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Convert measurement units

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5100365
\(6\,\text{lb}\,12\,\text{oz}\) of candy is shared equally among \(18\) children. How many ounces does each child receive? Use \(1\,\text{lb} = 16\,\text{oz}\).

Hints

- The answer is requested in ounces. - Convert the pounds to ounces before adding the extra ounces. - Use division to share the total equally.

Solution

1. Convert the total amount to ounces: \(6 \times 16 + 12 = 108\), so there are \(108\,\text{oz}\) of candy. 2. Divide equally: \(108\,\text{oz} \div 18 = 6\,\text{oz}\).

Answer

Each child receives \(6\,\text{oz}\).
5102575
What fraction of the larger unit is the given measurement? Write each fraction in simplest form. a) \(250\,\text{m}\) of \(1\,\text{km}\) b) \(40\,\text{min}\) of \(1\,\text{hr}\) c) \(375\,\text{g}\) of \(1\,\text{kg}\) d) \(20\,\text{cm}\) of \(1\,\text{m}\) e) \(12\,\text{in.}\) of \(1\,\text{yd}\)

Hints

- Convert the part and the whole to the same unit. - Write the part as the numerator and the whole as the denominator. - Divide the numerator and denominator by common factors to simplify.

Solution

1. Convert the part and the whole to the same unit, then write \(\frac{\text{part}}{\text{whole}}\). 2. For a), \(1\,\text{km}=1000\,\text{m}\), so \(\frac{250}{1000}=\frac{1}{4}\). 3. For b), \(1\,\text{hr}=60\,\text{min}\), so \(\frac{40}{60}=\frac{2}{3}\). 4. For c), \(1\,\text{kg}=1000\,\text{g}\), so \(\frac{375}{1000}=\frac{3}{8}\). 5. For d), \(1\,\text{m}=100\,\text{cm}\), so \(\frac{20}{100}=\frac{1}{5}\). 6. For e), \(1\,\text{yd}=36\,\text{in.}\), so \(\frac{12}{36}=\frac{1}{3}\).

Answer

a) \(\frac{1}{4}\) b) \(\frac{2}{3}\) c) \(\frac{3}{8}\) d) \(\frac{1}{5}\) e) \(\frac{1}{3}\)
5102595
Convert each measurement to the specified unit. a) \(450\,\text{mm}\) to centimeters b) \(15\,\text{s}\) to minutes c) \(800\,\text{mg}\) to grams d) \(250\,\text{mL}\) to liters

Hints

- Identify how many of the smaller units equal one larger unit. - When you convert to a larger unit, divide the numerical value. - Recall the conversion factors for length, time, mass, and capacity.

Solution

1. Divide by \(10\): \(450 \div 10 = 45\), so \(450\,\text{mm}=45\,\text{cm}\). 2. Divide by \(60\): \(15 \div 60 = 0.25\), so \(15\,\text{s}=0.25\,\text{min}\). 3. Divide by \(1000\): \(800 \div 1000 = 0.8\), so \(800\,\text{mg}=0.8\,\text{g}\). 4. Divide by \(1000\): \(250 \div 1000 = 0.25\), so \(250\,\text{mL}=0.25\,\text{L}\).

Answer

a) \(45\,\text{cm}\) b) \(0.25\,\text{min}\) c) \(0.8\,\text{g}\) d) \(0.25\,\text{L}\)
5102615
Convert each value to the specified unit. Use a decimal when needed. a) \(45\,\text{min}\) to hours b) \(250\,\text{mL}\) to liters c) \(35\) cents to dollars

Hints

- Determine how many smaller units make one target unit. - Divide by \(60\), \(1000\), or \(100\), as appropriate. - A decimal less than \(1\) is expected when the original amount is less than one full target unit.

Solution

1. \(45 \div 60 = 0.75\), so \(45\,\text{min}=0.75\,\text{h}\). 2. \(250 \div 1000 = 0.25\), so \(250\,\text{mL}=0.25\,\text{L}\). 3. \(35 \div 100 = 0.35\), so \(35\) cents equals \(\$0.35\).

Answer

a) \(0.75\,\text{h}\) b) \(0.25\,\text{L}\) c) \(\$0.35\)
5102625
Divide and give each result in the requested units. a) \(11\,\text{m}\div4\): as a decimal in meters and in meters and centimeters b) \(3\,\text{kg}\div5\): as a decimal in kilograms and in grams c) \(9\,\text{L}\div2\): as a decimal in liters and in liters and milliliters

Hints

- Perform each division first. - Recall how many centimeters are in a meter. - Recall how many grams are in a kilogram and how many milliliters are in a liter. - Use the decimal part to find the amount in the smaller unit.

Solution

1. For a), \(11\div4=2.75\), so the result is \(2.75\,\text{m}\). Since \(0.75\,\text{m}=75\,\text{cm}\), this is also \(2\,\text{m}\ 75\,\text{cm}\). 2. For b), \(3\div5=0.6\), so the result is \(0.6\,\text{kg}\). Since \(1\,\text{kg}=1000\,\text{g}\), \(0.6\times1000=600\), so the result is \(600\,\text{g}\). 3. For c), \(9\div2=4.5\), so the result is \(4.5\,\text{L}\). Since \(0.5\,\text{L}=500\,\text{mL}\), this is also \(4\,\text{L}\ 500\,\text{mL}\).

Answer

a) \(2.75\,\text{m}\); \(2\,\text{m}\ 75\,\text{cm}\) b) \(0.6\,\text{kg}\); \(600\,\text{g}\) c) \(4.5\,\text{L}\); \(4\,\text{L}\ 500\,\text{mL}\)
5102775
Convert each measurement to the indicated unit. Write each result as a fraction in simplest form. a) \(450\,\text{m}\) in kilometers b) \(12\,\text{min}\) in hours c) \(18\,\text{in.}\) in yards d) \(800\,\text{mL}\) in liters

Hints

- Identify how many of the smaller units make one larger unit. - Write the given amount over the conversion factor. - Simplify each fraction completely.

Solution

1. For a), \(1000\,\text{m}=1\,\text{km}\), so \(450\,\text{m}=\frac{450}{1000}\,\text{km}=\frac{9}{20}\,\text{km}\). 2. For b), \(60\,\text{min}=1\,\text{hr}\), so \(12\,\text{min}=\frac{12}{60}\,\text{hr}=\frac{1}{5}\,\text{hr}\). 3. For c), \(36\,\text{in.}=1\,\text{yd}\), so \(18\,\text{in.}=\frac{18}{36}\,\text{yd}=\frac{1}{2}\,\text{yd}\). 4. For d), \(1000\,\text{mL}=1\,\text{L}\), so \(800\,\text{mL}=\frac{800}{1000}\,\text{L}=\frac{4}{5}\,\text{L}\).

Answer

a) \(\frac{9}{20}\,\text{km}\) b) \(\frac{1}{5}\,\text{hr}\) c) \(\frac{1}{2}\,\text{yd}\) d) \(\frac{4}{5}\,\text{L}\)
5111225
Match each everyday object with a reasonable volume. Use the relative sizes to justify your choices mentally. Objects: - a full cleaning bucket - a large grain silo - a carton of orange juice - a grain of rice - a USB flash drive Volumes: \(30\,\text{mm}^3\), \(8\,\text{cm}^3\), \(1000\,\text{cm}^3\), \(10\,\text{L}\), \(400\,\text{m}^3\)

Hints

- Which object is smallest, and which is largest? - Think about how cubic millimeters compare with cubic centimeters. - Recall the relationship between liters and cubic decimeters. - Which units are commonly used for household containers and for large structures?

Solution

1. A grain of rice is the smallest object, so \(30\,\text{mm}^3\) is reasonable. 2. A USB flash drive is small but much larger than a grain of rice, so \(8\,\text{cm}^3\) is reasonable. 3. A typical juice carton holds \(1\,\text{L}=1000\,\text{cm}^3\). 4. A full cleaning bucket can reasonably hold about \(10\,\text{L}\). 5. A large grain silo is the largest object, so \(400\,\text{m}^3\) is reasonable.

Answer

- grain of rice \(\rightarrow30\,\text{mm}^3\) - USB flash drive \(\rightarrow8\,\text{cm}^3\) - carton of orange juice \(\rightarrow1000\,\text{cm}^3\) - full cleaning bucket \(\rightarrow10\,\text{L}\) - large grain silo \(\rightarrow400\,\text{m}^3\)
5113745
A class is making fruit punch. They combine: - \(1\frac{1}{4}\,\text{L}\) of apple juice - \(0.8\,\text{L}\) of sparkling water - \(750\,\text{mL}\) of orange juice - \(\frac{1}{5}\,\text{L}\) of berry syrup - \(50\,\text{mL}\) of lemon juice Will all the punch fit in a bowl that holds \(3\,\text{L}\)? Find the total volume and explain your decision.

Hints

- Convert every amount to the same unit. - Use \(1000\,\text{mL}=1\,\text{L}\). - Compare the total with the bowl’s capacity.

Solution

1. Convert every amount to liters: \(1\frac{1}{4}\,\text{L}=1.25\,\text{L}\), \(750\,\text{mL}=0.75\,\text{L}\), \(\frac{1}{5}\,\text{L}=0.2\,\text{L}\), and \(50\,\text{mL}=0.05\,\text{L}\). 2. Add: \(1.25+0.8+0.75+0.2+0.05=3.05\). The total volume is \(3.05\,\text{L}\). 3. Since \(3.05\,\text{L}>3\,\text{L}\), the bowl is too small.

Answer

The total volume is \(3.05\,\text{L}\). It will not all fit in a \(3\,\text{L}\) bowl.
5116415
Convert each measurement to the indicated unit. a) \(\frac{2}{5}\,\text{m}\) to centimeters b) \(\frac{3}{4}\,\text{L}\) to milliliters c) \(\frac{9}{10}\,\text{kg}\) to grams d) \(\frac{1}{8}\,\text{km}\) to meters

Hints

- Identify how many smaller units are in one larger unit. - Multiply the conversion factor by the fraction. - You may divide the conversion factor by the denominator first, then multiply by the numerator.

Solution

1. Since \(1\,\text{m}=100\,\text{cm}\), \(\frac{2}{5}\times100=40\), so \(\frac{2}{5}\,\text{m}=40\,\text{cm}\). 2. Since \(1\,\text{L}=1000\,\text{mL}\), \(\frac{3}{4}\times1000=750\), so \(\frac{3}{4}\,\text{L}=750\,\text{mL}\). 3. Since \(1\,\text{kg}=1000\,\text{g}\), \(\frac{9}{10}\times1000=900\), so \(\frac{9}{10}\,\text{kg}=900\,\text{g}\). 4. Since \(1\,\text{km}=1000\,\text{m}\), \(\frac{1}{8}\times1000=125\), so \(\frac{1}{8}\,\text{km}=125\,\text{m}\).

Answer

a) \(40\,\text{cm}\) b) \(750\,\text{mL}\) c) \(900\,\text{g}\) d) \(125\,\text{m}\)
5160495
A sporting goods store measures the mass and diameter of several balls. Convert each mass from kilograms (\(\text{kg}\)) to grams (\(\text{g}\)) and each diameter from meters (\(\text{m}\)) to centimeters (\(\text{cm}\)). <table> <tr> <th>Ball</th> <th>Basketball</th> <th>Soccer ball</th> <th>Handball</th> <th>Tennis ball</th> </tr> <tr> <th>Mass</th> <td>\(0.625\,\text{kg}\)</td> <td>\(0.430\,\text{kg}\)</td> <td>\(0.375\,\text{kg}\)</td> <td>\(0.058\,\text{kg}\)</td> </tr> <tr> <th>Diameter</th> <td>\(0.24\,\text{m}\)</td> <td>\(0.22\,\text{m}\)</td> <td>\(0.19\,\text{m}\)</td> <td>\(0.07\,\text{m}\)</td> </tr> </table>

Hints

- How many grams are in \(1\,\text{kg}\)? - How many centimeters are in \(1\,\text{m}\)? - Think about how the decimal point shifts when you convert to a smaller unit. - You can also use fractions or a place-value chart to think about each conversion.

Solution

1. Convert the masses from kilograms to grams by multiplying by \(1000\): - Basketball: \(0.625 \times 1000 = 625\,\text{g}\) - Soccer ball: \(0.430 \times 1000 = 430\,\text{g}\) - Handball: \(0.375 \times 1000 = 375\,\text{g}\) - Tennis ball: \(0.058 \times 1000 = 58\,\text{g}\) 2. Convert the diameters from meters to centimeters by multiplying by \(100\): - Basketball: \(0.24 \times 100 = 24\,\text{cm}\) - Soccer ball: \(0.22 \times 100 = 22\,\text{cm}\) - Handball: \(0.19 \times 100 = 19\,\text{cm}\) - Tennis ball: \(0.07 \times 100 = 7\,\text{cm}\)

Answer

Masses: Basketball \(625\,\text{g}\), soccer ball \(430\,\text{g}\), handball \(375\,\text{g}\), tennis ball \(58\,\text{g}\). Diameters: Basketball \(24\,\text{cm}\), soccer ball \(22\,\text{cm}\), handball \(19\,\text{cm}\), tennis ball \(7\,\text{cm}\).
5160505
Convert each mass to grams (\(\text{g}\)): a) \(1.500\,\text{kg}\) b) \(0.725\,\text{kg}\) c) \(0.040\,\text{kg}\) d) \(3.005\,\text{kg}\)

Hints

- Recall the conversion factor between kilograms and grams. - What happens to the decimal point when you multiply by \(1000\)? - Picture each number in a place-value chart.

Solution

To convert kilograms to grams, multiply by \(1000\), because \(1\,\text{kg}=1000\,\text{g}\). 1. a) \(1.500 \times 1000 = 1500\,\text{g}\) 2. b) \(0.725 \times 1000 = 725\,\text{g}\) 3. c) \(0.040 \times 1000 = 40\,\text{g}\) 4. d) \(3.005 \times 1000 = 3005\,\text{g}\)

Answer

a) \(1500\,\text{g}\) b) \(725\,\text{g}\) c) \(40\,\text{g}\) d) \(3005\,\text{g}\)
5166985
Convert each amount to liters (\(\text{L}\)). Write each answer as a decimal number. \(5000\,\text{mL}\); \(800\,\text{mL}\); \(1500\,\text{mL}\); \(200\,\text{mL}\); \(600\,\text{mL}\)

Hints

- Think about how many groups of \(1000\) are in each number. Each full group is \(1\,\text{L}\). - What happens to the remaining hundreds when you write the amount in liters? - Dividing by \(1000\) moves the decimal point three places to the left.

Solution

1. Use \(1000\,\text{mL}=1\,\text{L}\). 2. Divide each number of milliliters by \(1000\): \(5000 \div 1000=5.0\) \(800 \div 1000=0.8\) \(1500 \div 1000=1.5\) \(200 \div 1000=0.2\) \(600 \div 1000=0.6\)

Answer

\(5.0\,\text{L}\); \(0.8\,\text{L}\); \(1.5\,\text{L}\); \(0.2\,\text{L}\); \(0.6\,\text{L}\)
5168535
Write each mass as a decimal number of kilograms, \(\text{kg}\). Use three decimal places to show grams. a) \(3250\,\text{g}\) b) \(805\,\text{g}\) c) \(12{,}004\,\text{g}\)

Hints

- One kilogram equals \(1000\) grams. - The tenths, hundredths, and thousandths places represent hundreds, tens, and ones of grams. - Use zeros as placeholders when a place value is missing.

Solution

1. Since \(1000\,\text{g} = 1\,\text{kg}\), divide each number of grams by \(1000\). 2. a) \(3250\,\text{g} = 3.250\,\text{kg}\). 3. b) \(805\,\text{g} = 0.805\,\text{kg}\). 4. c) \(12{,}004\,\text{g} = 12.004\,\text{kg}\).

Answer

a) \(3.250\,\text{kg}\) b) \(0.805\,\text{kg}\) c) \(12.004\,\text{kg}\)
5168545
Write each mixed-unit mass as a decimal number of kilograms. a) \(5\,\text{kg}\ 70\,\text{g}\) b) \(18\,\text{kg}\ 205\,\text{g}\) c) \(0\,\text{kg}\ 9\,\text{g}\)

Hints

- Grams occupy three decimal places in a kilogram measurement. - Use a leading zero when the number of grams has fewer than three digits. - Keep meaningful trailing zeros that show the measurement in grams.

Solution

1. Write the kilograms as the whole-number part. 2. Since \(1\,\text{kg} = 1000\,\text{g}\), write the grams as three decimal places, using zeros as needed. 3. a) \(70\,\text{g} = 0.070\,\text{kg}\), so \(5\,\text{kg}\ 70\,\text{g} = 5.070\,\text{kg}\). 4. b) \(205\,\text{g} = 0.205\,\text{kg}\), so \(18\,\text{kg}\ 205\,\text{g} = 18.205\,\text{kg}\). 5. c) \(9\,\text{g} = 0.009\,\text{kg}\), so \(0\,\text{kg}\ 9\,\text{g} = 0.009\,\text{kg}\).

Answer

a) \(5.070\,\text{kg}\) b) \(18.205\,\text{kg}\) c) \(0.009\,\text{kg}\)
5205005
Fill in each missing length unit. Use millimeters, centimeters, meters, or kilometers. a) \(13\,\text{m}=1300\,\square\) b) \(5\,\text{m}=500\,\square\) c) \(8000\,\text{m}=8\,\square\) d) \(4\,\square=40\,\text{mm}\)

Hints

- Decide whether the numerical value became larger or smaller. - Recall the factors \(10\), \(100\), and \(1000\). - Match the factor to the pair of length units.

Solution

1. Since \(13\,\text{m}=1300\,\text{cm}\), the missing unit is centimeters. 2. Since \(5\,\text{m}=500\,\text{cm}\), the missing unit is centimeters. 3. Since \(8000\,\text{m}=8\,\text{km}\), the missing unit is kilometers. 4. Since \(4\,\text{cm}=40\,\text{mm}\), the missing unit is centimeters.

Answer

a) centimeters b) centimeters c) kilometers d) centimeters
5205015
Convert each measurement to the indicated unit. Write the missing numerical value. a) \(12\,\text{m}=\square\,\text{cm}\) b) \(900\,\text{cm}=\square\,\text{m}\) c) \(7\,\text{km}=\square\,\text{m}\) d) \(250\,\text{mm}=\square\,\text{cm}\)

Hints

- Decide whether the target unit is larger or smaller than the original unit. - Use \(10\,\text{mm}=1\,\text{cm}\), \(100\,\text{cm}=1\,\text{m}\), and \(1000\,\text{m}=1\,\text{km}\). - Converting to a smaller unit increases the numerical value.

Solution

1. \(12 \times 100 = 1200\), so \(12\,\text{m}=1200\,\text{cm}\). 2. \(900 \div 100 = 9\), so \(900\,\text{cm}=9\,\text{m}\). 3. \(7 \times 1000 = 7000\), so \(7\,\text{km}=7000\,\text{m}\). 4. \(250 \div 10 = 25\), so \(250\,\text{mm}=25\,\text{cm}\).

Answer

a) \(1200\) b) \(9\) c) \(7000\) d) \(25\)
5205025
Write \(<\), \(>\), or \(=\) in each blank. a) \(400\,\text{m}\ \square\ 4\,\text{km}\) b) \(300\,\text{cm}\ \square\ 3\,\text{m}\) c) \(25\,\text{cm}\ \square\ 250\,\text{mm}\) d) \(5\,\text{m}\ \square\ 550\,\text{cm}\)

Hints

- Convert both measurements in each comparison to the same unit. - Choosing the smaller unit can help you avoid decimals. - Compare the numerical values only after the units match.

Solution

1. \(4\,\text{km}=4000\,\text{m}\), so \(400\,\text{m}<4\,\text{km}\). 2. \(3\,\text{m}=300\,\text{cm}\), so the measurements are equal. 3. \(25\,\text{cm}=250\,\text{mm}\), so the measurements are equal. 4. \(5\,\text{m}=500\,\text{cm}\), and \(500<550\), so \(5\,\text{m}<550\,\text{cm}\).

Answer

a) \(<\) b) \(=\) c) \(=\) d) \(<\)
5205035
Convert each length to the unit in parentheses. a) \(5200\,\text{mm}\) (centimeters) b) \(14\,\text{m}\ 5\,\text{cm}\) (centimeters) c) \(3200\,\text{cm}\) (meters) d) \(9\,\text{km}\ 75\,\text{m}\) (meters) e) \(63\,\text{cm}\) (millimeters)

Hints

- Decide whether to multiply or divide based on the target unit. - Use factors of \(10\), \(100\), and \(1000\). - For a mixed measurement, convert the larger-unit part first and then add the remaining amount.

Solution

1. \(5200 \div 10 = 520\), so \(5200\,\text{mm}=520\,\text{cm}\). 2. \(14\,\text{m}=1400\,\text{cm}\). Then \(1400+5=1405\), so the total is \(1405\,\text{cm}\). 3. \(3200 \div 100 = 32\), so \(3200\,\text{cm}=32\,\text{m}\). 4. \(9\,\text{km}=9000\,\text{m}\). Then \(9000+75=9075\), so the total is \(9075\,\text{m}\). 5. \(63 \times 10 = 630\), so \(63\,\text{cm}=630\,\text{mm}\).

Answer

a) \(520\,\text{cm}\) b) \(1405\,\text{cm}\) c) \(32\,\text{m}\) d) \(9075\,\text{m}\) e) \(630\,\text{mm}\)
5205105
Julia says, “\(0.8\,\text{m}\) is shorter than \(0.75\,\text{m}\) because \(8\) is less than \(75\).” Convert both lengths to centimeters and explain why Julia is incorrect.

Hints

- Recall how many centimeters are in one meter. - You can write \(0.8\) as \(0.80\) without changing its value. - Compare the lengths after converting both to the same unit.

Solution

1. \(0.8 \times 100 = 80\), so \(0.8\,\text{m}=80\,\text{cm}\). 2. \(0.75 \times 100 = 75\), so \(0.75\,\text{m}=75\,\text{cm}\). 3. Since \(80>75\), \(0.8\,\text{m}>0.75\,\text{m}\). Julia compared the digits after the decimal point as whole numbers instead of comparing decimal place values.

Answer

Julia is incorrect. \(0.8\,\text{m}=80\,\text{cm}\) and \(0.75\,\text{m}=75\,\text{cm}\). Since \(80\,\text{cm}>75\,\text{cm}\), \(0.8\,\text{m}\) is longer.
5205115
Two measuring cups contain water. The first cup contains \(0.4\,\text{L}\), and the second contains \(0.35\,\text{L}\). Which cup contains more water? Justify your answer by converting both amounts to milliliters.

Hints

- Recall how many milliliters are in one liter. - Write both decimals to the same number of decimal places if that helps. - Compare the milliliter values.

Solution

1. First cup: \(0.4 \times 1000 = 400\), so \(0.4\,\text{L}=400\,\text{mL}\). 2. Second cup: \(0.35 \times 1000 = 350\), so \(0.35\,\text{L}=350\,\text{mL}\). 3. Since \(400>350\), the first cup contains more water.

Answer

The first cup contains more water because \(0.4\,\text{L}=400\,\text{mL}\), \(0.35\,\text{L}=350\,\text{mL}\), and \(400\,\text{mL}>350\,\text{mL}\).
5205125
Four lengths are shown. A: \(5.6\,\text{m}\) B: \(5.06\,\text{m}\) C: \(5.60\,\text{m}\) D: \(56\,\text{cm}\) Which two measurements represent the same length? Justify your choice by converting all four measurements to centimeters.

Hints

- Convert each measurement to centimeters. - Pay attention to the position of the zero in each decimal. - A trailing zero does not change a decimal’s value.

Solution

1. A: \(5.6\,\text{m}=560\,\text{cm}\). 2. B: \(5.06\,\text{m}=506\,\text{cm}\). 3. C: \(5.60\,\text{m}=560\,\text{cm}\). 4. D is \(56\,\text{cm}\). 5. Only A and C have the same converted value, \(560\,\text{cm}\).

Answer

A and C represent the same length. Both equal \(560\,\text{cm}\).
5205235
Fill in each missing number or unit so the equation is true. a) \(4.2\,\text{m}=4\,\text{m}\ \square\,\text{cm}\) b) \(520\,\text{mm}=52\,\square\) c) \(3\,\text{cm}\ 4\,\text{mm}=\square\,\text{mm}\) d) \(0.8\,\text{m}=\square\,\text{cm}\)

Hints

- Convert the measurements in each equation to compatible units. - Use \(10\,\text{mm}=1\,\text{cm}\) and \(100\,\text{cm}=1\,\text{m}\). - Converting from a larger unit to a smaller unit increases the numerical value.

Solution

1. \(0.2\,\text{m}=20\,\text{cm}\), so the missing number is \(20\). 2. \(520\,\text{mm}=52\,\text{cm}\), so the missing unit is centimeters. 3. \(3\,\text{cm}=30\,\text{mm}\), and \(30+4=34\), so the total is \(34\,\text{mm}\). 4. \(0.8 \times 100 = 80\), so \(0.8\,\text{m}=80\,\text{cm}\).

Answer

a) \(20\) b) centimeters c) \(34\) d) \(80\)
5205245
Fill in each missing value or unit. a) \(12.5\,\text{km}=12\,\text{km}\ \square\,\text{m}\) b) \(705\,\text{cm}=7\,\square\ 5\,\text{cm}\) c) \(2\,\text{m}\ 3\,\text{cm}=203\,\square\) d) \(1500\,\text{m}=1.5\,\square\)

Hints

- Use \(1000\,\text{m}=1\,\text{km}\) and \(100\,\text{cm}=1\,\text{m}\). - Separate a decimal amount into whole units and the remaining part. - Check that both sides of each equation represent the same length.

Solution

1. \(0.5\,\text{km}=500\,\text{m}\), so the missing value is \(500\). 2. \(700\,\text{cm}=7\,\text{m}\), so the missing unit is meters. 3. \(2\,\text{m}=200\,\text{cm}\), and \(200+3=203\), so the missing unit is centimeters. 4. \(1500\,\text{m}=1.5\,\text{km}\), so the missing unit is kilometers.

Answer

a) \(500\) b) meters c) centimeters d) kilometers
5205255
Fill in each missing number or unit. a) \(8\,\text{cm}\ 9\,\text{mm}=\square\,\text{mm}\) b) \(1.03\,\text{m}=1\,\text{m}\ 3\,\square\) c) \(600\,\text{cm}=\square\,\text{m}\) d) \(4500\,\text{m}=\square\,\text{km}\)

Hints

- Convert mixed measurements to one unit before adding. - Use \(10\,\text{mm}=1\,\text{cm}\), \(100\,\text{cm}=1\,\text{m}\), and \(1000\,\text{m}=1\,\text{km}\). - Converting to a larger unit requires division.

Solution

1. \(8\,\text{cm}=80\,\text{mm}\). Then \(80+9=89\), so the total is \(89\,\text{mm}\). 2. \(0.03\,\text{m}=3\,\text{cm}\), so the missing unit is centimeters. 3. \(600 \div 100=6\), so \(600\,\text{cm}=6\,\text{m}\). 4. \(4500 \div 1000=4.5\), so \(4500\,\text{m}=4.5\,\text{km}\).

Answer

a) \(89\) b) centimeters c) \(6\) d) \(4.5\)
5205315
Convert each length to the specified smaller unit. a) \(4.2\,\text{km}\) to meters b) \(17\,\text{m}\) to centimeters c) \(0.5\,\text{m}\) to centimeters d) \(240\,\text{cm}\) to millimeters

Hints

- Identify the conversion factor for each pair of units. - Converting to a smaller unit increases the numerical value. - Use factors of \(10\), \(100\), or \(1000\).

Solution

1. \(4.2 \times 1000=4200\), so \(4.2\,\text{km}=4200\,\text{m}\). 2. \(17 \times 100=1700\), so \(17\,\text{m}=1700\,\text{cm}\). 3. \(0.5 \times 100=50\), so \(0.5\,\text{m}=50\,\text{cm}\). 4. \(240 \times 10=2400\), so \(240\,\text{cm}=2400\,\text{mm}\).

Answer

a) \(4200\,\text{m}\) b) \(1700\,\text{cm}\) c) \(50\,\text{cm}\) d) \(2400\,\text{mm}\)
5205385
Write each length using mixed units. Example: \(125\,\text{cm}=1\,\text{m}\ 25\,\text{cm}\). a) \(842\,\text{cm}\) b) \(305\,\text{cm}\) c) \(2007\,\text{mm}\) d) \(95\,\text{mm}\)

Hints

- Separate the measurement into the greatest possible whole larger units and a remainder. - Use \(100\,\text{cm}=1\,\text{m}\), \(1000\,\text{mm}=1\,\text{m}\), and \(10\,\text{mm}=1\,\text{cm}\). - Check that the remainder is smaller than one of the larger units.

Solution

1. \(842\,\text{cm}=800\,\text{cm}+42\,\text{cm}=8\,\text{m}\ 42\,\text{cm}\). 2. \(305\,\text{cm}=300\,\text{cm}+5\,\text{cm}=3\,\text{m}\ 5\,\text{cm}\). 3. \(2007\,\text{mm}=2000\,\text{mm}+7\,\text{mm}=2\,\text{m}\ 7\,\text{mm}\). 4. \(95\,\text{mm}=90\,\text{mm}+5\,\text{mm}=9\,\text{cm}\ 5\,\text{mm}\).

Answer

a) \(8\,\text{m}\ 42\,\text{cm}\) b) \(3\,\text{m}\ 5\,\text{cm}\) c) \(2\,\text{m}\ 7\,\text{mm}\) d) \(9\,\text{cm}\ 5\,\text{mm}\)
5205565
Convert each length to the unit in parentheses. a) \(7.2\,\text{m}\) (centimeters) b) \(540\,\text{cm}\) (meters) c) \(20\,\text{cm}\ 8\,\text{mm}\) (millimeters) d) \(9000\,\text{mm}\) (centimeters)

Hints

- Identify whether the target unit is larger or smaller. - Use \(10\,\text{mm}=1\,\text{cm}\) and \(100\,\text{cm}=1\,\text{m}\). - For mixed measurements, convert both parts to the target unit before adding.

Solution

1. \(7.2 \times 100=720\), so \(7.2\,\text{m}=720\,\text{cm}\). 2. \(540 \div 100=5.4\), so \(540\,\text{cm}=5.4\,\text{m}\). 3. \(20\,\text{cm}=200\,\text{mm}\). Then \(200+8=208\), so the total is \(208\,\text{mm}\). 4. \(9000 \div 10=900\), so \(9000\,\text{mm}=900\,\text{cm}\).

Answer

a) \(720\,\text{cm}\) b) \(5.4\,\text{m}\) c) \(208\,\text{mm}\) d) \(900\,\text{cm}\)
5205675
Write \(<\), \(>\), or \(=\) in each box. a) \(230\,\text{mm}\ \square\ 2.3\,\text{cm}\) b) \(1.5\,\text{m}\ \square\ 150\,\text{cm}\) c) \(47\,\text{cm}\ \square\ 407\,\text{mm}\) d) \(0.8\,\text{km}\ \square\ 800\,\text{m}\)

Hints

- Convert both measurements in each comparison to the same unit. - A smaller unit can help avoid decimals. - Compare only after the units match.

Solution

1. \(2.3\,\text{cm}=23\,\text{mm}\), and \(230>23\), so the symbol is \(>\). 2. \(1.5\,\text{m}=150\,\text{cm}\), so the symbol is \(=\). 3. \(47\,\text{cm}=470\,\text{mm}\), and \(470>407\), so the symbol is \(>\). 4. \(0.8\,\text{km}=800\,\text{m}\), so the symbol is \(=\).

Answer

a) \(>\) b) \(=\) c) \(>\) d) \(=\)
5206265
Calculate each result and write it using mixed units of meters and centimeters. a) \(86\,\text{cm} + 45\,\text{cm}\) b) \(264\,\text{cm} - 0.84\,\text{m}\)

Hints

- Express each calculation in centimeters first. - Use \(100\,\text{cm} = 1\,\text{m}\). - Convert each final centimeter amount to meters and centimeters.

Solution

1. For a), add: \(86\,\text{cm} + 45\,\text{cm} = 131\,\text{cm}\). Convert: \(131\,\text{cm} = 1\,\text{m}\,31\,\text{cm}\). 2. For b), convert \(0.84\,\text{m} = 84\,\text{cm}\). Then subtract: \(264\,\text{cm} - 84\,\text{cm} = 180\,\text{cm}\). Convert: \(180\,\text{cm} = 1\,\text{m}\,80\,\text{cm}\).

Answer

a) \(1\,\text{m}\,31\,\text{cm}\) b) \(1\,\text{m}\,80\,\text{cm}\)
5208595
Fill in each missing unit or numerical value. a) \(5600\,\text{g}=5.6\,\square\) b) \(12\,\text{cm}=\square\,\text{mm}\) c) \(0.25\,\text{km}=250\,\square\) d) \(35\,\text{g}=\square\,\text{mg}\)

Hints

- Notice whether the numerical value becomes larger or smaller. - A smaller numerical value must be paired with a larger unit. - Identify the conversion factor for each pair of units.

Solution

1. \(5600\,\text{g}=5.6\,\text{kg}\), so the missing unit is kilograms. 2. \(12 \times 10=120\), so \(12\,\text{cm}=120\,\text{mm}\). 3. \(0.25 \times 1000=250\), so the missing unit is meters. 4. \(35 \times 1000=35{,}000\), so \(35\,\text{g}=35{,}000\,\text{mg}\).

Answer

a) kilograms b) \(120\) c) meters d) \(35{,}000\)
5208725
Rewrite each decimal mass without a decimal, using mixed units when needed. a) \(5.4\,\text{kg}\) b) \(0.025\,\text{kg}\) c) \(10.705\,\text{kg}\) d) \(0.008\,\text{g}\)

Hints

- Convert the decimal part to the next smaller unit. - Use a factor of \(1000\) between kilograms and grams and between grams and milligrams. - Omit a larger unit when its amount is zero.

Solution

1. \(0.4\,\text{kg}=400\,\text{g}\), so \(5.4\,\text{kg}=5\,\text{kg}\ 400\,\text{g}\). 2. \(0.025\,\text{kg}=25\,\text{g}\). 3. \(0.705\,\text{kg}=705\,\text{g}\), so \(10.705\,\text{kg}=10\,\text{kg}\ 705\,\text{g}\). 4. \(0.008\,\text{g}=8\,\text{mg}\).

Answer

a) \(5\,\text{kg}\ 400\,\text{g}\) b) \(25\,\text{g}\) c) \(10\,\text{kg}\ 705\,\text{g}\) d) \(8\,\text{mg}\)
5208765
Write \(<\), \(>\), or \(=\) in each box. Convert to a common unit first. a) \(2\,\text{km}\ 50\,\text{m}\ \square\ 2.5\,\text{km}\) b) \(0.08\,\text{km}\ \square\ 80\,\text{m}\) c) \(1205\,\text{m}\ \square\ 1\,\text{km}\ 25\,\text{m}\) d) \(3.007\,\text{km}\ \square\ 3007\,\text{m}\)

Hints

- Convert both sides to meters. - Use \(1000\,\text{m}=1\,\text{km}\). - Pay attention to all three decimal places when converting kilometers to meters.

Solution

1. \(2\,\text{km}\ 50\,\text{m}=2050\,\text{m}\), and \(2.5\,\text{km}=2500\,\text{m}\). Since \(2050<2500\), the symbol is \(<\). 2. \(0.08\,\text{km}=80\,\text{m}\), so the symbol is \(=\). 3. \(1\,\text{km}\ 25\,\text{m}=1025\,\text{m}\). Since \(1205>1025\), the symbol is \(>\). 4. \(3.007\,\text{km}=3007\,\text{m}\), so the symbol is \(=\).

Answer

a) \(<\) b) \(=\) c) \(>\) d) \(=\)
5209135
Convert each mass to the unit in parentheses. a) \(75\,\text{g}\) (kilograms) b) \(4000\,\text{g}\) (kilograms) c) \(850\,\text{g}\) (kilograms) d) \(1.2\,\text{g}\) (milligrams)

Hints

- Decide whether the target unit is larger or smaller. - Divide by \(1000\) to convert grams to kilograms. - Multiply by \(1000\) to convert grams to milligrams.

Solution

1. \(75 \div 1000=0.075\), so \(75\,\text{g}=0.075\,\text{kg}\). 2. \(4000 \div 1000=4\), so \(4000\,\text{g}=4\,\text{kg}\). 3. \(850 \div 1000=0.85\), so \(850\,\text{g}=0.85\,\text{kg}\). 4. \(1.2 \times 1000=1200\), so \(1.2\,\text{g}=1200\,\text{mg}\).

Answer

a) \(0.075\,\text{kg}\) b) \(4\,\text{kg}\) c) \(0.85\,\text{kg}\) d) \(1200\,\text{mg}\)
5209155
Write \(<\), \(>\), or \(=\) in each blank. a) \(0.3\,\text{kg}\ \square\ 30\,\text{g}\) b) \(\frac{1}{2}\,\text{kg}\ \square\ 500\,\text{g}\) c) \(1200\,\text{mg}\ \square\ 1.2\,\text{g}\) d) \(0.07\,\text{kg}\ \square\ 700\,\text{g}\)

Hints

- Convert both sides to the same unit. - Using the smaller unit can help avoid decimals. - Compare the numerical values only after the units match.

Solution

1. \(0.3\,\text{kg}=300\,\text{g}\), and \(300>30\), so the symbol is \(>\). 2. \(\frac{1}{2}\,\text{kg}=500\,\text{g}\), so the symbol is \(=\). 3. \(1200\,\text{mg}=1.2\,\text{g}\), so the symbol is \(=\). 4. \(0.07\,\text{kg}=70\,\text{g}\), and \(70<700\), so the symbol is \(<\).

Answer

a) \(>\) b) \(=\) c) \(=\) d) \(<\)
5209295
A storage bin contains \(9\,\text{lb}\) of coffee beans. The beans are packed into \(8\)-ounce bags. How many full bags can be filled? Use \(1\,\text{lb} = 16\,\text{oz}\).

Hints

- Express the total and the amount per bag in the same unit. - Convert pounds to ounces. - Divide to find how many times the smaller amount fits into the total. - Check by multiplying the number of bags by \(8\,\text{oz}\).

Solution

1. Convert the coffee beans to ounces: \(9 \times 16 = 144\), so there are \(144\,\text{oz}\). 2. Divide by the amount in one bag: \(144\,\text{oz} \div 8\,\text{oz} = 18\).

Answer

\(18\) full bags can be filled.
5209735
Convert each time measurement to the unit in parentheses. a) \(8\,\text{h}\) (minutes) b) \(360\,\text{s}\) (minutes) c) \(3\,\text{days}\) (hours) d) \(540\,\text{min}\) (hours)

Hints

- Decide whether the target unit is larger or smaller. - Use \(60\,\text{s}=1\,\text{min}\), \(60\,\text{min}=1\,\text{h}\), and \(24\,\text{h}=1\,\text{day}\). - Convert to a smaller unit by multiplying and to a larger unit by dividing.

Solution

1. \(8 \times 60=480\), so \(8\,\text{h}=480\,\text{min}\). 2. \(360 \div 60=6\), so \(360\,\text{s}=6\,\text{min}\). 3. \(3 \times 24=72\), so \(3\,\text{days}=72\,\text{h}\). 4. \(540 \div 60=9\), so \(540\,\text{min}=9\,\text{h}\).

Answer

a) \(480\,\text{min}\) b) \(6\,\text{min}\) c) \(72\,\text{h}\) d) \(9\,\text{h}\)
5209895
Write each amount entirely in cents. a) \(\$5.75\) b) \(\$12.05\) c) \(\$110.10\) d) \(\$2.04\)

Hints

- One dollar equals \(100\) cents. - Multiply the dollar amount by \(100\). - Pay attention to placeholder zeros in amounts such as \(\$12.05\).

Solution

1. \(\$5.75=575\) cents. 2. \(\$12.05=1205\) cents. 3. \(\$110.10=11{,}010\) cents. 4. \(\$2.04=204\) cents.

Answer

a) \(575\) cents b) \(1205\) cents c) \(11{,}010\) cents d) \(204\) cents
5210125
An intercity train trip takes exactly \(2\,\text{h}\ 17\,\text{min}\). Write the entire travel time in minutes.

Hints

- Use \(60\,\text{min}=1\,\text{h}\). - Convert the hours first, then add the remaining minutes. - Check that your total includes both parts of the travel time.

Solution

1. \(2\,\text{h}=2 \times 60\,\text{min}=120\,\text{min}\). 2. Add the remaining \(17\,\text{min}\): \(120+17=137\).

Answer

\(137\,\text{min}\)
5211125
Calculate: \(3\,\text{days}\,18\,\text{hr} \div 15\)

Hints

- Convert the days and hours to one smaller unit first. - How many hours are in one day? - Then divide the total number of hours.

Solution

1. Convert the duration to hours: \(3 \times 24 + 18 = 72 + 18 = 90\) hours. 2. Divide: \(90\,\text{hr} \div 15 = 6\,\text{hr}\).

Answer

\(6\,\text{hr}\)
5211365
Convert each measurement to the unit in parentheses. Use decimals when needed. a) \(45\,\text{mm}\) (centimeters); \(0.32\,\text{m}\) (centimeters); \(1\,\text{m}\ 5\,\text{cm}\) (centimeters) b) \(2700\,\text{g}\) (kilograms); \(4000\,\text{g}\) (kilograms); \(15\,\text{kg}\ 500\,\text{g}\) (kilograms)

Hints

- Identify the conversion factor for each pair of units. - Decide whether to multiply or divide. - For mixed units, convert the smaller-unit part to the target unit and add.

Solution

1. Lengths: \(45 \div 10=4.5\), so \(45\,\text{mm}=4.5\,\text{cm}\). \(0.32 \times 100=32\), so \(0.32\,\text{m}=32\,\text{cm}\). \(1\,\text{m}\ 5\,\text{cm}=100\,\text{cm}+5\,\text{cm}=105\,\text{cm}\). 2. Masses: \(2700 \div 1000=2.7\), so \(2700\,\text{g}=2.7\,\text{kg}\). \(4000\,\text{g}=4\,\text{kg}\). \(500\,\text{g}=0.5\,\text{kg}\), so \(15\,\text{kg}\ 500\,\text{g}=15.5\,\text{kg}\).

Answer

a) \(4.5\,\text{cm}\); \(32\,\text{cm}\); \(105\,\text{cm}\) b) \(2.7\,\text{kg}\); \(4\,\text{kg}\); \(15.5\,\text{kg}\)
5211375
Write \(<\), \(>\), or \(=\) in each blank. Convert one measurement to the other unit before comparing. a) \(2500\,\text{m}\ \square\ 2.5\,\text{km}\) b) \(75\,\text{min}\ \square\ 1.5\,\text{h}\) c) \(300\,\text{mg}\ \square\ 0.3\,\text{g}\) d) \(1\,\text{kg}\ 50\,\text{g}\ \square\ 1500\,\text{g}\)

Hints

- Measurements must use the same unit before you compare them. - One-half hour is \(30\) minutes. - Use \(1000\,\text{g}=1\,\text{kg}\).

Solution

1. \(2.5\,\text{km}=2500\,\text{m}\), so the symbol is \(=\). 2. \(1.5\,\text{h}=90\,\text{min}\), and \(75<90\), so the symbol is \(<\). 3. \(300\,\text{mg}=0.3\,\text{g}\), so the symbol is \(=\). 4. \(1\,\text{kg}\ 50\,\text{g}=1050\,\text{g}\), and \(1050<1500\), so the symbol is \(<\).

Answer

a) \(=\) b) \(<\) c) \(=\) d) \(<\)
5211445
Measurements can be written as decimals or with mixed units. Fill in each blank. a) \(5.4\,\text{m}=5\,\text{m}\ \square\,\text{cm}\) b) \(3.02\,\text{kg}=3\,\text{kg}\ \square\,\text{g}\) c) \(0.75\,\text{m}=\square\,\text{cm}\) d) \(8.1\,\text{cm}=8\,\text{cm}\ \square\,\text{mm}\)

Hints

- Interpret the decimal part in terms of the smaller unit. - Use \(100\,\text{cm}=1\,\text{m}\), \(1000\,\text{g}=1\,\text{kg}\), and \(10\,\text{mm}=1\,\text{cm}\). - A place-value chart can help you align the units.

Solution

1. \(0.4\,\text{m}=40\,\text{cm}\), so the missing number is \(40\). 2. \(0.02\,\text{kg}=20\,\text{g}\), so the missing number is \(20\). 3. \(0.75 \times 100=75\), so \(0.75\,\text{m}=75\,\text{cm}\). 4. \(0.1\,\text{cm}=1\,\text{mm}\), so the missing number is \(1\).

Answer

a) \(40\) b) \(20\) c) \(75\) d) \(1\)
5212275
Write \(<\), \(>\), or \(=\) to make each statement true. a) \(2\,\text{km}\ 50\,\text{m}\ \square\ 2005\,\text{m}\) b) \(0.12\,\text{m}\ \square\ 12\,\text{cm}\) c) \(60\,\text{mm}\ \square\ 0.06\,\text{m}\)

Hints

- Convert mixed units to one unit. - Use \(100\,\text{cm}=1\,\text{m}\) and \(1000\,\text{mm}=1\,\text{m}\). - Compare after both measurements use the same unit.

Solution

1. \(2\,\text{km}\ 50\,\text{m}=2050\,\text{m}\), and \(2050>2005\), so the symbol is \(>\). 2. \(0.12\,\text{m}=12\,\text{cm}\), so the symbol is \(=\). 3. \(0.06\,\text{m}=60\,\text{mm}\), so the symbol is \(=\).

Answer

a) \(>\) b) \(=\) c) \(=\)
5212325
A student writes \(6.5\,\text{m} + 250\,\text{cm} = 256.5\,\text{m}\). Explain the error and find the correct result in meters.

Hints

- Measurements must use the same unit before they are added. - Convert centimeters to meters because the answer is requested in meters. - Check whether the size of the result is reasonable.

Solution

1. The student added the numbers without first expressing both measurements in the same unit. 2. Convert: \(250\,\text{cm} = 2.5\,\text{m}\). 3. Add: \(6.5\,\text{m} + 2.5\,\text{m} = 9\,\text{m}\).

Answer

The error was adding measurements with different units without converting. The correct result is \(9\,\text{m}\).
5213035
Write \(<\), \(>\), or \(=\) in each box. a) \(12\,\text{cm}\ 5\,\text{mm}\ \square\ 12.05\,\text{cm}\) b) \(3\,\text{km}\ 20\,\text{m}\ \square\ 3.2\,\text{km}\) c) \(450\,\text{m}\ \square\ 0.45\,\text{km}\)

Hints

- Convert both sides to the same unit. - Converting to the smaller unit often avoids decimals. - Pay close attention to decimal place value.

Solution

1. \(12\,\text{cm}\ 5\,\text{mm}=12.5\,\text{cm}\), and \(12.5>12.05\), so the symbol is \(>\). 2. \(3\,\text{km}\ 20\,\text{m}=3020\,\text{m}\), while \(3.2\,\text{km}=3200\,\text{m}\). Since \(3020<3200\), the symbol is \(<\). 3. \(0.45\,\text{km}=450\,\text{m}\), so the symbol is \(=\).

Answer

a) \(>\) b) \(<\) c) \(=\)
5213045
Write \(<\), \(>\), or \(=\) in each box. a) \(4\,\text{kg}\ 50\,\text{g}\ \square\ 4.5\,\text{kg}\) b) \(0.8\,\text{kg}\ \square\ 80\,\text{g}\) c) \(2500\,\text{mg}\ \square\ 2.5\,\text{g}\)

Hints

- Convert both measurements to the same unit. - Use \(1000\,\text{g}=1\,\text{kg}\) and \(1000\,\text{mg}=1\,\text{g}\). - Compare the numerical values only after the units match.

Solution

1. \(4\,\text{kg}\ 50\,\text{g}=4.05\,\text{kg}\), and \(4.05<4.5\), so the symbol is \(<\). 2. \(0.8\,\text{kg}=800\,\text{g}\), and \(800>80\), so the symbol is \(>\). 3. \(2500\,\text{mg}=2.5\,\text{g}\), so the symbol is \(=\).

Answer

a) \(<\) b) \(>\) c) \(=\)
5213885
Match each object with the most reasonable area measurement. **Objects:** 1. One small square on millimeter graph paper 2. One key on a computer keyboard 3. A classroom whiteboard 4. A suburban home lot 5. A large farm 6. The state of West Virginia **Area measurements:** \(1\,\text{mm}^2\) | \(1.5\,\text{cm}^2\) | \(4\,\text{m}^2\) | \(6000\,\text{ft}^2\) | \(500\) acres | \(24{,}000\,\text{mi}^2\)

Hints

- Estimate each object's length and width. - Choose small units for small objects and large units for land areas. - Compare the objects from smallest to largest.

Solution

1. A \(1\,\text{mm}\times 1\,\text{mm}\) graph-paper square has area \(1\,\text{mm}^2\). 2. A keyboard key is about a square centimeter, so \(1.5\,\text{cm}^2\) is reasonable. 3. A classroom whiteboard can have an area of about \(4\,\text{m}^2\). 4. A suburban home lot can reasonably be about \(6000\,\text{ft}^2\). 5. Large farms are commonly measured in acres, so \(500\) acres is reasonable. 6. A state is measured in square miles, and \(24{,}000\,\text{mi}^2\) is the appropriate scale for West Virginia.

Answer

1. \(1\,\text{mm}^2\) 2. \(1.5\,\text{cm}^2\) 3. \(4\,\text{m}^2\) 4. \(6000\,\text{ft}^2\) 5. \(500\) acres 6. \(24{,}000\,\text{mi}^2\)
5214095
Write each mixed area measurement in the unit shown in parentheses. a) \(2\,\text{yd}^2\ 9\,\text{ft}^2\) \((\text{yd}^2)\) b) \(1\,\text{m}^2\ 20\,\text{cm}^2\) \((\text{cm}^2)\)

Hints

- Convert the part that is not already in the requested unit. - Add only after both parts use the same unit.

Solution

1. Since \(9\,\text{ft}^2=1\,\text{yd}^2\), \(2\,\text{yd}^2\ 9\,\text{ft}^2=3\,\text{yd}^2\). 2. Since \(1\,\text{m}^2=10{,}000\,\text{cm}^2\), adding \(20\,\text{cm}^2\) gives \(10{,}020\,\text{cm}^2\).

Answer

a) \(3\,\text{yd}^2\) b) \(10{,}020\,\text{cm}^2\)
5214165
Convert each area measurement to the requested unit. a) \(12\,\text{m}^2\) to square centimeters b) \(0.5\,\text{m}^2\) to square centimeters c) \(7\,\text{yd}^2\) to square feet d) \(13.4\,\text{cm}^2\) to square millimeters

Hints

- Use the area conversion factor for each pair of units. - Converting to a smaller unit makes the numerical value larger. - Use \(1\,\text{yd}^2=9\,\text{ft}^2\).

Solution

1. \(12\times 10{,}000=120{,}000\,\text{cm}^2\). 2. \(0.5\times 10{,}000=5000\,\text{cm}^2\). 3. \(7\times 9=63\,\text{ft}^2\). 4. \(13.4\times 100=1340\,\text{mm}^2\).

Answer

a) \(120{,}000\,\text{cm}^2\) b) \(5000\,\text{cm}^2\) c) \(63\,\text{ft}^2\) d) \(1340\,\text{mm}^2\)
5214185
Compare each pair of area measurements. Insert \(<\), \(>\), or \(=\). a) \(5\,\text{m}^2\ \underline{\hspace{0.6cm}}\ 45{,}000\,\text{cm}^2\) b) \(0.2\,\text{yd}^2\ \underline{\hspace{0.6cm}}\ 1.8\,\text{ft}^2\) c) \(850\,\text{mm}^2\ \underline{\hspace{0.6cm}}\ 8.5\,\text{cm}^2\) d) \(1\,\text{ft}^2\ \underline{\hspace{0.6cm}}\ 150\,\text{in.}^2\)

Hints

- Convert each pair to a common unit before comparing. - Use area conversion factors. - Check whether the converted numerical values are equal.

Solution

1. \(5\,\text{m}^2=50{,}000\,\text{cm}^2\), so \(5\,\text{m}^2>45{,}000\,\text{cm}^2\). 2. \(0.2\times 9=1.8\,\text{ft}^2\), so the measurements are equal. 3. \(8.5\,\text{cm}^2=850\,\text{mm}^2\), so the measurements are equal. 4. \(1\,\text{ft}^2=144\,\text{in.}^2\), so \(1\,\text{ft}^2<150\,\text{in.}^2\).

Answer

a) \(>\) b) \(=\) c) \(=\) d) \(<\)
5214325
Convert each area measurement to the requested unit. a) \(1400\,\text{mm}^2\) to \(\text{cm}^2\) b) \(252\,\text{ft}^2\) to \(\text{yd}^2\) c) \(8000\,\text{cm}^2\) to \(\text{m}^2\) d) \(720\,\text{in.}^2\) to \(\text{ft}^2\)

Hints

- Identify the area conversion factor for each part. - Converting to a larger unit requires division. - Record each result with the requested square unit.

Solution

1. \(1400\div 100=14\,\text{cm}^2\). 2. \(252\div 9=28\,\text{yd}^2\). 3. \(8000\div 10{,}000=0.8\,\text{m}^2\). 4. \(720\div 144=5\,\text{ft}^2\).

Answer

a) \(14\,\text{cm}^2\) b) \(28\,\text{yd}^2\) c) \(0.8\,\text{m}^2\) d) \(5\,\text{ft}^2\)
5214395
Convert each area measurement to the unit in parentheses. a) \(450\,\text{ft}^2\) \((\text{in.}^2)\) b) \(6\,\text{yd}^2\) \((\text{ft}^2)\) c) \(288\,\text{ft}^2\) \((\text{yd}^2)\)

Hints

- Decide whether the target unit is larger or smaller. - Use \(1\,\text{ft}^2=144\,\text{in.}^2\) and \(1\,\text{yd}^2=9\,\text{ft}^2\).

Solution

1. \(450\times 144=64{,}800\,\text{in.}^2\). 2. \(6\times 9=54\,\text{ft}^2\). 3. \(288\div 9=32\,\text{yd}^2\).

Answer

a) \(64{,}800\,\text{in.}^2\) b) \(54\,\text{ft}^2\) c) \(32\,\text{yd}^2\)
5214655
Calculate each result and write it in the smaller unit used in the expression. a) \(7\,\text{yd}^2+15\,\text{ft}^2\) b) \(12\,\text{ft}^2-400\,\text{in.}^2\) c) \(1\,\text{yd}^2+50\,\text{ft}^2\)

Hints

- Convert the larger unit to the smaller unit first. - Then add or subtract. - Use \(1\,\text{yd}^2=9\,\text{ft}^2\) and \(1\,\text{ft}^2=144\,\text{in.}^2\).

Solution

1. \(7\,\text{yd}^2=63\,\text{ft}^2\), so \(63+15=78\,\text{ft}^2\). 2. \(12\,\text{ft}^2=1728\,\text{in.}^2\), so \(1728-400=1328\,\text{in.}^2\). 3. \(1\,\text{yd}^2=9\,\text{ft}^2\), so \(9+50=59\,\text{ft}^2\).

Answer

a) \(78\,\text{ft}^2\) b) \(1328\,\text{in.}^2\) c) \(59\,\text{ft}^2\)
5214685
Convert each area measurement to square feet. a) \(7\,\text{yd}^2\ 12\,\text{ft}^2\) b) \(432\,\text{in.}^2\) c) \(2\,\text{yd}^2\ 5\,\text{ft}^2\) d) \(10\,\text{yd}^2\) e) \(6480\,\text{in.}^2\)

Hints

- Use \(1\,\text{yd}^2=9\,\text{ft}^2\). - Use \(1\,\text{ft}^2=144\,\text{in.}^2\). - Add mixed-unit parts after converting.

Solution

1. \(7\times 9+12=75\,\text{ft}^2\). 2. \(432\div 144=3\,\text{ft}^2\). 3. \(2\times 9+5=23\,\text{ft}^2\). 4. \(10\times 9=90\,\text{ft}^2\). 5. \(6480\div 144=45\,\text{ft}^2\).

Answer

a) \(75\,\text{ft}^2\) b) \(3\,\text{ft}^2\) c) \(23\,\text{ft}^2\) d) \(90\,\text{ft}^2\) e) \(45\,\text{ft}^2\)
5216735
Complete each statement about area-unit conversion. a) To convert \(1800\,\text{mm}^2\) to square centimeters, divide the numerical value by _____. The result is _____ \(\text{cm}^2\). b) To convert \(0.45\,\text{m}^2\) to square centimeters, multiply the numerical value by _____. The result is _____ \(\text{cm}^2\).

Hints

- Write the area conversion factor for each pair of units. - A larger target unit gives a smaller numerical value. - A smaller target unit gives a larger numerical value.

Solution

1. Since \(1\,\text{cm}^2=100\,\text{mm}^2\), divide by \(100\): \(1800\div 100=18\). 2. Since \(1\,\text{m}^2=10{,}000\,\text{cm}^2\), multiply by \(10{,}000\): \(0.45\times 10{,}000=4500\).

Answer

a) divide by \(100\); \(18\,\text{cm}^2\) b) multiply by \(10{,}000\); \(4500\,\text{cm}^2\)
5216755
A square cloth has an area of \(0.25\,\text{m}^2\). Find its area in square centimeters. Describe the conversion factor you use.

Hints

- Square the linear conversion factor from meters to centimeters. - Multiply the given area by the resulting area conversion factor.

Solution

1. Since \(1\,\text{m}=100\,\text{cm}\), \(1\,\text{m}^2=100\times 100=10{,}000\,\text{cm}^2\). 2. Multiply by \(10{,}000\): \(0.25\times 10{,}000=2500\,\text{cm}^2\).

Answer

\(2500\,\text{cm}^2\); multiply by \(10{,}000\).
5216825
Which area is greatest? Convert all three measurements to square inches before comparing. A: \(5\,\text{ft}^2\ 4\,\text{in.}^2\) B: \(540\,\text{in.}^2\) C: \(5\,\text{ft}^2\ 40\,\text{in.}^2\)

Hints

- Convert each mixed measurement to square inches. - Use \(1\,\text{ft}^2=144\,\text{in.}^2\). - Compare the converted values.

Solution

1. \(A=5\times 144+4=724\,\text{in.}^2\). 2. \(B=540\,\text{in.}^2\). 3. \(C=5\times 144+40=760\,\text{in.}^2\). 4. Since \(760>724>540\), measurement C is greatest.

Answer

C: \(5\,\text{ft}^2\ 40\,\text{in.}^2\)
5217785
Convert each measurement to the unit in parentheses. a) \(34\,\text{m}\) (centimeters) b) \(8000\,\text{g}\) (kilograms) c) \(4000\,\text{g}\) (kilograms) d) \(120\,\text{min}\) (hours) e) \(70\,\text{cm}\) (meters)

Hints

- Decide whether the target unit is larger or smaller. - Identify the conversion factor for each pair of units. - Time conversions may use \(60\) rather than a power of ten.

Solution

1. \(34 \times 100=3400\), so \(34\,\text{m}=3400\,\text{cm}\). 2. \(8000 \div 1000=8\), so \(8000\,\text{g}=8\,\text{kg}\). 3. \(4000 \div 1000=4\), so \(4000\,\text{g}=4\,\text{kg}\). 4. \(120 \div 60=2\), so \(120\,\text{min}=2\,\text{h}\). 5. \(70 \div 100=0.7\), so \(70\,\text{cm}=0.7\,\text{m}\).

Answer

a) \(3400\,\text{cm}\) b) \(8\,\text{kg}\) c) \(4\,\text{kg}\) d) \(2\,\text{h}\) e) \(0.7\,\text{m}\)
5217815
Convert each fractional measurement to the smaller unit. a) How many grams are in one-half kilogram? b) How many centimeters are in one-fifth meter? c) How many grams are in one-tenth kilogram?

Hints

- Identify how many smaller units make one larger unit. - Divide the full amount by the denominator of the fraction. - Use \(1000\,\text{g}=1\,\text{kg}\) and \(100\,\text{cm}=1\,\text{m}\).

Solution

1. \(1\,\text{kg}=1000\,\text{g}\), and \(1000 \div 2=500\), so one-half kilogram is \(500\,\text{g}\). 2. \(1\,\text{m}=100\,\text{cm}\), and \(100 \div 5=20\), so one-fifth meter is \(20\,\text{cm}\). 3. \(1\,\text{kg}=1000\,\text{g}\), and \(1000 \div 10=100\), so one-tenth kilogram is \(100\,\text{g}\).

Answer

a) \(500\,\text{g}\) b) \(20\,\text{cm}\) c) \(100\,\text{g}\)
5217825
Convert each fractional amount to the smaller unit. a) How many minutes are in one-tenth hour? b) How many cents are in one-half dollar? c) How many seconds are in one-fourth minute?

Hints

- Identify the number of smaller units in one whole larger unit. - Divide that amount by the denominator of the fraction. - Use \(60\) for time conversions and \(100\) cents for one dollar.

Solution

1. \(1\,\text{h}=60\,\text{min}\), and \(60 \div 10=6\), so one-tenth hour is \(6\,\text{min}\). 2. One dollar is \(100\) cents, and \(100 \div 2=50\), so one-half dollar is \(50\) cents. 3. \(1\,\text{min}=60\,\text{s}\), and \(60 \div 4=15\), so one-fourth minute is \(15\,\text{s}\).

Answer

a) \(6\,\text{min}\) b) \(50\) cents c) \(15\,\text{s}\)
5217875
Complete the table for each money amount. <table> <thead><tr><th>Decimal dollars</th><th>Dollars and cents</th><th>Cents only</th></tr></thead> <tbody> <tr><td>\(\$15.24\)</td><td>...</td><td>...</td></tr> <tr><td>...</td><td>\(\$6\) and \(2\) cents</td><td>...</td></tr> <tr><td>...</td><td>...</td><td>\(90\) cents</td></tr> <tr><td>\(\$0.05\)</td><td>...</td><td>...</td></tr> </tbody> </table>

Hints

- One dollar equals \(100\) cents. - Dollar amounts use two decimal places for cents. - A placeholder zero is needed for amounts less than \(10\) cents.

Solution

1. \(\$15.24\) is \(\$15\) and \(24\) cents, or \(1524\) cents. 2. \(\$6\) and \(2\) cents is \(\$6.02\), or \(602\) cents. 3. \(90\) cents is \(\$0.90\), or \(\$0\) and \(90\) cents. 4. \(\$0.05\) is \(\$0\) and \(5\) cents, or \(5\) cents.

Answer

<table> <thead><tr><th>Decimal dollars</th><th>Dollars and cents</th><th>Cents only</th></tr></thead> <tbody> <tr><td>\(\$15.24\)</td><td>\(\$15\) and \(24\) cents</td><td>\(1524\) cents</td></tr> <tr><td>\(\$6.02\)</td><td>\(\$6\) and \(2\) cents</td><td>\(602\) cents</td></tr> <tr><td>\(\$0.90\)</td><td>\(\$0\) and \(90\) cents</td><td>\(90\) cents</td></tr> <tr><td>\(\$0.05\)</td><td>\(\$0\) and \(5\) cents</td><td>\(5\) cents</td></tr> </tbody> </table>
5217885
Fill in the blanks so each chain is correct. a) \(\$24.08=\$\square\) and \(\square\) cents \(=\square\) cents b) \(\$\square=\$3\) and \(70\) cents \(=\square\) cents c) \(\$\square=\$\square\) and \(\square\) cents \(=1005\) cents d) \(\$0.40=\$\square\) and \(\square\) cents \(=\square\) cents

Hints

- Cents occupy two decimal places in a dollar amount. - Divide a cents-only amount by \(100\) to write it in dollars. - Use a zero in the hundredths place when the cents amount is less than \(10\).

Solution

1. \(\$24.08\) is \(\$24\) and \(8\) cents. In cents, \(24 \times 100+8=2408\). 2. \(\$3\) and \(70\) cents is \(\$3.70\), or \(370\) cents. 3. \(1005\) cents is \(\$10\) and \(5\) cents, or \(\$10.05\). 4. \(\$0.40\) is \(\$0\) and \(40\) cents, or \(40\) cents.

Answer

a) \(\$24.08=\$24\) and \(8\) cents \(=2408\) cents b) \(\$3.70=\$3\) and \(70\) cents \(=370\) cents c) \(\$10.05=\$10\) and \(5\) cents \(=1005\) cents d) \(\$0.40=\$0\) and \(40\) cents \(=40\) cents
5102585
Evaluate each expression and give the result in the smaller unit. a) \(\frac{3}{10}\,\text{km}+450\,\text{m}\) b) \(1\,\text{h}-\frac{1}{12}\,\text{h}\) c) \(\frac{1}{4}\,\text{t}-120\,\text{kg}\) d) \(\frac{7}{20}\,\text{L}+150\,\text{mL}\)

Hints

- Convert every measurement to the smaller unit first. - Add or subtract only after the units match. - Recall that \(1\,\text{km}=1000\,\text{m}\) and \(1\,\text{L}=1000\,\text{mL}\).

Solution

1. For a), \(\frac{3}{10}\,\text{km}=300\,\text{m}\), so \(300\,\text{m}+450\,\text{m}=750\,\text{m}\). 2. For b), \(1\,\text{h}=60\,\text{min}\) and \(\frac{1}{12}\,\text{h}=5\,\text{min}\), so \(60-5=55\,\text{min}\). 3. For c), \(\frac{1}{4}\,\text{t}=250\,\text{kg}\), so \(250-120=130\,\text{kg}\). 4. For d), \(\frac{7}{20}\,\text{L}=350\,\text{mL}\), so \(350+150=500\,\text{mL}\).

Answer

a) \(750\,\text{m}\) b) \(55\,\text{min}\) c) \(130\,\text{kg}\) d) \(500\,\text{mL}\)
5102605
Convert each measurement to the specified larger unit. Write each result as a fraction in simplest form and as a decimal. a) \(750\,\text{mL}\) to liters b) \(12\,\text{min}\) to hours c) \(50\,\text{cm}\) to meters d) \(1250\,\text{g}\) to kilograms

Hints

- Write each conversion as a fraction first. - Simplify the fraction by dividing the numerator and denominator by a common factor. - Divide the numerator by the denominator to write the decimal.

Solution

1. \(750\,\text{mL}=\frac{750}{1000}\,\text{L}=\frac{3}{4}\,\text{L}=0.75\,\text{L}\). 2. \(12\,\text{min}=\frac{12}{60}\,\text{h}=\frac{1}{5}\,\text{h}=0.2\,\text{h}\). 3. \(50\,\text{cm}=\frac{50}{100}\,\text{m}=\frac{1}{2}\,\text{m}=0.5\,\text{m}\). 4. \(1250\,\text{g}=\frac{1250}{1000}\,\text{kg}=\frac{5}{4}\,\text{kg}=1.25\,\text{kg}\).

Answer

a) \(\frac{3}{4}\,\text{L}\) and \(0.75\,\text{L}\) b) \(\frac{1}{5}\,\text{h}\) and \(0.2\,\text{h}\) c) \(\frac{1}{2}\,\text{m}\) and \(0.5\,\text{m}\) d) \(\frac{5}{4}\,\text{kg}\) and \(1.25\,\text{kg}\)
5102645
Compare the values. Insert \(<\), \(>\), or \(=\). First evaluate each quotient and express both sides in the same unit. a) \(5\,\text{m}\div4\ \square\ 120\,\text{cm}\) b) \(1\,\text{kg}\div8\ \square\ 125\,\text{g}\) c) \(7\,\text{km}\div10\ \square\ 0.75\,\text{km}\)

Hints

- Express both sides in the same unit before comparing. - Recall the decimal value of \(1\div8\). - Compare decimal values by writing the same number of decimal places. - You can also rewrite the right side as a fraction or quotient.

Solution

1. For a), \(5\,\text{m}\div4=1.25\,\text{m}=125\,\text{cm}\). Since \(125>120\), the correct symbol is \(>\). 2. For b), \(1\,\text{kg}\div8=0.125\,\text{kg}=125\,\text{g}\). The values are equal, so the symbol is \(=\). 3. For c), \(7\,\text{km}\div10=0.7\,\text{km}\). Since \(0.7<0.75\), the correct symbol is \(<\).

Answer

a) \(>\) b) \(=\) c) \(<\)
5102695
Evaluate both expressions in square feet and compare the results. Which result is greater? Expression A: \(6\,\text{yd}^2\div8\) Expression B: \(70\,\text{ft}^2\div10\)

Hints

- Remember that \(1\,\text{yd}=3\,\text{ft}\), so \(1\,\text{yd}^2=9\,\text{ft}^2\). - Convert both expressions to square feet before comparing. - Divide after completing the unit conversion.

Solution

1. Convert \(6\,\text{yd}^2\) to square feet. Since \(1\,\text{yd}^2=9\,\text{ft}^2\), \(6\,\text{yd}^2=54\,\text{ft}^2\). Then \(54\div8=6.75\), so Expression A is \(6.75\,\text{ft}^2\). 2. For Expression B, \(70\div10=7\), so the result is \(7\,\text{ft}^2\). 3. Since \(7>6.75\), Expression B is greater.

Answer

Expression B, \(7\,\text{ft}^2\), is greater than Expression A, \(6.75\,\text{ft}^2\).
5102785
Compare each pair of measurements. State which measurement is greater and justify your answer by converting both measurements to the same unit. a) \(\frac{2}{5}\,\text{L}\) or \(450\,\text{mL}\) b) \(18\,\text{min}\) or \(\frac{1}{4}\,\text{hr}\) c) \(\frac{7}{20}\,\text{kg}\) or \(320\,\text{g}\)

Hints

- Convert the larger unit to the smaller unit in each pair. - Multiply the fraction by the number of smaller units in one larger unit. - Once the units match, compare the numerical values.

Solution

1. For a), \(\frac{2}{5}\times1000\,\text{mL}=400\,\text{mL}\). Since \(450\,\text{mL}>400\,\text{mL}\), \(450\,\text{mL}\) is greater. 2. For b), \(\frac{1}{4}\times60\,\text{min}=15\,\text{min}\). Since \(18\,\text{min}>15\,\text{min}\), \(18\,\text{min}\) is greater. 3. For c), \(\frac{7}{20}\times1000\,\text{g}=350\,\text{g}\). Since \(350\,\text{g}>320\,\text{g}\), \(\frac{7}{20}\,\text{kg}\) is greater.

Answer

a) \(450\,\text{mL}\) is greater. b) \(18\,\text{min}\) is greater. c) \(\frac{7}{20}\,\text{kg}\) is greater.
5102795
Calculate each sum. Write each answer as a fraction in simplest form using the larger unit. a) \(25\,\text{cm} + \frac{1}{2}\,\text{m}\) b) \(150\,\text{g} + \frac{3}{4}\,\text{kg}\) c) \(10\,\text{min} + \frac{1}{6}\,\text{h}\)

Hints

- Convert both quantities in each part to the larger unit before adding. - After converting, use a common denominator when you add the fractions. - Simplify each final fraction.

Solution

1. For a), convert \(25\,\text{cm}\) to meters: \(25\,\text{cm} = \frac{25}{100}\,\text{m} = \frac{1}{4}\,\text{m}\). Then \(\frac{1}{4}\,\text{m} + \frac{1}{2}\,\text{m} = \frac{3}{4}\,\text{m}\). 2. For b), convert \(150\,\text{g}\) to kilograms: \(150\,\text{g} = \frac{150}{1000}\,\text{kg} = \frac{3}{20}\,\text{kg}\). Then \(\frac{3}{20}\,\text{kg} + \frac{3}{4}\,\text{kg} = \frac{18}{20}\,\text{kg} = \frac{9}{10}\,\text{kg}\). 3. For c), convert \(10\,\text{min}\) to hours: \(10\,\text{min} = \frac{10}{60}\,\text{h} = \frac{1}{6}\,\text{h}\). Then \(\frac{1}{6}\,\text{h} + \frac{1}{6}\,\text{h} = \frac{1}{3}\,\text{h}\).

Answer

a) \(\frac{3}{4}\,\text{m}\) b) \(\frac{9}{10}\,\text{kg}\) c) \(\frac{1}{3}\,\text{h}\)
5104725
Convert each measurement to the unit in parentheses. Then round to the nearest whole unit. a) \(0.0675\,\text{km}\) (meters) b) \(0.0184\,\text{km}\) (meters) c) \(0.0456\,\text{kg}\) (grams) d) \(0.00125\,\text{L}\) (milliliters)

Hints

- Convert first by multiplying by the correct power of \(10\). - Then inspect the tenths digit. - Round down for tenths digits \(0\) through \(4\), and round up for tenths digits \(5\) through \(9\).

Solution

1. \(0.0675 \times 1000 = 67.5\,\text{m}\), which rounds to \(68\,\text{m}\). 2. \(0.0184 \times 1000 = 18.4\,\text{m}\), which rounds to \(18\,\text{m}\). 3. \(0.0456 \times 1000 = 45.6\,\text{g}\), which rounds to \(46\,\text{g}\). 4. \(0.00125 \times 1000 = 1.25\,\text{mL}\), which rounds to \(1\,\text{mL}\).

Answer

a) \(68\,\text{m}\) b) \(18\,\text{m}\) c) \(46\,\text{g}\) d) \(1\,\text{mL}\)
5104805
Two packages are weighed. Package A weighs \(39\,\text{oz}\), and Package B weighs \(2.55\,\text{lb}\). a) What is each weight rounded to the nearest whole pound? b) Are the rounded weights different? Explain why this happens even though the actual weights are close.

Hints

- Convert both weights to pounds before comparing or rounding. - Find the midpoint between \(2\,\text{lb}\) and \(3\,\text{lb}\). - Consider what happens to values just below and just above that midpoint.

Solution

1. Convert Package A to pounds: \(39 \div 16 = 2.4375\,\text{lb}\). 2. Since \(2.4375 < 2.5\), Package A rounds to \(2\,\text{lb}\). 3. Since \(2.55 > 2.5\), Package B rounds to \(3\,\text{lb}\). 4. The rounded values differ because the actual weights lie on opposite sides of the rounding midpoint, \(2.5\,\text{lb}\).

Answer

a) Package A: \(2\,\text{lb}\); Package B: \(3\,\text{lb}\) b) Yes. Package A is below \(2.5\,\text{lb}\), while Package B is above \(2.5\,\text{lb}\), so they round in opposite directions.
5105385
Three friends compare their routes to school. Lucas walks \(\frac{3}{4}\) mile, Mia walks \(1200\) yards, and Ben walks \(\frac{4}{5}\) mile. Order the routes from shortest to longest.

Hints

- Express all three distances in the same unit. - Recall how many yards are in one mile. - Multiply the whole distance by each fraction.

Solution

1. Convert the mile distances to yards using \(1\,\text{mile}=1760\,\text{yd}\). 2. Lucas walks \(\frac{3}{4}\times1760=1320\,\text{yd}\). 3. Ben walks \(\frac{4}{5}\times1760=1408\,\text{yd}\). 4. Compare: \(1200<1320<1408\). Therefore, the order is Mia, Lucas, Ben.

Answer

Shortest to longest: Mia \((1200\,\text{yd})\), Lucas \((1320\,\text{yd})\), Ben \((1408\,\text{yd})\).
5105395
A cook needs \(\frac{3}{8}\) quart of cream for a soup recipe. The carton contains \(2\) cups of cream. How many fluid ounces of cream will remain after the recipe is made?

Hints

- Convert both amounts to fluid ounces. - Recall the number of fluid ounces in a quart and in a cup. - Subtract the amount used from the amount in the carton.

Solution

1. Convert both amounts to fluid ounces. One quart is \(32\) fluid ounces, so \(\frac{3}{8}\times32=12\) fluid ounces are needed. 2. Two cups equal \(16\) fluid ounces. 3. Subtract: \(16-12=4\). Therefore, \(4\) fluid ounces remain.

Answer

\(4\) fluid ounces remain.
5105435
Find what fraction the combined area of \(100\,\text{yd}^2\) and \(140\,\text{yd}^2\) is of a total area of \(3600\,\text{ft}^2\). Write the fraction in simplest form.

Hints

- Add the two smaller areas first. - Express all areas in the same unit. - Simplify the part-to-whole fraction.

Solution

1. Add the two areas: \(100+140=240\,\text{yd}^2\). 2. Convert the total area to square yards. Since \(1\,\text{yd}^2=9\,\text{ft}^2\), \(3600\,\text{ft}^2=400\,\text{yd}^2\). 3. The fraction is \(\frac{240}{400}=\frac{3}{5}\).

Answer

\(\frac{3}{5}\)
5105565
Four packages have weights written in different forms. - Package A: \(40\,\text{oz}\) - Package B: \(20\,\text{lb}\) - Package C: \(2.05\,\text{lb}\) - Package D: \(2\,\text{lb}\ 0.8\,\text{oz}\) a) Convert every weight to pounds. b) Order the packages from lightest to heaviest.

Hints

- Recall that \(16\,\text{oz}=1\,\text{lb}\). - Convert every measurement to pounds before comparing. - Write decimals to the same number of places when that helps you compare them.

Solution

1. Package A: \(40 \div 16 = 2.5\,\text{lb}\). 2. Package B is already \(20\,\text{lb}\). 3. Package C is already \(2.05\,\text{lb}\). 4. Package D: \(0.8 \div 16 = 0.05\,\text{lb}\), so \(2\,\text{lb}\ 0.8\,\text{oz}=2.05\,\text{lb}\). 5. Compare: \(2.05 < 2.5 < 20\). Packages C and D have equal weights.

Answer

a) Package A: \(2.5\,\text{lb}\); Package B: \(20\,\text{lb}\); Package C: \(2.05\,\text{lb}\); Package D: \(2.05\,\text{lb}\) b) Package C and Package D are tied for lightest, followed by Package A, then Package B.
5105575
Four students compare the weights of their lunch bags. - Alex: \(3.45\,\text{lb}\) - Mia: \(48.8\,\text{oz}\) - Jordan: \(3.6\,\text{lb}\) - Sam: \(3\frac{1}{2}\,\text{lb}\) Order the lunch bags from heaviest to lightest. Briefly explain why Mia’s bag is not the heaviest even though \(48.8\) is the largest numerical value shown.

Hints

- Convert \(3\frac{1}{2}\) to a decimal. - Use \(16\,\text{oz}=1\,\text{lb}\). - Compare the ones, tenths, and hundredths digits after all values use the same unit.

Solution

1. Convert every weight to pounds. Alex has \(3.45\,\text{lb}\). 2. Mia: \(48.8 \div 16 = 3.05\,\text{lb}\). 3. Jordan has \(3.6\,\text{lb}\). 4. Sam: \(3\frac{1}{2}\,\text{lb}=3.5\,\text{lb}\). 5. Compare: \(3.6 > 3.5 > 3.45 > 3.05\). The numerical value depends on the size of the unit, so \(48.8\,\text{oz}\) is only \(3.05\,\text{lb}\).

Answer

Jordan, Sam, Alex, Mia. Mia’s bag weighs \(3.05\,\text{lb}\); ounces are smaller than pounds, so the larger numeral does not mean the larger weight.
5105585
A construction site receives these materials: - Sand: \(0.8\,\text{ton}\) - Gravel: \(1700\,\text{lb}\) - Cement: \(0.09\,\text{ton}\) - Water: \(240\,\text{lb}\) a) Find the total weight in tons. b) Which material has the second-greatest weight? Use \(1\,\text{ton} = 2000\,\text{lb}\).

Hints

- Express all weights in the same unit. - Divide pounds by \(2000\) to convert to tons. - Add the converted values, then order the individual weights.

Solution

1. Convert each weight to tons: sand is \(0.8\,\text{ton}\); gravel is \(1700 \div 2000 = 0.85\,\text{ton}\); cement is \(0.09\,\text{ton}\); and water is \(240 \div 2000 = 0.12\,\text{ton}\). 2. Add: \(0.8 + 0.85 + 0.09 + 0.12 = 1.86\), so the total is \(1.86\,\text{tons}\). 3. Order the weights: \(0.85 > 0.8 > 0.12 > 0.09\). Sand has the second-greatest weight.

Answer

a) \(1.86\,\text{tons}\) b) Sand, at \(0.8\,\text{ton}\)
5106555
A shelf can safely hold at most \(30\,\text{lb}\). Four packages on the shelf weigh \(7 \frac{1}{2}\,\text{lb}\), \(4 \frac{1}{4}\,\text{lb}\), \(9 \frac{3}{4}\,\text{lb}\), and \(3 \frac{1}{2}\,\text{lb}\). Can another package weighing \(72\,\text{oz}\) be added safely? Explain.

Hints

- Add the weights already on the shelf first. - Find how much of the shelf's weight capacity remains. - How many ounces are in one pound? - Compare the new package's weight with the remaining capacity.

Solution

1. Add the four package weights: \(7 \frac{1}{2}+4 \frac{1}{4}+9 \frac{3}{4}+3 \frac{1}{2}=25\), so the shelf currently holds \(25\,\text{lb}\). 2. The remaining capacity is \(30\,\text{lb}-25\,\text{lb}=5\,\text{lb}\). 3. Convert the new package: \(72\,\text{oz}=4 \frac{1}{2}\,\text{lb}\). 4. Since \(4 \frac{1}{2}\,\text{lb}<5\,\text{lb}\), the package can be added safely.

Answer

Yes. The shelf has \(5\,\text{lb}\) of capacity left, and \(72\,\text{oz}=4 \frac{1}{2}\,\text{lb}\).
5106565
A beverage dispenser holds at most \(5 \frac{1}{4}\) gallons. It already contains \(1 \frac{1}{2}\) gallons of orange juice, \(\frac{3}{4}\) gallon of apple juice, \(1 \frac{1}{2}\) gallons of sparkling water, and \(\frac{3}{4}\) gallon of grape juice. Can \(13\) more cups of cherry juice be added without overflowing the dispenser? Explain.

Hints

- Add the amounts already in the dispenser. - Subtract that total from the dispenser's capacity. - How many cups are in one gallon? - Compare the available space with the amount to be added.

Solution

1. Add the amounts already in the dispenser: \(1 \frac{1}{2}+\frac{3}{4}+1 \frac{1}{2}+\frac{3}{4}=4 \frac{1}{2}\) gallons. 2. Find the remaining capacity: \(5 \frac{1}{4}-4 \frac{1}{2}=\frac{3}{4}\) gallon. 3. Convert the remaining capacity to cups. Since \(1\) gallon is \(16\) cups, \(\frac{3}{4}\) gallon is \(12\) cups. 4. Since \(13\) cups is more than \(12\) cups, the cherry juice would overflow the dispenser.

Answer

No. Only \(12\) cups of space remain, so adding \(13\) cups would overflow the dispenser.
5108825
A red blood cell has a diameter of about \(0.0075\,\text{mm}\). A small blood vessel has an inside diameter of about \(0.03\,\text{mm}\). a) Convert both measurements to micrometers \((\mu\text{m})\). Use \(1\,\text{mm} = 1000\,\mu\text{m}\). b) How many times as large is the blood vessel’s diameter as the blood cell’s diameter?

Hints

- Use the given relationship between millimeters and micrometers. - A “how many times as large” comparison can be found with division. - Compare the larger measurement with the smaller one after both use the same unit.

Solution

1. Convert the blood cell diameter: \(0.0075 \times 1000 = 7.5\), so it is \(7.5\,\mu\text{m}\). 2. Convert the blood vessel diameter: \(0.03 \times 1000 = 30\), so it is \(30\,\mu\text{m}\). 3. Find the multiplicative comparison: \(30 \div 7.5 = 4\).

Answer

a) The blood cell diameter is \(7.5\,\mu\text{m}\), and the blood vessel’s inside diameter is \(30\,\mu\text{m}\). b) The blood vessel’s diameter is \(4\) times as large.
5108995
A workshop receives a box containing \(400\) identical metal washers. The full box weighs \(1.58\,\text{kg}\), and the empty box weighs \(140\,\text{g}\). a) Find the weight of one washer in grams. b) How many grams do \(10\) washers weigh?

Hints

- Remove the empty-box weight before finding the weight of the washers. - Put all weights in the same unit before calculating. - Once you know the weight of one washer, scale to \(10\) washers.

Solution

1. Convert the full-box weight: \(1.58\,\text{kg}=1580\,\text{g}\). 2. Subtract the box weight: \(1580-140=1440\,\text{g}\) of washers. 3. Divide by the number of washers: \(1440\div400=3.6\,\text{g}\) per washer. 4. For b), \(3.6\times10=36\,\text{g}\).

Answer

a) \(3.6\,\text{g}\) b) \(36\,\text{g}\)
5109205
A room is \(12.5\) ft long. a) Find the room length in inches. b) A hallway runner is \(12\) ft \(8\) in long. Will it fit lengthwise in the room? Justify your answer by comparing the lengths in the same unit.

Hints

- Use the relationship between feet and inches to convert the room length. - Convert the runner length to the same unit before comparing. - Remember that \(1\,\text{ft}=12\,\text{in}\).

Solution

1. For a), use \(1\,\text{ft}=12\,\text{in}\): \(12.5\times12=150\). The room is \(150\) in long. 2. Convert the runner length to inches: \(12\times12+8=152\), so the runner is \(152\) in long. 3. Since \(152>150\), the runner is longer than the room and will not fit lengthwise.

Answer

a) \(150\) in b) No. The runner is \(152\) in long, which is \(2\) in longer than the \(150\)-in room.
5111315
A rectangular container has a volume of \(2.5\,\text{dm}^3\). a) Express this volume in liters, milliliters, and cubic centimeters. b) What do you notice when you compare the numerical values in liters and cubic decimeters, and in milliliters and cubic centimeters?

Hints

- Recall the relationship between capacity units and cubic volume units. - How many milliliters are in one liter? - What is the conversion factor from cubic decimeters to cubic centimeters?

Solution

1. Since \(1\,\text{dm}^3=1\,\text{L}\), \(2.5\,\text{dm}^3=2.5\,\text{L}\). 2. Since \(1\,\text{L}=1000\,\text{mL}\), \(2.5\,\text{L}=2500\,\text{mL}\). 3. Since \(1\,\text{dm}^3=1000\,\text{cm}^3\), \(2.5\,\text{dm}^3=2500\,\text{cm}^3\). 4. Liters and cubic decimeters have matching numerical values, and milliliters and cubic centimeters have matching numerical values.

Answer

a) \(2.5\,\text{L}\), \(2500\,\text{mL}\), and \(2500\,\text{cm}^3\) b) \(1\,\text{dm}^3=1\,\text{L}\) and \(1\,\text{cm}^3=1\,\text{mL}\), so each pair has the same numerical value.
5111345
Express \(0.045\,\text{m}^3\) first in cubic decimeters and then in cubic centimeters.

Hints

- Think about how many smaller unit cubes fit inside one larger unit cube. - Adjacent cubic metric units differ by a factor of \(1000\). - When the unit becomes smaller, the numerical value becomes larger.

Solution

1. Convert cubic meters to cubic decimeters by multiplying by \(1000\): \(0.045\times1000=45\). Therefore, \(0.045\,\text{m}^3=45\,\text{dm}^3\). 2. Convert cubic decimeters to cubic centimeters by multiplying by \(1000\) again: \(45\times1000=45{,}000\). Therefore, \(45\,\text{dm}^3=45{,}000\,\text{cm}^3\).

Answer

\(45\,\text{dm}^3\) and \(45{,}000\,\text{cm}^3\)
5111365
Find the sum \(V=1.2\,\text{dm}^3+800\,\text{cm}^3\). Give \(V\) in both liters and cubic centimeters.

Hints

- Convert both addends to the same unit first. - Which unit is more convenient for the addition? - Give the final result in both requested units.

Solution

1. Convert \(1.2\,\text{dm}^3\) to cubic centimeters: \(1.2\times1000=1200\,\text{cm}^3\). 2. Add: \(1200\,\text{cm}^3+800\,\text{cm}^3=2000\,\text{cm}^3\). 3. Since \(1000\,\text{cm}^3=1\,\text{L}\), \(2000\,\text{cm}^3=2\,\text{L}\).

Answer

\(V=2\,\text{L}=2000\,\text{cm}^3\)
5111375
A water bottle has a capacity of \(750\,\text{mL}\). a) How many cubic centimeters of water fit in the bottle? b) Assume that \(1\,\text{cm}^3\) of water has a mass of exactly \(1\,\text{g}\). Find the mass of the water in a full bottle in kilograms.

Hints

- How are milliliters and cubic centimeters related? - How many grams are in one kilogram? - Use the given mass of one cubic centimeter of water.

Solution

1. Since \(1\,\text{mL}=1\,\text{cm}^3\), the bottle holds \(750\,\text{cm}^3\) of water. 2. At \(1\,\text{g}\) per cubic centimeter, \(750\,\text{cm}^3\) of water has a mass of \(750\,\text{g}\). 3. Since \(1000\,\text{g}=1\,\text{kg}\), \(750\,\text{g}=0.75\,\text{kg}\).

Answer

a) \(750\,\text{cm}^3\) b) \(0.75\,\text{kg}\)
5112335
Four of these five measurements represent the same volume. One is different. \(2.5\,\text{dm}^3\); \(2500\,\text{mL}\); \(250\,\text{cm}^3\); \(0.0025\,\text{m}^3\); \(2500\,\text{cm}^3\) Identify the different measurement. Justify your answer by converting every value to cubic centimeters.

Hints

- How are milliliters and cubic centimeters related? - Recall the factors used to convert cubic metric units. - Convert every measurement to one common unit.

Solution

1. \(2.5\,\text{dm}^3=2500\,\text{cm}^3\). 2. \(2500\,\text{mL}=2500\,\text{cm}^3\). 3. \(250\,\text{cm}^3\) remains \(250\,\text{cm}^3\). 4. \(0.0025\,\text{m}^3=2500\,\text{cm}^3\). 5. The final value is already \(2500\,\text{cm}^3\). Four values equal \(2500\,\text{cm}^3\), but one equals \(250\,\text{cm}^3\).

Answer

\(250\,\text{cm}^3\) is different. Each of the other measurements equals \(2500\,\text{cm}^3\).
5113755
A relay course has six sections with these lengths: \(1.45\,\text{mi}\), \(\frac{3}{4}\,\text{mi}\), \(2.55\,\text{mi}\), \(0.25\,\text{mi}\), \(1\frac{1}{2}\,\text{mi}\), and \(880\,\text{yd}\). Find the total length. Group the values strategically and explain which values you combined.

Hints

- Convert the yard measurement to miles. - Convert the fractions to decimals. - Look for pairs that add to whole numbers.

Solution

1. Convert every length to miles. Since \(1760\,\text{yd}=1\,\text{mi}\), \(880\,\text{yd}=0.5\,\text{mi}\). Also, \(\frac{3}{4}\,\text{mi}=0.75\,\text{mi}\) and \(1\frac{1}{2}\,\text{mi}=1.5\,\text{mi}\). 2. Form convenient pairs: \(1.45+2.55=4\), \(0.75+0.25=1\), and \(1.5+0.5=2\). 3. Add the pair sums: \(4+1+2=7\).

Answer

The total length is \(7\,\text{mi}\). One efficient grouping is \((1.45+2.55)+(0.75+0.25)+(1.5+0.5)\).
5116425
Convert each measurement to the smaller unit. a) \(\frac{7}{12}\,\text{yd}\) to inches b) \(\frac{11}{16}\,\text{ft}^2\) to square inches c) \(\frac{17}{32}\,\text{gal}\) to fluid ounces d) \(\frac{5}{8}\,\text{lb}\) to ounces

Hints

- Use the correct conversion factor for each pair of units. - For square units, square the linear conversion factor. - Divide the conversion factor by the denominator before multiplying by the numerator when convenient.

Solution

1. Since \(1\,\text{yd}=36\,\text{in}\), \(\frac{7}{12}\times36=21\), so the result is \(21\,\text{in}\). 2. Since \(1\,\text{ft}^2=144\,\text{in}^2\), \(\frac{11}{16}\times144=99\), so the result is \(99\,\text{in}^2\). 3. Since \(1\,\text{gal}=128\) fluid ounces, \(\frac{17}{32}\times128=68\), so the result is \(68\) fluid ounces. 4. Since \(1\,\text{lb}=16\,\text{oz}\), \(\frac{5}{8}\times16=10\), so the result is \(10\,\text{oz}\).

Answer

a) \(21\,\text{in}\) b) \(99\,\text{in}^2\) c) \(68\) fluid ounces d) \(10\,\text{oz}\)
5116435
Compare the measurements. Insert \(<\), \(>\), or \(=\). a) \(\frac{3}{8}\,\text{km}\ \square\ 370\,\text{m}\) b) \(\frac{4}{5}\,\text{L}\ \square\ 800\,\text{mL}\) c) \(\frac{7}{16}\,\text{ft}^2\ \square\ 64\,\text{in}^2\) d) \(\frac{13}{20}\,\text{kg}\ \square\ 640\,\text{g}\)

Hints

- Convert both measurements in each comparison to the same unit. - It is often easiest to convert the fractional measurement to the smaller unit. - For square units, remember that \(1\,\text{ft}^2=144\,\text{in}^2\).

Solution

1. For a), \(\frac{3}{8}\times1000=375\), so \(\frac{3}{8}\,\text{km}=375\,\text{m}>370\,\text{m}\). 2. For b), \(\frac{4}{5}\times1000=800\), so \(\frac{4}{5}\,\text{L}=800\,\text{mL}\). 3. For c), \(1\,\text{ft}^2=144\,\text{in}^2\), and \(\frac{7}{16}\times144=63\). Thus, \(63\,\text{in}^2<64\,\text{in}^2\). 4. For d), \(\frac{13}{20}\times1000=650\), so \(\frac{13}{20}\,\text{kg}=650\,\text{g}>640\,\text{g}\).

Answer

a) \(>\) b) \(=\) c) \(<\) d) \(>\)
5116785
A case contains \(12\) identical juice pouches. The full case weighs \(2.7\,\text{kg}\), and the empty cardboard case weighs \(60\,\text{g}\). Find the weight of one juice pouch in grams.

Hints

- Put all weights in the same unit first. - Subtract the packaging weight before dividing. - Divide the net weight equally among the \(12\) pouches.

Solution

1. Convert the total weight: \(2.7\,\text{kg}=2700\,\text{g}\). 2. Subtract the empty case: \(2700-60=2640\,\text{g}\). 3. Divide by \(12\): \(2640\div12=220\,\text{g}\).

Answer

\(220\,\text{g}\)
5117705
Convert each area measurement to the requested unit. a) \(0.8\,\text{m}^2\) to square centimeters b) \(12{,}500\,\text{mm}^2\) to square centimeters c) \(5\,\text{yd}^2\) to square feet d) \(3400\,\text{cm}^2\) to square meters

Hints

- Use the area conversion factor, not the linear conversion factor. - Decide whether to multiply or divide based on whether the target unit is smaller or larger. - Remember that \(1\,\text{yd}^2=9\,\text{ft}^2\).

Solution

1. Since \(1\,\text{m}^2=10{,}000\,\text{cm}^2\), \(0.8\times 10{,}000=8000\,\text{cm}^2\). 2. Since \(1\,\text{cm}^2=100\,\text{mm}^2\), \(12{,}500\div 100=125\,\text{cm}^2\). 3. Since \(1\,\text{yd}^2=9\,\text{ft}^2\), \(5\times 9=45\,\text{ft}^2\). 4. Since \(1\,\text{m}^2=10{,}000\,\text{cm}^2\), \(3400\div 10{,}000=0.34\,\text{m}^2\).

Answer

a) \(8000\,\text{cm}^2\) b) \(125\,\text{cm}^2\) c) \(45\,\text{ft}^2\) d) \(0.34\,\text{m}^2\)
5117715
Order the area measurements from least to greatest: \(4500\,\text{in.}^2\), \(30\,\text{ft}^2\), \(31\,\text{ft}^2\), and \(4400\,\text{in.}^2\).

Hints

- Convert all four measurements to the same unit. - Use \(1\,\text{ft}^2=144\,\text{in.}^2\). - Then compare the numerical values.

Solution

1. Convert the square-foot measurements to square inches using \(1\,\text{ft}^2=144\,\text{in.}^2\). 2. \(30\times 144=4320\,\text{in.}^2\), and \(31\times 144=4464\,\text{in.}^2\). 3. Compare: \(4320<4400<4464<4500\).

Answer

\(30\,\text{ft}^2<4400\,\text{in.}^2<31\,\text{ft}^2<4500\,\text{in.}^2\)
5117725
Complete each equation with the missing number or unit. a) \(0.75\,\text{m}^2=7500\,\underline{\hspace{1cm}}\) b) \(252\,\text{ft}^2=\underline{\hspace{1cm}}\,\text{yd}^2\) c) \(12\,\text{cm}^2=1200\,\underline{\hspace{1cm}}\) d) \(\underline{\hspace{1cm}}\,\text{in.}^2=5\,\text{ft}^2\)

Hints

- Identify the conversion factor for each pair of area units. - Square units use squared conversion factors. - Check whether the missing value should be larger or smaller than the given value.

Solution

1. Since \(1\,\text{m}^2=10{,}000\,\text{cm}^2\), \(0.75\,\text{m}^2=7500\,\text{cm}^2\). 2. Since \(1\,\text{yd}^2=9\,\text{ft}^2\), \(252\div 9=28\,\text{yd}^2\). 3. Since \(1\,\text{cm}^2=100\,\text{mm}^2\), \(12\,\text{cm}^2=1200\,\text{mm}^2\). 4. Since \(1\,\text{ft}^2=144\,\text{in.}^2\), \(5\times 144=720\,\text{in.}^2\).

Answer

a) \(\text{cm}^2\) b) \(28\) c) \(\text{mm}^2\) d) \(720\)
5164475
A bakery orders flour in bags that each have a mass of \(25\,\text{kg}\). a) How many bags have a total mass of exactly \(1\) metric ton? b) What is the total mass of \(10\) bags in grams?

Hints

- Convert \(1\) metric ton to kilograms. - Divide the total kilograms by the mass of one bag. - For part b, find the kilograms first and then convert to grams.

Solution

1. One metric ton is \(1000\,\text{kg}\). Divide to find the number of bags: \(1000\,\text{kg} \div 25\,\text{kg} = 40\). 2. Ten bags have a mass of \(10 \times 25\,\text{kg} = 250\,\text{kg}\). 3. Convert to grams by multiplying \(250\,\text{kg}\) by \(1000\,\frac{\text{g}}{\text{kg}}\). The total mass is \(250{,}000\,\text{g}\).

Answer

a) \(40\) bags b) \(250{,}000\,\text{g}\)
5166995
Compare the capacities. Insert \(<\), \(>\), or \(=\). a) \(0.5\,\text{L}\ \_\_\_\ 500\,\text{mL}\) b) \(1200\,\text{mL}\ \_\_\_\ 1.5\,\text{L}\) c) \(1\,\text{L}\ 20\,\text{mL}\ \_\_\_\ 1.2\,\text{L}\) d) \(0.3\,\text{L}\ \_\_\_\ 30\,\text{mL}\) e) \(250\,\text{mL}\ \_\_\_\ 0.25\,\text{L}\)

Hints

- Comparisons are easier when both quantities use the same unit. Try converting everything to milliliters. - Pay close attention to the milliliters in the mixed-unit amount. - How many milliliters are in \(0.1\,\text{L}\)? Use that as a benchmark.

Solution

1. Convert both quantities in each comparison to the same unit. 2. a) \(0.5\,\text{L}=500\,\text{mL}\), so the quantities are equal. 3. b) \(1.5\,\text{L}=1500\,\text{mL}\). Since \(1200<1500\), \(1200\,\text{mL}<1.5\,\text{L}\). 4. c) \(1\,\text{L}\ 20\,\text{mL}=1020\,\text{mL}\), and \(1.2\,\text{L}=1200\,\text{mL}\). Since \(1020<1200\), \(1\,\text{L}\ 20\,\text{mL}<1.2\,\text{L}\). 5. d) \(0.3\,\text{L}=300\,\text{mL}\). Since \(300>30\), \(0.3\,\text{L}>30\,\text{mL}\). 6. e) \(0.25\,\text{L}=250\,\text{mL}\), so the quantities are equal.

Answer

a) \(=\) b) \(<\) c) \(<\) d) \(>\) e) \(=\)
5167485
Lucas attends school on \(180\) days each year. His one-way walk to school is \(950\,\text{m}\). How many kilometers does he walk altogether traveling to and from school during the year?

Hints

- Remember that Lucas walks the route twice each school day. - First find his walking distance for one day. - Convert the final distance from meters to kilometers.

Solution

1. Find the round-trip distance for one day: \(950 \times 2 = 1900\,\text{m}\). 2. Find the distance for \(180\) school days: \(1900 \times 180 = 342{,}000\,\text{m}\). 3. Convert meters to kilometers: \(342{,}000\,\text{m} = 342\,\text{km}\).

Answer

Lucas walks \(342\,\text{km}\) altogether.
5168245
Fill in each missing measurement. a) \(\frac{3}{4}\,\text{hr}=\_\_\_\,\text{min}\) b) \(130\,\text{min}=\_\_\_\,\text{hr}\ \_\_\_\,\text{min}\) c) \(1\,\text{hr}\ 25\,\text{min}=\_\_\_\,\text{min}\) d) \(0.5\,\text{L}=\_\_\_\,\text{mL}\) e) \(1.25\,\text{L}=\_\_\_\,\text{mL}\)

Hints

- How many minutes are in \(1\) hour? - How many milliliters are in \(1\) liter? - To convert minutes to hours and minutes, find how many groups of \(60\) fit in the total. - Half a liter is half of \(1000\,\text{mL}\).

Solution

1. a) Three-fourths of \(60\) minutes is \(\frac{3}{4} \times 60=45\) minutes. 2. b) Since \(130=2 \times 60+10\), \(130\,\text{min}=2\,\text{hr}\ 10\,\text{min}\). 3. c) \(1\,\text{hr}\ 25\,\text{min}=60\,\text{min}+25\,\text{min}=85\,\text{min}\). 4. d) Since \(1\,\text{L}=1000\,\text{mL}\), \(0.5\,\text{L}=500\,\text{mL}\). 5. e) \(1.25\,\text{L}=1000\,\text{mL}+250\,\text{mL}=1250\,\text{mL}\).

Answer

a) \(45\,\text{min}\) b) \(2\,\text{hr}\ 10\,\text{min}\) c) \(85\,\text{min}\) d) \(500\,\text{mL}\) e) \(1250\,\text{mL}\)
5168555
Complete the table. <table> <tr> <td>Fraction form</td> <td>Mass in grams</td> <td>Decimal form in kilograms</td> </tr> <tr> <td>\(\frac{1}{2}\,\text{kg}\)</td> <td>\(500\,\text{g}\)</td> <td>\(0.500\,\text{kg}\)</td> </tr> <tr> <td>\(\frac{1}{4}\,\text{kg}\)</td> <td>...</td> <td>...</td> </tr> <tr> <td>\(\frac{3}{4}\,\text{kg}\)</td> <td>...</td> <td>...</td> </tr> </table>

Hints

- One kilogram equals \(1000\) grams. - Find one-fourth of \(1000\) grams. - Use the grams value to write the decimal through the thousandths place.

Solution

1. Since \(1\,\text{kg} = 1000\,\text{g}\), \(\frac{1}{4}\,\text{kg} = 1000\,\text{g} \div 4 = 250\,\text{g}\). In decimal form, this is \(0.250\,\text{kg}\). 2. Three-fourths of a kilogram is \(3 \times 250\,\text{g} = 750\,\text{g}\). In decimal form, this is \(0.750\,\text{kg}\).

Answer

For \(\frac{1}{4}\,\text{kg}\): \(250\,\text{g}\) and \(0.250\,\text{kg}\). For \(\frac{3}{4}\,\text{kg}\): \(750\,\text{g}\) and \(0.750\,\text{kg}\).
5171045
A bakery uses exactly \(500\,\text{g}\) of butter for one sheet cake. A bulk box contains \(10\,\text{kg}\) of butter. a) How many sheet cakes can be made with one box? b) The bakery uses \(12\) boxes in one week. How many sheet cakes are made that week?

Hints

- Convert the box's mass to grams. - Determine how many \(500\)-gram portions are in \(1\,\text{kg}\), then scale to \(10\,\text{kg}\). - Multiply the cakes per box by \(12\).

Solution

1. Convert the butter in one box: \(10\,\text{kg} = 10{,}000\,\text{g}\). 2. Two \(500\,\text{g}\) portions make \(1000\,\text{g}\), or \(1\,\text{kg}\). Therefore, \(10\,\text{kg}\) provides \(10 \times 2 = 20\) cake portions. 3. Find the weekly number of cakes: \(20 \times 12 = 240\).

Answer

a) One box makes \(20\) sheet cakes. b) The bakery makes \(240\) sheet cakes in one week.
5186285
A school festival needs at least \(45\,\text{L}\) of apple juice. The school buys four \(10\)-liter containers that cost \(\$8\) each and three \(2000\,\text{mL}\) bottles that cost \(\$2\) each. Does the school have enough juice, and what is the total cost?

Hints

- Convert the milliliter bottles to liters first. - Add the capacities after all amounts use the same unit. - Multiply each price by the number of containers or bottles.

Solution

1. Convert each small bottle to liters: \(2000\,\text{mL}=2\,\text{L}\). 2. The four large containers hold \(4 \times 10=40\,\text{L}\). The three small bottles hold \(3 \times 2=6\,\text{L}\). 3. The total amount is \(40+6=46\,\text{L}\). Since \(46\ge 45\), the school has enough juice. 4. The total cost is \(4 \times \$8+3 \times \$2=\$32+\$6=\$38\).

Answer

Yes. The school buys \(46\,\text{L}\) of juice for a total cost of \(\$38\).
5193485
An elephant in a national park lives to be \(70\) years old. Determine whether the elephant has lived more than \(600{,}000\) hours by its 70th birthday. Assume every year has exactly \(365\) days.

Hints

- How many days are in each year under the problem's assumption? - How many hours are in one day? - First find the total number of days. - Then convert the days to hours and compare.

Solution

1. Find the number of days in \(70\) years: \(70 \times 365 = 25{,}550\) days. 2. Convert days to hours: \(25{,}550 \times 24 = 613{,}200\) hours. 3. Compare: \(613{,}200 > 600{,}000\), so the elephant has lived more than \(600{,}000\) hours.

Answer

Yes. The elephant has lived \(613{,}200\) hours, which is more than \(600{,}000\) hours.
5193495
A high-speed train travels \(2850\,\text{km}\) on its route. A student claims that this distance is greater than \(250\) million centimeters. Is the student correct? Show how you know.

Hints

- How many meters are in one kilometer? - How many centimeters are in one meter? - You may convert in two steps or use one combined conversion factor. - Compare the converted distance with \(250\) million centimeters.

Solution

1. Convert kilometers to meters: \(2850 \times 1000 = 2{,}850{,}000\), so the distance is \(2{,}850{,}000\,\text{m}\). 2. Convert meters to centimeters: \(2{,}850{,}000 \times 100 = 285{,}000{,}000\), so the distance is \(285{,}000{,}000\,\text{cm}\). 3. Compare: \(285{,}000{,}000\,\text{cm} > 250{,}000{,}000\,\text{cm}\).

Answer

Yes. The student is correct because \(2850\,\text{km} = 285{,}000{,}000\,\text{cm}\).
5193505
An adult blue whale eats an average of \(3500\,\text{kg}\) of krill each day. Does it eat more than \(1250\,\text{t}\) of krill in \(365\) days? Show how you know.

Hints

- Use \(1000\,\text{kg} = 1\,\text{t}\). - Find the whale’s total food for \(365\) days. - Express both amounts in the same unit before comparing.

Solution

1. Find the yearly amount: \(3500\,\text{kg} \times 365 = 1{,}277{,}500\,\text{kg}\). 2. Convert the comparison amount: \(1250\,\text{t} = 1250 \times 1000\,\text{kg} = 1{,}250{,}000\,\text{kg}\). 3. Compare: \(1{,}277{,}500\,\text{kg} > 1{,}250{,}000\,\text{kg}\).

Answer

Yes. The whale eats about \(1{,}277{,}500\,\text{kg}\), which is more than \(1250\,\text{t}\).
5204775
Lucas takes exactly \(2000\) steps on a walk. His average step length is \(60\,\text{cm}\). Find the total distance first in centimeters, then convert it to meters and kilometers.

Hints

- Use multiplication to combine many equal step lengths. - How many centimeters are in one meter? - How many meters are in one kilometer?

Solution

1. Find the total distance in centimeters: \(2000 \times 60\,\text{cm} = 120{,}000\,\text{cm}\). 2. Convert centimeters to meters: \(120{,}000 \div 100 = 1200\), so the distance is \(1200\,\text{m}\). 3. Convert meters to kilometers: \(1200 \div 1000 = 1.2\), so the distance is \(1.2\,\text{km}\).

Answer

\(120{,}000\,\text{cm} = 1200\,\text{m} = 1.2\,\text{km}\)
5204785
Mia and Tom each walk exactly \(400\,\text{m}\). Mia’s average step length is \(50\,\text{cm}\), and Tom’s average step length is \(80\,\text{cm}\). How many steps does each person take?

Hints

- Use the same unit for the total distance and each step length. - Think of dividing the full distance into equal step-length parts. - Divide the total distance by each person’s step length.

Solution

1. Convert the distance to centimeters: \(400\,\text{m} = 40{,}000\,\text{cm}\). 2. Find Mia’s number of steps: \(40{,}000 \div 50 = 800\). 3. Find Tom’s number of steps: \(40{,}000 \div 80 = 500\).

Answer

Mia takes \(800\) steps, and Tom takes \(500\) steps.
5204845
Order the lengths from least to greatest using \(<\). First convert every measurement to centimeters. \(30\,\text{cm}\); \(250\,\text{mm}\); \(0.5\,\text{m}\); \(12\,\text{cm}\); \(0.002\,\text{km}\)

Hints

- Measurements are easier to compare when they use the same unit. - Use \(10\,\text{mm}=1\,\text{cm}\), \(100\,\text{cm}=1\,\text{m}\), and \(100{,}000\,\text{cm}=1\,\text{km}\). - After converting, compare the numerical values from left to right by place value.

Solution

1. Convert to centimeters: \(30\,\text{cm}=30\,\text{cm}\), \(250\,\text{mm}=25\,\text{cm}\), \(0.5\,\text{m}=50\,\text{cm}\), \(12\,\text{cm}=12\,\text{cm}\), and \(0.002\,\text{km}=200\,\text{cm}\). 2. Compare the values: \(12 < 25 < 30 < 50 < 200\). 3. Write the original measurements in that order.

Answer

\(12\,\text{cm} < 250\,\text{mm} < 30\,\text{cm} < 0.5\,\text{m} < 0.002\,\text{km}\)
5204895
A stack of \(100\) sheets of printer paper is exactly \(1\,\text{cm}\) high. a) How thick is one sheet of paper in millimeters? b) About how many sheets are in a stack that is \(20\,\text{cm}\) high? c) A student claims, “A package of \(500\) sheets is \(25\,\text{cm}\) high.” Use a calculation to decide whether the claim is reasonable.

Hints

- Convert \(1\,\text{cm}\) to millimeters first. - Think of each centimeter as one group of \(100\) sheets. - Determine how many groups of \(100\) are in \(500\).

Solution

1. Since \(1\,\text{cm}=10\,\text{mm}\), one sheet is \(10\,\text{mm}\div 100=0.1\,\text{mm}\) thick. 2. Each centimeter of height represents \(100\) sheets, so a \(20\,\text{cm}\) stack contains \(20\times 100=2000\) sheets. 3. Five hundred sheets are \(5\) groups of \(100\) sheets. Their height is \(5\times 1\,\text{cm}=5\,\text{cm}\). 4. The claim is not reasonable because \(5\,\text{cm}\neq 25\,\text{cm}\).

Answer

a) \(0.1\,\text{mm}\) b) About \(2000\) sheets c) The claim is not reasonable; \(500\) sheets would be \(5\,\text{cm}\) high.
5205045
Which measurements describe the same length? Find the three matching pairs. A: \(400\,\text{cm}\) B: \(40\,\text{m}\) C: \(40\,\text{cm}\) D: \(400\,\text{mm}\) E: \(4000\,\text{mm}\) F: \(4000\,\text{cm}\)

Hints

- Convert all measurements to one common unit. - Work through the letters one at a time. - Match measurements that have the same converted value.

Solution

1. Convert each measurement to centimeters. 2. A is \(400\,\text{cm}\). 3. B: \(40\,\text{m}=4000\,\text{cm}\). 4. C is \(40\,\text{cm}\). 5. D: \(400\,\text{mm}=40\,\text{cm}\). 6. E: \(4000\,\text{mm}=400\,\text{cm}\). 7. F is \(4000\,\text{cm}\). 8. Therefore, the pairs are A and E, B and F, and C and D.

Answer

A and E; B and F; C and D
5205055
Fill in each missing number or unit so the equation is true. a) \(7\,\text{m}\ 4\,\text{cm}=\square\,\text{cm}\) b) \(85\,\text{mm}=8\,\text{cm}\ \square\,\text{mm}\) c) \(1200\,\text{m}=1\,\text{km}\ \square\,\text{m}\) d) \(4500\,\text{cm}=45\,\square\) e) \(3\,\text{cm}\ 2\,\text{mm}=\square\,\text{mm}\)

Hints

- Convert the larger-unit part to the smaller unit before adding. - A mixed measurement can be separated into full larger units and a remainder. - When a unit is missing, use the change in numerical value to identify it.

Solution

1. \(7\,\text{m}=700\,\text{cm}\), and \(700+4=704\). 2. \(8\,\text{cm}=80\,\text{mm}\), leaving \(85-80=5\,\text{mm}\). 3. \(1\,\text{km}=1000\,\text{m}\), leaving \(1200-1000=200\,\text{m}\). 4. \(4500\,\text{cm}=45\,\text{m}\), so the missing unit is meters. 5. \(3\,\text{cm}=30\,\text{mm}\), and \(30+2=32\,\text{mm}\).

Answer

a) \(704\) b) \(5\) c) \(200\) d) meters e) \(32\)
5205335
Complete the conversion chain. \(0.05\,\text{km}=\square\,\text{m}=\square\,\text{cm}=\square\,\text{mm}\)

Hints

- Convert one step at a time. - The kilometer-to-meter step uses a factor of \(1000\). - The meter-to-centimeter and centimeter-to-millimeter steps use factors of \(100\) and \(10\).

Solution

1. \(0.05 \times 1000=50\), so \(0.05\,\text{km}=50\,\text{m}\). 2. \(50 \times 100=5000\), so \(50\,\text{m}=5000\,\text{cm}\). 3. \(5000 \times 10=50{,}000\), so \(5000\,\text{cm}=50{,}000\,\text{mm}\).

Answer

\(0.05\,\text{km}=50\,\text{m}=5000\,\text{cm}=50{,}000\,\text{mm}\)
5205395
Write each length without a decimal. Begin with the greatest reasonable unit, then use the greatest possible smaller unit. a) \(7.2\,\text{m}\) b) \(14.06\,\text{m}\) c) \(5.125\,\text{km}\) d) \(0.68\,\text{m}\)

Hints

- Separate the whole-number part from the decimal part. - Convert only the decimal part to a smaller unit. - Use \(100\,\text{cm}=1\,\text{m}\) and \(1000\,\text{m}=1\,\text{km}\).

Solution

1. \(0.2\,\text{m}=20\,\text{cm}\), so \(7.2\,\text{m}=7\,\text{m}\ 20\,\text{cm}\). 2. \(0.06\,\text{m}=6\,\text{cm}\), so \(14.06\,\text{m}=14\,\text{m}\ 6\,\text{cm}\). 3. \(0.125\,\text{km}=125\,\text{m}\), so \(5.125\,\text{km}=5\,\text{km}\ 125\,\text{m}\). 4. \(0.68\,\text{m}=68\,\text{cm}\).

Answer

a) \(7\,\text{m}\ 20\,\text{cm}\) b) \(14\,\text{m}\ 6\,\text{cm}\) c) \(5\,\text{km}\ 125\,\text{m}\) d) \(68\,\text{cm}\)
5205405
Rewrite each length without a decimal. Use mixed units when needed. a) \(10.04\,\text{km}\) b) \(2.5\,\text{cm}\) c) \(4070\,\text{mm}\) d) \(0.009\,\text{m}\)

Hints

- Pay close attention to zeros in the decimal. - Convert the decimal part to a smaller unit. - A unit may be omitted from the mixed form when its amount is zero.

Solution

1. \(0.04\,\text{km}=40\,\text{m}\), so \(10.04\,\text{km}=10\,\text{km}\ 40\,\text{m}\). 2. \(0.5\,\text{cm}=5\,\text{mm}\), so \(2.5\,\text{cm}=2\,\text{cm}\ 5\,\text{mm}\). 3. \(4070\,\text{mm}=4000\,\text{mm}+70\,\text{mm}=4\,\text{m}\ 7\,\text{cm}\). 4. \(0.009\,\text{m}=9\,\text{mm}\).

Answer

a) \(10\,\text{km}\ 40\,\text{m}\) b) \(2\,\text{cm}\ 5\,\text{mm}\) c) \(4\,\text{m}\ 7\,\text{cm}\) d) \(9\,\text{mm}\)
5205495
Pair measurements that have the same weight. Which measurement has no partner? Explain. \(12.5\,\text{lb}\); \(200\,\text{oz}\); \(125\,\text{lb}\); \(2000\,\text{oz}\); \(20\,\text{oz}\)

Hints

- Convert all measurements to the same unit. - Use \(16\,\text{oz}=1\,\text{lb}\). - Compare the converted values and match equal weights.

Solution

1. Convert the ounce measurements to pounds using \(16\,\text{oz}=1\,\text{lb}\). 2. \(200 \div 16=12.5\), so \(200\,\text{oz}=12.5\,\text{lb}\). 3. \(2000 \div 16=125\), so \(2000\,\text{oz}=125\,\text{lb}\). 4. \(20 \div 16=1.25\), so \(20\,\text{oz}=1.25\,\text{lb}\). 5. The pairs are \(12.5\,\text{lb}\) with \(200\,\text{oz}\), and \(125\,\text{lb}\) with \(2000\,\text{oz}\). The unpaired measurement is \(20\,\text{oz}\).

Answer

\(20\,\text{oz}\) has no partner. The other pairs are \(12.5\,\text{lb}=200\,\text{oz}\) and \(125\,\text{lb}=2000\,\text{oz}\).
5205575
Evaluate each expression. First convert every measurement to the smallest unit used in that part. a) \(12\,\text{m}\ 8\,\text{cm}+30\,\text{cm}\) b) \(2\,\text{km}\ 45\,\text{m}-120\,\text{m}\) c) \((40\,\text{cm}\ 6\,\text{mm})\times 5\)

Hints

- Identify the smallest unit used in each part. - Convert mixed measurements to one numerical value before calculating. - Include the correct unit in each answer.

Solution

1. Convert to centimeters: \(12\,\text{m}=1200\,\text{cm}\). Then \(1200+8+30=1238\), so the result is \(1238\,\text{cm}\). 2. Convert to meters: \(2\,\text{km}=2000\,\text{m}\). Then \(2000+45-120=1925\), so the result is \(1925\,\text{m}\). 3. Convert to millimeters: \(40\,\text{cm}=400\,\text{mm}\). Then \((400+6)\times 5=406\times 5=2030\), so the result is \(2030\,\text{mm}\).

Answer

a) \(1238\,\text{cm}\) b) \(1925\,\text{m}\) c) \(2030\,\text{mm}\)
5205585
In each group, one length does not match the others. Find it and justify your choice by converting to a common unit. a) \(3.5\,\text{m}\); \(350\,\text{cm}\); \(3500\,\text{mm}\); \(0.035\,\text{km}\) b) \(142\,\text{cm}\); \(1.42\,\text{m}\); \(1420\,\text{mm}\); \(14.2\,\text{cm}\)

Hints

- Choose one common unit for each group. - Convert every measurement before comparing. - Watch the position of the decimal point and the number of zeros.

Solution

1. For a), convert to meters: \(3.5\,\text{m}\), \(350\,\text{cm}=3.5\,\text{m}\), \(3500\,\text{mm}=3.5\,\text{m}\), and \(0.035\,\text{km}=35\,\text{m}\). The outlier is \(0.035\,\text{km}\). 2. For b), convert to centimeters: \(142\,\text{cm}\), \(1.42\,\text{m}=142\,\text{cm}\), \(1420\,\text{mm}=142\,\text{cm}\), and \(14.2\,\text{cm}\). The outlier is \(14.2\,\text{cm}\).

Answer

a) \(0.035\,\text{km}\) b) \(14.2\,\text{cm}\)
5205685
Write \(<\), \(>\), or \(=\) in each box. a) \(6\,\text{m}\ 50\,\text{cm}\ \square\ 6.05\,\text{m}\) b) \(1200\,\text{mm}\ \square\ 120\,\text{cm}\) c) \(0.04\,\text{km}\ \square\ 400\,\text{m}\) d) \(9\,\text{cm}\ 2\,\text{mm}\ \square\ 9.2\,\text{cm}\)

Hints

- Convert mixed units to one unit before comparing. - Pay attention to decimal place value. - Use centimeters or millimeters to compare lengths without introducing more decimals.

Solution

1. \(6\,\text{m}\ 50\,\text{cm}=6.5\,\text{m}\), and \(6.5>6.05\), so the symbol is \(>\). 2. \(120\,\text{cm}=1200\,\text{mm}\), so the symbol is \(=\). 3. \(0.04\,\text{km}=40\,\text{m}\), and \(40<400\), so the symbol is \(<\). 4. \(9\,\text{cm}\ 2\,\text{mm}=92\,\text{mm}\), and \(9.2\,\text{cm}=92\,\text{mm}\), so the symbol is \(=\).

Answer

a) \(>\) b) \(=\) c) \(<\) d) \(=\)
5205695
Order the lengths from least to greatest using \(<\). \(75\,\text{cm}\); \(0.8\,\text{m}\); \(720\,\text{mm}\); \(71\,\text{cm}\)

Hints

- Convert all values to one common unit. - Record each converted value beside its original measurement. - Use the original forms in the final inequality.

Solution

1. Convert all measurements to millimeters: \(75\,\text{cm}=750\,\text{mm}\), \(0.8\,\text{m}=800\,\text{mm}\), \(720\,\text{mm}=720\,\text{mm}\), and \(71\,\text{cm}=710\,\text{mm}\). 2. Compare: \(710<720<750<800\). 3. Write the original measurements in that order.

Answer

\(71\,\text{cm}<720\,\text{mm}<75\,\text{cm}<0.8\,\text{m}\)
5205745
A package currently weighs \(16\,\text{lb}\,7\,\text{oz}\). Two more items are added, and each weighs \(1\,\text{lb}\,14\,\text{oz}\). Is the package now heavier or lighter than \(20\,\text{lb}\)? Use \(1\,\text{lb} = 16\,\text{oz}\).

Hints

- Express all weights in ounces. - Find the combined weight of the two added items. - Compare the final weight with \(20\,\text{lb}\).

Solution

1. Convert the package’s starting weight to ounces: \(16 \times 16 + 7 = 263\,\text{oz}\). 2. Convert one added item: \(1 \times 16 + 14 = 30\,\text{oz}\). Two items weigh \(2 \times 30 = 60\,\text{oz}\). 3. Add: \(263 + 60 = 323\,\text{oz}\). 4. Convert back: \(323\,\text{oz} = 20\,\text{lb}\,3\,\text{oz}\), which is heavier than \(20\,\text{lb}\).

Answer

The package weighs \(20\,\text{lb}\,3\,\text{oz}\), so it is heavier than \(20\,\text{lb}\).
5205755
A large rain barrel contains \(155.75\,\text{L}\) of water. A storm adds \(45\,\text{L}\,350\,\text{mL}\). The barrel can hold \(200\,\text{L}\). Will the barrel overflow?

Hints

- Convert the added liters and milliliters to one decimal number of liters. - Add the new water to the amount already in the barrel. - Compare the total with the barrel’s capacity.

Solution

1. Convert the added water to liters: \(45\,\text{L}\,350\,\text{mL} = 45.35\,\text{L}\). 2. Add the amounts: \(155.75\,\text{L} + 45.35\,\text{L} = 201.10\,\text{L}\). 3. Since \(201.10\,\text{L} > 200\,\text{L}\), the barrel will overflow.

Answer

Yes. The barrel would contain \(201.10\,\text{L}\), which is \(1.10\,\text{L}\) more than its capacity.
5205765
A rope is \(12.4\,\text{m}\) long. First, a \(3.85\,\text{m}\) piece is cut off. Then a piece measuring \(4\,\text{m}\,60\,\text{cm}\) is cut off. Is the remaining piece longer or shorter than \(4\,\text{m}\)?

Hints

- Express both cut pieces in the same unit. - Find the total length removed. - Subtract from the original length and compare the result with \(4\,\text{m}\).

Solution

1. Convert the mixed measurement to meters: \(4\,\text{m}\,60\,\text{cm} = 4.6\,\text{m}\). 2. Add the lengths removed: \(3.85\,\text{m} + 4.6\,\text{m} = 8.45\,\text{m}\). 3. Subtract from the original length: \(12.4\,\text{m} - 8.45\,\text{m} = 3.95\,\text{m}\). 4. Since \(3.95\,\text{m} < 4\,\text{m}\), the remaining piece is shorter than \(4\,\text{m}\).

Answer

The remaining piece is \(3.95\,\text{m}\) long, so it is shorter than \(4\,\text{m}\).
5205865
Find the remaining length and write the result as a decimal number of meters: \(3.5\,\text{m} - 120\,\text{cm} - 450\,\text{mm}\).

Hints

- Convert all measurements to meters first. - How many centimeters are in one meter? - How many millimeters are in one meter? - Line up decimal points when subtracting.

Solution

1. Convert each length to meters: \(120\,\text{cm} = 1.20\,\text{m}\) and \(450\,\text{mm} = 0.45\,\text{m}\). 2. Subtract the first length: \(3.5\,\text{m} - 1.20\,\text{m} = 2.30\,\text{m}\). 3. Subtract the second length: \(2.30\,\text{m} - 0.45\,\text{m} = 1.85\,\text{m}\).

Answer

\(1.85\,\text{m}\)
5205875
Calculate and write the result as a decimal number of kilograms: \(1.25\,\text{kg} + 850\,\text{g} - 0.4\,\text{kg} - 325\,\text{g}\).

Hints

- The answer is requested in kilograms. - Use \(1000\,\text{g} = 1\,\text{kg}\). - Convert every amount to kilograms before calculating. - Keep the decimal places aligned.

Solution

1. Convert the gram amounts to kilograms: \(850\,\text{g} = 0.850\,\text{kg}\) and \(325\,\text{g} = 0.325\,\text{kg}\). 2. Add the positive amounts: \(1.25\,\text{kg} + 0.850\,\text{kg} = 2.100\,\text{kg}\). 3. Subtract: \(2.100\,\text{kg} - 0.4\,\text{kg} - 0.325\,\text{kg} = 1.375\,\text{kg}\).

Answer

\(1.375\,\text{kg}\)
5205905
Convert \(4\,\text{kg}\ 20\,\text{g}\), \(420\,\text{g}\), \(4002\,\text{g}\), and \(0.42\,\text{kg}\) to grams. Then order the original measurements from least to greatest.

Hints

- Use \(1000\,\text{g}=1\,\text{kg}\). - Convert every measurement to grams before comparing. - Check whether any two measurements are equal.

Solution

1. \(4\,\text{kg}\ 20\,\text{g}=4000\,\text{g}+20\,\text{g}=4020\,\text{g}\). 2. \(420\,\text{g}\) remains \(420\,\text{g}\). 3. \(4002\,\text{g}\) remains \(4002\,\text{g}\). 4. \(0.42\,\text{kg}=420\,\text{g}\). 5. Therefore, \(420=420<4002<4020\).

Answer

Conversions: \(4020\,\text{g}\), \(420\,\text{g}\), \(4002\,\text{g}\), \(420\,\text{g}\) Order: \(420\,\text{g}=0.42\,\text{kg}<4002\,\text{g}<4\,\text{kg}\ 20\,\text{g}\)
5205915
Convert \(5\,\text{m}\ 8\,\text{cm}\), \(580\,\text{cm}\), \(508\,\text{mm}\), and \(5.8\,\text{m}\) to centimeters. Then order the original measurements from greatest to least.

Hints

- Use \(10\,\text{mm}=1\,\text{cm}\) and \(100\,\text{cm}=1\,\text{m}\). - Convert each measurement to centimeters. - Greatest to least means the values decrease from left to right.

Solution

1. \(5\,\text{m}\ 8\,\text{cm}=500\,\text{cm}+8\,\text{cm}=508\,\text{cm}\). 2. \(580\,\text{cm}\) remains \(580\,\text{cm}\). 3. \(508\,\text{mm}=50.8\,\text{cm}\). 4. \(5.8\,\text{m}=580\,\text{cm}\). 5. Compare: \(580=580>508>50.8\).

Answer

Conversions: \(508\,\text{cm}\), \(580\,\text{cm}\), \(50.8\,\text{cm}\), \(580\,\text{cm}\) Order: \(580\,\text{cm}=5.8\,\text{m}>5\,\text{m}\ 8\,\text{cm}>508\,\text{mm}\)
5205925
Convert \(2\,\text{L}\ 500\,\text{mL}\), \(2050\,\text{mL}\), \(2.5\,\text{L}\), and \(25\,\text{mL}\) to milliliters. Then order the original measurements from least to greatest.

Hints

- Use \(1000\,\text{mL}=1\,\text{L}\). - Convert all measurements to milliliters. - Least to greatest means the values increase from left to right.

Solution

1. \(2\,\text{L}\ 500\,\text{mL}=2000\,\text{mL}+500\,\text{mL}=2500\,\text{mL}\). 2. \(2050\,\text{mL}\) remains \(2050\,\text{mL}\). 3. \(2.5\,\text{L}=2500\,\text{mL}\). 4. \(25\,\text{mL}\) remains \(25\,\text{mL}\). 5. Compare: \(25<2050<2500=2500\).

Answer

Conversions: \(2500\,\text{mL}\), \(2050\,\text{mL}\), \(2500\,\text{mL}\), \(25\,\text{mL}\) Order: \(25\,\text{mL}<2050\,\text{mL}<2\,\text{L}\ 500\,\text{mL}=2.5\,\text{L}\)
5205955
Evaluate each expression. Use the smaller unit shown in each part. Decide whether the result is a measurement or a unitless number. a) \(450\,\text{g}+1.2\,\text{kg}\) b) \(3\,\text{m}\times 25\) c) \(72\,\text{cm}\div 8\,\text{cm}\) d) \(\$2.00-45\) cents

Hints

- Convert measurements to the same unit before adding or subtracting. - Multiplying a measurement by a number keeps the measurement unit. - Dividing two measurements with the same unit produces a unitless ratio.

Solution

1. \(1.2\,\text{kg}=1200\,\text{g}\). Then \(450+1200=1650\), so the result is \(1650\,\text{g}\). 2. \(3\times 25=75\). Multiplying a measurement by a number keeps the unit, so the result is \(75\,\text{m}\). 3. \(72\div 8=9\). The centimeter units cancel because one length is divided by another length, so the result is the unitless number \(9\). 4. \(\$2.00=200\) cents. Then \(200-45=155\), so the result is \(155\) cents.

Answer

a) \(1650\,\text{g}\) b) \(75\,\text{m}\) c) \(9\) d) \(155\) cents
5205975
Evaluate the left side of each comparison, then write \(<\), \(>\), or \(=\). a) \(5\times 200\,\text{mg}\ \square\ 1\,\text{g}\) b) \(1\,\text{km}-350\,\text{m}\ \square\ 700\,\text{m}\) c) \(48\,\text{h}\div 2\ \square\ 1\,\text{day}\) d) \(\$3.50+150\) cents \(\square\ \$5.00\)

Hints

- Evaluate the left side first. - Convert both sides to the same unit before comparing. - Recall that \(1000\,\text{mg}=1\,\text{g}\) and \(24\,\text{h}=1\,\text{day}\).

Solution

1. \(5\times 200\,\text{mg}=1000\,\text{mg}=1\,\text{g}\), so the symbol is \(=\). 2. \(1\,\text{km}=1000\,\text{m}\). Then \(1000-350=650\), and \(650<700\), so the symbol is \(<\). 3. \(48\,\text{h}\div 2=24\,\text{h}=1\,\text{day}\), so the symbol is \(=\). 4. \(150\) cents is \(\$1.50\). Then \(\$3.50+\$1.50=\$5.00\), so the symbol is \(=\).

Answer

a) \(=\) b) \(<\) c) \(=\) d) \(=\)
5206065
Convert all addends to grams, write the sum as a product, and calculate: \(250\,\text{g} + 250\,\text{g} + 250{,}000\,\text{mg} + 0.25\,\text{kg} + 250\,\text{g}\).

Hints

- Convert every amount to grams. - Compare the converted addends. - Count how many equal addends there are. - Replace repeated addition with multiplication.

Solution

1. Convert: \(250{,}000\,\text{mg} = 250\,\text{g}\) and \(0.25\,\text{kg} = 250\,\text{g}\). 2. All five addends equal \(250\,\text{g}\). 3. Write the product: \(5 \times 250\,\text{g}\). 4. Calculate: \(5 \times 250\,\text{g} = 1250\,\text{g} = 1.25\,\text{kg}\).

Answer

\(5 \times 250\,\text{g} = 1250\,\text{g} = 1.25\,\text{kg}\)
5206075
Write the sum as a product, calculate its value, and give the result in meters: \(45\,\text{cm} + 0.45\,\text{m} + 450\,\text{mm} + 45\,\text{cm}\).

Hints

- Check whether the different measurements represent equal lengths. - Convert all terms to one unit. - Count the equal addends and write a multiplication expression. - Convert the final result to meters.

Solution

1. Convert each measurement to centimeters: \(0.45\,\text{m} = 45\,\text{cm}\) and \(450\,\text{mm} = 45\,\text{cm}\). 2. The sum has four equal addends: \(45\,\text{cm} + 45\,\text{cm} + 45\,\text{cm} + 45\,\text{cm}\). 3. Write and evaluate the product: \(4 \times 45\,\text{cm} = 180\,\text{cm}\). 4. Convert to meters: \(180\,\text{cm} = 1.8\,\text{m}\).

Answer

\(4 \times 45\,\text{cm} = 180\,\text{cm} = 1.8\,\text{m}\)
5206275
Calculate efficiently and give the result in meters and centimeters: \(6\,\text{m}\,80\,\text{cm} + 14\,\text{m}\,36\,\text{cm} + 20\,\text{cm} + 65\,\text{cm}\).

Hints

- Look for centimeter amounts that combine to make a full meter. - You may change the order of addends. - Use \(100\,\text{cm} = 1\,\text{m}\).

Solution

1. Combine measurements that make a full meter: \(80\,\text{cm} + 20\,\text{cm} = 100\,\text{cm} = 1\,\text{m}\). 2. Combine the remaining centimeters: \(36\,\text{cm} + 65\,\text{cm} = 101\,\text{cm} = 1\,\text{m}\,1\,\text{cm}\). 3. Add the meter amounts: \(6\,\text{m} + 14\,\text{m} + 1\,\text{m} + 1\,\text{m}\,1\,\text{cm} = 22\,\text{m}\,1\,\text{cm}\).

Answer

\(22\,\text{m}\,1\,\text{cm}\)
5206285
Calculate each result and write it in mixed units of meters and centimeters. a) \(18\,\text{m}\,42\,\text{cm} - 5.9\,\text{m}\) b) \(2\,\text{m}\,5\,\text{cm} + 740\,\text{mm} + 30\,\text{cm}\)

Hints

- Express all measurements in one unit before calculating. - Pay attention to the relationships among millimeters, centimeters, and meters. - Convert the final result back to meters and centimeters.

Solution

1. For a), convert to centimeters: \(18\,\text{m}\,42\,\text{cm} = 1842\,\text{cm}\) and \(5.9\,\text{m} = 590\,\text{cm}\). Subtract: \(1842 - 590 = 1252\), so the result is \(12\,\text{m}\,52\,\text{cm}\). 2. For b), convert to centimeters: \(2\,\text{m}\,5\,\text{cm} = 205\,\text{cm}\) and \(740\,\text{mm} = 74\,\text{cm}\). Add: \(205 + 74 + 30 = 309\), so the result is \(3\,\text{m}\,9\,\text{cm}\).

Answer

a) \(12\,\text{m}\,52\,\text{cm}\) b) \(3\,\text{m}\,9\,\text{cm}\)
5206295
Calculate each result and briefly describe what happens to the units. a) \(4.2\,\text{m} \div 7\) b) \(4.2\,\text{m} \div 30\,\text{cm}\) c) \(4.2\,\text{m} - 50\,\text{cm}\) What general rule can you state about dividing one measurement by another measurement of the same kind, as in part b)?

Hints

- Convert to centimeters so each calculation uses compatible units. - Compare dividing a measurement by a number with dividing a measurement by another measurement. - For subtraction, the units must match first.

Solution

1. Convert \(4.2\,\text{m}\) to \(420\,\text{cm}\). 2. For a), \(420\,\text{cm} \div 7 = 60\,\text{cm}\). Dividing a length by a number gives another length. 3. For b), \(420\,\text{cm} \div 30\,\text{cm} = 14\). The matching units cancel, so the result is a count with no measurement unit. 4. For c), \(420\,\text{cm} - 50\,\text{cm} = 370\,\text{cm}\), or \(3.7\,\text{m}\). 5. In general, dividing two measurements of the same kind after expressing them in the same unit gives a unitless number that tells how many times one measurement fits into the other.

Answer

a) \(60\,\text{cm}\), or \(0.6\,\text{m}\) b) \(14\) c) \(370\,\text{cm}\), or \(3.7\,\text{m}\) Rule: Dividing one measurement by another measurement of the same kind gives a unitless number after the units are made the same.
5206305
A ribbon is \(2.4\,\text{m}\) long. a) The ribbon is cut into \(12\) equal pieces. Find the length of each piece in centimeters and explain your method. b) How many \(15\,\text{cm}\) pieces can be cut from the entire ribbon? c) Give another everyday example in which dividing one length by another length gives a count.

Hints

- Convert the entire ribbon to centimeters first. - In part a), divide a length by a number of pieces. - In part b), find how many times the smaller length fits into the larger length. - For part c), think about building, crafts, or sports.

Solution

1. Convert the ribbon length: \(2.4\,\text{m} = 240\,\text{cm}\). 2. For a), divide the total length by the number of equal pieces: \(240\,\text{cm} \div 12 = 20\,\text{cm}\). 3. For b), divide the total length by the length of one piece: \(240\,\text{cm} \div 15\,\text{cm} = 16\). 4. For c), one valid example is finding how many \(50\,\text{cm}\)-wide pavers fit along a \(10\,\text{m}\) walkway.

Answer

a) \(20\,\text{cm}\) b) \(16\) pieces c) Answers will vary. Example: finding how many \(50\,\text{cm}\)-wide pavers fit along a \(10\,\text{m}\) walkway.
5206505
A class hikes for four days. The distances are \(12\,\text{km}\,450\,\text{m}\), \(15\,\text{km}\,720\,\text{m}\), \(8\,\text{km}\,900\,\text{m}\), and \(11\,\text{km}\,130\,\text{m}\). How far does the class hike altogether? Give the result in kilometers and meters.

Hints

- Add kilometers and meters separately. - Use \(1000\,\text{m} = 1\,\text{km}\). - Regroup any extra meters as kilometers.

Solution

1. Add the meter amounts: \(450 + 720 + 900 + 130 = 2200\), so the meters total \(2\,\text{km}\,200\,\text{m}\). 2. Add the kilometer amounts: \(12 + 15 + 8 + 11 = 46\), so they total \(46\,\text{km}\). 3. Combine the results: \(46\,\text{km} + 2\,\text{km}\,200\,\text{m} = 48\,\text{km}\,200\,\text{m}\).

Answer

\(48\,\text{km}\,200\,\text{m}\)
5206515
A courier records four driving segments during the morning. <table> <tr><th>Segment</th><th>Distance</th></tr> <tr><td>Warehouse to Stop A</td><td>\(8\,\text{mi}\,1200\,\text{ft}\)</td></tr> <tr><td>Stop A to Stop B</td><td>\(2600\,\text{ft}\)</td></tr> <tr><td>Stop B to Stop C</td><td>\(2\,\text{mi}\,375\,\text{ft}\)</td></tr> <tr><td>Stop C back to the warehouse</td><td>\(1800\,\text{ft}\)</td></tr> </table> Find the total distance driven. Give the result in miles and feet. Use \(1\,\text{mi} = 5280\,\text{ft}\).

Hints

- Convert each mixed distance to feet. - Add all four segments. - Divide the total number of feet by \(5280\) to convert back to miles and feet.

Solution

1. Convert each distance to feet: \(8\,\text{mi}\,1200\,\text{ft} = 43{,}440\,\text{ft}\) and \(2\,\text{mi}\,375\,\text{ft} = 10{,}935\,\text{ft}\). 2. Add all four distances: \(43{,}440 + 2600 + 10{,}935 + 1800 = 58{,}775\), so the total is \(58{,}775\,\text{ft}\). 3. Convert back to mixed units: \(58{,}775\,\text{ft} = 11\,\text{mi}\,695\,\text{ft}\).

Answer

\(11\,\text{mi}\,695\,\text{ft}\)
5206655
A truck carries two heavy crates. The first weighs \(2\,\text{tons}\,900\,\text{lb}\), and the second weighs \(1750\,\text{lb}\). First estimate the total weight. Then find the exact total in pounds. Use \(1\,\text{ton} = 2000\,\text{lb}\).

Hints

- Convert tons and pounds to pounds. - Round to numbers that are easy to add mentally. - Compare the exact total with the estimate.

Solution

1. Convert the first crate: \(2 \times 2000 + 900 = 4900\,\text{lb}\). 2. Estimate by rounding: \(4900\,\text{lb} \approx 5000\,\text{lb}\) and \(1750\,\text{lb} \approx 1800\,\text{lb}\), so the total is about \(6800\,\text{lb}\). 3. Add exactly: \(4900\,\text{lb} + 1750\,\text{lb} = 6650\,\text{lb}\).

Answer

Estimate: about \(6800\,\text{lb}\) Exact: \(6650\,\text{lb}\)
5206665
A garden fence requires \(12\) rolls of wire. Each roll is \(15\,\text{ft}\,11\,\text{in}\) long. First estimate the total length. Then find the exact total in feet and inches.

Hints

- Round the length of one roll to a nearby whole number of feet for the estimate. - Convert feet and inches to inches before multiplying. - Convert the exact total back to feet and inches.

Solution

1. Estimate by rounding each roll to \(16\,\text{ft}\): \(16 \times 12 = 192\), so the total is about \(192\,\text{ft}\). 2. Convert one roll to inches: \(15 \times 12 + 11 = 191\,\text{in}\). 3. Multiply: \(191 \times 12 = 2292\,\text{in}\). 4. Convert back to feet and inches: \(2292 \div 12 = 191\), so the exact total is \(191\,\text{ft}\).

Answer

Estimate: about \(192\,\text{ft}\) Exact: \(191\,\text{ft}\)
5206675
A ribbon \(15\,\text{ft}\,6\,\text{in}\) long is cut into \(6\) equal pieces. First estimate the length of one piece. Then find the exact length.

Hints

- Use a nearby total length that is easy to divide by \(6\). - Convert the exact length to inches before dividing. - Convert the quotient back to feet and inches.

Solution

1. Estimate by using \(15\,\text{ft} = 180\,\text{in}\): \(180 \div 6 = 30\,\text{in}\), or about \(2\,\text{ft}\,6\,\text{in}\). 2. Convert the exact length to inches: \(15 \times 12 + 6 = 186\,\text{in}\). 3. Divide: \(186 \div 6 = 31\,\text{in}\). 4. Convert back to feet and inches: \(31\,\text{in} = 2\,\text{ft}\,7\,\text{in}\).

Answer

Estimate: about \(2\,\text{ft}\,6\,\text{in}\) Exact: \(2\,\text{ft}\,7\,\text{in}\)
5206935
Evaluate and write the result as a decimal number of kilograms: \(12.5\,\text{kg} - (4\,\text{kg}\,250\,\text{g} - 1\,\text{kg}\,75\,\text{g})\).

Hints

- Evaluate the parentheses first. - Convert the mixed kilogram-and-gram amounts to one unit. - Use \(1000\,\text{g} = 1\,\text{kg}\). - Align decimal points when subtracting.

Solution

1. Convert the measurements in parentheses to grams: \(4\,\text{kg}\,250\,\text{g} = 4250\,\text{g}\) and \(1\,\text{kg}\,75\,\text{g} = 1075\,\text{g}\). 2. Evaluate the parentheses: \(4250\,\text{g} - 1075\,\text{g} = 3175\,\text{g}\). 3. Convert: \(3175\,\text{g} = 3.175\,\text{kg}\). 4. Subtract: \(12.5\,\text{kg} - 3.175\,\text{kg} = 9.325\,\text{kg}\).

Answer

\(9.325\,\text{kg}\)
5206945
Evaluate and write the result as a decimal number of liters: \(10\,\text{L} - [3.4\,\text{L} + (1\,\text{L}\,250\,\text{mL} - 800\,\text{mL})]\).

Hints

- Begin with the innermost parentheses. - Express liters and milliliters in one unit. - Use \(1000\,\text{mL} = 1\,\text{L}\). - Follow the grouping symbols from inside to outside.

Solution

1. Evaluate the inner parentheses: \(1\,\text{L}\,250\,\text{mL} - 800\,\text{mL} = 1250\,\text{mL} - 800\,\text{mL} = 450\,\text{mL}\). 2. Convert: \(450\,\text{mL} = 0.45\,\text{L}\). 3. Evaluate the brackets: \(3.4\,\text{L} + 0.45\,\text{L} = 3.85\,\text{L}\). 4. Subtract: \(10\,\text{L} - 3.85\,\text{L} = 6.15\,\text{L}\).

Answer

\(6.15\,\text{L}\)
5207095
Leo wants to hang a picture. The hook is \(7\,\text{ft}\,1\,\text{in}\) above the floor. Leo is \(5\,\text{ft}\,2\,\text{in}\) tall and can reach \(10\,\text{in}\) above the top of his head. He stands on a \(16\)-inch stool. Can Leo reach the hook?

Hints

- Convert all measurements to inches. - Add Leo’s height, extra reach, and stool height. - Compare the total reach with the hook height.

Solution

1. Convert Leo’s height to inches: \(5 \times 12 + 2 = 62\,\text{in}\). 2. Add his reach and the stool height: \(62 + 10 + 16 = 88\,\text{in}\). 3. Convert the hook height to inches: \(7 \times 12 + 1 = 85\,\text{in}\). 4. Since \(88\,\text{in} \ge 85\,\text{in}\), Leo can reach the hook.

Answer

Yes. Leo can reach \(88\,\text{in}\), which is \(3\,\text{in}\) above the hook.
5207105
Maya wants to touch a ceiling that is \(8\,\text{ft}\,2\,\text{in}\) high. She is \(4\,\text{ft}\,10\,\text{in}\) tall and can reach \(13\,\text{in}\) above the top of her head. Each ladder step raises her \(8\,\text{in}\). What is the lowest step Maya must stand on to reach the ceiling with her fingertips?

Hints

- Convert all heights to inches. - Find Maya’s reach without the ladder. - Divide the remaining height by the rise per step, then decide whether to round up.

Solution

1. Convert Maya’s height to inches: \(4 \times 12 + 10 = 58\,\text{in}\). 2. Add her extra reach: \(58 + 13 = 71\,\text{in}\). 3. Convert the ceiling height: \(8 \times 12 + 2 = 98\,\text{in}\). 4. Find the missing height: \(98 - 71 = 27\,\text{in}\). 5. Divide by the rise per step: \(27 \div 8 = 3.375\). Three steps are not enough, so Maya must stand on the fourth step.

Answer

Maya must stand on at least the fourth step.
5207115
An apple hangs from a branch \(9\,\text{ft}\,4\,\text{in}\) above the ground. Tom is \(5\,\text{ft}\,1\,\text{in}\) tall, can reach \(14\,\text{in}\) above his head, and has a \(24\)-inch stool. Marie is \(5\,\text{ft}\,4\,\text{in}\) tall, can reach \(15\,\text{in}\) above her head, and has a \(36\)-inch step ladder. Who can reach the apple? Find each person’s maximum reach.

Hints

- Find each person’s maximum reach separately. - Convert feet and inches to one unit before adding. - Compare each total with the branch height.

Solution

1. Tom’s height is \(5 \times 12 + 1 = 61\,\text{in}\). His maximum reach is \(61 + 14 + 24 = 99\,\text{in}\), or \(8\,\text{ft}\,3\,\text{in}\). Since \(99 < 112\), Tom cannot reach the apple. 2. Marie’s height is \(5 \times 12 + 4 = 64\,\text{in}\). Her maximum reach is \(64 + 15 + 36 = 115\,\text{in}\), or \(9\,\text{ft}\,7\,\text{in}\). Since \(115 > 112\), Marie can reach the apple.

Answer

Tom can reach \(8\,\text{ft}\,3\,\text{in}\) and cannot reach the apple. Marie can reach \(9\,\text{ft}\,7\,\text{in}\) and can reach the apple.
5207165
Estimate first, then evaluate exactly: \((15\,\text{m} - 700\,\text{cm}) \times 12 + 400\,\text{cm} \times 5\).

Hints

- Convert all measurements to meters. - Evaluate the parentheses before multiplication and addition. - Use nearby easy numbers for the estimate.

Solution

1. Estimate by using \(700\,\text{cm} = 7\,\text{m}\) and rounding \(12\) to \(10\): \((15 - 7) \times 10 + 4 \times 5 = 100\), so an estimate is \(100\,\text{m}\). 2. Convert centimeters to meters: \(700\,\text{cm} = 7\,\text{m}\) and \(400\,\text{cm} = 4\,\text{m}\). 3. Evaluate the parentheses: \(15\,\text{m} - 7\,\text{m} = 8\,\text{m}\). 4. Multiply: \(8\,\text{m} \times 12 = 96\,\text{m}\) and \(4\,\text{m} \times 5 = 20\,\text{m}\). 5. Add: \(96\,\text{m} + 20\,\text{m} = 116\,\text{m}\).

Answer

Estimate: about \(100\,\text{m}\) Exact: \(116\,\text{m}\)
5207175
Estimate first, then evaluate exactly: \((2\,\text{km} + 400\,\text{m}) \div 8 + 70\,\text{m} \times 6\).

Hints

- Convert kilometers to meters first. - Evaluate division and multiplication before adding. - Use compatible numbers for the estimate.

Solution

1. Estimate: \(2400\,\text{m} \div 8 \approx 300\,\text{m}\) and \(70\,\text{m} \times 6 \approx 400\,\text{m}\), so the result should be about \(700\,\text{m}\). 2. Convert: \(2\,\text{km} = 2000\,\text{m}\). 3. Evaluate the parentheses: \(2000\,\text{m} + 400\,\text{m} = 2400\,\text{m}\). 4. Divide and multiply: \(2400\,\text{m} \div 8 = 300\,\text{m}\) and \(70\,\text{m} \times 6 = 420\,\text{m}\). 5. Add: \(300\,\text{m} + 420\,\text{m} = 720\,\text{m}\).

Answer

Estimate: about \(700\,\text{m}\) Exact: \(720\,\text{m}\)
5207185
Estimate first, then evaluate exactly: \(25\,\text{cm} \times 4 + (1.8\,\text{m} - 60\,\text{cm}) \div 3\).

Hints

- Convert the meter measurement to centimeters. - Evaluate the parentheses first, then multiplication and division, and finally addition. - Use nearby easy measurements for the estimate.

Solution

1. Estimate by using \(1.8\,\text{m} \approx 2\,\text{m}\): \(25\,\text{cm} \times 4 = 100\,\text{cm}\), and \((200\,\text{cm} - 60\,\text{cm}) \div 3\) is about \(50\,\text{cm}\). The total should be about \(150\,\text{cm}\). 2. Multiply: \(25\,\text{cm} \times 4 = 100\,\text{cm}\). 3. Convert and evaluate the parentheses: \(1.8\,\text{m} = 180\,\text{cm}\), so \(180\,\text{cm} - 60\,\text{cm} = 120\,\text{cm}\). 4. Divide: \(120\,\text{cm} \div 3 = 40\,\text{cm}\). 5. Add: \(100\,\text{cm} + 40\,\text{cm} = 140\,\text{cm}\), or \(1.4\,\text{m}\).

Answer

Estimate: about \(150\,\text{cm}\) Exact: \(140\,\text{cm}\), or \(1.4\,\text{m}\)
5207275
During a school fundraiser walk, Team A travels \(6.2\,\text{km}\). Team B travels \(450\,\text{m}\) farther than Team A. Team C travels half as far as Team A. Give all three distances in kilometers.

Hints

- Express distances in the same unit before adding or dividing. - “Half as far” means divide by \(2\). - Convert each final distance back to kilometers.

Solution

1. Team A travels \(6.2\,\text{km}\). 2. Convert Team A’s distance to meters: \(6.2\,\text{km} = 6200\,\text{m}\). Team B travels \(6200 + 450 = 6650\,\text{m} = 6.65\,\text{km}\). 3. Team C travels half of Team A’s distance: \(6200 \div 2 = 3100\,\text{m} = 3.1\,\text{km}\).

Answer

Team A: \(6.2\,\text{km}\) Team B: \(6.65\,\text{km}\) Team C: \(3.1\,\text{km}\)
5207285
Three crates have different masses. Find the missing masses for Crates 2 and 3 in grams. <table> <tr><th>Crate</th><th>Mass</th></tr> <tr><td>Crate 1</td><td>\(3.6\,\text{kg}\)</td></tr> <tr><td>Crate 2</td><td>\(150\,\text{g}\) less than Crate 1</td></tr> <tr><td>Crate 3</td><td>Half the mass of Crate 1</td></tr> </table>

Hints

- Convert Crate 1’s mass to grams. - “Less than” means subtract. - Find half by dividing by \(2\).

Solution

1. Convert Crate 1’s mass: \(3.6\,\text{kg} = 3600\,\text{g}\). 2. Crate 2 has a mass of \(3600\,\text{g} - 150\,\text{g} = 3450\,\text{g}\). 3. Crate 3 has a mass of \(3600\,\text{g} \div 2 = 1800\,\text{g}\).

Answer

Crate 2: \(3450\,\text{g}\) Crate 3: \(1800\,\text{g}\)
5207295
A bucket contains \(4.8\,\text{L}\) of rainwater. A watering can holds \(1500\,\text{mL}\) more than the bucket. A small pitcher holds exactly one-third as much water as the bucket. How many liters of water are in the three containers altogether?

Hints

- Find the amount in each container separately. - Convert liters to milliliters before calculating. - One-third means divide by \(3\). - Add all amounts and convert back to liters.

Solution

1. Convert the bucket amount: \(4.8\,\text{L} = 4800\,\text{mL}\). 2. The watering can holds \(4800 + 1500 = 6300\,\text{mL}\). 3. The pitcher holds \(4800 \div 3 = 1600\,\text{mL}\). 4. Add: \(4800 + 6300 + 1600 = 12{,}700\,\text{mL}\). 5. Convert: \(12{,}700\,\text{mL} = 12.7\,\text{L}\).

Answer

The three containers hold \(12.7\,\text{L}\) altogether.
5207305
A hiking trail is \(15\,\text{km}\,600\,\text{m}\) long and is divided into equal sections of \(1300\,\text{m}\). a) Determine whether the trail has \(8\), \(10\), or \(12\) sections. b) A hiker’s average step length is \(65\,\text{cm}\). How many steps are needed for one section?

Hints

- Convert the trail length and section length to the same unit. - Divide the total length by one section’s length. - Convert the section length to centimeters before using the step length.

Solution

1. Convert the total trail length to meters: \(15\,\text{km}\,600\,\text{m} = 15{,}600\,\text{m}\). 2. Divide by the section length: \(15{,}600 \div 1300 = 12\), so the trail has \(12\) sections. 3. Convert one section to centimeters: \(1300\,\text{m} = 130{,}000\,\text{cm}\). 4. Divide by the step length: \(130{,}000 \div 65 = 2000\).

Answer

a) \(12\) sections b) \(2000\) steps per section
5208575
Convert each mass to the unit in parentheses. a) \(8\,\text{kg}\ 150\,\text{g}\) (grams) b) \(450\,\text{g}\) (kilograms) c) \(1.5\,\text{g}\) (milligrams) d) \(7200\,\text{mg}\) (grams)

Hints

- Decide whether the target unit is larger or smaller. - Use a conversion factor of \(1000\) between kilograms and grams and between grams and milligrams. - For a mixed measurement, convert the larger-unit part and then add.

Solution

1. \(8\,\text{kg}=8000\,\text{g}\). Then \(8000+150=8150\), so the total is \(8150\,\text{g}\). 2. \(450 \div 1000=0.45\), so \(450\,\text{g}=0.45\,\text{kg}\). 3. \(1.5 \times 1000=1500\), so \(1.5\,\text{g}=1500\,\text{mg}\). 4. \(7200 \div 1000=7.2\), so \(7200\,\text{mg}=7.2\,\text{g}\).

Answer

a) \(8150\,\text{g}\) b) \(0.45\,\text{kg}\) c) \(1500\,\text{mg}\) d) \(7.2\,\text{g}\)
5208585
Convert each length to the unit in parentheses. a) \(2.75\,\text{km}\) (meters) b) \(80\,\text{cm}\) (meters) c) \(4\,\text{m}\ 3\,\text{cm}\) (centimeters) d) \(5\,\text{cm}\) (millimeters)

Hints

- Use conversion factors of \(10\), \(100\), or \(1000\). - Multiplying by a power of ten moves the decimal point to the right. - Dividing by a power of ten moves the decimal point to the left.

Solution

1. \(2.75 \times 1000=2750\), so \(2.75\,\text{km}=2750\,\text{m}\). 2. \(80 \div 100=0.8\), so \(80\,\text{cm}=0.8\,\text{m}\). 3. \(4\,\text{m}=400\,\text{cm}\). Then \(400+3=403\), so the total is \(403\,\text{cm}\). 4. \(5 \times 10=50\), so \(5\,\text{cm}=50\,\text{mm}\).

Answer

a) \(2750\,\text{m}\) b) \(0.8\,\text{m}\) c) \(403\,\text{cm}\) d) \(50\,\text{mm}\)
5208745
A student converted each mass to mixed units. Check each result. Identify the incorrect results and write the correct conversion. a) \(4.5\,\text{kg}=4\,\text{kg}\ 5\,\text{g}\) b) \(0.025\,\text{kg}=25\,\text{g}\) c) \(7.08\,\text{g}=7\,\text{g}\ 80\,\text{mg}\) d) \(1.3\,\text{kg}=1\,\text{kg}\ 3\,\text{g}\)

Hints

- Check each conversion independently. - Multiply the decimal part by \(1000\) to convert to the next smaller mass unit. - Pay attention to tenths, hundredths, and thousandths.

Solution

1. For a), \(0.5\,\text{kg}=500\,\text{g}\), not \(5\,\text{g}\). The correct conversion is \(4\,\text{kg}\ 500\,\text{g}\). 2. For b), \(0.025 \times 1000=25\), so \(0.025\,\text{kg}=25\,\text{g}\). This is correct. 3. For c), \(0.08 \times 1000=80\), so \(7.08\,\text{g}=7\,\text{g}\ 80\,\text{mg}\). This is correct. 4. For d), \(0.3\,\text{kg}=300\,\text{g}\), not \(3\,\text{g}\). The correct conversion is \(1\,\text{kg}\ 300\,\text{g}\).

Answer

a) Incorrect. \(4.5\,\text{kg}=4\,\text{kg}\ 500\,\text{g}\) b) Correct. c) Correct. d) Incorrect. \(1.3\,\text{kg}=1\,\text{kg}\ 300\,\text{g}\)
5208755
Complete the table for each mass. <table> <tr> <td>Mass in grams</td> <td>\(7040\,\text{g}\)</td> <td></td> <td></td> <td>\(12{,}300\,\text{g}\)</td> </tr> <tr> <td>Mass in kilograms and grams</td> <td></td> <td>\(5\,\text{kg}\ 8\,\text{g}\)</td> <td></td> <td></td> </tr> <tr> <td>Decimal mass in kilograms</td> <td></td> <td></td> <td>\(0.015\,\text{kg}\)</td> <td></td> </tr> </table>

Hints

- Use \(1000\,\text{g}=1\,\text{kg}\). - Pay attention to placeholder zeros when the gram part is less than \(100\). - Dividing grams by \(1000\) gives kilograms.

Solution

1. \(7040\,\text{g}=7\,\text{kg}\ 40\,\text{g}=7.04\,\text{kg}\). 2. \(5\,\text{kg}\ 8\,\text{g}=5008\,\text{g}=5.008\,\text{kg}\). 3. \(0.015\,\text{kg}=15\,\text{g}\), so the mixed-unit form is simply \(15\,\text{g}\). 4. \(12{,}300\,\text{g}=12\,\text{kg}\ 300\,\text{g}=12.3\,\text{kg}\).

Answer

<table> <tr> <td>Mass in grams</td> <td>\(7040\,\text{g}\)</td> <td>\(5008\,\text{g}\)</td> <td>\(15\,\text{g}\)</td> <td>\(12{,}300\,\text{g}\)</td> </tr> <tr> <td>Mass in kilograms and grams</td> <td>\(7\,\text{kg}\ 40\,\text{g}\)</td> <td>\(5\,\text{kg}\ 8\,\text{g}\)</td> <td>\(15\,\text{g}\)</td> <td>\(12\,\text{kg}\ 300\,\text{g}\)</td> </tr> <tr> <td>Decimal mass in kilograms</td> <td>\(7.04\,\text{kg}\)</td> <td>\(5.008\,\text{kg}\)</td> <td>\(0.015\,\text{kg}\)</td> <td>\(12.3\,\text{kg}\)</td> </tr> </table>
5208775
A student wrote three chains of equivalent masses, but each row contains exactly one error. Find and correct the incorrect measurement. a) \(4500\,\text{g}=4\,\text{kg}\ 500\,\text{g}=45\,\text{kg}\) b) \(3\,\text{kg}\ 20\,\text{g}=3020\,\text{g}=3.2\,\text{kg}\) c) \(6\,\text{kg}\ 4\,\text{g}=6.004\,\text{kg}=604\,\text{g}\)

Hints

- Check each conversion independently. - Track the number of zeros and the decimal point carefully. - Convert mixed units to a single unit before comparing equivalent forms.

Solution

1. In a), \(45\,\text{kg}\) is incorrect. \(4500\,\text{g}=4.5\,\text{kg}\). 2. In b), \(3.2\,\text{kg}\) is incorrect. \(3020\,\text{g}=3.02\,\text{kg}\). 3. In c), \(604\,\text{g}\) is incorrect. \(6\,\text{kg}\ 4\,\text{g}=6004\,\text{g}\).

Answer

a) Replace \(45\,\text{kg}\) with \(4.5\,\text{kg}\). b) Replace \(3.2\,\text{kg}\) with \(3.02\,\text{kg}\). c) Replace \(604\,\text{g}\) with \(6004\,\text{g}\).
5208785
Order the lengths from least to greatest. \(40.5\,\text{cm}\); \(4005\,\text{mm}\); \(4\,\text{m}\ 5\,\text{cm}\); \(0.0041\,\text{km}\); \(420\,\text{cm}\); \(4.5\,\text{m}\)

Hints

- Choose one common unit for all six lengths. - Use \(10\,\text{mm}=1\,\text{cm}\), \(1000\,\text{mm}=1\,\text{m}\), and \(1{,}000{,}000\,\text{mm}=1\,\text{km}\). - Record the converted values before ordering the original forms.

Solution

1. Convert to millimeters: \(40.5\,\text{cm}=405\,\text{mm}\); \(4005\,\text{mm}=4005\,\text{mm}\); \(4\,\text{m}\ 5\,\text{cm}=4050\,\text{mm}\); \(0.0041\,\text{km}=4100\,\text{mm}\); \(420\,\text{cm}=4200\,\text{mm}\); and \(4.5\,\text{m}=4500\,\text{mm}\). 2. Compare: \(405<4005<4050<4100<4200<4500\).

Answer

\(40.5\,\text{cm}<4005\,\text{mm}<4\,\text{m}\ 5\,\text{cm}<0.0041\,\text{km}<420\,\text{cm}<4.5\,\text{m}\)
5208795
Order the masses from least to greatest. \(850\,\text{g}\); \(0.805\,\text{kg}\); \(8\,\text{kg}\); \(8050\,\text{g}\); \(8.5\,\text{kg}\); \(85\,\text{g}\)

Hints

- Convert every mass to grams. - Use \(1000\,\text{g}=1\,\text{kg}\). - Compare the numerical values after the units match.

Solution

1. Convert to grams: \(850\,\text{g}=850\,\text{g}\); \(0.805\,\text{kg}=805\,\text{g}\); \(8\,\text{kg}=8000\,\text{g}\); \(8050\,\text{g}=8050\,\text{g}\); \(8.5\,\text{kg}=8500\,\text{g}\); and \(85\,\text{g}=85\,\text{g}\). 2. Compare: \(85<805<850<8000<8050<8500\).

Answer

\(85\,\text{g}<0.805\,\text{kg}<850\,\text{g}<8\,\text{kg}<8050\,\text{g}<8.5\,\text{kg}\)
5208805
Order the capacities from least to greatest. \(1250\,\text{mL}\); \(1.2\,\text{L}\); \(0.125\,\text{L}\); \(1025\,\text{mL}\); \(1.205\,\text{L}\); \(12\,\text{L}\)

Hints

- Convert every liter measurement to milliliters. - Use \(1000\,\text{mL}=1\,\text{L}\). - Compare close values carefully by writing all milliliter amounts.

Solution

1. Convert to milliliters: \(1250\,\text{mL}\); \(1.2\,\text{L}=1200\,\text{mL}\); \(0.125\,\text{L}=125\,\text{mL}\); \(1025\,\text{mL}\); \(1.205\,\text{L}=1205\,\text{mL}\); and \(12\,\text{L}=12{,}000\,\text{mL}\). 2. Compare: \(125<1025<1200<1205<1250<12{,}000\).

Answer

\(0.125\,\text{L}<1025\,\text{mL}<1.2\,\text{L}<1.205\,\text{L}<1250\,\text{mL}<12\,\text{L}\)
5208835
Calculate and give each result in kilograms. a) \(14.8\,\text{kg} + 350\,\text{g}\) b) \(2.05\,\text{t} - 80\,\text{kg}\) c) \(1.2\,\text{kg} + 400\,\text{g} + 0.85\,\text{kg}\)

Hints

- Use \(1000\,\text{g} = 1\,\text{kg}\) and \(1000\,\text{kg} = 1\,\text{t}\). - Convert every value to kilograms before calculating. - Align decimal points when adding.

Solution

1. Convert to kilograms: \(350\,\text{g} = 0.35\,\text{kg}\), \(2.05\,\text{t} = 2050\,\text{kg}\), and \(400\,\text{g} = 0.4\,\text{kg}\). 2. For a), \(14.8\,\text{kg} + 0.35\,\text{kg} = 15.15\,\text{kg}\). 3. For b), \(2050\,\text{kg} - 80\,\text{kg} = 1970\,\text{kg}\). 4. For c), \(1.2\,\text{kg} + 0.4\,\text{kg} + 0.85\,\text{kg} = 2.45\,\text{kg}\).

Answer

a) \(15.15\,\text{kg}\) b) \(1970\,\text{kg}\) c) \(2.45\,\text{kg}\)
5208845
Calculate and give each result in kilograms. a) \(12\frac{1}{2}\,\text{kg} - 4.25\,\text{kg} - 750\,\text{g}\) b) \(3\,\text{t} - (1200\,\text{kg} + 400\,\text{kg} \times 2)\)

Hints

- Rewrite familiar fractions as decimals when useful. - Follow the grouping symbols and order of operations. - Convert tons to kilograms before calculating.

Solution

1. For a), convert \(12\frac{1}{2}\,\text{kg} = 12.5\,\text{kg}\) and \(750\,\text{g} = 0.75\,\text{kg}\). Then \(12.5 - 4.25 - 0.75 = 7.5\), so the result is \(7.5\,\text{kg}\). 2. For b), convert \(3\,\text{t} = 3000\,\text{kg}\). 3. Evaluate the parentheses: \(1200\,\text{kg} + 400\,\text{kg} \times 2 = 1200\,\text{kg} + 800\,\text{kg} = 2000\,\text{kg}\). 4. Subtract: \(3000\,\text{kg} - 2000\,\text{kg} = 1000\,\text{kg}\).

Answer

a) \(7.5\,\text{kg}\) b) \(1000\,\text{kg}\)
5208855
Calculate and give each final result in kilograms. a) \(5.4\,\text{kg} + (8.2\,\text{kg} - 6500\,\text{g}) \times 5\) b) \(0.6\,\text{t} - 150\frac{1}{2}\,\text{kg} - 4500\,\text{g}\)

Hints

- Evaluate the parentheses before multiplying. - The factor \(5\) applies to the entire result inside the parentheses. - Convert tons and grams to kilograms before subtracting.

Solution

1. For a), convert \(6500\,\text{g} = 6.5\,\text{kg}\). 2. Evaluate: \(8.2\,\text{kg} - 6.5\,\text{kg} = 1.7\,\text{kg}\), then \(1.7\,\text{kg} \times 5 = 8.5\,\text{kg}\). 3. Add: \(5.4\,\text{kg} + 8.5\,\text{kg} = 13.9\,\text{kg}\). 4. For b), convert \(0.6\,\text{t} = 600\,\text{kg}\), \(150\frac{1}{2}\,\text{kg} = 150.5\,\text{kg}\), and \(4500\,\text{g} = 4.5\,\text{kg}\). 5. Subtract: \(600\,\text{kg} - 150.5\,\text{kg} - 4.5\,\text{kg} = 445\,\text{kg}\).

Answer

a) \(13.9\,\text{kg}\) b) \(445\,\text{kg}\)
5209005
Evaluate: \(12\,\text{kg}\,750\,\text{g} \div 250\,\text{g} + 8\,\text{g}\,40\,\text{mg} \div 20\,\text{mg} - 15\).

Hints

- When one measurement is divided by another in the same unit, the units cancel. - Convert each mixed measurement to the smaller unit used in its division. - Complete the divisions before addition and subtraction.

Solution

1. Convert for each division: \(12\,\text{kg}\,750\,\text{g} = 12{,}750\,\text{g}\) and \(8\,\text{g}\,40\,\text{mg} = 8040\,\text{mg}\). 2. Divide: \(12{,}750\,\text{g} \div 250\,\text{g} = 51\) and \(8040\,\text{mg} \div 20\,\text{mg} = 402\). 3. Add and subtract: \(51 + 402 - 15 = 438\).

Answer

\(438\)
5209025
Evaluate: \(15\,\text{kg} - (3\,\text{kg}\,450\,\text{g} + 720\,\text{g} \times 5)\).

Hints

- Evaluate the parentheses first. - Inside the parentheses, multiply before adding. - Converting all masses to grams can simplify the calculation.

Solution

1. Multiply inside the parentheses: \(720\,\text{g} \times 5 = 3600\,\text{g}\). 2. Convert and add inside the parentheses: \(3450\,\text{g} + 3600\,\text{g} = 7050\,\text{g}\). 3. Convert the starting mass: \(15\,\text{kg} = 15{,}000\,\text{g}\). 4. Subtract: \(15{,}000\,\text{g} - 7050\,\text{g} = 7950\,\text{g}\). 5. Convert: \(7950\,\text{g} = 7\,\text{kg}\,950\,\text{g}\).

Answer

\(7\,\text{kg}\,950\,\text{g}\)
5209045
Two boards have a total length of \(4.50\,\text{m}\). One board is \(70\,\text{cm}\) shorter than the other. Find the length of each board in meters.

Hints

- Express the total and difference in the same unit. - Remove the extra length so the two parts would be equal. - Divide the remaining total into two equal lengths.

Solution

1. Convert the difference: \(70\,\text{cm} = 0.70\,\text{m}\). 2. Subtract the difference from the total: \(4.50\,\text{m} - 0.70\,\text{m} = 3.80\,\text{m}\). 3. The result is twice the shorter board’s length, so \(3.80\,\text{m} \div 2 = 1.90\,\text{m}\). 4. Add the difference to find the longer board: \(1.90\,\text{m} + 0.70\,\text{m} = 2.60\,\text{m}\).

Answer

The shorter board is \(1.90\,\text{m}\), and the longer board is \(2.60\,\text{m}\).
5209145
Convert each mass to the unit in parentheses. Pay attention to fractions and mixed units. a) \(\frac{3}{4}\,\text{kg}\) (grams) b) \(5\,\text{kg}\ 20\,\text{g}\) (kilograms) c) \(12\,\text{kg}\ 500\,\text{g}\) (kilograms) d) \(2\,\text{kg}\ \frac{1}{2}\,\text{g}\) (grams)

Hints

- Convert the fraction to a smaller unit when helpful. - For mixed units, convert each part to the target unit and add. - Use \(1000\,\text{g}=1\,\text{kg}\).

Solution

1. \(\frac{1}{4}\,\text{kg}=250\,\text{g}\), so \(\frac{3}{4}\,\text{kg}=750\,\text{g}\). 2. \(20\,\text{g}=0.02\,\text{kg}\), so the total is \(5.02\,\text{kg}\). 3. \(500\,\text{g}=0.5\,\text{kg}\), so the total is \(12.5\,\text{kg}\). 4. \(2\,\text{kg}=2000\,\text{g}\). Adding \(\frac{1}{2}\,\text{g}\) gives \(2000.5\,\text{g}\).

Answer

a) \(750\,\text{g}\) b) \(5.02\,\text{kg}\) c) \(12.5\,\text{kg}\) d) \(2000.5\,\text{g}\)
5209245
A charter bus is preparing for a class trip with \(48\) students and \(2\) teachers. Use an average person weight of \(120\,\text{lb}\). Each person may also bring a suitcase weighing up to \(30\,\text{lb}\) and a carry-on weighing \(10\,\text{lb}\). a) If everyone uses the full luggage allowance, what total weight do the people and luggage add to the bus? b) The unloaded bus weighs \(13\) tons. What is the fully loaded bus’s total weight in tons? Use \(1\,\text{ton} = 2000\,\text{lb}\). c) Ten people each have a suitcase that is \(4\,\text{lb}\) over the limit. By how many pounds does this increase the answer to part a)?

Hints

- Find the total number of people first. - Find the weight added by one person and that person’s luggage. - Convert between tons and pounds for part b). - For part c), calculate only the new excess weight.

Solution

1. There are \(48 + 2 = 50\) people. 2. Each person and full luggage allowance contribute \(120 + 30 + 10 = 160\,\text{lb}\). 3. For a), \(50 \times 160\,\text{lb} = 8000\,\text{lb}\). 4. For b), the unloaded bus weighs \(13 \times 2000 = 26{,}000\,\text{lb}\). The loaded weight is \(26{,}000 + 8000 = 34{,}000\,\text{lb}\), which is \(34{,}000 \div 2000 = 17\) tons. 5. For c), the extra luggage adds \(10 \times 4\,\text{lb} = 40\,\text{lb}\).

Answer

a) \(8000\,\text{lb}\) b) \(17\) tons c) \(40\,\text{lb}\)
5209525
Evaluate and give the result in kilograms: \((1\,\text{kg}\,450\,\text{g} + 850\,\text{g} - 700\,\text{g}) \times 6\).

Hints

- First convert all measurements to the same unit. - Remember to evaluate the expression inside the parentheses first. - How many grams are in \(1\,\text{kg}\)?

Solution

1. Convert the mixed measurement to grams: \(1\,\text{kg}\,450\,\text{g} = 1450\,\text{g}\). 2. Evaluate inside the parentheses: \(1450\,\text{g} + 850\,\text{g} - 700\,\text{g} = 1600\,\text{g}\). 3. Multiply: \(1600\,\text{g} \times 6 = 9600\,\text{g}\). 4. Convert to kilograms: \(9600\,\text{g} = 9.6\,\text{kg}\).

Answer

\(9.6\,\text{kg}\)
5209535
Evaluate: \(7.2\,\text{kg} - (2\frac{1}{4}\,\text{kg} + 1\,\text{kg}\,850\,\text{g} + 450\,\text{g})\).

Hints

- Convert the fraction of a kilogram to grams. - Convert every measurement to the same unit before calculating. - Evaluate the expression inside the parentheses first.

Solution

1. Convert each measurement to grams: \(7.2\,\text{kg} = 7200\,\text{g}\), \(2\frac{1}{4}\,\text{kg} = 2250\,\text{g}\), and \(1\,\text{kg}\,850\,\text{g} = 1850\,\text{g}\). 2. Add inside the parentheses: \(2250\,\text{g} + 1850\,\text{g} + 450\,\text{g} = 4550\,\text{g}\). 3. Subtract: \(7200\,\text{g} - 4550\,\text{g} = 2650\,\text{g}\). 4. Convert the result: \(2650\,\text{g} = 2\,\text{kg}\,650\,\text{g} = 2.65\,\text{kg}\).

Answer

\(2\,\text{kg}\,650\,\text{g}\), or \(2.65\,\text{kg}\)
5209545
Find the sum: \(4\,\text{kg}\,200\,\text{g} \div 60 + 12\,\text{kg} \div 80\).

Hints

- Perform each division before adding. - Convert the kilogram measurements to grams before dividing. - Look for common factors of \(10\) to simplify each division.

Solution

1. Convert the first measurement to grams and divide: \(4\,\text{kg}\,200\,\text{g} = 4200\,\text{g}\), so \(4200\,\text{g} \div 60 = 70\,\text{g}\). 2. Convert the second measurement to grams and divide: \(12\,\text{kg} = 12{,}000\,\text{g}\), so \(12{,}000\,\text{g} \div 80 = 150\,\text{g}\). 3. Add the two quotients: \(70\,\text{g} + 150\,\text{g} = 220\,\text{g}\).

Answer

\(220\,\text{g}\)
5209745
Write each time interval entirely in the smaller unit shown. a) \(4\,\text{min}\ 12\,\text{s}\) b) \(2\,\text{h}\ 45\,\text{min}\) c) \(1\,\text{day}\ 10\,\text{h}\)

Hints

- Convert the larger-unit part to the smaller unit first. - Then add the remaining smaller-unit amount. - Use \(60\,\text{s}=1\,\text{min}\), \(60\,\text{min}=1\,\text{h}\), and \(24\,\text{h}=1\,\text{day}\).

Solution

1. \(4\,\text{min}=240\,\text{s}\). Then \(240+12=252\), so the interval is \(252\,\text{s}\). 2. \(2\,\text{h}=120\,\text{min}\). Then \(120+45=165\), so the interval is \(165\,\text{min}\). 3. \(1\,\text{day}=24\,\text{h}\). Then \(24+10=34\), so the interval is \(34\,\text{h}\).

Answer

a) \(252\,\text{s}\) b) \(165\,\text{min}\) c) \(34\,\text{h}\)
5209885
Write each mixed measurement entirely in the smaller unit. a) \(2\,\text{kg}\ 450\,\text{g}\) b) \(15\,\text{kg}\ 30\,\text{g}\) c) \(9\,\text{km}\ 75\,\text{m}\) d) \(4\,\text{m}\ 25\,\text{cm}\)

Hints

- Convert the larger-unit part to the smaller unit. - Then add the remaining amount. - Use the factors \(100\) or \(1000\), depending on the units.

Solution

1. \(2\,\text{kg}=2000\,\text{g}\). Then \(2000+450=2450\), so the mass is \(2450\,\text{g}\). 2. \(15\,\text{kg}=15{,}000\,\text{g}\). Then \(15{,}000+30=15{,}030\), so the mass is \(15{,}030\,\text{g}\). 3. \(9\,\text{km}=9000\,\text{m}\). Then \(9000+75=9075\), so the length is \(9075\,\text{m}\). 4. \(4\,\text{m}=400\,\text{cm}\). Then \(400+25=425\), so the length is \(425\,\text{cm}\).

Answer

a) \(2450\,\text{g}\) b) \(15{,}030\,\text{g}\) c) \(9075\,\text{m}\) d) \(425\,\text{cm}\)
5210135
A satellite needs exactly \(1\,\text{day}\ 5\,\text{h}\ 12\,\text{min}\) to complete a set of measurements. Find the total time in minutes.

Hints

- Use \(24\,\text{h}=1\,\text{day}\). - Convert the total hours to minutes using \(60\,\text{min}=1\,\text{h}\). - Add the remaining minutes at the end.

Solution

1. \(1\,\text{day}=24\,\text{h}\). Adding \(5\,\text{h}\) gives \(29\,\text{h}\). 2. \(29 \times 60=1740\), so \(29\,\text{h}=1740\,\text{min}\). 3. Add \(12\,\text{min}\): \(1740+12=1752\).

Answer

\(1752\,\text{min}\)
5210255
Write each time measurement using mixed units. a) \(450\,\text{s}\) b) \(190\,\text{min}\) c) \(2.5\,\text{h}\) d) \(62\,\text{h}\)

Hints

- Divide by \(60\) when separating seconds into minutes or minutes into hours. - The quotient gives the larger units, and the remainder gives the smaller units. - Use \(24\,\text{h}=1\,\text{day}\).

Solution

1. \(450 \div 60=7\) remainder \(30\), so \(450\,\text{s}=7\,\text{min}\ 30\,\text{s}\). 2. \(190 \div 60=3\) remainder \(10\), so \(190\,\text{min}=3\,\text{h}\ 10\,\text{min}\). 3. \(0.5\,\text{h}=30\,\text{min}\), so \(2.5\,\text{h}=2\,\text{h}\ 30\,\text{min}\). 4. \(62 \div 24=2\) remainder \(14\), so \(62\,\text{h}=2\,\text{days}\ 14\,\text{h}\).

Answer

a) \(7\,\text{min}\ 30\,\text{s}\) b) \(3\,\text{h}\ 10\,\text{min}\) c) \(2\,\text{h}\ 30\,\text{min}\) d) \(2\,\text{days}\ 14\,\text{h}\)
5210265
Which time measurements are equivalent? Match each letter to a number. A: \(130\,\text{s}\) B: \(2\,\text{h}\ 15\,\text{min}\) C: \(85\,\text{min}\) 1: \(135\,\text{min}\) 2: \(2\,\text{min}\ 10\,\text{s}\) 3: \(1\,\text{h}\ 25\,\text{min}\)

Hints

- Convert both measurements in a possible pair to the same unit. - You can also divide a large number of seconds or minutes by \(60\) and use the remainder.

Solution

1. A: \(130\,\text{s}=2\,\text{min}\ 10\,\text{s}\), so A matches 2. 2. B: \(2\,\text{h}=120\,\text{min}\), and \(120+15=135\), so B matches 1. 3. C: \(85\,\text{min}=1\,\text{h}\ 25\,\text{min}\), so C matches 3.

Answer

A–2; B–1; C–3
5210275
A computer needs \(3000\,\text{min}\) to complete a large update. Write this time in days, hours, and minutes.

Hints

- First convert minutes to hours. - Then determine how many full \(24\)-hour days fit in that number of hours. - Use each remainder as the amount in the smaller unit.

Solution

1. \(3000 \div 60=50\), so the update takes \(50\,\text{h}\). 2. \(50 \div 24=2\) remainder \(2\), so \(50\,\text{h}=2\,\text{days}\ 2\,\text{h}\). 3. There is no remainder from the minutes-to-hours conversion, so there are \(0\,\text{min}\).

Answer

\(2\,\text{days}\ 2\,\text{h}\ 0\,\text{min}\)
5210535
Evaluate each expression. a) \(2\,\text{h}-\frac{1}{4}\,\text{h}-55\,\text{min}\) b) \(3\times(12\,\text{min}\ 45\,\text{s})\) c) \(5\,\text{kg}\div 125\,\text{g}\) d) \(3\,\text{km}\ 50\,\text{m}-1\,\text{km}\ 200\,\text{m}\)

Hints

- Convert measurements to the smaller unit before calculating. - Convert the fractional hour to minutes before subtracting. - When dividing two masses, convert both to the same unit first. - For the length subtraction, convert both mixed lengths to meters.

Solution

1. Convert to minutes: \(120\,\text{min}-15\,\text{min}-55\,\text{min}=50\,\text{min}\). 2. \(3\times 12\,\text{min}=36\,\text{min}\), and \(3\times 45\,\text{s}=135\,\text{s}=2\,\text{min}\ 15\,\text{s}\). The total is \(38\,\text{min}\ 15\,\text{s}\). 3. \(5\,\text{kg}=5000\,\text{g}\). Then \(5000\,\text{g}\div 125\,\text{g}=40\). 4. Convert to meters: \(3050\,\text{m}-1200\,\text{m}=1850\,\text{m}=1\,\text{km}\ 850\,\text{m}\).

Answer

a) \(50\,\text{min}\) b) \(38\,\text{min}\ 15\,\text{s}\) c) \(40\) d) \(1\,\text{km}\ 850\,\text{m}\)
5210545
A delivery van carries packages that each weigh \(5\,\text{lb}\,8\,\text{oz}\). a) What is the total weight of \(18\) packages? b) The van can carry at most \(1320\,\text{lb}\) of cargo. What is the greatest number of these packages it can carry?

Hints

- Convert the ounces in one package’s weight to a fraction or decimal part of a pound. - For part b), divide the maximum cargo weight by the weight of one package. - Think about how many times one package’s weight fits into the cargo limit.

Solution

1. Convert one package’s weight to pounds: \(8\,\text{oz} = 0.5\,\text{lb}\), so each package weighs \(5.5\,\text{lb}\). 2. a) Multiply: \(18 \times 5.5\,\text{lb} = 99\,\text{lb}\). 3. b) Divide the cargo limit by the weight of one package: \(1320\,\text{lb} \div 5.5\,\text{lb} = 240\). 4. The van can carry exactly \(240\) packages without exceeding the limit.

Answer

a) \(99\,\text{lb}\) b) \(240\) packages
5210555
Evaluate both sides and write \(<\), \(>\), or \(=\). Show your calculations. a) \(10\,\text{h}\div 15\ \square\ 45\,\text{min}\) b) \(0.5\,\text{km}\times 8\ \square\ 3\,\text{km}\ 900\,\text{m}\) c) \(4\,\text{kg}-1\,\text{kg}\ 200\,\text{g}\ \square\ 14\times 200\,\text{g}\)

Hints

- Evaluate the left side first. - Convert both values to the same unit before comparing. - Use \(1000\,\text{m}=1\,\text{km}\) when converting between kilometers and meters.

Solution

1. \(10\,\text{h}=600\,\text{min}\). Then \(600\div 15=40\,\text{min}\), and \(40<45\), so the symbol is \(<\). 2. \(0.5\,\text{km}=500\,\text{m}\). Then \(500\times 8=4000\,\text{m}=4\,\text{km}\), which is greater than \(3\,\text{km}\ 900\,\text{m}\), so the symbol is \(>\). 3. Left side: \(4000\,\text{g}-1200\,\text{g}=2800\,\text{g}\). Right side: \(14\times 200\,\text{g}=2800\,\text{g}\). The symbol is \(=\).

Answer

a) \(<\) b) \(>\) c) \(=\)
5210825
An elephant at a zoo is exactly \(2000\) days old. How many years and days is that? Use \(365\) days for one year.

Hints

- Use \(365\) days for one year. - Divide the total number of days by \(365\). - The remainder is the number of extra days.

Solution

1. \(2000 \div 365=5\) remainder \(175\). 2. Check: \(5 \times 365=1825\), and \(2000-1825=175\). 3. Therefore, the elephant is \(5\) years and \(175\) days old.

Answer

\(5\) years and \(175\) days
5210845
A record says that an olive tree was planted \(1150\) months ago. A gardener claims, “The tree is already more than \(95\) years old.” Is the gardener correct? Justify your answer with a calculation.

Hints

- Use \(12\) months for one year. - You may convert \(95\) years to months or convert \(1150\) months to years. - Compare the values after they use the same unit.

Solution

1. Convert \(95\) years to months: \(95 \times 12=1140\) months. 2. Since \(1150>1140\), the tree is older than \(95\) years. 3. In mixed units, \(1150 \div 12=95\) remainder \(10\), so the age is \(95\) years and \(10\) months.

Answer

Yes. The tree is \(95\) years and \(10\) months old, which is more than \(95\) years.
5211135
Find the value of the quotient: \(4\,\text{hr}\,30\,\text{min} \div 18\,\text{min}\)

Hints

- What happens to the unit when one duration is divided by another duration in the same unit? - Convert both quantities to minutes first. - How many minutes are in one hour?

Solution

1. Convert the first duration to minutes: \(4 \times 60 + 30 = 240 + 30 = 270\) minutes. 2. Divide durations with the same unit: \(270\,\text{min} \div 18\,\text{min} = 15\). 3. The minutes cancel, so the quotient has no unit.

Answer

\(15\)
5211145
Find the sum: \(1\,\text{day}\,14\,\text{hr}\,45\,\text{min} + 19\,\text{hr}\,30\,\text{min}\)

Hints

- Add one unit at a time, beginning with minutes. - Regroup \(60\) minutes as \(1\) hour and \(24\) hours as \(1\) day. - Check whether each unit needs to be regrouped.

Solution

1. Add the minutes: \(45 + 30 = 75\) minutes, which is \(1\) hour \(15\) minutes. 2. Add the hours, including the regrouped hour: \(14 + 19 + 1 = 34\) hours, which is \(1\) day \(10\) hours. 3. Add the days: \(1 + 1 = 2\) days. 4. The sum is \(2\) days, \(10\) hours, and \(15\) minutes.

Answer

\(2\,\text{days}\,10\,\text{hr}\,15\,\text{min}\)
5211385
Write each mixed measurement as a decimal in the larger unit. Example: \(3\,\text{m}\ 4\,\text{cm}=3.04\,\text{m}\). a) \(8\,\text{km}\ 20\,\text{m}\) b) \(5\,\text{kg}\ 7\,\text{g}\) c) \(12\,\text{m}\ 40\,\text{cm}\) d) \(\$2\) and \(5\) cents

Hints

- The number of decimal places depends on the conversion factor. - Use three decimal places when converting meters to kilometers or grams to kilograms. - Use two decimal places for cents in a dollar amount.

Solution

1. \(20\,\text{m}=0.02\,\text{km}\), so the total is \(8.02\,\text{km}\). 2. \(7\,\text{g}=0.007\,\text{kg}\), so the total is \(5.007\,\text{kg}\). 3. \(40\,\text{cm}=0.4\,\text{m}\), so the total is \(12.4\,\text{m}\). 4. Five cents is \(\$0.05\), so the total is \(\$2.05\).

Answer

a) \(8.02\,\text{km}\) b) \(5.007\,\text{kg}\) c) \(12.4\,\text{m}\) d) \(\$2.05\)
5211435
Fill in each missing number or unit. a) \(4\,\text{km}\ 12\,\text{m}=\square\,\text{m}\) b) \(7\,\text{kg}\ \square\,\text{g}=7005\,\text{g}\) c) \(2\,\text{m}\ 3\,\text{cm}\ 5\,\text{mm}=\square\,\text{mm}\) d) \(12\,\text{kg}\ 80\,\text{g}=12{,}080\,\square\)

Hints

- Convert every part of a mixed measurement to the same unit. - Use subtraction to find a missing remainder. - Track the zeros carefully when converting kilometers to meters or kilograms to grams.

Solution

1. \(4\,\text{km}=4000\,\text{m}\). Then \(4000+12=4012\), so the missing number is \(4012\). 2. \(7\,\text{kg}=7000\,\text{g}\). The difference \(7005-7000=5\), so the missing number is \(5\). 3. \(2\,\text{m}=2000\,\text{mm}\) and \(3\,\text{cm}=30\,\text{mm}\). Then \(2000+30+5=2035\). 4. \(12\,\text{kg}=12{,}000\,\text{g}\). Adding \(80\,\text{g}\) gives \(12{,}080\,\text{g}\), so the missing unit is grams.

Answer

a) \(4012\) b) \(5\) c) \(2035\) d) grams
5211455
Find each missing value. a) \(3\,\text{days}\ 4\,\text{h}=\square\,\text{h}\) b) \(2\,\text{h}\ 15\,\text{min}=\square\,\text{min}\) c) \(\square\,\text{h}\ 20\,\text{min}=200\,\text{min}\) d) \(1.5\,\text{h}=\square\,\text{min}\)

Hints

- Time units do not use powers of ten. - Use \(60\) between seconds and minutes or minutes and hours, and \(24\) between hours and days. - One-half hour is \(30\) minutes.

Solution

1. \(3\,\text{days}=72\,\text{h}\). Then \(72+4=76\), so the missing value is \(76\). 2. \(2\,\text{h}=120\,\text{min}\). Then \(120+15=135\), so the missing value is \(135\). 3. Subtract the \(20\,\text{min}\): \(200-20=180\,\text{min}\). Since \(180 \div 60=3\), the missing value is \(3\). 4. \(1.5\,\text{h}=1\,\text{h}+0.5\,\text{h}=60\,\text{min}+30\,\text{min}=90\,\text{min}\).

Answer

a) \(76\) b) \(135\) c) \(3\) d) \(90\)
5211465
Evaluate each expression. Convert units when needed. a) \(3.4\,\text{kg}+850\,\text{g}-1.02\,\text{kg}+470\,\text{g}\) b) \(12\,\text{m}\ 5\,\text{cm}-4.8\,\text{m}+72\,\text{cm}\) c) \((4.2\,\text{kg}+580\,\text{g})\times 15\)

Hints

- Convert all terms in an expression to one unit. - Align decimal points carefully. - A mixed meter-and-centimeter measurement can be written as a decimal in meters.

Solution

1. Convert to grams: \(3400+850-1020+470=3700\). The result is \(3700\,\text{g}=3.7\,\text{kg}\). 2. Convert to meters: \(12.05-4.8+0.72=7.25+0.72=7.97\). The result is \(7.97\,\text{m}\). 3. Convert to grams: \((4200+580)\times 15=4780\times 15=71{,}700\,\text{g}=71.7\,\text{kg}\).

Answer

a) \(3.7\,\text{kg}\) b) \(7.97\,\text{m}\) c) \(71.7\,\text{kg}\)
5211475
Evaluate each expression. Write time answers in hours and minutes. a) \(4\,\text{h}\ 12\,\text{min}-155\,\text{min}+1\,\text{h}\ 48\,\text{min}\) b) \(18\,\text{km}\div 25\,\text{m}\) c) \(144\,\text{kg}\div 12-36\,\text{kg}\div 12\)

Hints

- Convert hours to minutes before adding or subtracting. - Dividing two measurements with the same unit gives a unitless number. - In c), divide each term before subtracting.

Solution

1. Convert to minutes: \(252-155+108=97+108=205\,\text{min}\). Since \(205\,\text{min}=3\,\text{h}\ 25\,\text{min}\), that is the result. 2. \(18\,\text{km}=18{,}000\,\text{m}\). Then \(18{,}000\,\text{m}\div 25\,\text{m}=720\). 3. \(144\,\text{kg}\div 12=12\,\text{kg}\), and \(36\,\text{kg}\div 12=3\,\text{kg}\). Then \(12-3=9\,\text{kg}\).

Answer

a) \(3\,\text{h}\ 25\,\text{min}\) b) \(720\) c) \(9\,\text{kg}\)
5211595
Evaluate each expression and write the result as a decimal in the larger unit. a) \(\$18.60-475\) cents \(+\$1.15\) b) \(6.3\,\text{kg}-(2100\,\text{g}+850\,\text{g})\)

Hints

- Convert all terms to the larger unit before calculating. - Evaluate the parentheses first. - Use \(100\) cents for one dollar and \(1000\,\text{g}=1\,\text{kg}\).

Solution

1. \(475\) cents is \(\$4.75\). Then \(\$18.60-\$4.75=\$13.85\), and \(\$13.85+\$1.15=\$15.00\). 2. Inside the parentheses, \(2100\,\text{g}+850\,\text{g}=2950\,\text{g}=2.95\,\text{kg}\). Then \(6.3-2.95=3.35\), so the result is \(3.35\,\text{kg}\).

Answer

a) \(\$15.00\) b) \(3.35\,\text{kg}\)
5211605
Evaluate each expression and write the result as a decimal. a) \((15\,\text{m}\,60\,\text{cm} + 4.4\,\text{m}) \div 5\) b) \(0.85\,\text{km} - 320\,\text{m} + 1.2\,\text{km}\)

Hints

- Use the parentheses to determine the order of operations. - Express all measurements in a common unit before calculating. - Use \(100\,\text{cm} = 1\,\text{m}\) and \(1000\,\text{m} = 1\,\text{km}\).

Solution

1. For a), convert \(15\,\text{m}\,60\,\text{cm} = 15.6\,\text{m}\). Then \(15.6\,\text{m} + 4.4\,\text{m} = 20.0\,\text{m}\), and \(20.0\,\text{m} \div 5 = 4.0\,\text{m}\). 2. For b), convert to meters: \(0.85\,\text{km} = 850\,\text{m}\) and \(1.2\,\text{km} = 1200\,\text{m}\). 3. Calculate: \(850\,\text{m} - 320\,\text{m} + 1200\,\text{m} = 1730\,\text{m}\). 4. Convert to kilometers: \(1730\,\text{m} = 1.73\,\text{km}\).

Answer

a) \(4.0\,\text{m}\) b) \(1.73\,\text{km}\)
5211685
A wooden fence has \(4\) sections, each \(6\,\text{ft}\) long. The fence will be made from vertical boards that are each \(6\,\text{in}\) wide, placed directly next to one another with no gaps. How many boards are needed altogether?

Hints

- Find the total length of all fence sections. - Express the total length and board width in the same unit. - Divide the fence length by one board’s width.

Solution

1. Find the total fence length: \(4 \times 6\,\text{ft} = 24\,\text{ft}\). 2. Convert to inches: \(24 \times 12 = 288\), so the fence is \(288\,\text{in}\) long. 3. Divide by the width of one board: \(288 \div 6 = 48\).

Answer

\(48\) boards are needed.
5211815
Find the missing measurement \(\square\) that makes the equation true: \(8\,\text{kg}\,450\,\text{g} - \square + 1250\,\text{g} = 7\,\text{kg}\,100\,\text{g}\).

Hints

- Convert every measurement to grams first. - Combine the known measurements on the left side. - Work backward to find the amount that must be subtracted.

Solution

1. Convert the mixed measurements to grams: \(8\,\text{kg}\,450\,\text{g} = 8450\,\text{g}\) and \(7\,\text{kg}\,100\,\text{g} = 7100\,\text{g}\). 2. Rewrite the equation: \(8450\,\text{g} - \square + 1250\,\text{g} = 7100\,\text{g}\). 3. Combine the known amounts on the left: \(8450\,\text{g} + 1250\,\text{g} = 9700\,\text{g}\). 4. Find the missing measurement: \(9700\,\text{g} - 7100\,\text{g} = 2600\,\text{g}\). 5. Check: \(8450\,\text{g} - 2600\,\text{g} + 1250\,\text{g} = 7100\,\text{g}\).

Answer

\(\square = 2600\,\text{g}\), or \(2\,\text{kg}\,600\,\text{g}\)
5211825
Find the missing length: \(3.4\,\text{km} - 850\,\text{m} + \square = 4\,\text{km}\,20\,\text{m}\).

Hints

- Express all lengths in meters. - Calculate the known part of the left side first. - Find how much is needed to reach the right side.

Solution

1. Convert to meters: \(3.4\,\text{km} = 3400\,\text{m}\) and \(4\,\text{km}\,20\,\text{m} = 4020\,\text{m}\). 2. Subtract: \(3400\,\text{m} - 850\,\text{m} = 2550\,\text{m}\). 3. Solve \(2550\,\text{m} + \square = 4020\,\text{m}\) by subtracting: \(4020 - 2550 = 1470\). 4. Therefore, \(\square = 1470\,\text{m} = 1\,\text{km}\,470\,\text{m}\).

Answer

\(\square = 1470\,\text{m}\), or \(1\,\text{km}\,470\,\text{m}\)
5212095
Complete the missing table entries and write each mass as a decimal in kilograms. <table> <tr> <th>\(100\,\text{kg}\)</th> <th>\(10\,\text{kg}\)</th> <th>\(1\,\text{kg}\)</th> <th>\(100\,\text{g}\)</th> <th>\(10\,\text{g}\)</th> <th>\(1\,\text{g}\)</th> <th>Decimal in kilograms</th> </tr> <tr><td>4</td><td>0</td><td>7</td><td>2</td><td>0</td><td>0</td><td>\(?\,\text{kg}\)</td></tr> <tr><td></td><td>5</td><td>0</td><td>3</td><td>0</td><td>0</td><td>\(?\,\text{kg}\)</td></tr> <tr><td>0</td><td>0</td><td>0</td><td>8</td><td>1</td><td>5</td><td>\(?\,\text{kg}\)</td></tr> <tr><td>1</td><td>2</td><td>0</td><td>4</td><td>5</td><td>0</td><td>\(?\,\text{kg}\)</td></tr> </table>

Hints

- Each column gives one place in the decimal representation. - The kilogram column is the ones place. - The \(100\)-gram, \(10\)-gram, and \(1\)-gram columns are the tenths, hundredths, and thousandths places in kilograms.

Solution

1. Row 1 represents \(400\,\text{kg}+7\,\text{kg}+200\,\text{g}=407.2\,\text{kg}\). 2. Row 2 represents \(50\,\text{kg}+300\,\text{g}=50.3\,\text{kg}\). The missing first entry is \(0\). 3. Row 3 represents \(815\,\text{g}=0.815\,\text{kg}\). 4. Row 4 represents \(120\,\text{kg}+450\,\text{g}=120.45\,\text{kg}\).

Answer

<table> <tr> <th>\(100\,\text{kg}\)</th> <th>\(10\,\text{kg}\)</th> <th>\(1\,\text{kg}\)</th> <th>\(100\,\text{g}\)</th> <th>\(10\,\text{g}\)</th> <th>\(1\,\text{g}\)</th> <th>Decimal in kilograms</th> </tr> <tr><td>4</td><td>0</td><td>7</td><td>2</td><td>0</td><td>0</td><td>\(407.2\,\text{kg}\)</td></tr> <tr><td>0</td><td>5</td><td>0</td><td>3</td><td>0</td><td>0</td><td>\(50.3\,\text{kg}\)</td></tr> <tr><td>0</td><td>0</td><td>0</td><td>8</td><td>1</td><td>5</td><td>\(0.815\,\text{kg}\)</td></tr> <tr><td>1</td><td>2</td><td>0</td><td>4</td><td>5</td><td>0</td><td>\(120.45\,\text{kg}\)</td></tr> </table>
5212105
Enter each length in a place-value table with columns for kilometers, hundreds of meters, tens of meters, meters, tenths of a meter, hundredths of a meter, and thousandths of a meter. Then write each length as a decimal in meters. a) \(2\,\text{km}\ 45\,\text{m}\ 3\,\text{cm}\) b) \(80\,\text{m}\ 70\,\text{cm}\ 5\,\text{mm}\) c) \(6\,\text{km}\ 90\,\text{cm}\)

Hints

- Place a zero in any place that is not represented in the mixed measurement. - The meter column is the ones place. - Centimeters occupy the hundredths place in meters, and millimeters occupy the thousandths place.

Solution

1. For a), the place values are \(2\mid0\mid4\mid5\mid0\mid3\mid0\), giving \(2045.03\,\text{m}\). 2. For b), the place values are \(0\mid0\mid8\mid0\mid7\mid0\mid5\), giving \(80.705\,\text{m}\). 3. For c), the place values are \(6\mid0\mid0\mid0\mid9\mid0\mid0\), giving \(6000.9\,\text{m}\).

Answer

<table> <tr><th>Part</th><th>km</th><th>\(100\,\text{m}\)</th><th>\(10\,\text{m}\)</th><th>\(1\,\text{m}\)</th><th>\(0.1\,\text{m}\)</th><th>\(0.01\,\text{m}\)</th><th>\(0.001\,\text{m}\)</th><th>Decimal in meters</th></tr> <tr><td>a)</td><td>2</td><td>0</td><td>4</td><td>5</td><td>0</td><td>3</td><td>0</td><td>\(2045.03\,\text{m}\)</td></tr> <tr><td>b)</td><td>0</td><td>0</td><td>8</td><td>0</td><td>7</td><td>0</td><td>5</td><td>\(80.705\,\text{m}\)</td></tr> <tr><td>c)</td><td>6</td><td>0</td><td>0</td><td>0</td><td>9</td><td>0</td><td>0</td><td>\(6000.9\,\text{m}\)</td></tr> </table>
5212115
Three packages are weighed. Package A weighs \(1\,\text{kg}\ 25\,\text{g}\). Package B weighs \(1.205\,\text{kg}\). Package C weighs \(1250\,\text{g}\). a) Enter each mass in a place-value table with columns for \(1\,\text{kg}\), \(100\,\text{g}\), \(10\,\text{g}\), and \(1\,\text{g}\). b) Write all three masses as decimals in kilograms. c) Which package is heaviest?

Hints

- Use the gram digits to fill the hundreds, tens, and ones columns. - Write all decimal masses to three decimal places before comparing. - Compare from left to right by place value.

Solution

1. Package A has entries \(1\mid0\mid2\mid5\), so its decimal mass is \(1.025\,\text{kg}\). 2. Package B has entries \(1\mid2\mid0\mid5\), so its decimal mass is \(1.205\,\text{kg}\). 3. Package C has entries \(1\mid2\mid5\mid0\), so its decimal mass is \(1.250\,\text{kg}\). 4. Since \(1.025<1.205<1.250\), Package C is heaviest.

Answer

a) A: \(1\mid0\mid2\mid5\); B: \(1\mid2\mid0\mid5\); C: \(1\mid2\mid5\mid0\) b) A: \(1.025\,\text{kg}\); B: \(1.205\,\text{kg}\); C: \(1.250\,\text{kg}\) c) Package C is heaviest.
5212285
Evaluate both sides. Then write \(<\), \(>\), or \(=\). a) \(1\,\text{km}-450\,\text{m}\ \square\ 500\,\text{m}+50\,\text{m}\) b) \(0.38\,\text{m}+22\,\text{cm}\ \square\ 0.6\,\text{m}-5\,\text{cm}\)

Hints

- Evaluate the left and right sides separately. - Use the same unit for every measurement on one side. - Convert both final values to the same unit before comparing.

Solution

1. In a), the left side is \(1000\,\text{m}-450\,\text{m}=550\,\text{m}\). The right side is \(500\,\text{m}+50\,\text{m}=550\,\text{m}\). Therefore, the correct symbol is \(=\). 2. In b), \(0.38\,\text{m}=38\,\text{cm}\), so the left side is \(38\,\text{cm}+22\,\text{cm}=60\,\text{cm}\). Also, \(0.6\,\text{m}=60\,\text{cm}\), so the right side is \(60\,\text{cm}-5\,\text{cm}=55\,\text{cm}\). Therefore, the correct symbol is \(>\).

Answer

a) \(=\) b) \(>\)
5212315
A student wrote: \(32\,\text{kg} \div 800\,\text{g} = 40\,\text{g}\). Find the error, explain it briefly, and give the correct result.

Hints

- Decide whether the quotient represents a weight or a number of equal groups. - Convert both measurements to the same unit before dividing. - Think about what happens to identical units in the dividend and divisor.

Solution

1. Convert the measurements to the same unit: \(32\,\text{kg} = 32{,}000\,\text{g}\). 2. Divide: \(32{,}000\,\text{g} \div 800\,\text{g} = 40\). 3. The unit cancels because a mass is divided by a mass measured in the same unit. The quotient tells how many groups of \(800\,\text{g}\) fit into \(32\,\text{kg}\), so the result is a number, not a measurement in grams.

Answer

The error is writing \(\text{g}\) after the quotient. The correct calculation is \(32{,}000\,\text{g} \div 800\,\text{g} = 40\), so the correct result is \(40\).
5212335
Check the calculation for an error and correct it: \(4 \times 15\,\text{min} + 2\,\text{hr} = 62\,\text{min}\)

Hints

- Which operation should be completed first? - How many minutes are in one hour? - Are all terms written in the same time unit?

Solution

1. Multiply first: \(4 \times 15\,\text{min} = 60\,\text{min}\). 2. Convert the hours to minutes: \(2\,\text{hr} = 120\,\text{min}\). 3. Add quantities in the same unit: \(60\,\text{min} + 120\,\text{min} = 180\,\text{min}\). 4. The incorrect work treated \(2\) hours as if it were \(2\) minutes. The corrected result is \(180\) minutes, or \(3\) hours.

Answer

The error is adding \(2\) hours as though it were \(2\) minutes. The correct result is \(180\,\text{min}\), or \(3\,\text{hr}\).
5212505
Sarah's goal is to read for exactly \(7\) hours this week. Her reading times from Monday through Saturday are shown below. <table> <tr> <th>Day</th> <th>Reading Time</th> </tr> <tr> <td>Monday</td> <td>\(45\,\text{min}\)</td> </tr> <tr> <td>Tuesday</td> <td>\(1\,\text{hr}\,10\,\text{min}\)</td> </tr> <tr> <td>Wednesday</td> <td>\(55\,\text{min}\)</td> </tr> <tr> <td>Thursday</td> <td>\(1\,\text{hr}\,20\,\text{min}\)</td> </tr> <tr> <td>Friday</td> <td>\(40\,\text{min}\)</td> </tr> <tr> <td>Saturday</td> <td>\(1\,\text{hr}\,5\,\text{min}\)</td> </tr> </table> How long must Sarah read on Sunday to reach her goal?

Hints

- Convert every reading time to minutes. - How many minutes are in \(7\) hours? - Add the times from Monday through Saturday. - Subtract the total from the weekly goal.

Solution

1. Convert the reading times to minutes and add: \(45 + 70 + 55 + 80 + 40 + 65 = 355\) minutes. 2. Convert the goal to minutes: \(7 \times 60 = 420\) minutes. 3. Find the remaining time: \(420 - 355 = 65\) minutes. 4. Convert \(65\) minutes to \(1\) hour \(5\) minutes.

Answer

Sarah must read for \(1\,\text{hr}\,5\,\text{min}\) on Sunday.
5212545
Rewrite each measurement without a decimal, using mixed units when needed. a) \(12.050\,\text{km}\) b) \(4.07\,\text{m}\) c) \(0.625\,\text{kg}\) d) \(30.2\,\text{kg}\)

Hints

- Keep the whole-number part in the original unit. - Convert the decimal part to a smaller unit. - Pay close attention to zeros immediately after the decimal point.

Solution

1. \(0.050\,\text{km}=50\,\text{m}\), so \(12.050\,\text{km}=12\,\text{km}\ 50\,\text{m}\). 2. \(0.07\,\text{m}=7\,\text{cm}\), so \(4.07\,\text{m}=4\,\text{m}\ 7\,\text{cm}\). 3. \(0.625\,\text{kg}=625\,\text{g}\). 4. \(0.2\,\text{kg}=200\,\text{g}\), so \(30.2\,\text{kg}=30\,\text{kg}\ 200\,\text{g}\).

Answer

a) \(12\,\text{km}\ 50\,\text{m}\) b) \(4\,\text{m}\ 7\,\text{cm}\) c) \(625\,\text{g}\) d) \(30\,\text{kg}\ 200\,\text{g}\)
5212555
Rewrite each measurement without a decimal, using mixed units when needed. a) \(4.5\,\text{h}\) b) \(1.5\,\text{min}\) c) \(8.005\,\text{L}\) d) \(0.45\,\text{L}\)

Hints

- Time conversions use \(60\), not a power of ten. - Use \(60\,\text{min}=1\,\text{h}\), \(60\,\text{s}=1\,\text{min}\), and \(1000\,\text{mL}=1\,\text{L}\). - Omit the larger unit when its amount is zero.

Solution

1. \(0.5\,\text{h}=30\,\text{min}\), so \(4.5\,\text{h}=4\,\text{h}\ 30\,\text{min}\). 2. \(0.5\,\text{min}=30\,\text{s}\), so \(1.5\,\text{min}=1\,\text{min}\ 30\,\text{s}\). 3. \(0.005\,\text{L}=5\,\text{mL}\), so \(8.005\,\text{L}=8\,\text{L}\ 5\,\text{mL}\). 4. \(0.45\,\text{L}=450\,\text{mL}\).

Answer

a) \(4\,\text{h}\ 30\,\text{min}\) b) \(1\,\text{min}\ 30\,\text{s}\) c) \(8\,\text{L}\ 5\,\text{mL}\) d) \(450\,\text{mL}\)
5212565
Complete the missing equivalent forms. a) \(6.305\,\text{km}=\square\,\text{km}\ \square\,\text{m}=\square\,\text{m}\) b) \(15\,\text{kg}\ 400\,\text{g}=\square\,\text{kg}=\square\,\text{g}\) c) \(0.08\,\text{m}=\square\,\text{cm}\)

Hints

- Identify the conversion factor in each part. - Keep track of whether you are converting to a larger or smaller unit. - Use place value to preserve zeros correctly.

Solution

1. \(6.305\,\text{km}=6\,\text{km}\ 305\,\text{m}=6305\,\text{m}\). 2. \(15\,\text{kg}\ 400\,\text{g}=15.4\,\text{kg}=15{,}400\,\text{g}\). 3. \(0.08\,\text{m}=8\,\text{cm}\).

Answer

a) \(6\,\text{km}\ 305\,\text{m}\); \(6305\,\text{m}\) b) \(15.4\,\text{kg}\); \(15{,}400\,\text{g}\) c) \(8\,\text{cm}\)
5212665
Ms. Wagner takes the bus between Oakville and Riverton for work. Morning trip: Oakville departs \(7{:}12\) a.m. – Riverton arrives \(7{:}58\) a.m. Return trip: Riverton departs \(4{:}35\) p.m. – Oakville arrives \(5{:}19\) p.m. a) Find the travel time for each trip. b) How much total time does Ms. Wagner spend on the bus during a \(5\)-day workweek? Give the answer in hours and minutes. c) How long is Ms. Wagner in Riverton on one workday?

Hints

- Count to the next hour when that makes an elapsed time easier. - One hour has \(60\) minutes. - Multiply the daily bus time by the number of workdays. - The time in Riverton is from the morning arrival to the afternoon departure.

Solution

1. The morning trip lasts \(46\) minutes, from \(7{:}12\) a.m. to \(7{:}58\) a.m. The return trip lasts \(44\) minutes, from \(4{:}35\) p.m. to \(5{:}19\) p.m. 2. The daily travel time is \(46 + 44 = 90\) minutes. For \(5\) days, the total is \(5 \times 90 = 450\) minutes, or \(7\) hours \(30\) minutes. 3. The time from arrival in Riverton at \(7{:}58\) a.m. to departure at \(4{:}35\) p.m. is \(8\) hours \(37\) minutes.

Answer

a) Morning: \(46\,\text{min}\); return: \(44\,\text{min}\) b) \(7\,\text{hr}\,30\,\text{min}\) c) \(8\,\text{hr}\,37\,\text{min}\)
5212675
Liam attends a full-day school. His schedule is: - Leaves home: \(7{:}35\) a.m. - School begins: \(8{:}10\) a.m. - School ends: \(3{:}45\) p.m. - Arrives home: \(4{:}25\) p.m. a) How long is Liam away from home each day? b) How much time does he spend traveling to and from school each day? c) Find his total time at school, including breaks, during a \(5\)-day school week.

Hints

- Break each elapsed time into smaller steps if needed. - Include both the trip to school and the trip home. - Convert one school day to minutes before multiplying by \(5\).

Solution

1. The time from \(7{:}35\) a.m. to \(4{:}25\) p.m. is \(8\) hours \(50\) minutes. 2. The trip to school takes \(35\) minutes, and the trip home takes \(40\) minutes. The total travel time is \(75\) minutes, or \(1\) hour \(15\) minutes. 3. One school day lasts from \(8{:}10\) a.m. to \(3{:}45\) p.m., or \(7\) hours \(35\) minutes. In minutes, that is \(7 \times 60 + 35 = 455\) minutes. 4. For \(5\) days, the total is \(5 \times 455 = 2275\) minutes, or \(37\) hours \(55\) minutes.

Answer

a) \(8\,\text{hr}\,50\,\text{min}\) b) \(75\,\text{min}\), or \(1\,\text{hr}\,15\,\text{min}\) c) \(37\,\text{hr}\,55\,\text{min}\)
5212685
An express train takes \(52\) minutes to travel from City A to City B. A local train takes \(1\,\text{hr}\,14\,\text{min}\) for the same trip because it makes more stops. a) How many minutes faster is the express train on one trip? b) A commuter makes a round trip on \(5\) days each week. How much time would the commuter save in one week by switching from the local train to the express train? Give the answer in hours and minutes.

Hints

- Convert the longer trip time completely to minutes. - The commuter travels the route twice each day. - How many one-way trips are made in a \(5\)-day week?

Solution

1. Convert the local train's time to minutes: \(1\,\text{hr}\,14\,\text{min} = 74\,\text{min}\). 2. The savings on one trip is \(74 - 52 = 22\) minutes. 3. A round trip saves \(2 \times 22 = 44\) minutes per day. 4. Over \(5\) days, the savings is \(5 \times 44 = 220\) minutes. 5. Convert \(220\) minutes to \(3\) hours \(40\) minutes.

Answer

a) \(22\,\text{min}\) b) \(3\,\text{hr}\,40\,\text{min}\)
5212715
Two packages have a total mass of \(15.6\,\text{kg}\). The first package has a mass that is \(2\,\text{kg}\,400\,\text{g}\) greater than the second package. Find the mass of each package.

Hints

- Convert both measurements to the same unit. - Subtract the difference from the total so the remaining amount represents two equal parts. - Divide the remaining amount by \(2\). - Add the difference back to one part to find the greater mass.

Solution

1. Convert both given measurements to grams: \(15.6\,\text{kg} = 15{,}600\,\text{g}\) and \(2\,\text{kg}\,400\,\text{g} = 2400\,\text{g}\). 2. Remove the difference from the total: \(15{,}600\,\text{g} - 2400\,\text{g} = 13{,}200\,\text{g}\). 3. Divide the remaining mass equally: \(13{,}200\,\text{g} \div 2 = 6600\,\text{g}\). This is the second package’s mass. 4. Add the difference to find the first package’s mass: \(6600\,\text{g} + 2400\,\text{g} = 9000\,\text{g}\). 5. Convert the results: \(9000\,\text{g} = 9\,\text{kg}\) and \(6600\,\text{g} = 6.6\,\text{kg}\).

Answer

The first package has a mass of \(9\,\text{kg}\), and the second package has a mass of \(6.6\,\text{kg}\).
5213055
Write \(<\), \(>\), or \(=\) in each box. a) \(1\,\text{h}\ 45\,\text{min}\ \square\ 100\,\text{min}\) b) \(3\,\text{min}\ 12\,\text{s}\ \square\ 192\,\text{s}\) c) \(2\,\text{days}\ 4\,\text{h}\ \square\ 50\,\text{h}\)

Hints

- Convert each mixed time to the smaller unit. - Use \(60\,\text{min}=1\,\text{h}\), \(60\,\text{s}=1\,\text{min}\), and \(24\,\text{h}=1\,\text{day}\). - Compare after both sides use the same unit.

Solution

1. \(1\,\text{h}\ 45\,\text{min}=60\,\text{min}+45\,\text{min}=105\,\text{min}\). Since \(105>100\), the symbol is \(>\). 2. \(3\,\text{min}\ 12\,\text{s}=180\,\text{s}+12\,\text{s}=192\,\text{s}\), so the symbol is \(=\). 3. \(2\,\text{days}\ 4\,\text{h}=48\,\text{h}+4\,\text{h}=52\,\text{h}\). Since \(52>50\), the symbol is \(>\).

Answer

a) \(>\) b) \(=\) c) \(>\)
5213375
Evaluate the expression and write the result in meters and centimeters: \(15.6\,\text{m} \div 4 + 85\,\text{cm} \times 7\).

Hints

- Evaluate the division and multiplication before adding. - Convert both results to the same unit before addition. - Convert the final centimeter amount to meters and centimeters.

Solution

1. Divide the first term: \(15.6\,\text{m} \div 4 = 3.9\,\text{m} = 390\,\text{cm}\). 2. Multiply the second term: \(85\,\text{cm} \times 7 = 595\,\text{cm}\). 3. Add in centimeters: \(390\,\text{cm} + 595\,\text{cm} = 985\,\text{cm}\). 4. Convert to mixed units: \(985\,\text{cm} = 9\,\text{m}\,85\,\text{cm}\).

Answer

\(9\,\text{m}\,85\,\text{cm}\)
5213385
Evaluate and write the result using mixed units: \(2.5\,\text{kg} \times 9 - 14\,\text{kg}\,400\,\text{g} \div 6\).

Hints

- Converting decimal kilograms to grams may make the multiplication easier. - Perform the multiplication and division before the subtraction. - How many grams are in \(1\,\text{kg}\)?

Solution

1. Multiply the first measurement: \(2.5\,\text{kg} \times 9 = 22.5\,\text{kg} = 22{,}500\,\text{g}\). 2. Convert the second measurement to grams: \(14\,\text{kg}\,400\,\text{g} = 14{,}400\,\text{g}\). 3. Divide: \(14{,}400\,\text{g} \div 6 = 2400\,\text{g}\). 4. Subtract: \(22{,}500\,\text{g} - 2400\,\text{g} = 20{,}100\,\text{g}\). 5. Write the result using mixed units: \(20{,}100\,\text{g} = 20\,\text{kg}\,100\,\text{g}\).

Answer

\(20\,\text{kg}\,100\,\text{g}\)
5213395
Calculate and give the result in mixed time units: \(1\,\text{hr}\,45\,\text{min} + (3\,\text{hr}\,20\,\text{min}) \div 8\)

Hints

- One hour equals \(60\) minutes. - Complete the division before the addition. - Regroup if the final number of minutes is at least \(60\).

Solution

1. Convert the duration being divided to minutes: \(3 \times 60 + 20 = 200\) minutes. 2. Divide: \(200\,\text{min} \div 8 = 25\,\text{min}\). 3. Add: \(1\,\text{hr}\,45\,\text{min} + 25\,\text{min} = 1\,\text{hr}\,70\,\text{min}\). 4. Regroup \(70\) minutes as \(1\) hour \(10\) minutes. The result is \(2\) hours \(10\) minutes.

Answer

\(2\,\text{hr}\,10\,\text{min}\)
5213605
A coffee roaster packages coffee in \(1\)-pound bags and \(8\)-ounce bags. The roaster will package \(30\,\text{lb}\) of coffee and use the same number of each bag size. How many bags will be filled altogether?

Hints

- Convert the \(8\)-ounce bag size to pounds. - Group one large bag and one small bag together. - Find the weight of one such pair. - Determine how many pairs fit into the total weight.

Solution

1. Convert the smaller bag size to pounds: \(8\,\text{oz} = 0.5\,\text{lb}\). 2. Group one bag of each size: \(1\,\text{lb} + 0.5\,\text{lb} = 1.5\,\text{lb}\) per pair. 3. Find the number of pairs: \(30\,\text{lb} \div 1.5\,\text{lb} = 20\). 4. Each pair contains two bags, so \(20 \times 2 = 40\) bags will be filled.

Answer

A total of \(40\) bags will be filled: \(20\) one-pound bags and \(20\) eight-ounce bags.
5213905
Match each object to the comparison square whose area is closest. **Objects:** a) A US dime b) A sheet of US letter-size paper c) A classroom whiteboard d) A small neighborhood park **Comparison squares:** - side length \(2\,\text{cm}\) - side length \(10\,\text{in.}\) - side length \(6\,\text{ft}\) - side length \(100\,\text{yd}\)

Hints

- Find each comparison square's area. - Estimate each object's dimensions. - Match each estimate to the nearest square area.

Solution

1. The comparison-square areas are \(4\,\text{cm}^2\), \(100\,\text{in.}^2\), \(36\,\text{ft}^2\), and \(10{,}000\,\text{yd}^2\). 2. A dime covers only a few square centimeters, so the \(2\,\text{cm}\) square is closest. 3. US letter paper measures \(8.5\,\text{in.}\times 11\,\text{in.}\), with area \(93.5\,\text{in.}^2\), so the \(10\,\text{in.}\) square is closest. 4. A classroom whiteboard can be about \(4\,\text{ft}\times 8\,\text{ft}=32\,\text{ft}^2\), so the \(6\,\text{ft}\) square is closest. 5. A small neighborhood park can be on the scale of a \(100\,\text{yd}\times 100\,\text{yd}\) square.

Answer

a) side length \(2\,\text{cm}\) b) side length \(10\,\text{in.}\) c) side length \(6\,\text{ft}\) d) side length \(100\,\text{yd}\)
5213945
A child's bedroom has an area of \(12{,}960\,\text{in.}^2\). 1. Convert the area to square feet and then to square yards. 2. Which of the three units—square inches, square feet, or square yards—is best for describing the room in a real-estate listing? Briefly explain.

Hints

- Use \(1\,\text{ft}^2=144\,\text{in.}^2\). - Use \(1\,\text{yd}^2=9\,\text{ft}^2\). - Consider which unit is commonly used for room size.

Solution

1. Since \(1\,\text{ft}^2=144\,\text{in.}^2\), \(12{,}960\div 144=90\,\text{ft}^2\). Since \(1\,\text{yd}^2=9\,\text{ft}^2\), \(90\div 9=10\,\text{yd}^2\). 2. Square feet is the best unit because room sizes in US real-estate listings are commonly described in square feet, and \(90\) is easy to interpret.

Answer

1. \(90\,\text{ft}^2\) and \(10\,\text{yd}^2\) 2. Square feet is the most appropriate unit.
5213955
A rectangular community sports field is \(330\,\text{ft}\) long and \(132\,\text{ft}\) wide. 1. Find its area in square feet. 2. Convert the area to square yards. 3. How many acres is the field? Is acres a reasonable unit for comparing this field with larger properties?

Hints

- Use the rectangle area formula. - Use \(1\,\text{yd}^2=9\,\text{ft}^2\). - Compare the square-foot area with \(43{,}560\,\text{ft}^2\) per acre.

Solution

1. The area is \(330\times 132=43{,}560\,\text{ft}^2\). 2. Since \(1\,\text{yd}^2=9\,\text{ft}^2\), \(43{,}560\div 9=4840\,\text{yd}^2\). 3. Since \(1\) acre is \(43{,}560\,\text{ft}^2\), the field is exactly \(1\) acre. Acres is reasonable when comparing the field with larger land areas, although square feet or square yards is more direct for one field.

Answer

1. \(43{,}560\,\text{ft}^2\) 2. \(4840\,\text{yd}^2\) 3. \(1\) acre; yes, acres is useful for comparison with larger properties.
5213995
A large poster is estimated by covering it with square sticky notes. Each sticky note has an area of about \(9\,\text{in.}^2\), and \(40\) notes cover the poster. Estimate the poster's area in square inches and then convert it to square feet.

Hints

- Multiply the number of sticky notes by the area of one note. - Use \(1\,\text{ft}^2=144\,\text{in.}^2\) to convert.

Solution

1. The estimated area is \(40\times 9=360\,\text{in.}^2\). 2. Since \(1\,\text{ft}^2=144\,\text{in.}^2\), \(360\div 144=2.5\,\text{ft}^2\).

Answer

Approximately \(360\,\text{in.}^2\), or \(2.5\,\text{ft}^2\)
5214075
Convert each mixed area measurement to the unit in parentheses. a) \(5\,\text{yd}^2\ 12\,\text{ft}^2\) \((\text{ft}^2)\) b) \(9\,\text{ft}^2\ 4\,\text{in.}^2\) \((\text{in.}^2)\) c) \(14\,\text{m}^2\ 50\,\text{cm}^2\) \((\text{cm}^2)\)

Hints

- Convert the larger unit to the requested smaller unit first. - Add only after both parts use the same unit. - Use area conversion factors, not linear conversion factors.

Solution

1. \(5\,\text{yd}^2=5\times 9=45\,\text{ft}^2\). Adding \(12\,\text{ft}^2\) gives \(57\,\text{ft}^2\). 2. \(9\,\text{ft}^2=9\times 144=1296\,\text{in.}^2\). Adding \(4\,\text{in.}^2\) gives \(1300\,\text{in.}^2\). 3. \(14\,\text{m}^2=14\times 10{,}000=140{,}000\,\text{cm}^2\). Adding \(50\,\text{cm}^2\) gives \(140{,}050\,\text{cm}^2\).

Answer

a) \(57\,\text{ft}^2\) b) \(1300\,\text{in.}^2\) c) \(140{,}050\,\text{cm}^2\)
5214085
Write each length as a decimal in the unit in parentheses. a) \(6\,\text{m}\ 30\,\text{cm}\) (meters) b) \(25\,\text{cm}\) (meters) c) \(8\,\text{km}\ 75\,\text{m}\) (kilometers)

Hints

- Convert the smaller-unit part to the larger target unit. - Use the conversion factor to determine the number of decimal places. - Add the converted amount to the whole larger units.

Solution

1. \(30\,\text{cm}=0.3\,\text{m}\), so the total is \(6.3\,\text{m}\). 2. \(25 \div 100=0.25\), so \(25\,\text{cm}=0.25\,\text{m}\). 3. \(75\,\text{m}=0.075\,\text{km}\), so the total is \(8.075\,\text{km}\).

Answer

a) \(6.3\,\text{m}\) b) \(0.25\,\text{m}\) c) \(8.075\,\text{km}\)
5214295
A teacher says, “Our classroom has an area of \(650{,}000\,\text{cm}^2\).” A student thinks the number is too large for one room. Convert the area to square meters. Is the measurement reasonable for a classroom? Explain.

Hints

- Square the length conversion factor when converting area units. - Think about reasonable classroom length and width measurements. - A large numerical value can result from using a small unit.

Solution

1. Since \(1\,\text{m}=100\,\text{cm}\), one square meter contains \(100\times 100=10{,}000\,\text{cm}^2\). 2. Convert the area: \(650{,}000\div 10{,}000=65\). Therefore, \(650{,}000\,\text{cm}^2=65\,\text{m}^2\). 3. A room measuring about \(8\,\text{m}\times 8\,\text{m}\) has an area close to \(65\,\text{m}^2\), so the measurement is reasonable. The original number looks large because square centimeters are small units.

Answer

The measurement is reasonable. \(650{,}000\,\text{cm}^2=65\,\text{m}^2\), which is a plausible classroom area.
5214305
Two students compare area measurements from homework. Lucas says, “The front of my smartphone has an area of \(12{,}500\,\text{mm}^2\).” Julia says, “My postcard has an area of \(1.5\,\text{m}^2\).” One student made a major error. Convert both measurements to square centimeters and decide who is incorrect.

Hints

- Convert both areas to square centimeters. - Remember that area conversion factors are squared. - Compare each result with the size of an object you can hold.

Solution

1. Since \(1\,\text{cm}^2=100\,\text{mm}^2\), \(12{,}500\div 100=125\). Lucas's measurement is \(125\,\text{cm}^2\), which is reasonable for a smartphone. 2. Since \(1\,\text{m}^2=10{,}000\,\text{cm}^2\), \(1.5\times 10{,}000=15{,}000\). Julia's measurement is \(15{,}000\,\text{cm}^2\). 3. A typical postcard has an area near \(150\,\text{cm}^2\), not \(15{,}000\,\text{cm}^2\). Julia's value is about \(100\) times too large.

Answer

Julia is incorrect. Her value equals \(15{,}000\,\text{cm}^2\), which is far too large for a postcard. Lucas's value of \(125\,\text{cm}^2\) is reasonable.
5214345
Check each area conversion. Correct any conversion that is wrong. a) \(300\,\text{mm}^2=30\,\text{cm}^2\) b) \(15\,\text{yd}^2=1.5\,\text{ft}^2\) c) \(2000\,\text{cm}^2=0.2\,\text{m}^2\)

Hints

- Recalculate each conversion independently. - Use area conversion factors, not linear conversion factors. - Decide whether the numerical value should increase or decrease.

Solution

1. Statement a is incorrect. Since \(1\,\text{cm}^2=100\,\text{mm}^2\), \(300\div 100=3\,\text{cm}^2\). 2. Statement b is incorrect. Since \(1\,\text{yd}^2=9\,\text{ft}^2\), \(15\times 9=135\,\text{ft}^2\). 3. Statement c is correct because \(2000\div 10{,}000=0.2\,\text{m}^2\).

Answer

a) Incorrect; \(300\,\text{mm}^2=3\,\text{cm}^2\) b) Incorrect; \(15\,\text{yd}^2=135\,\text{ft}^2\) c) Correct
5214405
Write each mixed area measurement using the smaller of the two units. a) \(5\,\text{yd}^2\ 20\,\text{ft}^2\) b) \(12\,\text{ft}^2\ 7\,\text{in.}^2\) c) \(4\,\text{m}^2\ 9\,\text{cm}^2\) d) \(1\,\text{m}^2\ 1500\,\text{cm}^2\)

Hints

- Identify the smaller unit in each pair. - Convert the larger-unit part to the smaller unit. - Add the two values after converting.

Solution

1. \(5\,\text{yd}^2=45\,\text{ft}^2\), so the total is \(45+20=65\,\text{ft}^2\). 2. \(12\,\text{ft}^2=12\times 144=1728\,\text{in.}^2\), so the total is \(1728+7=1735\,\text{in.}^2\). 3. \(4\,\text{m}^2=40{,}000\,\text{cm}^2\). Adding \(9\,\text{cm}^2\) gives \(40{,}009\,\text{cm}^2\). 4. \(1\,\text{m}^2=10{,}000\,\text{cm}^2\). Adding \(1500\,\text{cm}^2\) gives \(11{,}500\,\text{cm}^2\).

Answer

a) \(65\,\text{ft}^2\) b) \(1735\,\text{in.}^2\) c) \(40{,}009\,\text{cm}^2\) d) \(11{,}500\,\text{cm}^2\)
5214465
Convert each area measurement to both requested units. a) \(810\,\text{ft}^2\) to square yards and square inches b) \(15{,}000\,\text{cm}^2\) to square meters and square millimeters c) \(400\,\text{yd}^2\) to square feet and square inches d) \(20{,}736\,\text{in.}^2\) to square feet and square yards

Hints

- Write the conversion factors for each requested unit. - Use multiplication for smaller target units and division for larger target units. - Convert in two steps when helpful.

Solution

1. \(810\div 9=90\,\text{yd}^2\), and \(810\times 144=116{,}640\,\text{in.}^2\). 2. \(15{,}000\div 10{,}000=1.5\,\text{m}^2\), and \(15{,}000\times 100=1{,}500{,}000\,\text{mm}^2\). 3. \(400\times 9=3600\,\text{ft}^2\), and \(3600\times 144=518{,}400\,\text{in.}^2\). 4. \(20{,}736\div 144=144\,\text{ft}^2\), and \(144\div 9=16\,\text{yd}^2\).

Answer

a) \(90\,\text{yd}^2\) and \(116{,}640\,\text{in.}^2\) b) \(1.5\,\text{m}^2\) and \(1{,}500{,}000\,\text{mm}^2\) c) \(3600\,\text{ft}^2\) and \(518{,}400\,\text{in.}^2\) d) \(144\,\text{ft}^2\) and \(16\,\text{yd}^2\)
5214475
Order the area measurements from least to greatest. Convert all measurements to square inches to compare. \(3\,\text{yd}^2\), \(2800\,\text{in.}^2\), \(31\,\text{ft}^2\), \(2\,\text{yd}^2\ 90\,\text{in.}^2\)

Hints

- Convert every measurement to square inches. - Use \(1\,\text{ft}^2=144\,\text{in.}^2\) and \(1\,\text{yd}^2=1296\,\text{in.}^2\). - Then order the numerical values.

Solution

1. \(3\,\text{yd}^2=3\times 1296=3888\,\text{in.}^2\). 2. \(2800\,\text{in.}^2\) is already in the target unit. 3. \(31\,\text{ft}^2=31\times 144=4464\,\text{in.}^2\). 4. \(2\,\text{yd}^2\ 90\,\text{in.}^2=2\times 1296+90=2682\,\text{in.}^2\). 5. Therefore, \(2682<2800<3888<4464\).

Answer

\(2\,\text{yd}^2\ 90\,\text{in.}^2<2800\,\text{in.}^2<3\,\text{yd}^2<31\,\text{ft}^2\)
5214485
A farmer owns three fields. The first is \(15\) acres, the second is \(217{,}800\,\text{ft}^2\), and the third is \(435{,}600\,\text{ft}^2\). Find the total area in acres.

Hints

- Express each square-foot area as a multiple of \(43{,}560\,\text{ft}^2\). - Use \(1\) acre \(=43{,}560\,\text{ft}^2\). - Add the converted areas.

Solution

1. Since \(1\) acre is \(43{,}560\,\text{ft}^2\) and \(43{,}560\times 5=217{,}800\), the second field is \(5\) acres. 2. Since \(43{,}560\times 10=435{,}600\), the third field is \(10\) acres. 3. The total area is \(15+5+10=30\) acres.

Answer

\(30\) acres
5214665
What number belongs in each box? a) \(5\,\text{ft}^2-\square\,\text{in.}^2=420\,\text{in.}^2\) b) \(2\,\text{yd}^2+\square\,\text{ft}^2=26\,\text{ft}^2\)

Hints

- Convert the known larger-unit measurement to the unit on the right. - Then find the missing difference or addend.

Solution

1. \(5\,\text{ft}^2=720\,\text{in.}^2\). Therefore, \(720-420=300\), so the missing number is \(300\). 2. \(2\,\text{yd}^2=18\,\text{ft}^2\). Therefore, \(26-18=8\), so the missing number is \(8\).

Answer

a) \(300\) b) \(8\)
5214675
Compare the values. Insert \(<\), \(>\), or \(=\). a) \(4\,\text{ft}^2+9\,\text{in.}^2\ \underline{\hspace{0.6cm}}\ 490\,\text{in.}^2\) b) \(10\,\text{yd}^2-10\,\text{ft}^2\ \underline{\hspace{0.6cm}}\ 80\,\text{ft}^2\)

Hints

- Convert the entire left side to the unit used on the right. - Calculate first, then compare.

Solution

1. \(4\,\text{ft}^2+9\,\text{in.}^2=576\,\text{in.}^2+9\,\text{in.}^2=585\,\text{in.}^2\), and \(585>490\). 2. \(10\,\text{yd}^2-10\,\text{ft}^2=90\,\text{ft}^2-10\,\text{ft}^2=80\,\text{ft}^2\), so the values are equal.

Answer

a) \(>\) b) \(=\)
5214705
Which area measurements are equal? Find the three matching pairs. A: \(2\,\text{yd}^2\ 4\,\text{ft}^2\) B: \(24\,\text{ft}^2\) C: \(3168\,\text{in.}^2\) D: \(2\,\text{yd}^2\ 6\,\text{ft}^2\) E: \(240\,\text{ft}^2\) F: \(34{,}560\,\text{in.}^2\)

Hints

- Convert every measurement to square feet. - Use \(1\,\text{ft}^2=144\,\text{in.}^2\) and look for multiples of \(144\). - Match values with equal numerical results.

Solution

1. Convert all measurements to square feet. 2. \(A=2\times 9+4=22\,\text{ft}^2\). 3. \(B=24\,\text{ft}^2\). 4. Since \(20\times 144=2880\) and \(2\times 144=288\), \(C=3168\,\text{in.}^2=22\,\text{ft}^2\). 5. \(D=2\times 9+6=24\,\text{ft}^2\). 6. \(E=240\,\text{ft}^2\). 7. Since \(24\times 144=3456\), \(240\times 144=34{,}560\), so \(F=240\,\text{ft}^2\).

Answer

A and C B and D E and F
5214725
Write each area measurement in mixed units. a) \(4005\,\text{in.}^2\) as square feet and square inches b) \(10{,}200\,\text{in.}^2\) as square feet and square inches c) \(807\,\text{ft}^2\) as square yards and square feet d) \(30{,}050\,\text{cm}^2\) as square meters and square centimeters

Hints

- Find the greatest whole-number multiple of the conversion factor that does not exceed the given area. - Express the remainder in the original smaller unit. - Use \(144\), \(9\), or \(10{,}000\) as appropriate.

Solution

1. Since \(20\times 144=2880\) and \(7\times 144=1008\), \(27\times 144=3888\). The remainder is \(4005-3888=117\), so \(4005\,\text{in.}^2=27\,\text{ft}^2\ 117\,\text{in.}^2\). 2. Since \(7\times 144=1008\), \(70\times 144=10{,}080\). The remainder is \(10{,}200-10{,}080=120\), so \(10{,}200\,\text{in.}^2=70\,\text{ft}^2\ 120\,\text{in.}^2\). 3. \(807\div 9=89\) remainder \(6\), so \(807\,\text{ft}^2=89\,\text{yd}^2\ 6\,\text{ft}^2\). 4. \(30{,}050\div 10{,}000=3\) remainder \(50\), so \(30{,}050\,\text{cm}^2=3\,\text{m}^2\ 50\,\text{cm}^2\).

Answer

a) \(27\,\text{ft}^2\ 117\,\text{in.}^2\) b) \(70\,\text{ft}^2\ 120\,\text{in.}^2\) c) \(89\,\text{yd}^2\ 6\,\text{ft}^2\) d) \(3\,\text{m}^2\ 50\,\text{cm}^2\)
5214735
A property has an area of \(56{,}343\,\text{ft}^2\). a) Write the area in square yards and square feet. b) Write the area in acres, square yards, and square feet.

Hints

- Divide by \(9\) to convert square feet to square yards with a remainder. - For part b, first remove one acre, or \(43{,}560\,\text{ft}^2\). - Convert the remaining square feet to square yards.

Solution

1. Since \(1\,\text{yd}^2=9\,\text{ft}^2\), \(56{,}343\div 9=6260\) remainder \(3\). Therefore, the area is \(6260\,\text{yd}^2\ 3\,\text{ft}^2\). 2. One acre is \(43{,}560\,\text{ft}^2\). The remaining area is \(56{,}343-43{,}560=12{,}783\,\text{ft}^2\). 3. \(12{,}783\div 9=1420\) remainder \(3\), so the complete mixed-unit form is \(1\) acre \(1420\,\text{yd}^2\ 3\,\text{ft}^2\).

Answer

a) \(6260\,\text{yd}^2\ 3\,\text{ft}^2\) b) \(1\) acre \(1420\,\text{yd}^2\ 3\,\text{ft}^2\)
5214805
Find the missing area needed to reach each target. a) \(5\,\text{ft}^2+\underline{\hspace{1cm}}=1\,\text{yd}^2\) b) \(62\,\text{in.}^2+\underline{\hspace{1cm}}=1\,\text{ft}^2\) c) \(450\,\text{cm}^2+\underline{\hspace{1cm}}=0.05\,\text{m}^2\) d) \(8.5\,\text{cm}^2+\underline{\hspace{1cm}}=0.01\,\text{m}^2\)

Hints

- Convert each target to the smaller unit in the equation. - Subtract the starting area from the converted target. - Keep the answer in the smaller unit.

Solution

1. \(1\,\text{yd}^2=9\,\text{ft}^2\), so \(9-5=4\,\text{ft}^2\). 2. \(1\,\text{ft}^2=144\,\text{in.}^2\), so \(144-62=82\,\text{in.}^2\). 3. \(0.05\,\text{m}^2=500\,\text{cm}^2\), so \(500-450=50\,\text{cm}^2\). 4. \(0.01\,\text{m}^2=100\,\text{cm}^2\), so \(100-8.5=91.5\,\text{cm}^2\).

Answer

a) \(4\,\text{ft}^2\) b) \(82\,\text{in.}^2\) c) \(50\,\text{cm}^2\) d) \(91.5\,\text{cm}^2\)
5214885
Calculate the quotient: \(2\,\text{yd}^2\ 42\,\text{ft}^2\div 3\,\text{ft}^2\).

Hints

- Convert the dividend to one unit. - What happens to the units when one area is divided by another area in the same unit?

Solution

1. Convert the mixed area to square feet: \(2\,\text{yd}^2=18\,\text{ft}^2\), so the total is \(18+42=60\,\text{ft}^2\). 2. Divide: \(60\,\text{ft}^2\div 3\,\text{ft}^2=20\). The area units cancel.

Answer

\(20\)
5215065
Ms. Rivera is painting her garage floor. The total floor area is \(180\,\text{ft}^2\ 20\,\text{in.}^2\). She has painted \(90\,\text{ft}^2\ 50\,\text{in.}^2\). What area remains? Write the result in square feet and square inches.

Hints

- Compare the square-inch parts before subtracting. - Regroup \(1\,\text{ft}^2\) as \(144\,\text{in.}^2\) when needed. - Subtract the square inches and square feet separately.

Solution

1. Since \(20\,\text{in.}^2\) is less than \(50\,\text{in.}^2\), regroup \(1\,\text{ft}^2\) as \(144\,\text{in.}^2\). The total area becomes \(179\,\text{ft}^2\ 164\,\text{in.}^2\). 2. Subtract the square inches: \(164-50=114\,\text{in.}^2\). 3. Subtract the square feet: \(179-90=89\,\text{ft}^2\). Therefore, the remaining area is \(89\,\text{ft}^2\ 114\,\text{in.}^2\).

Answer

\(89\,\text{ft}^2\ 114\,\text{in.}^2\)
5215075
A farmer has a field measuring \(4\) acres \(21{,}780\,\text{ft}^2\). The field will be divided equally into \(6\) paddocks. What is the area of each paddock in square feet?

Hints

- Convert the entire field to square feet. - Use \(1\) acre \(=43{,}560\,\text{ft}^2\). - Divide the total by \(6\).

Solution

1. Convert the acres to square feet: \(4\times 43{,}560=174{,}240\,\text{ft}^2\). 2. Add the remaining area: \(174{,}240+21{,}780=196{,}020\,\text{ft}^2\). 3. Divide equally: \(196{,}020\div 6=32{,}670\,\text{ft}^2\).

Answer

\(32{,}670\,\text{ft}^2\) per paddock
5215085
A wall area of \(7\,\text{yd}^2\ 9\,\text{ft}^2\) will be completely covered with posters. Each poster covers exactly \(6\,\text{ft}^2\). Assume the posters can be arranged with no gaps or waste. How many posters are needed?

Hints

- Convert the wall area to square feet. - Divide the total area by the area of one poster.

Solution

1. Convert the wall area to square feet: \(7\times 9+9=72\,\text{ft}^2\). 2. Divide by the area of one poster: \(72\,\text{ft}^2\div 6\,\text{ft}^2=12\).

Answer

\(12\) posters
5215155
Convert each area measurement to the unit in parentheses, then round to the nearest whole unit. a) \(548\,\text{ft}^2\) \((\text{yd}^2)\) b) \(12{,}750\,\text{mm}^2\) \((\text{cm}^2)\) c) \(142.8\,\text{in.}^2\) \((\text{ft}^2)\)

Hints

- Convert before rounding. - Use the correct area conversion factor. - Look at the decimal part or use benchmark values to round to the nearest whole unit.

Solution

1. \(548\div 9\approx 60.89\,\text{yd}^2\), which rounds to \(61\,\text{yd}^2\). 2. \(12{,}750\div 100=127.5\,\text{cm}^2\), which rounds to \(128\,\text{cm}^2\). 3. Since \(1\,\text{ft}^2=144\,\text{in.}^2\), \(142.8\,\text{in.}^2\) is slightly less than \(1\,\text{ft}^2\) and greater than \(0.5\,\text{ft}^2\). Therefore, it rounds to \(1\,\text{ft}^2\).

Answer

a) \(61\,\text{yd}^2\) b) \(128\,\text{cm}^2\) c) \(1\,\text{ft}^2\)
5215165
Solve by converting and rounding to the requested whole unit. a) A sports field has an area of \(7420\,\text{ft}^2\). Round the area to the nearest square yard. b) A nature preserve has an area of \(28{,}490\,\text{yd}^2\). Round the area to the nearest acre.

Hints

- Convert before rounding. - Use \(1\,\text{yd}^2=9\,\text{ft}^2\). - For acres, compare the area with nearby whole-number multiples of \(4840\,\text{yd}^2\).

Solution

1. \(7420\div 9\approx 824.44\,\text{yd}^2\), which rounds to \(824\,\text{yd}^2\). 2. Since \(1\) acre is \(4840\,\text{yd}^2\), \(5\) acres is \(24{,}200\,\text{yd}^2\) and \(6\) acres is \(29{,}040\,\text{yd}^2\). The given area is \(4290\,\text{yd}^2\) above \(5\) acres but only \(550\,\text{yd}^2\) below \(6\) acres, so it rounds to \(6\) acres.

Answer

a) \(824\,\text{yd}^2\) b) \(6\) acres
5215175
Complete the table. Convert each area to the target unit, then round to the nearest whole unit. <table> <tr> <th>Area</th> <th>Nearest square yard</th> <th>Nearest acre</th> </tr> <tr> <td>\(184{,}300\,\text{ft}^2\)</td> <td></td> <td></td> </tr> <tr> <td>\(75{,}550\,\text{ft}^2\)</td> <td></td> <td></td> </tr> </table>

Hints

- Convert each square-foot value before rounding. - Use \(1\,\text{yd}^2=9\,\text{ft}^2\) and \(1\) acre \(=43{,}560\,\text{ft}^2\). - Compare with nearby whole-number acre values.

Solution

1. For \(184{,}300\,\text{ft}^2\), \(184{,}300\div 9\approx 20{,}477.78\,\text{yd}^2\), which rounds to \(20{,}478\,\text{yd}^2\). Four acres is \(174{,}240\,\text{ft}^2\), and five acres is \(217{,}800\,\text{ft}^2\). The given area is closer to four acres, so it rounds to \(4\) acres. 2. For \(75{,}550\,\text{ft}^2\), \(75{,}550\div 9\approx 8394.44\,\text{yd}^2\), which rounds to \(8394\,\text{yd}^2\). One acre is \(43{,}560\,\text{ft}^2\), and two acres is \(87{,}120\,\text{ft}^2\). The given area is closer to two acres, so it rounds to \(2\) acres.

Answer

First row: \(20{,}478\,\text{yd}^2\) and \(4\) acres Second row: \(8394\,\text{yd}^2\) and \(2\) acres
5215255
Calculate each expression. When the result has an area unit, write it in the smallest area unit used in that expression. a) \(5\,\text{yd}^2+35\,\text{ft}^2\) b) \((2\,\text{yd}^2\ 5\,\text{ft}^2)\times 4\) c) \(1\text{ acre}-40{,}000\,\text{ft}^2\) d) \(12\,\text{ft}^2\div 48\,\text{in.}^2\)

Hints

- Convert to the smallest unit before calculating. - Use the appropriate area conversion factor. - When equal area units are divided, the units cancel.

Solution

1. \(5\,\text{yd}^2=45\,\text{ft}^2\), so \(45+35=80\,\text{ft}^2\). 2. \(2\,\text{yd}^2\ 5\,\text{ft}^2=18+5=23\,\text{ft}^2\). Then \(23\times 4=92\,\text{ft}^2\). 3. \(1\) acre is \(43{,}560\,\text{ft}^2\), so \(43{,}560-40{,}000=3560\,\text{ft}^2\). 4. \(12\,\text{ft}^2=1728\,\text{in.}^2\), so \(1728\div 48=36\). The units cancel.

Answer

a) \(80\,\text{ft}^2\) b) \(92\,\text{ft}^2\) c) \(3560\,\text{ft}^2\) d) \(36\)
5215265
Calculate. a) \(12\,\text{m}\,40\,\text{cm} - 85\,\text{cm}\) b) \(7 \times 3\,\text{m}\,20\,\text{cm}\) c) \(2\,\text{km} - 750\,\text{m}\) d) \(1\,\text{m} \div 5\,\text{mm}\)

Hints

- Choose a common unit that avoids decimals in each part. - Use the metric relationships among millimeters, centimeters, meters, and kilometers. - When one length is divided by another length in the same unit, the result is a count.

Solution

1. For a), convert to centimeters: \(1240\,\text{cm} - 85\,\text{cm} = 1155\,\text{cm} = 11\,\text{m}\,55\,\text{cm}\). 2. For b), convert to centimeters: \(7 \times 320\,\text{cm} = 2240\,\text{cm} = 22\,\text{m}\,40\,\text{cm}\). 3. For c), convert to meters: \(2000\,\text{m} - 750\,\text{m} = 1250\,\text{m} = 1\,\text{km}\,250\,\text{m}\). 4. For d), convert to millimeters: \(1000\,\text{mm} \div 5\,\text{mm} = 200\).

Answer

a) \(11\,\text{m}\,55\,\text{cm}\) b) \(22\,\text{m}\,40\,\text{cm}\) c) \(1\,\text{km}\,250\,\text{m}\) d) \(200\)
5215275
Calculate each expression. a) \(3\,\text{yd}^2\ 5\,\text{ft}^2-12\,\text{ft}^2\) b) \(1\text{ acre}-1\,\text{ft}^2\) c) \(4\,\text{ft}^2\ 4\,\text{in.}^2\div 5\,\text{in.}^2\) d) \(12\times(2\,\text{ft}^2\ 50\,\text{in.}^2)\)

Hints

- Convert mixed areas to one unit before calculating. - Use \(1\,\text{yd}^2=9\,\text{ft}^2\), \(1\,\text{ft}^2=144\,\text{in.}^2\), and \(1\) acre \(=43{,}560\,\text{ft}^2\). - Decide whether the final result should retain an area unit.

Solution

1. \(3\,\text{yd}^2\ 5\,\text{ft}^2=27+5=32\,\text{ft}^2\). Then \(32-12=20\,\text{ft}^2\). 2. \(1\) acre is \(43{,}560\,\text{ft}^2\), so the result is \(43{,}559\,\text{ft}^2\). 3. \(4\,\text{ft}^2\ 4\,\text{in.}^2=4\times 144+4=580\,\text{in.}^2\). Then \(580\div 5=116\). 4. \(2\,\text{ft}^2\ 50\,\text{in.}^2=338\,\text{in.}^2\). Then \(12\times 338=4056\,\text{in.}^2\). Since \(28\times 144=4032\), the remainder is \(24\,\text{in.}^2\), so the mixed-unit result is \(28\,\text{ft}^2\ 24\,\text{in.}^2\).

Answer

a) \(20\,\text{ft}^2\) b) \(43{,}559\,\text{ft}^2\) c) \(116\) d) \(4056\,\text{in.}^2\), or \(28\,\text{ft}^2\ 24\,\text{in.}^2\)
5215505
A farmer has \(4\) acres \(21{,}780\,\text{ft}^2\) of land. Corn is planted on \(1\) acre. The remaining land is divided into \(4\) equal vegetable plots. Find the area of one vegetable plot in square feet.

Hints

- Convert all land areas to square feet. - Subtract the corn area from the total. - Divide the remainder by \(4\).

Solution

1. The total land area is \(4\times 43{,}560+21{,}780=196{,}020\,\text{ft}^2\). 2. The corn area is \(43{,}560\,\text{ft}^2\). 3. The remaining area is \(196{,}020-43{,}560=152{,}460\,\text{ft}^2\). 4. Each vegetable plot has area \(152{,}460\div 4=38{,}115\,\text{ft}^2\).

Answer

\(38{,}115\,\text{ft}^2\)
5216745
Determine whether each statement is true. Correct any false statement. a) “\(288\,\text{in.}^2\) is equal to \(2\,\text{ft}^2\).” b) “To convert \(3\,\text{yd}^2\) to square inches, multiply by \(9\) and then by \(12\).”

Hints

- Check each statement with a complete conversion. - Remember that linear conversion factors must be squared for area units.

Solution

1. Statement a is true because \(2\times 144=288\,\text{in.}^2\). 2. Statement b is false. First convert square yards to square feet by multiplying by \(9\), then convert square feet to square inches by multiplying by \(144\), not \(12\). Therefore, \(3\times 9\times 144=3888\,\text{in.}^2\).

Answer

a) True b) False; multiply by \(9\) and then by \(144\). The result is \(3888\,\text{in.}^2\).
5216835
Calculate the sum. Give your answer in square feet. \(4\,\text{yd}^2\ 12\,\text{ft}^2+88\,\text{ft}^2+2\,\text{yd}^2\ 5\,\text{ft}^2\)

Hints

- Recall the relationship between square yards and square feet. - Convert each mixed area to square feet. - Add only after all three measurements use the same unit.

Solution

1. Convert the first mixed area: \(4\,\text{yd}^2\ 12\,\text{ft}^2=4\times 9+12=48\,\text{ft}^2\). 2. The second addend is already \(88\,\text{ft}^2\). 3. Convert the third mixed area: \(2\,\text{yd}^2\ 5\,\text{ft}^2=2\times 9+5=23\,\text{ft}^2\). 4. Add: \(48+88+23=159\,\text{ft}^2\).

Answer

\(159\,\text{ft}^2\)
5216985
Calculate each expression. Write each area answer in mixed units when possible. a) \((5\,\text{yd}^2\ 3\,\text{ft}^2)\times 8\) b) \((24\,\text{ft}^2\ 48\,\text{in.}^2)\div 6\)

Hints

- Convert each mixed area completely to its smaller unit first. - Multiply or divide the numerical value. - Convert the result back to mixed units.

Solution

1. For a), convert to square feet: \(5\,\text{yd}^2\ 3\,\text{ft}^2=5\times 9+3=48\,\text{ft}^2\). Then \(48\times 8=384\,\text{ft}^2\). Since \(384=42\times 9+6\), the result is \(42\,\text{yd}^2\ 6\,\text{ft}^2\). 2. For b), convert to square inches: \(24\,\text{ft}^2\ 48\,\text{in.}^2=24\times 144+48=3504\,\text{in.}^2\). Then \(3504\div 6=584\,\text{in.}^2\). Since \(584=4\times 144+8\), the result is \(4\,\text{ft}^2\ 8\,\text{in.}^2\).

Answer

a) \(42\,\text{yd}^2\ 6\,\text{ft}^2\) b) \(4\,\text{ft}^2\ 8\,\text{in.}^2\)
5217055
For each measurement, find how much more area is needed to make \(1\,\text{ft}^2\). Give each answer in square inches. a) \(75\,\text{in.}^2\) b) \(0.5\,\text{ft}^2\) c) \(30\,\text{in.}^2+0.25\,\text{ft}^2\)

Hints

- Convert every given area to square inches. - Use \(1\,\text{ft}^2=144\,\text{in.}^2\). - Subtract each converted area from \(144\,\text{in.}^2\).

Solution

1. The target area is \(1\,\text{ft}^2=144\,\text{in.}^2\). 2. For a), \(144-75=69\,\text{in.}^2\). 3. For b), \(0.5\times 144=72\,\text{in.}^2\), so \(144-72=72\,\text{in.}^2\). 4. For c), \(0.25\times 144=36\,\text{in.}^2\). The given area is \(30+36=66\,\text{in.}^2\), so \(144-66=78\,\text{in.}^2\).

Answer

a) \(69\,\text{in.}^2\) b) \(72\,\text{in.}^2\) c) \(78\,\text{in.}^2\)
5217795
Find each missing value. a) \(2.5\,\text{kg}=\square\,\text{g}\) b) \(3\,\text{h}\ 15\,\text{min}=\square\,\text{min}\) c) \(12\,\text{cm}\ 5\,\text{mm}=\square\,\text{mm}\) d) \(1.25\,\text{m}=\square\,\text{cm}\)

Hints

- For mixed units, convert the larger part first and add the remainder. - Multiplying by \(10\), \(100\), or \(1000\) shifts the decimal point to the right. - Use \(60\,\text{min}=1\,\text{h}\).

Solution

1. \(2.5 \times 1000=2500\), so the missing value is \(2500\). 2. \(3\,\text{h}=180\,\text{min}\). Then \(180+15=195\). 3. \(12\,\text{cm}=120\,\text{mm}\). Then \(120+5=125\). 4. \(1.25 \times 100=125\), so the missing value is \(125\).

Answer

a) \(2500\) b) \(195\) c) \(125\) d) \(125\)
5217835
Convert each fractional measurement to the smaller unit. a) One-fifth liter in milliliters b) Two-fifths kilometer in meters c) One-eighth kilogram in grams

Hints

- Use \(1000\) smaller units for each whole measurement. - Find one-fifth before finding two-fifths. - Repeated halving can help you find one-eighth of \(1000\).

Solution

1. \(1\,\text{L}=1000\,\text{mL}\), and \(1000 \div 5=200\), so one-fifth liter is \(200\,\text{mL}\). 2. One-fifth kilometer is \(1000 \div 5=200\,\text{m}\). Two-fifths is \(2 \times 200=400\,\text{m}\). 3. \(1\,\text{kg}=1000\,\text{g}\), and \(1000 \div 8=125\), so one-eighth kilogram is \(125\,\text{g}\).

Answer

a) \(200\,\text{mL}\) b) \(400\,\text{m}\) c) \(125\,\text{g}\)
5217895
Form three groups of three equivalent money amounts. \(\$3.20\); \(\$3\) and \(2\) cents; \(32\) cents; \(\$3\) and \(20\) cents; \(\$3.02\); \(320\) cents; \(\$0.32\); \(302\) cents; \(\$0\) and \(32\) cents

Hints

- Convert every amount to cents. - Compare the resulting numerical values. - Pay attention to the difference between \(2\) cents and \(20\) cents.

Solution

1. \(\$3.20=\$3\) and \(20\) cents \(=320\) cents. 2. \(\$3.02=\$3\) and \(2\) cents \(=302\) cents. 3. \(\$0.32=\$0\) and \(32\) cents \(=32\) cents.

Answer

Group 1: \(\$3.20\), \(\$3\) and \(20\) cents, \(320\) cents Group 2: \(\$3.02\), \(\$3\) and \(2\) cents, \(302\) cents Group 3: \(\$0.32\), \(\$0\) and \(32\) cents, \(32\) cents
5313525
How much time remains in a \(24\)-hour day after \(5\,\text{hr}\,18\,\text{min}\) have passed?

Hints

- How many hours are in a full day? - Regroup one hour as \(60\) minutes before subtracting. - Subtract the minutes and hours.

Solution

1. A full day has \(24\) hours. 2. Regroup \(24\) hours as \(23\) hours \(60\) minutes. 3. Subtract: \(23\,\text{hr}\,60\,\text{min} - 5\,\text{hr}\,18\,\text{min} = 18\,\text{hr}\,42\,\text{min}\).

Answer

\(18\,\text{hr}\,42\,\text{min}\) remain.
5313595
A runner starts at exactly \(10{:}45{:}15\) a.m. (hours:minutes:seconds) and crosses the finish line at \(11{:}02{:}40\) a.m. How long is the run? Give the answer in minutes and seconds.

Hints

- Find the difference between the seconds first. - How many minutes pass before the next full hour? - Add the minutes after the hour.

Solution

1. Subtract the seconds: \(40 - 15 = 25\) seconds. 2. From \(10{:}45\) a.m. to \(11{:}00\) a.m. is \(15\) minutes, and from \(11{:}00\) a.m. to \(11{:}02\) a.m. is \(2\) minutes. 3. The total elapsed time is \(17\) minutes \(25\) seconds.

Answer

The run lasts \(17\,\text{min}\,25\,\text{s}\).
5313655
Solve these time conversion and addition problems: a) How many hours and minutes are \(315\,\text{min}\)? b) Find \(1\,\text{hr}\,45\,\text{min} + 230\,\text{min}\). Give the result in hours and minutes.

Hints

- Use \(60\) minutes in one hour. - Divide the total number of minutes by \(60\). - When adding time, regroup every \(60\) minutes as one hour.

Solution

1. For a), divide by \(60\): \(315 \div 60 = 5\) remainder \(15\). Therefore, \(315\,\text{min} = 5\,\text{hr}\,15\,\text{min}\). 2. For b), convert \(230\) minutes: \(230 \div 60 = 3\) remainder \(50\), so \(230\,\text{min} = 3\,\text{hr}\,50\,\text{min}\). 3. Add: \(1\,\text{hr}\,45\,\text{min} + 3\,\text{hr}\,50\,\text{min} = 4\,\text{hr}\,95\,\text{min} = 5\,\text{hr}\,35\,\text{min}\).

Answer

a) \(5\,\text{hr}\,15\,\text{min}\) b) \(5\,\text{hr}\,35\,\text{min}\)
5355835
A square metal sheet has side length \(36\,\text{in.}\). A square opening with side length \(12\,\text{in.}\) is centered in the sheet. Find the area of the remaining metal in square feet.
Figure for problem 535583

Hints

- Convert the side lengths to feet before finding the areas. - Find the area of each square. - Subtract the opening area from the outer-square area.

Solution

1. Convert the side lengths: \(36\,\text{in.}=3\,\text{ft}\) and \(12\,\text{in.}=1\,\text{ft}\). 2. The area of the outer square is \(3\,\text{ft}\times 3\,\text{ft}=9\,\text{ft}^2\). 3. The area of the opening is \(1\,\text{ft}\times 1\,\text{ft}=1\,\text{ft}^2\). 4. The remaining area is \(9-1=8\,\text{ft}^2\).

Answer

\(8\,\text{ft}^2\)
5359005
A rectangular field is \(660\,\text{ft}\) long and \(330\,\text{ft}\) wide. What is its area in acres?
Figure for problem 535900

Hints

- Find the rectangular area in square feet first. - Use \(1\) acre \(=43{,}560\,\text{ft}^2\) and look for a whole-number multiple.

Solution

1. Find the area in square feet: \(660\,\text{ft}\times 330\,\text{ft}=217{,}800\,\text{ft}^2\). 2. Since \(1\) acre is \(43{,}560\,\text{ft}^2\) and \(43{,}560\times 5=217{,}800\), the field has an area of \(5\) acres.

Answer

\(5\) acres
5104735
Three containers hold different amounts of water: Container A: \(0.0024\,\text{m}^3\) Container B: \(2350\,\text{mL}\) Container C: \(2.6\,\text{dm}^3\) Round each amount to the nearest whole liter. Which container has the greatest rounded volume?

Hints

- Convert all three amounts to liters before rounding. - One cubic decimeter equals one liter. - How many milliliters equal one liter?

Solution

1. Container A: \(0.0024\,\text{m}^3=2.4\,\text{L}\), which rounds to \(2\,\text{L}\). 2. Container B: \(2350\,\text{mL}=2.35\,\text{L}\), which rounds to \(2\,\text{L}\). 3. Container C: Since \(1\,\text{dm}^3=1\,\text{L}\), \(2.6\,\text{dm}^3=2.6\,\text{L}\), which rounds to \(3\,\text{L}\). 4. Since \(3\,\text{L}>2\,\text{L}\), Container C has the greatest rounded volume.

Answer

Container C, with a rounded volume of \(3\,\text{L}\).
5106575
A flatbed truck can carry at most \(12 \frac{1}{8}\) tons. It is already carrying loads of \(2 \frac{3}{8}\) tons, \(3 \frac{1}{2}\) tons, \(1 \frac{3}{4}\) tons, and \(2 \frac{1}{2}\) tons. A pallet weighing \(4200\,\text{lb}\) is ready to be added. Find the truck's remaining capacity and decide whether the pallet can be loaded safely.

Hints

- Add the loads already on the truck using a common denominator. - Find the remaining payload capacity. - How many pounds are in one US ton? - Convert the pallet weight to tons before comparing.

Solution

1. Add the current loads: \(2 \frac{3}{8}+3 \frac{1}{2}+1 \frac{3}{4}+2 \frac{1}{2}=10 \frac{1}{8}\) tons. 2. Find the remaining capacity: \(12 \frac{1}{8}-10 \frac{1}{8}=2\) tons. 3. Convert the pallet weight. Since \(1\) ton is \(2000\,\text{lb}\), \(4200\,\text{lb}=2.1\) tons. 4. Since \(2.1\) tons is greater than the remaining \(2\) tons, the pallet cannot be loaded safely.

Answer

The truck has \(2\) tons of capacity left. The \(4200\,\text{lb}\) pallet weighs \(2.1\) tons, so it cannot be loaded safely.
5108845
Air-quality measurements classify particles by size. A \(\text{PM}_{10}\) particle can have a diameter of up to \(0.01\,\text{mm}\), while a \(\text{PM}_{2.5}\) particle can have a diameter of up to \(0.0025\,\text{mm}\). a) Using the maximum diameters, how many \(\text{PM}_{2.5}\) particles placed end to end would match the diameter of one \(\text{PM}_{10}\) particle? b) A soot nanoparticle has a diameter of \(50\,\text{nm}\). How many such nanoparticles placed end to end would match the maximum diameter of one \(\text{PM}_{2.5}\) particle? Use \(1\,\text{mm} = 1{,}000{,}000\,\text{nm}\).

Hints

- For a comparison by length, divide the larger diameter by the smaller diameter. - Use the same unit before comparing or dividing measurements. - Apply the given millimeter-to-nanometer relationship before solving part b).

Solution

1. For a), divide the maximum diameters: \(0.01 \div 0.0025 = 4\). 2. For b), convert the maximum \(\text{PM}_{2.5}\) diameter to nanometers: \(0.0025 \times 1{,}000{,}000 = 2500\,\text{nm}\). 3. Divide by the nanoparticle diameter: \(2500 \div 50 = 50\).

Answer

a) \(4\) particles b) \(50\) nanoparticles
5110655
Evaluate the expression. Give the result in cubic decimeters. \(1.2\,\text{m}^3-450\,\text{dm}^3+12{,}000\,\text{cm}^3\)

Hints

- Convert every volume to the requested unit first. - How many cubic decimeters are in one cubic meter? - How many cubic centimeters are in one cubic decimeter? - Then calculate from left to right.

Solution

1. Convert each volume to cubic decimeters: \(1.2\,\text{m}^3=1200\,\text{dm}^3\) and \(12{,}000\,\text{cm}^3=12\,\text{dm}^3\). 2. Substitute the converted values: \(1200\,\text{dm}^3-450\,\text{dm}^3+12\,\text{dm}^3\). 3. Calculate from left to right: \(1200-450=750\), and \(750+12=762\). 4. The result is \(762\,\text{dm}^3\).

Answer

\(762\,\text{dm}^3\)
5110665
What number makes the equation true? \(3500\,\text{cm}^3+\Box\,\text{dm}^3=0.02\,\text{m}^3\)

Hints

- Convert the known volumes to one common unit. - How are cubic centimeters, cubic decimeters, and cubic meters related? - Think of the equation as a balance: both sides must have the same total volume.

Solution

1. Convert the known volumes to cubic decimeters. 2. Since \(3500\,\text{cm}^3=3.5\,\text{dm}^3\) and \(0.02\,\text{m}^3=20\,\text{dm}^3\), the equation becomes \(3.5+\Box=20\). 3. Subtract to find the missing number: \(20-3.5=16.5\). 4. The missing number is \(16.5\).

Answer

\(16.5\)
5111075
Fill in each missing value. a) \(4.5\,\text{L}=\dots\,\text{mL}\) b) \(250\,\text{cm}^3=\dots\,\text{dm}^3\) c) \(0.03\,\text{m}^3=\dots\,\text{L}\) d) \(12\,\text{dm}^3+5\,\text{cm}^3=\dots\,\text{cm}^3\)

Hints

- Determine how many smaller units make one larger unit. - Recall the relationship between liters and cubic decimeters. - When converting from a larger unit to a smaller unit, the numerical value increases. - In part d), convert both quantities to the same unit before adding.

Solution

1. Since \(1\,\text{L}=1000\,\text{mL}\), \(4.5\times1000=4500\,\text{mL}\). 2. Since \(1000\,\text{cm}^3=1\,\text{dm}^3\), \(250\div1000=0.25\,\text{dm}^3\). 3. Since \(1\,\text{m}^3=1000\,\text{L}\), \(0.03\times1000=30\,\text{L}\). 4. Convert \(12\,\text{dm}^3\) to \(12{,}000\,\text{cm}^3\). Then \(12{,}000+5=12{,}005\,\text{cm}^3\).

Answer

a) \(4500\,\text{mL}\) b) \(0.25\,\text{dm}^3\) c) \(30\,\text{L}\) d) \(12{,}005\,\text{cm}^3\)
5111095
Find the sum. Give the result in liters. \(0.4\,\text{m}^3+120\,\text{dm}^3+5000\,\text{cm}^3\)

Hints

- Quantities with different units cannot be added directly. - Convert each term to liters first. - One liter equals one cubic decimeter.

Solution

1. Convert the first term: \(0.4\,\text{m}^3=400\,\text{L}\). 2. Convert the second term: \(120\,\text{dm}^3=120\,\text{L}\). 3. Convert the third term: \(5000\,\text{cm}^3=5\,\text{L}\). 4. Add the volumes: \(400\,\text{L}+120\,\text{L}+5\,\text{L}=525\,\text{L}\).

Answer

\(525\,\text{L}\)
5111115
Compare each pair of volumes. Write \(<\), \(>\), or \(=\) in the blank. a) \(450\,\text{mL}\quad\dots\quad0.45\,\text{dm}^3\) b) \(0.02\,\text{m}^3\quad\dots\quad2000\,\text{cm}^3\) c) \(1500\,\text{mm}^3\quad\dots\quad1.5\,\text{cm}^3\) d) \(12\,\text{L}\quad\dots\quad1200\,\text{cm}^3\)

Hints

- Convert both quantities in each pair to the same unit. - One liter equals one cubic decimeter. - One milliliter equals one cubic centimeter. - Keep track of the order \(\text{m}^3\), \(\text{dm}^3\), \(\text{cm}^3\), and \(\text{mm}^3\).

Solution

1. For part a, \(0.45\,\text{dm}^3=450\,\text{cm}^3=450\,\text{mL}\), so the volumes are equal. 2. For part b, \(0.02\,\text{m}^3=20{,}000\,\text{cm}^3\), and \(20{,}000>2000\). 3. For part c, \(1500\,\text{mm}^3=1.5\,\text{cm}^3\), so the volumes are equal. 4. For part d, \(12\,\text{L}=12{,}000\,\text{cm}^3\), and \(12{,}000>1200\).

Answer

a) \(=\) b) \(>\) c) \(=\) d) \(>\)
5111145
Complete the table so that all three entries in each row represent the same volume. <table> <tr> <th>\(\text{m}^3\)</th> <th>\(\text{dm}^3\) (or \(\text{L}\))</th> <th>\(\text{cm}^3\) (or \(\text{mL}\))</th> </tr> <tr> <td>\(1.2\)</td> <td>a) \(\dots\)</td> <td>b) \(\dots\)</td> </tr> <tr> <td>c) \(\dots\)</td> <td>d) \(\dots\)</td> <td>\(750\)</td> </tr> <tr> <td>e) \(\dots\)</td> <td>\(40\)</td> <td>f) \(\dots\)</td> </tr> </table>

Hints

- Recall the conversion factor between adjacent cubic metric units. - How many smaller cubes fit inside one cube of the next larger unit? - Convert step by step from cubic meters to cubic decimeters and then to cubic centimeters.

Solution

1. First row: \(1.2\,\text{m}^3=1200\,\text{dm}^3=1{,}200{,}000\,\text{cm}^3\). 2. Second row: \(750\,\text{cm}^3=0.75\,\text{dm}^3=0.00075\,\text{m}^3\). 3. Third row: \(40\,\text{dm}^3=0.04\,\text{m}^3=40{,}000\,\text{cm}^3\).

Answer

a) \(1200\) b) \(1{,}200{,}000\) c) \(0.00075\) d) \(0.75\) e) \(0.04\) f) \(40{,}000\)
5111155
Lucas says, “Since \(10\,\text{mm}=1\,\text{cm}\), it must be true that \(100\,\text{mm}^3=1\,\text{cm}^3\).” Explain Lucas's error and give the correct value of \(100\,\text{mm}^3\) in cubic centimeters.

Hints

- Picture a cube with an edge length of \(1\,\text{cm}\). How many \(1\,\text{mm}\) cubes fit along one edge? - How many layers of the smaller cubes are there? - What happens to volume when each of three edge lengths is multiplied by \(10\)?

Solution

1. A cube with volume \(1\,\text{cm}^3\) has dimensions \(1\,\text{cm}\times1\,\text{cm}\times1\,\text{cm}\). 2. In millimeters, those dimensions are \(10\,\text{mm}\times10\,\text{mm}\times10\,\text{mm}\), so \(1\,\text{cm}^3=1000\,\text{mm}^3\). 3. The volume conversion factor is \(10\times10\times10=1000\), not \(100\). 4. Therefore, \(100\,\text{mm}^3\div1000=0.1\,\text{cm}^3\).

Answer

Lucas applied a two-dimensional conversion factor to a three-dimensional measurement. Since \(1\,\text{cm}^3=1000\,\text{mm}^3\), the correct conversion is \(100\,\text{mm}^3=0.1\,\text{cm}^3\).
5111195
Convert each measurement to the unit shown in parentheses. a) \(1.5\,\text{m}^3\) (\(\text{dm}^3\)) b) \(450\,\text{cm}^3\) (\(\text{dm}^3\)) c) \(0.02\,\text{dm}^3\) (\(\text{mm}^3\)) d) \(2\,\text{L}\ 5\,\text{mL}\) (\(\text{cm}^3\))

Hints

- Decide how many smaller units make one larger unit. - Adjacent cubic metric units differ by a factor of \(1000\). - One cubic decimeter equals one liter. - Decide whether the numerical value should increase or decrease.

Solution

1. For part a, multiply by \(1000\): \(1.5\times1000=1500\,\text{dm}^3\). 2. For part b, divide by \(1000\): \(450\div1000=0.45\,\text{dm}^3\). 3. From cubic decimeters to cubic millimeters, the factor is \(1000\times1000=1{,}000{,}000\). Therefore, \(0.02\times1{,}000{,}000=20{,}000\,\text{mm}^3\). 4. Since \(1\,\text{L}=1000\,\text{cm}^3\) and \(1\,\text{mL}=1\,\text{cm}^3\), \(2\times1000+5=2005\,\text{cm}^3\).

Answer

a) \(1500\,\text{dm}^3\) b) \(0.45\,\text{dm}^3\) c) \(20{,}000\,\text{mm}^3\) d) \(2005\,\text{cm}^3\)
5111235
Determine which measurements represent equal volumes. Then arrange all the measurements in increasing order, beginning with the smallest. \(2.5\,\text{dm}^3\); \(0.025\,\text{m}^3\); \(2500\,\text{cm}^3\); \(25\,\text{L}\)

Hints

- Convert all measurements to the same unit, such as cubic decimeters. - How many liters are in one cubic meter? - Adjacent cubic metric units differ by a factor of \(1000\).

Solution

1. Convert all measurements to cubic decimeters or liters. 2. \(2.5\,\text{dm}^3\) remains \(2.5\,\text{dm}^3\). 3. \(0.025\,\text{m}^3=25\,\text{dm}^3\). 4. \(2500\,\text{cm}^3=2.5\,\text{dm}^3\). 5. \(25\,\text{L}=25\,\text{dm}^3\). 6. Therefore, \(2.5\,\text{dm}^3=2500\,\text{cm}^3<0.025\,\text{m}^3=25\,\text{L}\).

Answer

\(2.5\,\text{dm}^3=2500\,\text{cm}^3<0.025\,\text{m}^3=25\,\text{L}\)
5111245
Two water tanks are being compared. Tank A has a volume of \(0.06\,\text{m}^3\). Tank B has a volume of \(55{,}000\,\text{cm}^3\). Which tank can hold more water? Find the difference in liters.

Hints

- Convert both measurements to liters. - One cubic decimeter equals one liter. - How many cubic centimeters are in one cubic decimeter? - How many liters are in one cubic meter?

Solution

1. Convert Tank A to liters: \(0.06\,\text{m}^3=60\,\text{L}\). 2. Convert Tank B to liters: \(55{,}000\,\text{cm}^3=55\,\text{L}\). 3. Since \(60\,\text{L}>55\,\text{L}\), Tank A holds more. 4. The difference is \(60\,\text{L}-55\,\text{L}=5\,\text{L}\).

Answer

Tank A holds more water. The difference is \(5\,\text{L}\).
5111275
Fill in each missing number or unit so that the equation is true. a) \(3.5\,\text{m}^3=\dots\,\text{dm}^3\) b) \(7500\,\text{mL}=\dots\,\text{L}\) c) \(0.2\,\text{dm}^3=\dots\,\text{mm}^3\) d) \(450{,}000\,\text{cm}^3=0.45\,\dots\)

Hints

- Decide whether each conversion uses a larger or smaller unit. - Determine whether to multiply or divide by \(1000\). - Adjacent cubic metric units differ by a factor of \(1000\).

Solution

1. For part a, \(3.5\times1000=3500\), so \(3.5\,\text{m}^3=3500\,\text{dm}^3\). 2. For part b, \(7500\div1000=7.5\), so \(7500\,\text{mL}=7.5\,\text{L}\). 3. For part c, \(1\,\text{dm}^3=1{,}000{,}000\,\text{mm}^3\). Therefore, \(0.2\times1{,}000{,}000=200{,}000\,\text{mm}^3\). 4. For part d, \(450{,}000\,\text{cm}^3\div1{,}000{,}000=0.45\,\text{m}^3\), so the missing unit is \(\text{m}^3\).

Answer

a) \(3500\,\text{dm}^3\) b) \(7.5\,\text{L}\) c) \(200{,}000\,\text{mm}^3\) d) \(0.45\,\text{m}^3\)
5111295
Evaluate the expression. Give the result in liters. \(1.5\,\text{m}^3-(400\,\text{dm}^3+250{,}000\,\text{cm}^3)\)

Hints

- Evaluate the parentheses first. - Convert each term to liters before calculating. - How many liters are in one cubic meter?

Solution

1. Convert each volume to liters: \(1.5\,\text{m}^3=1500\,\text{L}\), \(400\,\text{dm}^3=400\,\text{L}\), and \(250{,}000\,\text{cm}^3=250\,\text{L}\). 2. Evaluate the parentheses: \(400\,\text{L}+250\,\text{L}=650\,\text{L}\). 3. Subtract: \(1500\,\text{L}-650\,\text{L}=850\,\text{L}\).

Answer

\(850\,\text{L}\)
5111305
Fill in each missing number so that the equation is true. a) \(0.25\,\text{L}+\dots\,\text{cm}^3=1\,\text{dm}^3\) b) \(0.002\,\text{m}^3-1500\,\text{cm}^3=\dots\,\text{L}\)

Hints

- Convert all quantities in each equation to the unit shown next to the blank. - Recall the factor of \(1000\) between adjacent cubic metric units. - How many cubic centimeters are in one-quarter liter?

Solution

1. For part a, \(1\,\text{dm}^3=1000\,\text{cm}^3\) and \(0.25\,\text{L}=250\,\text{cm}^3\). The missing amount is \(1000-250=750\,\text{cm}^3\). 2. For part b, \(0.002\,\text{m}^3=2\,\text{L}\) and \(1500\,\text{cm}^3=1.5\,\text{L}\). The difference is \(2-1.5=0.5\,\text{L}\).

Answer

a) \(750\) b) \(0.5\)
5111355
Compare \(2500\,\text{mL}\) and \(0.03\,\text{m}^3\). Which volume is greater? Also find the difference in liters.

Hints

- Convert both measurements to the same unit before comparing. - One liter equals one cubic decimeter. - How many liters are in one cubic meter?

Solution

1. Convert \(2500\,\text{mL}\) to liters: \(2500\,\text{mL}=2.5\,\text{L}\). 2. Convert \(0.03\,\text{m}^3\) to liters: \(0.03\,\text{m}^3=30\,\text{L}\). 3. Since \(30\,\text{L}>2.5\,\text{L}\), \(0.03\,\text{m}^3\) is greater. 4. The difference is \(30\,\text{L}-2.5\,\text{L}=27.5\,\text{L}\).

Answer

\(0.03\,\text{m}^3\) is greater. The difference is \(27.5\,\text{L}\).
5111405
Evaluate each expression, paying attention to the units. For parts a) and b), give the result in the larger of the two units shown. a) \(1.5\,\text{dm}^3+800\,\text{cm}^3\) b) \(0.04\,\text{m}^3-15\,\text{dm}^3\) c) \(2.4\,\text{L}\div6\)

Hints

- Decide how many smaller units make one larger unit. - It may be easier to calculate first in the smaller unit. - One cubic decimeter equals one liter.

Solution

1. For part a, \(1.5\,\text{dm}^3=1500\,\text{cm}^3\). The sum is \(1500+800=2300\,\text{cm}^3=2.3\,\text{dm}^3\). 2. For part b, \(0.04\,\text{m}^3=40\,\text{dm}^3\). The difference is \(40-15=25\,\text{dm}^3=0.025\,\text{m}^3\). 3. For part c, \(2.4\div6=0.4\), so the result is \(0.4\,\text{L}\).

Answer

a) \(2.3\,\text{dm}^3\) b) \(0.025\,\text{m}^3\) c) \(0.4\,\text{L}\)
5111435
Write each decimal volume as a sum using whole-number amounts of the listed units. a) \(4.007\,\text{m}^3\), using \(\text{m}^3\) and \(\text{dm}^3\) b) \(12.65\,\text{dm}^3\), using \(\text{dm}^3\) and \(\text{cm}^3\) c) \(2.3004\,\text{m}^3\), using \(\text{m}^3\), \(\text{dm}^3\), and \(\text{cm}^3\) d) \(5.08\,\text{L}\), using \(\text{L}\) and \(\text{mL}\)

Hints

- Keep the whole-number part in the original unit. - Convert the decimal part to the next smaller unit by multiplying by \(1000\). - Repeat the process when the converted value still has a decimal part.

Solution

1. For part a, \(0.007\,\text{m}^3=7\,\text{dm}^3\), so \(4.007\,\text{m}^3=4\,\text{m}^3+7\,\text{dm}^3\). 2. For part b, \(0.65\,\text{dm}^3=650\,\text{cm}^3\), so \(12.65\,\text{dm}^3=12\,\text{dm}^3+650\,\text{cm}^3\). 3. For part c, \(0.3004\,\text{m}^3=300.4\,\text{dm}^3=300\,\text{dm}^3+400\,\text{cm}^3\). Therefore, \(2.3004\,\text{m}^3=2\,\text{m}^3+300\,\text{dm}^3+400\,\text{cm}^3\). 4. For part d, \(0.08\,\text{L}=80\,\text{mL}\), so \(5.08\,\text{L}=5\,\text{L}+80\,\text{mL}\).

Answer

a) \(4\,\text{m}^3+7\,\text{dm}^3\) b) \(12\,\text{dm}^3+650\,\text{cm}^3\) c) \(2\,\text{m}^3+300\,\text{dm}^3+400\,\text{cm}^3\) d) \(5\,\text{L}+80\,\text{mL}\)
5111445
Write each sum as a decimal in the largest unit shown. a) \(7\,\text{m}^3+50\,\text{dm}^3\) b) \(2\,\text{dm}^3+4\,\text{cm}^3\) c) \(10\,\text{L}+5\,\text{mL}\) d) \(1\,\text{m}^3+20\,\text{dm}^3+300\,\text{cm}^3\)

Hints

- Convert every smaller-unit amount to the largest unit shown. - A factor of \(1000\) means that three decimal places represent the next smaller cubic unit. - Keep placeholder zeros when a smaller-unit amount does not fill all three decimal places.

Solution

1. For part a, \(50\,\text{dm}^3=0.05\,\text{m}^3\), so the total is \(7.05\,\text{m}^3\). 2. For part b, \(4\,\text{cm}^3=0.004\,\text{dm}^3\), so the total is \(2.004\,\text{dm}^3\). 3. For part c, \(5\,\text{mL}=0.005\,\text{L}\), so the total is \(10.005\,\text{L}\). 4. For part d, \(20\,\text{dm}^3=0.02\,\text{m}^3\) and \(300\,\text{cm}^3=0.0003\,\text{m}^3\). The total is \(1+0.02+0.0003=1.0203\,\text{m}^3\).

Answer

a) \(7.05\,\text{m}^3\) b) \(2.004\,\text{dm}^3\) c) \(10.005\,\text{L}\) d) \(1.0203\,\text{m}^3\)
5111455
Compare each pair. Write \(<\), \(>\), or \(=\). a) \(2.05\,\text{m}^3\quad\Box\quad2\,\text{m}^3+500\,\text{dm}^3\) b) \(3.4\,\text{dm}^3\quad\Box\quad3\,\text{dm}^3+40\,\text{cm}^3\) c) \(0.008\,\text{L}\quad\Box\quad80\,\text{mL}\) d) \(1\,\text{cm}^3+2\,\text{mm}^3\quad\Box\quad1.002\,\text{cm}^3\)

Hints

- Convert both quantities in each pair to the same unit. - Add the parts of each sum after converting them. - Adjacent cubic metric units differ by a factor of \(1000\).

Solution

1. For part a, \(2.05\,\text{m}^3=2050\,\text{dm}^3\), while \(2\,\text{m}^3+500\,\text{dm}^3=2500\,\text{dm}^3\). Therefore, \(2050<2500\). 2. For part b, \(3.4\,\text{dm}^3=3400\,\text{cm}^3\), while \(3\,\text{dm}^3+40\,\text{cm}^3=3040\,\text{cm}^3\). Therefore, \(3400>3040\). 3. For part c, \(0.008\,\text{L}=8\,\text{mL}\), and \(8<80\). 4. For part d, \(2\,\text{mm}^3=0.002\,\text{cm}^3\), so \(1\,\text{cm}^3+2\,\text{mm}^3=1.002\,\text{cm}^3\).

Answer

a) \(<\) b) \(>\) c) \(<\) d) \(=\)
5111485
Jonas says, “My new school backpack can hold \(150{,}000\,\text{mL}\) of lemonade.” Decide whether this is realistic for a normal school backpack. First convert the volume to cubic centimeters and then to liters. Compare your result with a typical school backpack capacity of about \(20\) to \(30\) liters.

Hints

- How are milliliters and cubic centimeters related? - How many milliliters are in one liter? - Compare the converted capacity with the stated range for a typical backpack.

Solution

1. Since \(1\,\text{mL}=1\,\text{cm}^3\), \(150{,}000\,\text{mL}=150{,}000\,\text{cm}^3\). 2. Since \(1000\,\text{cm}^3=1\,\text{L}\), \(150{,}000\div1000=150\,\text{L}\). 3. A typical school backpack holds about \(20\) to \(30\) liters. A capacity of \(150\) liters is about \(5\) to \(7.5\) times as large, so Jonas's claim is not realistic.

Answer

No. \(150{,}000\,\text{mL}=150{,}000\,\text{cm}^3=150\,\text{L}\), which is far greater than the typical \(20\)- to \(30\)-liter capacity of a school backpack.
5111515
Two containers are being compared. Container A holds \(2.05\,\text{L}\) of water. Container B holds \(1\,\text{L}+800\,\text{mL}\) of juice. a) Which container holds more liquid? b) Find the total amount of liquid in both containers in liters, written as a decimal. c) Write the total as a sum of whole liters and milliliters.

Hints

- Convert both amounts to the same unit before comparing. - Align decimal points when adding decimal numbers. - Convert the decimal part of the total liters to milliliters.

Solution

1. Container B holds \(1\,\text{L}+800\,\text{mL}=1.8\,\text{L}\). Since \(2.05>1.8\), Container A holds more. 2. The total is \(2.05\,\text{L}+1.8\,\text{L}=3.85\,\text{L}\). 3. Since \(0.85\,\text{L}=850\,\text{mL}\), \(3.85\,\text{L}=3\,\text{L}+850\,\text{mL}\).

Answer

a) Container A b) \(3.85\,\text{L}\) c) \(3\,\text{L}+850\,\text{mL}\)
5111525
Convert each volume to the requested unit. a) \(4\frac{3}{4}\,\text{m}^3\) to \(\text{dm}^3\) b) \(0.2\,\text{dm}^3\) to \(\text{cm}^3\) c) \(1\frac{1}{2}\,\text{cm}^3\) to \(\text{mm}^3\) d) \(2\frac{1}{4}\,\text{L}\) to \(\text{mL}\)

Hints

- Adjacent cubic metric units differ by a factor of \(1000\). - One liter equals one cubic decimeter. - Convert each mixed number to a decimal or improper fraction before multiplying.

Solution

1. For part a, \(4\frac{3}{4}=4.75\), and \(4.75\times1000=4750\,\text{dm}^3\). 2. For part b, \(0.2\times1000=200\,\text{cm}^3\). 3. For part c, \(1\frac{1}{2}=1.5\), and \(1.5\times1000=1500\,\text{mm}^3\). 4. For part d, \(2\frac{1}{4}=2.25\), and \(2.25\times1000=2250\,\text{mL}\).

Answer

a) \(4750\,\text{dm}^3\) b) \(200\,\text{cm}^3\) c) \(1500\,\text{mm}^3\) d) \(2250\,\text{mL}\)
5111565
Three containers hold water. Container A holds \(0.05\,\text{m}^3\), Container B holds \(48\,\text{L}\), and Container C holds \(52{,}000\,\text{cm}^3\). Order the containers from least to greatest amount of water. Justify your order by converting every volume to liters.

Hints

- How are liters and cubic decimeters related? - How many liters are in one cubic meter? - Convert every amount to the same unit before comparing.

Solution

1. Container A: \(0.05\,\text{m}^3=50\,\text{L}\). 2. Container B: \(48\,\text{L}\). 3. Container C: \(52{,}000\,\text{cm}^3=52\,\text{L}\). 4. Since \(48\,\text{L}<50\,\text{L}<52\,\text{L}\), the order is Container B, Container A, Container C.

Answer

Container B \(<\) Container A \(<\) Container C, because \(48\,\text{L}<50\,\text{L}<52\,\text{L}\).
5111575
A large tank has a capacity of exactly \(1\,\text{m}^3\). It already contains \(650\,\text{dm}^3\) of water and \(120\,\text{L}\) of oil. How many cubic decimeters of space remain before the tank is full?

Hints

- Express the tank's capacity in the requested unit. - Add the volumes already in the tank. - Subtract the filled volume from the total capacity.

Solution

1. Convert the total capacity: \(1\,\text{m}^3=1000\,\text{dm}^3\). 2. Convert the oil volume: \(120\,\text{L}=120\,\text{dm}^3\). 3. The filled volume is \(650\,\text{dm}^3+120\,\text{dm}^3=770\,\text{dm}^3\). 4. The remaining volume is \(1000\,\text{dm}^3-770\,\text{dm}^3=230\,\text{dm}^3\).

Answer

\(230\,\text{dm}^3\) of space remains.
5111635
Evaluate each expression. Give the answer in the unit shown in parentheses. a) \(1.2\,\text{L}+300\,\text{mL}+500\,\text{mL}\) (liters) b) \(20\,\text{L}-15\,\text{L}\) (milliliters)

Hints

- Convert all values in an expression to one common unit before calculating. - Give the final result in the unit requested.

Solution

1. For part a, \(300\,\text{mL}=0.3\,\text{L}\) and \(500\,\text{mL}=0.5\,\text{L}\). Then \(1.2+0.3+0.5=2\,\text{L}\). 2. For part b, \(20\,\text{L}-15\,\text{L}=5\,\text{L}\). Since \(1\,\text{L}=1000\,\text{mL}\), \(5\,\text{L}=5000\,\text{mL}\).

Answer

a) \(2\,\text{L}\) b) \(5000\,\text{mL}\)
5111685
Consider two cubes. a) Find the volume of a small cube with an edge length of \(10\,\text{cm}\). b) How many of these small cubes fit inside a large cube with an edge length of \(1\,\text{m}\)? c) Use your result from part b) to state the conversion from \(1\,\text{m}^3\) to cubic centimeters.

Hints

- Use the cube-volume formula. - Convert the large cube's edge length to centimeters. - Compare the large cube's volume with the small cube's volume. - Remember that all three dimensions change when converting cubic units.

Solution

1. The small cube has volume \(10\times10\times10=1000\,\text{cm}^3\). 2. The large cube has an edge length of \(1\,\text{m}=100\,\text{cm}\). Its volume is \(100\times100\times100=1{,}000{,}000\,\text{cm}^3\). 3. The number of small cubes is \(1{,}000{,}000\div1000=1000\). 4. Therefore, \(1\,\text{m}^3=1{,}000{,}000\,\text{cm}^3\).

Answer

a) \(1000\,\text{cm}^3\) b) \(1000\) small cubes c) \(1\,\text{m}^3=1{,}000{,}000\,\text{cm}^3\)
5118225
A water dispenser at a recreation center starts with \(18.5\,\text{L}\) of water. During the morning, people fill \(60\) cups with \(250\,\text{mL}\) of water each. How many milliliters of water remain in the dispenser?

Hints

- First find the total amount of water removed. - Keep track of the unit requested for the answer. - How are liters and milliliters related?

Solution

1. The total amount removed is \(60\times250\,\text{mL}=15{,}000\,\text{mL}\). 2. Convert the amount removed to liters: \(15{,}000\,\text{mL}=15\,\text{L}\). 3. The amount remaining is \(18.5\,\text{L}-15\,\text{L}=3.5\,\text{L}\). 4. Convert to milliliters: \(3.5\,\text{L}=3500\,\text{mL}\).

Answer

\(3500\,\text{mL}\) of water remains.
5118235
Two pitchers are in a kitchen. Pitcher A contains \(2500\,\text{mL}\) of water. Pitcher B contains \(2.45\,\text{L}\) and has a maximum capacity of \(2.6\,\text{L}\). Then \(125\,\text{mL}\) of water is poured from Pitcher A into Pitcher B. Will Pitcher B overflow? Justify your answer with a calculation.

Hints

- How much space remains in Pitcher B before any water is added? - Convert the amounts to the same unit. - Compare the new total with the pitcher's capacity.

Solution

1. Convert the added amount to liters: \(125\,\text{mL}=0.125\,\text{L}\). 2. The new amount in Pitcher B is \(2.45\,\text{L}+0.125\,\text{L}=2.575\,\text{L}\). 3. Since \(2.575\,\text{L}<2.6\,\text{L}\), Pitcher B does not overflow.

Answer

No. Pitcher B will contain \(2.575\,\text{L}\), which is less than its \(2.6\,\text{L}\) capacity.
5169575
Two ropes are cut into equal-length pieces. Rope A is \(25.68\,\text{m}\) long and is cut into \(8\) equal pieces. Rope B is \(19.26\,\text{m}\) long and is cut into \(6\) equal pieces. Find the length of one piece from each rope in centimeters. Which rope makes longer pieces, or are the pieces equal in length?

Hints

- Work with one rope at a time. - Convert each rope's length to centimeters before dividing. - Compare the two final piece lengths.

Solution

1. Convert Rope A to centimeters: \(25.68\,\text{m} = 2568\,\text{cm}\). 2. Divide Rope A into \(8\) equal pieces: \(2568 \div 8 = 321\,\text{cm}\). 3. Convert Rope B to centimeters: \(19.26\,\text{m} = 1926\,\text{cm}\). 4. Divide Rope B into \(6\) equal pieces: \(1926 \div 6 = 321\,\text{cm}\). 5. Both ropes make pieces of the same length.

Answer

Both ropes make pieces that are \(321\,\text{cm}\) long.
5171645
A student records the time spent on homework and studying each school day: <table> <tr><td>Math</td><td>\(25\,\text{minutes}\)</td></tr> <tr><td>Language arts</td><td>\(25\,\text{minutes}\)</td></tr> <tr><td>Science</td><td>\(15\,\text{minutes}\)</td></tr> <tr><td>Spanish</td><td>\(10\,\text{minutes}\)</td></tr> <tr><td>Independent reading</td><td>\(15\,\text{minutes}\)</td></tr> </table> a) Find the total study time per day in minutes. b) Find the total study time in a \(5\)-day school week. Give the answer in hours and minutes. c) A school year has \(40\) school weeks. Find the total study time for the school year in hours.

Hints

- Add all the times for one day first. - Use \(60\) minutes per hour when converting. - Multiply the daily total by the number of school days in a week. - Then scale the weekly total to \(40\) weeks.

Solution

1. Add the daily times: \(25 + 25 + 15 + 10 + 15 = 90\) minutes. 2. Find the weekly time: \(90 \times 5 = 450\) minutes. 3. Convert the weekly time: \(450 \div 60 = 7\) remainder \(30\), so the weekly time is \(7\) hours \(30\) minutes. 4. Find the yearly time: \(450 \times 40 = 18{,}000\) minutes. 5. Convert to hours: \(18{,}000 \div 60 = 300\) hours.

Answer

a) The daily study time is \(90\) minutes. b) The weekly study time is \(7\) hours \(30\) minutes. c) The yearly study time is \(300\) hours.
5172325
A leaky faucet loses exactly one drop every second. How many weeks, days, hours, minutes, and seconds will pass before \(10{,}000{,}000\) drops have fallen? Use \(1\) week \(= 7\) days.

Hints

- At one drop each second, the number of drops equals the number of seconds. - Convert one time unit at a time, starting with seconds. - What does each remainder represent in the smaller unit? - Remember that one week has \(7\) days.

Solution

1. At one drop per second, \(10{,}000{,}000\) drops take \(10{,}000{,}000\) seconds. 2. Convert seconds to minutes: \(10{,}000{,}000 \div 60 = 166{,}666\) remainder \(40\), so the time is \(166{,}666\) minutes and \(40\) seconds. 3. Convert minutes to hours: \(166{,}666 \div 60 = 2777\) remainder \(46\), so the time is \(2777\) hours, \(46\) minutes, and \(40\) seconds. 4. Convert hours to days: \(2777 \div 24 = 115\) remainder \(17\), so the time is \(115\) days, \(17\) hours, \(46\) minutes, and \(40\) seconds. 5. Convert days to weeks: \(115 \div 7 = 16\) remainder \(3\).

Answer

It takes \(16\) weeks, \(3\) days, \(17\) hours, \(46\) minutes, and \(40\) seconds.
5204795
A hiking trail is \(3\,\text{km}\) long. Mrs. Smith estimates that she takes \(4000\) steps to complete the trail. a) If her estimate is exact, what is her average step length in centimeters? b) Her small dog travels the same distance using exactly twice as many steps. How many steps does the dog take, and what is the dog’s average step length?

Hints

- Convert the trail length to centimeters first. - Divide the total length by the number of equal steps. - In part b), connect twice as many steps with the length of each step.

Solution

1. Convert the trail length: \(3\,\text{km} = 3000\,\text{m} = 300{,}000\,\text{cm}\). 2. Find Mrs. Smith’s step length: \(300{,}000 \div 4000 = 75\), so it is \(75\,\text{cm}\). 3. The dog takes twice as many steps: \(4000 \times 2 = 8000\) steps. 4. Find the dog’s step length: \(300{,}000 \div 8000 = 37.5\), so it is \(37.5\,\text{cm}\).

Answer

a) \(75\,\text{cm}\) b) The dog takes \(8000\) steps, and its average step length is \(37.5\,\text{cm}\).
5206195
Roadside marker posts will be placed along both sides of a straight highway from mile marker \(7\) to mile marker \(12\). On each side, the posts are spaced every \(200\,\text{ft}\), with posts at both endpoints. How many posts are needed altogether? Use \(1\,\text{mi} = 5280\,\text{ft}\).

Hints

- Find the distance between the two mile markers. - Convert the distance to feet before dividing by the spacing. - The number of posts on one side is one more than the number of spaces. - Double the result for both sides of the highway.

Solution

1. Find the highway length: \(12\,\text{mi} - 7\,\text{mi} = 5\,\text{mi}\). 2. Convert to feet: \(5 \times 5280 = 26{,}400\), so the section is \(26{,}400\,\text{ft}\) long. 3. Find the number of spaces on one side: \(26{,}400 \div 200 = 132\). 4. Include both endpoint posts: \(132 + 1 = 133\) posts on one side. 5. Account for both sides of the highway: \(133 \times 2 = 266\).

Answer

\(266\) marker posts are needed altogether.
5206955
Evaluate the expression and write the result as a decimal number of meters: \(5\,\text{m}\,4\,\text{cm} - 4 \times (15\,\text{cm} + 22\,\text{mm})\).

Hints

- Evaluate the parentheses before multiplying and subtracting. - Convert all lengths to the smallest unit used. - Complete the arithmetic before converting the final result to meters.

Solution

1. Evaluate the parentheses in millimeters: \(15\,\text{cm} + 22\,\text{mm} = 150\,\text{mm} + 22\,\text{mm} = 172\,\text{mm}\). 2. Multiply: \(4 \times 172\,\text{mm} = 688\,\text{mm}\). 3. Convert the first length to millimeters: \(5\,\text{m}\,4\,\text{cm} = 5000\,\text{mm} + 40\,\text{mm} = 5040\,\text{mm}\). 4. Subtract: \(5040\,\text{mm} - 688\,\text{mm} = 4352\,\text{mm}\). 5. Convert to meters: \(4352\,\text{mm} = 4.352\,\text{m}\).

Answer

\(4.352\,\text{m}\)
5209265
A cargo plane can carry at most \(25\,\text{t}\) of payload. For a relief flight, workers load \(840\) crates with a mass of \(25\,\text{kg}\) each. Twelve support staff members also fly, with a combined person-and-equipment mass of \(120\,\text{kg}\) each. a) Find the current payload mass in kilograms. b) How many tons of payload capacity remain? c) By how many kilograms would the remaining capacity decrease if improved packaging added \(500\,\text{g}\) to every crate?

Hints

- Find the crate mass and staff mass separately. - Express capacity and payload in the same unit before subtracting. - Remaining capacity is the difference between the maximum and current payload. - Convert \(500\,\text{g}\) to kilograms before multiplying.

Solution

1. Find the crates’ mass: \(840 \times 25\,\text{kg} = 21{,}000\,\text{kg}\). 2. Find the support staff mass: \(12 \times 120\,\text{kg} = 1440\,\text{kg}\). 3. For a), add: \(21{,}000 + 1440 = 22{,}440\), so the payload mass is \(22{,}440\,\text{kg}\). 4. For b), convert the capacity: \(25\,\text{t} = 25{,}000\,\text{kg}\). The remaining capacity is \(25{,}000 - 22{,}440 = 2560\,\text{kg} = 2.56\,\text{t}\). 5. For c), \(500\,\text{g} = 0.5\,\text{kg}\), so the added packaging mass is \(840 \times 0.5\,\text{kg} = 420\,\text{kg}\).

Answer

a) \(22{,}440\,\text{kg}\) b) \(2.56\,\text{t}\) c) The remaining capacity would decrease by \(420\,\text{kg}\).
5211835
Find the missing duration \(\square\) that makes the equation true: \(1\,\text{hr}\,25\,\text{min} + 140\,\text{s} + \square = 2\,\text{hr}\)

Hints

- How many seconds are in one minute and one hour? - Express every duration in seconds. - How much time is missing from the known total to reach two hours?

Solution

1. Convert the known duration to seconds: \(1\,\text{hr}\,25\,\text{min} = 60 \times 60 + 25 \times 60 = 5100\,\text{s}\). 2. Convert the target to seconds: \(2\,\text{hr} = 2 \times 3600 = 7200\,\text{s}\). 3. Add the known times: \(5100 + 140 = 5240\,\text{s}\). 4. Find the missing time: \(7200 - 5240 = 1960\,\text{s}\). 5. Convert if desired: \(1960\,\text{s} = 32\,\text{min}\,40\,\text{s}\).

Answer

\(\square = 1960\,\text{s}\), or \(32\,\text{min}\,40\,\text{s}\)
5212515
Ms. Weber works exactly \(35\) hours each week. She takes a \(30\)-minute unpaid lunch break each day. Her times for Monday through Thursday are shown below. <table> <tr> <th>Day</th> <th>Start</th> <th>End</th> </tr> <tr> <td>Monday</td> <td>\(8{:}00\) a.m.</td> <td>\(4{:}00\) p.m.</td> </tr> <tr> <td>Tuesday</td> <td>\(8{:}30\) a.m.</td> <td>\(3{:}30\) p.m.</td> </tr> <tr> <td>Wednesday</td> <td>\(7{:}45\) a.m.</td> <td>\(4{:}15\) p.m.</td> </tr> <tr> <td>Thursday</td> <td>\(9{:}00\) a.m.</td> <td>\(3{:}30\) p.m.</td> </tr> </table> On Friday, Ms. Weber starts at \(8{:}00\) a.m. She also takes a \(30\)-minute unpaid lunch break. What time can she leave after completing exactly \(35\) work hours?

Hints

- The lunch break does not count as work time. - Add the work time from Monday through Thursday. - How many work hours remain for Friday? - Add Friday's lunch break when finding the clock time she can leave.

Solution

1. Subtract the \(30\)-minute lunch break from each day's elapsed time. The work times are \(7\,\text{hr}\,30\,\text{min}\), \(6\,\text{hr}\,30\,\text{min}\), \(8\,\text{hr}\), and \(6\,\text{hr}\). 2. The total from Monday through Thursday is \(28\) hours. 3. She needs \(35 - 28 = 7\) more work hours on Friday. 4. Including Friday's lunch break, she must be present for \(7\,\text{hr}\,30\,\text{min}\). 5. Starting at \(8{:}00\) a.m., she can leave at \(3{:}30\) p.m.

Answer

Ms. Weber can leave at \(3{:}30\) p.m.
5214315
A city record lists a small park's area as \(0.04\,\text{km}^2\). A news article says, “The park covers only \(400\,\text{m}^2\).” Convert the official area to square meters and explain the article's error. How many times as large is the actual park as the reported area?

Hints

- Square the conversion factor from kilometers to meters. - Multiply \(0.04\) by the number of square meters in one square kilometer. - Divide the actual area by the reported area to find the scale factor.

Solution

1. Since \(1\,\text{km}=1000\,\text{m}\), \(1\,\text{km}^2=1{,}000{,}000\,\text{m}^2\). 2. Convert the official area: \(0.04\times 1{,}000{,}000=40{,}000\). Therefore, the park covers \(40{,}000\,\text{m}^2\). 3. Compare the areas: \(40{,}000\div 400=100\). The actual park is \(100\) times as large as the area stated in the article.

Answer

The article is incorrect. \(0.04\,\text{km}^2=40{,}000\,\text{m}^2\), so the actual park is \(100\) times as large as the reported \(400\,\text{m}^2\).
5110675
Evaluate the expression. Give the result in cubic centimeters. \(0.5\,\text{L}+450\,\text{mL}+250\,\text{cm}^3-0.001\,\text{m}^3\)

Hints

- One liter equals one cubic decimeter. - How many cubic centimeters equal one milliliter? - Be careful when converting cubic meters to cubic centimeters. - Convert all quantities before performing the operations.

Solution

1. Convert each volume to cubic centimeters: \(0.5\,\text{L}=500\,\text{cm}^3\), \(450\,\text{mL}=450\,\text{cm}^3\), and \(0.001\,\text{m}^3=1000\,\text{cm}^3\). 2. Substitute the converted values: \(500\,\text{cm}^3+450\,\text{cm}^3+250\,\text{cm}^3-1000\,\text{cm}^3\). 3. Add the positive terms: \(500+450+250=1200\). 4. Subtract: \(1200-1000=200\). 5. The result is \(200\,\text{cm}^3\).

Answer

\(200\,\text{cm}^3\)
5111125
Find each value and give the result in the requested unit. a) \(0.8\,\text{m}^3+150\,\text{dm}^3=\dots\,\text{dm}^3\) b) \(4200\,\text{cm}^3-1.2\,\text{dm}^3=\dots\,\text{L}\) c) How many times does a volume of \(500\,\text{mm}^3\) fit into a volume of \(0.1\,\text{dm}^3\)?

Hints

- In parts a and b, convert all quantities to the unit requested in the answer. - Cubic metric units change by a factor of \(1000\) at each adjacent unit step. - In part c, first convert \(0.1\,\text{dm}^3\) to cubic millimeters.

Solution

1. For part a, \(0.8\,\text{m}^3=800\,\text{dm}^3\). Then \(800+150=950\,\text{dm}^3\). 2. For part b, \(4200\,\text{cm}^3=4.2\,\text{L}\) and \(1.2\,\text{dm}^3=1.2\,\text{L}\). Then \(4.2-1.2=3\,\text{L}\). 3. For part c, \(0.1\,\text{dm}^3=100{,}000\,\text{mm}^3\). Then \(100{,}000\div500=200\).

Answer

a) \(950\,\text{dm}^3\) b) \(3\,\text{L}\) c) \(200\) times
5111165
Evaluate each expression. Give the answer to part a) in cubic decimeters and the answer to part b) in cubic centimeters. a) \(0.75\,\text{m}^3+250\,\text{dm}^3-400\,\text{L}\) b) \(1\frac{3}{4}\,\text{L}-850\,\text{cm}^3+0.1\,\text{dm}^3\)

Hints

- Identify the requested unit for each part. - Convert every term to that unit before calculating. - One liter has the same volume as one cubic decimeter. - Write the mixed number as an equivalent decimal or improper fraction if helpful.

Solution

1. For part a, \(0.75\,\text{m}^3=750\,\text{dm}^3\) and \(400\,\text{L}=400\,\text{dm}^3\). Then \(750+250-400=600\,\text{dm}^3\). 2. For part b, \(1\frac{3}{4}\,\text{L}=1750\,\text{cm}^3\) and \(0.1\,\text{dm}^3=100\,\text{cm}^3\). Then \(1750-850+100=1000\,\text{cm}^3\).

Answer

a) \(600\,\text{dm}^3\) b) \(1000\,\text{cm}^3\)
5111175
Write \(<\), \(>\), or \(=\) in each box. Show your work. a) \(12.5\,\text{L}+500\,\text{mL}\quad\Box\quad0.014\,\text{m}^3\) b) \(2\frac{1}{5}\,\text{dm}^3-200\,\text{cm}^3\quad\Box\quad2000\,\text{mL}\)

Hints

- Convert both sides of each comparison to the same unit. - What decimal is equivalent to \(\frac{1}{5}\)? - Recall the relationships among liters, milliliters, and cubic decimeters.

Solution

1. For part a, the left side is \(12.5\,\text{L}+0.5\,\text{L}=13\,\text{L}\). The right side is \(0.014\,\text{m}^3=14\,\text{L}\). Therefore, \(13\,\text{L}<14\,\text{L}\). 2. For part b, \(2\frac{1}{5}\,\text{dm}^3=2.2\,\text{dm}^3\) and \(200\,\text{cm}^3=0.2\,\text{dm}^3\), so the left side is \(2\,\text{dm}^3\). The right side is \(2000\,\text{mL}=2\,\text{L}=2\,\text{dm}^3\). Therefore, the two sides are equal.

Answer

a) \(<\) b) \(=\)
5111185
Evaluate each expression. Give each final answer in liters. a) \(15\times(200\,\text{mL}+0.3\,\text{L})\) b) \((0.008\,\text{m}^3-3\,\text{dm}^3)\div4\)

Hints

- Evaluate the parentheses first. - Choose a common unit before adding or subtracting. - Convert the final value to liters if needed. - In part b, divide the remaining volume into \(4\) equal parts.

Solution

1. For part a, \(200\,\text{mL}=0.2\,\text{L}\), so the amount in parentheses is \(0.2\,\text{L}+0.3\,\text{L}=0.5\,\text{L}\). Then \(15\times0.5\,\text{L}=7.5\,\text{L}\). 2. For part b, \(0.008\,\text{m}^3=8\,\text{dm}^3\). The amount in parentheses is \(8\,\text{dm}^3-3\,\text{dm}^3=5\,\text{dm}^3\). Then \(5\,\text{dm}^3\div4=1.25\,\text{dm}^3=1.25\,\text{L}\).

Answer

a) \(7.5\,\text{L}\) b) \(1.25\,\text{L}\)
5111335
A dump truck can carry at most \(12\,\text{m}^3\) of gravel. Three job sites ordered these amounts: Site A: \(11{,}500\,\text{dm}^3\) Site B: \(12{,}500{,}000\,\text{cm}^3\) Site C: \(11{,}000\,\text{L}\) For which sites is one truckload enough? Justify your answer by converting every amount to cubic meters.

Hints

- Convert all amounts to the same unit before comparing them. - How do you convert cubic decimeters to cubic meters? - How many unit steps separate cubic centimeters from cubic meters? - How are liters related to cubic meters?

Solution

1. Site A: \(11{,}500\,\text{dm}^3=11.5\,\text{m}^3\). Since \(11.5<12\), one truckload is enough. 2. Site B: \(12{,}500{,}000\,\text{cm}^3=12.5\,\text{m}^3\). Since \(12.5>12\), one truckload is not enough. 3. Site C: \(11{,}000\,\text{L}=11{,}000\,\text{dm}^3=11\,\text{m}^3\). Since \(11<12\), one truckload is enough. 4. Therefore, one truckload is enough for Sites A and C.

Answer

One truckload is enough for Site A, which ordered \(11.5\,\text{m}^3\), and Site C, which ordered \(11\,\text{m}^3\). It is not enough for Site B, which ordered \(12.5\,\text{m}^3\).
5111425
Find the missing value \(x\) in each equation. a) \(x+450\,\text{cm}^3=2\,\text{dm}^3\) b) \(0.5\,\text{m}^3-x=320\,\text{L}\) c) \(x\times5=1.5\,\text{L}\)

Hints

- Identify the inverse operation needed to isolate \(x\). - Convert all quantities in an equation to the same unit before calculating. - Check each answer by substituting it into the original equation.

Solution

1. For part a, \(2\,\text{dm}^3=2000\,\text{cm}^3\). Therefore, \(x=2000-450=1550\,\text{cm}^3\). 2. For part b, \(0.5\,\text{m}^3=500\,\text{L}\). Therefore, \(x=500-320=180\,\text{L}\). 3. For part c, divide by \(5\): \(x=1.5\,\text{L}\div5=0.3\,\text{L}\).

Answer

a) \(x=1550\,\text{cm}^3\) b) \(x=180\,\text{L}\) c) \(x=0.3\,\text{L}\)
5111535
A rectangular aquarium has a total capacity of \(120\,\text{L}\). a) Express this volume in cubic decimeters. b) The aquarium is currently \(\frac{3}{4}\) full. Find the volume of water in cubic centimeters.

Hints

- How are liters and cubic decimeters related? - Find three-fourths of \(120\) before converting units. - How many cubic centimeters are in one cubic decimeter?

Solution

1. Since \(1\,\text{L}=1\,\text{dm}^3\), \(120\,\text{L}=120\,\text{dm}^3\). 2. The water volume is \(\frac{3}{4}\) of \(120\,\text{L}\): \(120\div4\times3=90\,\text{L}\). 3. Since \(1\,\text{L}=1000\,\text{cm}^3\), \(90\times1000=90{,}000\,\text{cm}^3\).

Answer

a) \(120\,\text{dm}^3\) b) \(90{,}000\,\text{cm}^3\)
5111545
Compare each pair of volumes. Write \(<\), \(>\), or \(=\). a) \(\frac{1}{4}\,\text{m}^3\quad\Box\quad250\,\text{L}\) b) \(3\frac{1}{5}\,\text{dm}^3\quad\Box\quad320\,\text{cm}^3\) c) \(0.8\,\text{L}\quad\Box\quad850\,\text{mL}\) d) \(\frac{1}{20}\,\text{cm}^3\quad\Box\quad50\,\text{mm}^3\)

Hints

- Convert both volumes in each pair to the same unit. - Convert the fractions or mixed numbers to equivalent decimals if helpful. - Check whether the numerical value should increase or decrease when the unit changes.

Solution

1. For part a, \(\frac{1}{4}\,\text{m}^3=0.25\,\text{m}^3=250\,\text{L}\), so the volumes are equal. 2. For part b, \(3\frac{1}{5}\,\text{dm}^3=3.2\,\text{dm}^3=3200\,\text{cm}^3\). Since \(3200>320\), the left volume is greater. 3. For part c, \(0.8\,\text{L}=800\,\text{mL}\). Since \(800<850\), the left volume is less. 4. For part d, \(\frac{1}{20}\,\text{cm}^3=0.05\,\text{cm}^3=50\,\text{mm}^3\), so the volumes are equal.

Answer

a) \(=\) b) \(>\) c) \(<\) d) \(=\)
5111605
Write each volume as a fraction of the indicated larger unit. Write each fraction in simplest form. a) \(15\,\text{cm}^3\) in \(\text{dm}^3\) b) \(800\,\text{mL}\) in \(\text{L}\) c) \(125\,\text{dm}^3\) in \(\text{m}^3\) d) \(600\,\text{mm}^3\) in \(\text{cm}^3\) e) \(350\,\text{mL}\) in \(\text{L}\)

Hints

- Write the conversion as a fraction with the larger unit's conversion factor in the denominator. - Most adjacent cubic metric units and the milliliter-liter pair use a factor of \(1000\). - Reduce each fraction using a common factor.

Solution

1. For part a, \(15\,\text{cm}^3=\frac{15}{1000}\,\text{dm}^3=\frac{3}{200}\,\text{dm}^3\). 2. For part b, \(800\,\text{mL}=\frac{800}{1000}\,\text{L}=\frac{4}{5}\,\text{L}\). 3. For part c, \(125\,\text{dm}^3=\frac{125}{1000}\,\text{m}^3=\frac{1}{8}\,\text{m}^3\). 4. For part d, \(600\,\text{mm}^3=\frac{600}{1000}\,\text{cm}^3=\frac{3}{5}\,\text{cm}^3\). 5. For part e, \(350\,\text{mL}=\frac{350}{1000}\,\text{L}=\frac{7}{20}\,\text{L}\).

Answer

a) \(\frac{3}{200}\,\text{dm}^3\) b) \(\frac{4}{5}\,\text{L}\) c) \(\frac{1}{8}\,\text{m}^3\) d) \(\frac{3}{5}\,\text{cm}^3\) e) \(\frac{7}{20}\,\text{L}\)
5111655
Evaluate each expression. a) \(6\times750\,\text{mL}\) (give the result in liters) b) \(3.6\,\text{dm}^3\div9\) c) \(2\frac{1}{2}\,\text{m}^3\div0.5\,\text{m}^3\) d) \(1.2\,\text{L}-\frac{1}{4}\,\text{L}\)

Hints

- When dividing two measurements with the same unit, the units cancel. - Convert fractions or mixed numbers to decimals if that makes the calculation easier. - Check whether a particular unit is requested for the answer.

Solution

1. For part a, \(6\times750\,\text{mL}=4500\,\text{mL}=4.5\,\text{L}\). 2. For part b, \(3.6\,\text{dm}^3\div9=0.4\,\text{dm}^3\). 3. For part c, \(2\frac{1}{2}\,\text{m}^3=2.5\,\text{m}^3\). Then \(2.5\,\text{m}^3\div0.5\,\text{m}^3=5\). The matching units cancel. 4. For part d, \(\frac{1}{4}\,\text{L}=0.25\,\text{L}\). Then \(1.2\,\text{L}-0.25\,\text{L}=0.95\,\text{L}\).

Answer

a) \(4.5\,\text{L}\) b) \(0.4\,\text{dm}^3\) c) \(5\) d) \(0.95\,\text{L}\)
5111665
Solve each volume problem. a) Order the volumes from least to greatest: \(0.2\,\text{m}^3\), \(220\,\text{L}\), and \(21{,}000\,\text{cm}^3\). b) Evaluate \((4.8\,\text{L}-800\,\text{cm}^3)\div2\). c) Find the missing value in cubic decimeters: \(0.05\,\text{m}^3+\dots\,\text{dm}^3=100\,\text{L}\).

Hints

- Convert values to a common unit before comparing or combining them. - Evaluate the parentheses before dividing. - One cubic decimeter equals one liter.

Solution

1. For part a, convert to liters: \(0.2\,\text{m}^3=200\,\text{L}\) and \(21{,}000\,\text{cm}^3=21\,\text{L}\). Therefore, \(21{,}000\,\text{cm}^3<0.2\,\text{m}^3<220\,\text{L}\). 2. For part b, \(4.8\,\text{L}=4800\,\text{cm}^3\). Then \((4800-800)\,\text{cm}^3\div2=4000\,\text{cm}^3\div2=2000\,\text{cm}^3=2\,\text{L}\). 3. For part c, \(0.05\,\text{m}^3=50\,\text{dm}^3\) and \(100\,\text{L}=100\,\text{dm}^3\). The missing value is \(100-50=50\,\text{dm}^3\).

Answer

a) \(21{,}000\,\text{cm}^3<0.2\,\text{m}^3<220\,\text{L}\) b) \(2\,\text{L}\) c) \(50\,\text{dm}^3\)
5112345
Order these volumes from least to greatest: \(0.4\,\text{L}\); \(45\,\text{cm}^3\); \(4000\,\text{mm}^3\); \(0.05\,\text{dm}^3\); \(0.0003\,\text{m}^3\)

Hints

- Choose one unit and convert every measurement to it. - Pay careful attention to factors of \(1000\) between cubic units. - One liter equals one cubic decimeter.

Solution

1. Convert every volume to cubic centimeters. 2. \(4000\,\text{mm}^3=4\,\text{cm}^3\). 3. \(45\,\text{cm}^3\) remains \(45\,\text{cm}^3\). 4. \(0.05\,\text{dm}^3=50\,\text{cm}^3\). 5. \(0.0003\,\text{m}^3=300\,\text{cm}^3\). 6. \(0.4\,\text{L}=400\,\text{cm}^3\). 7. Therefore, \(4000\,\text{mm}^3<45\,\text{cm}^3<0.05\,\text{dm}^3<0.0003\,\text{m}^3<0.4\,\text{L}\).

Answer

\(4000\,\text{mm}^3<45\,\text{cm}^3<0.05\,\text{dm}^3<0.0003\,\text{m}^3<0.4\,\text{L}\)
5118365
Evaluate each expression. Convert to a common unit when needed and give each answer in a reasonable form. a) \(450\,\text{mL}+0.75\,\text{L}\) b) \(0.004\,\text{m}^3-1200\,\text{cm}^3\) c) \(1.2\,\text{dm}^3\times5\) d) \(6\,\text{L}\div15\) e) \(2\,\text{m}^3\div400\,\text{L}\)

Hints

- Use the conversion factors for metric volume units. - One cubic decimeter equals one liter. - Convert quantities to the same unit before adding, subtracting, or dividing. - Decide whether a quotient should have a unit or be a unitless number.

Solution

1. For part a, \(0.75\,\text{L}=750\,\text{mL}\). Then \(450+750=1200\,\text{mL}=1.2\,\text{L}\). 2. For part b, \(0.004\,\text{m}^3=4000\,\text{cm}^3\). Then \(4000-1200=2800\,\text{cm}^3=2.8\,\text{dm}^3\). 3. For part c, \(1.2\times5=6\), so the result is \(6\,\text{dm}^3\). 4. For part d, \(6\,\text{L}=6000\,\text{mL}\). Then \(6000\div15=400\,\text{mL}\). 5. For part e, \(2\,\text{m}^3=2000\,\text{L}\). Then \(2000\,\text{L}\div400\,\text{L}=5\). The units cancel.

Answer

a) \(1.2\,\text{L}\) b) \(2.8\,\text{dm}^3\) c) \(6\,\text{dm}^3\) d) \(400\,\text{mL}\) e) \(5\)
5118375
Write \(<\), \(>\), or \(=\), or find the missing value so that each statement is true. a) \(0.2\,\text{m}^3\quad\Box\quad20{,}000\,\text{cm}^3\) b) \(4.5\,\text{L}-800\,\text{mL}=\Box\,\text{dm}^3\) c) \(\frac{3}{4}\,\text{m}^3+\Box\,\text{L}=1\,\text{m}^3\) d) \(12\times250\,\text{mL}\quad\Box\quad0.003\,\text{m}^3\)

Hints

- Convert quantities to the same unit before comparing them. - How many milliliters are in one liter, and how many liters are in one cubic meter? - Convert \(\frac{3}{4}\) of a cubic meter directly to liters.

Solution

1. For part a, \(0.2\,\text{m}^3=200{,}000\,\text{cm}^3\). Since \(200{,}000>20{,}000\), the correct symbol is \(>\). 2. For part b, \(800\,\text{mL}=0.8\,\text{L}\). Then \(4.5-0.8=3.7\,\text{L}=3.7\,\text{dm}^3\). 3. For part c, \(\frac{3}{4}\,\text{m}^3=750\,\text{L}\) and \(1\,\text{m}^3=1000\,\text{L}\). The missing amount is \(1000-750=250\,\text{L}\). 4. For part d, \(12\times250\,\text{mL}=3000\,\text{mL}=3\,\text{L}\), and \(0.003\,\text{m}^3=3\,\text{L}\). The volumes are equal.

Answer

a) \(>\) b) \(3.7\,\text{dm}^3\) c) \(250\,\text{L}\) d) \(=\)
5118385
Evaluate each expression step by step. Give the final answer in the unit shown in parentheses. a) \(\left(\frac{1}{4}\,\text{m}^3+150\,\text{L}\right)\times2-0.3\,\text{m}^3\) (liters) b) \(0.08\,\text{dm}^3\div4+15\,\text{mL}\) (\(\text{cm}^3\)) c) \(\frac{2}{5}\,\text{L}\times10-3500\,\text{mL}\) (liters)

Hints

- Follow the order of operations: parentheses first, then multiplication or division before addition or subtraction. - Convert intermediate values to the requested unit. - Milliliters and cubic centimeters represent equal volumes.

Solution

1. For part a, \(\frac{1}{4}\,\text{m}^3=250\,\text{L}\). The parentheses equal \(250\,\text{L}+150\,\text{L}=400\,\text{L}\). Then \(400\times2=800\,\text{L}\), and \(0.3\,\text{m}^3=300\,\text{L}\). The result is \(800-300=500\,\text{L}\). 2. For part b, \(0.08\,\text{dm}^3=80\,\text{cm}^3\). Then \(80\div4=20\,\text{cm}^3\). Since \(15\,\text{mL}=15\,\text{cm}^3\), the result is \(20+15=35\,\text{cm}^3\). 3. For part c, \(\frac{2}{5}\,\text{L}=0.4\,\text{L}\). Then \(0.4\times10=4\,\text{L}\), and \(3500\,\text{mL}=3.5\,\text{L}\). The result is \(4-3.5=0.5\,\text{L}\).

Answer

a) \(500\,\text{L}\) b) \(35\,\text{cm}^3\) c) \(0.5\,\text{L}\)

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