- 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.

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 \(\frac{12}{36}=\frac{1}{3}\).

a) \(\frac{1}{4}\) b) \(\frac{2}{3}\) c) \(\frac{3}{8}\) d) \(\frac{1}{5}\) e) \(\frac{1}{3}\)

- 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.

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 \(250\,\text{mL}=0.25\,\text{L}\).

a) \(45\,\text{cm}\) b) \(0.25\,\text{min}\) c) \(0.8\,\text{g}\) d) \(0.25\,\text{L}\)

- 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.

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. \(1250\,\text{g}=\frac{1250}{1000}\,\text{kg}=\frac{5}{4}\,\text{kg}=1.25\,\text{kg}\).

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 \(1.25\,\text{kg}\)

- Decide whether the target unit is larger or smaller than the starting unit. - Converting to a larger unit makes the numerical value smaller; converting to a smaller unit makes it larger. - Check whether your result has a reasonable size before finalizing it.

1. For a), \(45 \div 60=0.75\), so \(45\,\text{min}=0.75\,\text{h}\). 2. For b), \(250 \div 1000=0.25\), so \(250\,\text{mL}=0.25\,\text{L}\). 3. For c), \(1.35\times1000=1350\), so \(1.35\,\text{km}=1350\,\text{m}\).

a) \(0.75\,\text{h}\) b) \(0.25\,\text{L}\) c) \(1350\,\text{m}\)

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

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 \(800\,\text{mL}=\frac{800}{1000}\,\text{L}=\frac{4}{5}\,\text{L}\).

a) \(\frac{9}{20}\,\text{km}\) b) \(\frac{1}{5}\,\text{hr}\) c) \(\frac{1}{2}\,\text{yd}\) d) \(\frac{4}{5}\,\text{L}\)

- 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.

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 \(\frac{1}{8}\,\text{km}=125\,\text{m}\).

a) \(40\,\text{cm}\) b) \(750\,\text{mL}\) c) \(900\,\text{g}\) d) \(125\,\text{m}\)

- Use the relationship between hours and minutes for parts a)-c). - Use the relationship between liters and milliliters for parts d)-e). - For a mixed-unit time, separate the whole hours from the extra minutes.

1. For a), three-fourths of \(60\) minutes is \(\frac{3}{4}\times60=45\) minutes. 2. For b), \(130=2\times60+10\), so \(130\,\text{min}=2\,\text{hr}\ 10\,\text{min}\). 3. For c), \(1\,\text{hr}\ 25\,\text{min}=60\,\text{min}+25\,\text{min}=85\,\text{min}\). 4. For d), \(0.5\,\text{L}=500\,\text{mL}\). 5. For e), \(1.25\,\text{L}=1250\,\text{mL}\).

a) \(45\,\text{min}\) b) \(2\,\text{hr}\ 10\,\text{min}\) c) \(85\,\text{min}\) d) \(500\,\text{mL}\) e) \(1250\,\text{mL}\)

- The number of decimal places depends on the conversion factor between the units. - Use three decimal places when the smaller unit is one-thousandth of the larger unit. - Keep placeholder zeros so the decimal represents the correct smaller-unit amount.

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. \(5\,\text{mL}=0.005\,\text{L}\), so the total is \(3.005\,\text{L}\).

a) \(8.02\,\text{km}\) b) \(5.007\,\text{kg}\) c) \(12.4\,\text{m}\) d) \(3.005\,\text{L}\)

- 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.

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. \(0.45\,\text{L}=450\,\text{mL}\).

a) \(4\,\text{h}\ 30\,\text{min}\) b) \(1\,\text{min}\ 30\,\text{s}\) c) \(8\,\text{L}\ 5\,\text{mL}\) d) \(450\,\text{mL}\)

- Start with the number of smaller units in one whole larger unit. - The fraction tells what part of that full conversion amount is needed. - Check that each answer uses the smaller unit requested in its part.

1. For a), \(1\,\text{h}=60\,\text{min}\), and \(\frac{1}{10}\times60=6\), so the amount is \(6\,\text{min}\). 2. For b), \(1\,\text{L}=1000\,\text{mL}\), and \(\frac{3}{4}\times1000=750\), so the amount is \(750\,\text{mL}\). 3. For c), \(1\,\text{kg}=1000\,\text{g}\), and \(\frac{2}{5}\times1000=400\), so the amount is \(400\,\text{g}\).

a) \(6\,\text{min}\) b) \(750\,\text{mL}\) c) \(400\,\text{g}\)

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

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 \(125\,\text{g}\).

a) \(200\,\text{mL}\) b) \(400\,\text{m}\) c) \(125\,\text{g}\)

- Convert every measurement to the smaller unit first. - Add or subtract only after the units match. - Use the conversion relationship supplied in the problem when you reach the metric-ton part.

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), \(\frac{1}{4}\) of \(1000\,\text{kg}\) is \(250\,\text{kg}\), so \(250\,\text{kg}-120\,\text{kg}=130\,\text{kg}\). 4. For d), \(\frac{7}{20}\,\text{L}=350\,\text{mL}\), so \(350\,\text{mL}+150\,\text{mL}=500\,\text{mL}\).

a) \(750\,\text{m}\) b) \(55\,\text{min}\) c) \(130\,\text{kg}\) d) \(500\,\text{mL}\)

- 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.

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 \(4\,\text{L}\ 500\,\text{mL}\).

a) \(2.75\,\text{m}\); \(2\,\text{m}\ 75\,\text{cm}\) b) \(0.6\,\text{kg}\); \(600\,\text{g}\) c) \(4.5\,\text{L}\); \(4\,\text{L}\ 500\,\text{mL}\)

- 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.

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.

a) \(450\,\text{mL}\) is greater. b) \(18\,\text{min}\) is greater. c) \(\frac{7}{20}\,\text{kg}\) is greater.

- Convert each measurement before rounding it. - Compare the converted decimal with the two nearest whole numbers. - Use the tenths digit to decide which whole number is nearer.

1. \(675 \div 1000=0.675\,\text{km}\), which rounds to \(1\,\text{km}\). 2. \(1840 \div 1000=1.84\,\text{km}\), which rounds to \(2\,\text{km}\). 3. \(1456 \div 1000=1.456\,\text{kg}\), which rounds to \(1\,\text{kg}\). 4. \(1250 \div 1000=1.25\,\text{L}\), which rounds to \(1\,\text{L}\).

a) \(1\,\text{km}\) b) \(2\,\text{km}\) c) \(1\,\text{kg}\) d) \(1\,\text{L}\)

- 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.

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 \(10\,\text{oz}\).

a) \(21\,\text{in}\) b) \(99\,\text{in}^2\) c) \(68\) fluid ounces d) \(10\,\text{oz}\)

- 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 \(1\,\text{ft}^2=144\,\text{in}^2\).

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 \(\frac{13}{20}\,\text{kg}=650\,\text{g}>640\,\text{g}\).

a) \(>\) b) \(=\) c) \(<\) d) \(>\)

- Convert measurements to compatible units before adding, subtracting, or dividing. - A requested output unit may require a conversion after the arithmetic. - Dividing two measurements with the same unit produces a unitless ratio.

1. \(1.2\,\text{kg}=1200\,\text{g}\). Then \(450+1200=1650\), so the result is \(1650\,\text{g}\). 2. \(3\,\text{m}\times25=75\,\text{m}\). Since \(1\,\text{m}=100\,\text{cm}\), this is \(7500\,\text{cm}\). 3. \(72\,\text{cm}=720\,\text{mm}\). Then \(720\,\text{mm}\div8\,\text{mm}=90\). The units cancel, so the result is unitless. 4. \(2\,\text{L}=2000\,\text{mL}\). Then \(2000-450=1550\), so the result is \(1550\,\text{mL}\).

a) \(1650\,\text{g}\) b) \(7500\,\text{cm}\) c) \(90\) d) \(1550\,\text{mL}\)

- Evaluate the left side first. - Convert both sides to the same unit before comparing. - Check the unit relationship that applies in each part before doing the arithmetic.

1. \(5\times200\,\text{mg}=1000\,\text{mg}=1\,\text{g}\), so the symbol is \(=\). 2. \(1\,\text{km}=1000\,\text{m}\). Then \(1000-350=650\), and \(650<700\), so the symbol is \(<\). 3. \(48\,\text{h}\div2=24\,\text{h}=1\,\text{day}\), so the symbol is \(=\). 4. \(1500\,\text{mL}=1.5\,\text{L}\). Then \(3.5\,\text{L}+1.5\,\text{L}=5\,\text{L}\), so the symbol is \(=\).

a) \(=\) b) \(<\) c) \(=\) d) \(=\)