Free Math Worksheets

- How many ounces are in \(1\) pound? - What fraction of a pound is each smaller weight? - The price changes in the same proportion as the weight.

1. Since \(1\,\text{lb} = 16\,\text{oz}\), \(8\,\text{oz}\) is half a pound. Its price is \(\$4.80 \div 2 = \$2.40\). 2. Four ounces is half of \(8\,\text{oz}\), so its price is \(\$2.40 \div 2 = \$1.20\). 3. Two ounces is half of \(4\,\text{oz}\), so its price is \(\$1.20 \div 2 = \$0.60\).

\(8\,\text{oz}\) cost \(\$2.40\). \(4\,\text{oz}\) cost \(\$1.20\). \(2\,\text{oz}\) cost \(\$0.60\).

- Convert \(1\,\text{kg}\) to grams. - Determine how many groups of \(125\,\text{g}\) make the target mass. - You can use multiplication or repeated addition.

1. Convert the target mass: \(1\,\text{kg} = 1000\,\text{g}\). 2. Use multiplication to find how many groups of \(125\,\text{g}\) make \(1000\,\text{g}\): \(8 \times 125\,\text{g} = 1000\,\text{g}\). 3. About \(8\) lemons have a total mass of \(1\,\text{kg}\).

About \(8\) lemons.

- How many ounces are in \(1\) pound? - What fraction of a pound is \(8\,\text{oz}\)? - A half-pound costs half as much as a full pound at the same rate.

1. Since \(1\,\text{lb} = 16\,\text{oz}\), \(8\,\text{oz}\) is half a pound. Divide each price by \(2\). 2. Zucchini: \(\$1.80 \div 2 = \$0.90\). 3. Bell peppers: \(\$3.40 \div 2 = \$1.70\). 4. Carrots: \(\$1.20 \div 2 = \$0.60\). 5. Eggplant: \(\$2.60 \div 2 = \$1.30\).

Zucchini: \(\$0.90\) Bell peppers: \(\$1.70\) Carrots: \(\$0.60\) Eggplant: \(\$1.30\)

- How many \(4\)-ounce portions are in \(1\) pound? - Once you know the price of \(4\,\text{oz}\), scale it to other amounts. - Half a pound is \(8\,\text{oz}\).

1. Since \(1\,\text{lb} = 16\,\text{oz}\), \(4\,\text{oz}\) is one fourth of a pound. The price is \(\$9.00 \div 4 = \$2.25\). 2. Twelve ounces is three \(4\)-ounce portions, so \(3 \times \$2.25 = \$6.75\). 3. Half a pound is \(8\,\text{oz}\), so the price is \(\$9.00 \div 2 = \$4.50\).

a) \(\$2.25\) b) \(\$6.75\) c) \(\$4.50\)

- Express each target amount as a sum or difference of the two known amounts. - Add the prices when quantities are combined. - Subtract the smaller price when the smaller quantity is removed.

1. For \(36\,\text{fl oz}\), add the prices for \(32\,\text{fl oz}\) and \(4\,\text{fl oz}\): \(\$3.60 + \$0.45 = \$4.05\). 2. For \(28\,\text{fl oz}\), subtract the price of \(4\,\text{fl oz}\) from the price of \(32\,\text{fl oz}\): \(\$3.60 - \$0.45 = \$3.15\).

a) \(\$4.05\) b) \(\$3.15\)

- Convert the ribbon to centimeters first. - Divide the total number of centimeters by \(4\). - Check whether the division has a remainder.

1. Convert the ribbon's length: \(3\,\text{m} = 300\,\text{cm}\). 2. Divide the total length equally among four children: \(300 \div 4 = 75\). 3. Each child receives \(75\,\text{cm}\), with no ribbon left over.

Each child receives \(75\,\text{cm}\) of ribbon, and there is no remainder.

- How many days are in one week? - Multiply the daily amount by the number of days. - Recall that \(1000\,\text{kg}\) equals \(1\) metric ton.

1. Estimate the weekly amount: \(150\,\text{kg} \times 7 \approx 1050\,\text{kg}\). 2. Since \(1000\,\text{kg}\) is \(1\) metric ton, the estimate is \(1\) metric ton \(50\,\text{kg}\).

The elephant eats about \(1050\,\text{kg}\), or about \(1\) metric ton \(50\,\text{kg}\), in one week.

- Convert \(4\) metric tons to kilograms. - Divide by the mass of one newborn elephant. - Check by multiplication.

1. Convert the adult elephant's mass: \(4\) metric tons \(= 4000\,\text{kg}\). 2. Use the approximate masses to compare: \(4000\,\text{kg} \div 100\,\text{kg} \approx 40\).

\(40\) newborn elephants have about the same total mass as one adult elephant.

- How much length does one unit interval use? - How many unit intervals are needed to reach \(2500\)? - Convert millimeters to meters at the end.

1. Multiply the number of unit intervals by the length of each interval: \(2500\times4\,\text{mm}=10{,}000\,\text{mm}\). 2. Convert millimeters to meters: \(10{,}000\,\text{mm}\div1000=10\,\text{m}\).

The number line must be at least \(10\,\text{m}\) long.

- How many seconds make \(1\) minute? - Add the minutes and seconds separately. - Regroup any total of at least \(60\) seconds as minutes and seconds.

1. Add \(18+37=55\) seconds and \(6+9=15\) minutes. The result is \(15\,\text{min}\ 55\,\text{s}\). 2. Add \(44+26=70\) seconds. Since \(70\) seconds is \(1\) minute \(10\) seconds, add \(12+5+1=18\) minutes. The result is \(18\,\text{min}\ 10\,\text{s}\). 3. Add \(52+39=91\) seconds. Since \(91\) seconds is \(1\) minute \(31\) seconds, add \(15+11+1=27\) minutes. The result is \(27\,\text{min}\ 31\,\text{s}\).

1. \(15\,\text{min}\ 55\,\text{s}\) 2. \(18\,\text{min}\ 10\,\text{s}\) 3. \(27\,\text{min}\ 31\,\text{s}\)

- One hour is \(60\) minutes, and one minute is \(60\) seconds. - Regroup any total of at least \(60\) minutes. - When subtracting, regroup one larger unit if the smaller-unit amount is not large enough.

1. a) Add \(35+45=80\) minutes. Regroup \(80\) minutes as \(1\) hour \(20\) minutes. Then \(14+8+1=23\) hours, so the result is \(23\,\text{hr}\ 20\,\text{min}\). 2. b) Regroup \(12\,\text{min}\ 15\,\text{s}\) as \(11\,\text{min}\ 75\,\text{s}\). Then subtract to get \(4\,\text{min}\ 35\,\text{s}\).

a) \(23\,\text{hr}\ 20\,\text{min}\) b) \(4\,\text{min}\ 35\,\text{s}\)

- How many ounces are in \(1\,\text{lb}\)? - How many pounds are in \(1\,\text{ton}\)? - Add the smaller units first and regroup when possible. - You can also convert everything to the smallest unit, add, and convert back.

1. a) \(12\,\text{oz}+8\,\text{oz}=20\,\text{oz}=1\,\text{lb}\ 4\,\text{oz}\). 2. b) Add \(12+3=15\) pounds and \(12+4=16\) ounces. Since \(16\,\text{oz}=1\,\text{lb}\), the result is \(16\,\text{lb}\). 3. c) Add \(4+5=9\) tons and \(1200+1100=2300\) pounds. Since \(2300\,\text{lb}=1\,\text{ton}\ 300\,\text{lb}\), the result is \(10\,\text{tons}\ 300\,\text{lb}\).

a) \(1\,\text{lb}\ 4\,\text{oz}\) b) \(16\,\text{lb}\) c) \(10\,\text{tons}\ 300\,\text{lb}\)

- Compare the backpack before and after the books were removed. - Express both weights in the same unit first. - Think about which part of the original weight is no longer included. - Check that the books weigh less than the full backpack.

1. Convert both weights to ounces: \(7\,\text{lb}\ 4\,\text{oz} = 116\,\text{oz}\) and \(4\,\text{lb}\ 12\,\text{oz} = 76\,\text{oz}\). 2. Find the difference: \(116\,\text{oz} - 76\,\text{oz} = 40\,\text{oz}\). 3. Convert the difference to pounds and ounces: \(40\,\text{oz} = 2\,\text{lb}\ 8\,\text{oz}\).

\(2\,\text{lb}\ 8\,\text{oz}\)

- Picture the scarf as one long segment. Which part is finished, and which part is still missing? - First express both lengths in the same unit. - Decide whether centimeters or meters will give the clearest answer. - Check that the finished part and the missing part add to the full scarf length.

1. Convert the total length to centimeters: \(2\,\text{m} = 200\,\text{cm}\). 2. Subtract the finished length from the total length: \(200\,\text{cm} - 78\,\text{cm} = 122\,\text{cm}\).

\(122\,\text{cm}\)

- Recall how many milliliters equal one liter. - Determine how many \(200\)-milliliter cups make \(1\) liter. - Multiply the cups per liter by \(40\).

1. Convert the aquarium's capacity: \(40\,\text{L} = 40{,}000\,\text{mL}\). 2. Five \(200\,\text{mL}\) cups hold \(5 \times 200\,\text{mL} = 1000\,\text{mL}\), or \(1\,\text{L}\). 3. Multiply the \(5\) cups needed per liter by \(40\) liters: \(5 \times 40 = 200\) cups.

Maya needs \(200\) full cups of water.

- Divide the total value by the value of one token. - Divide the number of tokens by the number packed in each box. - Find the total weight in grams before converting to kilograms.

1. a) Divide the total value by the value of one token: \(6000\div3=2000\). The park buys \(2000\) tokens. 2. b) Divide the number of tokens by the number in each box: \(2000\div5=400\). The park needs \(400\) boxes. 3. c) Find the total weight in grams: \(2000\times4\,\text{g}=8000\,\text{g}\). Since \(1000\,\text{g}=1\,\text{kg}\), the tokens weigh \(8\,\text{kg}\).

a) \(2000\) tokens b) \(400\) boxes c) \(8\,\text{kg}\)

- Convert the full weeks to days first. - Then add the extra days. - One week has \(7\) days.

1. Convert the full weeks to days: \(4 \times 7 = 28\) days. 2. Add the extra days: \(28 + 2 = 30\) days.

The captain is at sea for \(30\) days altogether.

- Express both vacation lengths in days. - How many days are in \(3\) weeks? - Subtract to find how much longer one vacation is.

1. Convert Anton's vacation to days: \(3 \times 7 = 21\) days. 2. Compare the vacations: Lena has \(25\) days, and Anton has \(21\) days. 3. Find the difference: \(25 - 21 = 4\) days.

Lena has the longer vacation. It is \(4\) days longer than Anton's vacation.

- How many centimeters are in one meter? - Find the height of the stack and the estimated height of the building in the same unit. - Divide the total distance by the length of one step.

1. Using the estimated book thickness, the stack height is \(1000 \times 1\,\text{cm} \approx 1000\,\text{cm} = 10\,\text{m}\). 2. Using the estimated story height, the building height is \(3 \times 3\,\text{m} \approx 9\,\text{m}\). The book stack would be about \(1\,\text{m}\) taller. 3. Using the estimated step length, the number of steps is \(1000 \div 50 \approx 20\). 4. The distance is exactly \(1000\,\text{cm} = 10\,\text{m}\).

a) Yes. The stack would be about \(10\,\text{m}\) high, which is about \(1\,\text{m}\) taller than the building. b) About \(20\) steps. The distance is \(10\,\text{m}\).

- Convert each height to inches before you calculate. - Work one comparison at a time. - Compare all three heights after you find the missing heights.

1. Convert Leo's height to inches: \(4\,\text{ft}\ 8\,\text{in} = 56\,\text{in}\). 2. Find Mia's height: \(56\,\text{in} - 2\,\text{in} = 54\,\text{in}\), or \(4\,\text{ft}\ 6\,\text{in}\). 3. Find Felix's height: \(54\,\text{in} + 5\,\text{in} = 59\,\text{in}\), or \(4\,\text{ft}\ 11\,\text{in}\). 4. Compare the heights: \(59\,\text{in} > 56\,\text{in} > 54\,\text{in}\), so Felix is the tallest.

Mia is \(4\,\text{ft}\ 6\,\text{in}\) tall, and Felix is \(4\,\text{ft}\ 11\,\text{in}\) tall. Felix is the tallest.

- Standing on the books makes Jonah taller. Add the height of the books first. - He is still shorter than Lucas, so account for the remaining difference. - Convert the final number of inches to feet and inches.

1. Find Jonah's height while he is standing on the books: \(52\,\text{in} + 6\,\text{in} = 58\,\text{in}\). 2. Add the remaining difference: \(58\,\text{in} + 2\,\text{in} = 60\,\text{in}\). 3. Convert inches to feet and inches: \(60\,\text{in} = 5\,\text{ft}\ 0\,\text{in}\).

Lucas is \(5\,\text{ft}\ 0\,\text{in}\) tall.

- Find the sister's height first. - The word “add” tells you how to find the father's height. - Work through the people in the same order as the information in the problem.

1. Find Julia's height: \(34\,\text{in} + 4\,\text{in} = 38\,\text{in}\). 2. Find their father's height: \(34\,\text{in} + 38\,\text{in} = 72\,\text{in}\), or \(6\,\text{ft}\). 3. Find their mother's height: \(72\,\text{in} - 5\,\text{in} = 67\,\text{in}\), or \(5\,\text{ft}\ 7\,\text{in}\).

Julia is \(38\,\text{in}\) tall, their father is \(6\,\text{ft}\) tall, and their mother is \(5\,\text{ft}\ 7\,\text{in}\) tall.

- How many ounces are in \(1\) pound? - For each package size, decide how many equal packages make the target weight. - You can use repeated addition or division.

1. Convert the target weight: \(1\,\text{lb} = 16\,\text{oz}\). 2. For chocolate bars, calculate \(16 \div 2 = 8\). She needs \(8\) bars. 3. For yogurt cups, calculate \(16 \div 4 = 4\). She needs \(4\) cups. 4. For bags of oats, calculate \(16 \div 8 = 2\). She needs \(2\) bags.

a) \(8\) chocolate bars b) \(4\) yogurt cups c) \(2\) bags of oats

- Remember that \(1\,\text{kg} = 1000\,\text{g}\). - Find each total mass in grams first. - Break apart a multiplication fact if that helps, but keep the unit attached to the result.

1. For part a, \(8 \times 125\,\text{g} = 1000\,\text{g}\). Since \(1000\,\text{g} = 1\,\text{kg}\), the apples weigh \(1\,\text{kg}\). 2. For part b, convert \(1\,\text{kg} = 1000\,\text{g}\), then calculate \(1000 \div 200 = 5\). Five peaches weigh \(1\,\text{kg}\). 3. For part c, the apples weigh \(6 \times 125\,\text{g} = 750\,\text{g}\), and the peaches weigh \(4 \times 200\,\text{g} = 800\,\text{g}\). Since \(800\,\text{g} > 750\,\text{g}\), the \(4\) peaches are heavier.

a) \(1000\,\text{g} = 1\,\text{kg}\) b) \(5\) peaches c) The \(4\) peaches are heavier.

- Add the weights of all the packed items first. - How much more is needed to reach \(1000\,\text{g}\)? - Add the hundreds, tens, and ones carefully.

1. Add the item weights: \(450\,\text{g} + 320\,\text{g} + 180\,\text{g}\). 2. Calculate the total: \(450 + 320 = 770\), and \(770 + 180 = 950\). The items weigh \(950\,\text{g}\). 3. Convert \(1\,\text{kg} = 1000\,\text{g}\), then find the difference: \(1000\,\text{g} - 950\,\text{g} = 50\,\text{g}\).

The items weigh \(950\,\text{g}\) altogether. Another \(50\,\text{g}\) would make exactly \(1\,\text{kg}\).

- How many grams are in \(1\) kilogram? - Express every weight in grams before adding. - Convert the final total back to kilograms and grams.

1. Convert the empty backpack's weight to grams: \(1\,\text{kg}\ 200\,\text{g} = 1200\,\text{g}\). 2. Add all the weights: \(1200\,\text{g} + 1500\,\text{g} + 350\,\text{g} + 750\,\text{g} = 3800\,\text{g}\). 3. Convert the total to kilograms and grams: \(3800\,\text{g} = 3\,\text{kg}\ 800\,\text{g}\).

The packed backpack weighs \(3\,\text{kg}\ 800\,\text{g}\).

- What does “too heavy” mean when you compare two weights? - Express the \(3\,\text{kg}\) limit in grams. - Add the weights of the backpack and all its contents first.

1. Find the packed weight: \(1100\,\text{g} + 1600\,\text{g} + 800\,\text{g} = 3500\,\text{g}\). 2. Convert the limit to grams: \(3\,\text{kg} = 3000\,\text{g}\). 3. Compare the weights: \(3500\,\text{g} > 3000\,\text{g}\). 4. Find how far the backpack is over the limit: \(3500\,\text{g} - 3000\,\text{g} = 500\,\text{g}\).

Yes. The backpack weighs \(3500\,\text{g}\), or \(3\,\text{kg}\ 500\,\text{g}\), so it is \(500\,\text{g}\) over the limit.

- Convert the full backpack weight to grams first. - Add the weights of all the items whose weights are known. - What operation finds a missing part when you know the whole and the other parts?

1. Convert the total weight to grams: \(4\,\text{kg} = 4000\,\text{g}\). 2. Convert the textbook weight to grams: \(1\,\text{kg}\ 300\,\text{g} = 1300\,\text{g}\). 3. Add the known parts: \(1250\,\text{g} + 1300\,\text{g} + 300\,\text{g} = 2850\,\text{g}\). 4. Subtract the known weight from the total: \(4000\,\text{g} - 2850\,\text{g} = 1150\,\text{g}\).

The water bottle weighs \(1150\,\text{g}\).

- Convert all masses to grams. - One kilogram equals \(1000\) grams. - After ordering the masses, subtract to find the difference.

1. Convert each mass to grams: Max has \(3250\,\text{g}\), Mia has \(2900\,\text{g}\), and Leo has \(3040\,\text{g}\). 2. Order the masses: \(2900 < 3040 < 3250\). 3. Find the difference between Max's and Mia's backpacks: \(3250 - 2900 = 350\,\text{g}\).

a) Mia, Leo, Max b) Max's backpack is \(350\,\text{g}\) heavier than Mia's.

- First find the new total weight. - How many grams are in \(4\,\text{kg}\)? - Compare the new total with the limit.

1. Find the new weight: \(3580\,\text{g} + 470\,\text{g} = 4050\,\text{g}\). 2. Convert the limit: \(4\,\text{kg} = 4000\,\text{g}\). Since \(4050\,\text{g} > 4000\,\text{g}\), the bag is above the limit. 3. Find the difference: \(4050\,\text{g} - 4000\,\text{g} = 50\,\text{g}\).

a) The bag weighs \(4050\,\text{g}\). b) It is above the limit by \(50\,\text{g}\).

- Convert the full bag weight to grams first. - Remember that Sarah removes two books. - Convert the final answer back to kilograms and grams.

1. Convert the bag's weight to grams: \(4\,\text{kg}\ 250\,\text{g} = 4250\,\text{g}\). 2. Find the combined weight of the books: \(850\,\text{g} + 850\,\text{g} = 1700\,\text{g}\). 3. Subtract the books' weight: \(4250\,\text{g} - 1700\,\text{g} = 2550\,\text{g}\). 4. Convert back to kilograms and grams: \(2550\,\text{g} = 2\,\text{kg}\ 550\,\text{g}\).

The school bag weighs \(2\,\text{kg}\ 550\,\text{g}\).

- Add the backpack and all its contents first. - Remember that \(16\,\text{oz} = 1\,\text{lb}\). - Then add the packed backpack's weight to Lucas's weight.

1. Add the backpack and its contents: \(2\,\text{lb}\,10\,\text{oz} + 6\,\text{lb}\,3\,\text{oz} + 1\,\text{lb}\,3\,\text{oz} = 9\,\text{lb}\,16\,\text{oz}\). 2. Convert \(16\,\text{oz}\) to \(1\,\text{lb}\), so the packed backpack weighs \(10\,\text{lb}\). 3. Add Lucas's weight: \(75\,\text{lb} + 10\,\text{lb} = 85\,\text{lb}\).

Lucas and the packed backpack weigh \(85\,\text{lb}\) altogether.

- Determine how many one-way trips each child walks per day. - Convert all distances to meters. - Find each child's distance for one day and then for five days.

1. Emma walks \(750 + 750 = 1500\,\text{m}\) each day. 2. In five days, Emma walks \(5 \times 1500 = 7500\,\text{m}\). 3. Convert Leo's one-way distance: \(1\,\text{km}\,200\,\text{m} = 1200\,\text{m}\). 4. In five days, Leo walks \(5 \times 1200 = 6000\,\text{m}\). 5. Since \(7500 > 6000\), Emma walks farther.

Emma walks farther. Emma walks \(7500\,\text{m}\), and Leo walks \(6000\,\text{m}\).

- Find half of the bottle's starting amount. - Keep track of both the amount consumed and the amount remaining. - Convert \(1\,\text{L}\) to milliliters before comparing.

1. Find the amount Sofia drinks during the first break: \(400\,\text{mL} \div 2 = 200\,\text{mL}\). 2. Find the amount remaining after the first break: \(400\,\text{mL} - 200\,\text{mL} = 200\,\text{mL}\). 3. Find the amount remaining after the second break: \(200\,\text{mL} - 150\,\text{mL} = 50\,\text{mL}\). 4. Find the amount she has consumed: \(200\,\text{mL} + 150\,\text{mL} = 350\,\text{mL}\). 5. Convert the daily goal: \(1\,\text{L} = 1000\,\text{mL}\). 6. Find the additional amount needed: \(1000\,\text{mL} - 350\,\text{mL} = 650\,\text{mL}\).

1. \(50\,\text{mL}\) remain in the bottle. 2. Sofia must drink \(650\,\text{mL}\) more.

- Convert the mixed kilogram-and-gram amount to grams. - Look for two item weights that add to the target amount. - There is more than one correct combination.

1. Convert the target reduction: \(1\,\text{kg}\ 500\,\text{g} = 1500\,\text{g}\). 2. One combination is \(1200\,\text{g} + 300\,\text{g} = 1500\,\text{g}\), so Lucas can remove the atlas and pencil case. 3. Another combination is \(800\,\text{g} + 700\,\text{g} = 1500\,\text{g}\), so he can remove the book and water bottle.

Solution 1: Remove the atlas and pencil case. Solution 2: Remove the book and water bottle.

- Convert \(2\,\text{kg}\) to grams and compare it with the total weight of all the groceries. - Find how much capacity remains when only the potatoes are in the bag. - Find how many grams over the limit the bag would be with every item included.

1. Find the current weight: \(1500\,\text{g} + 150\,\text{g} = 1650\,\text{g}\). 2. With the carrots, the total would be \(1650\,\text{g} + 500\,\text{g} = 2150\,\text{g}\), which is \(150\,\text{g}\) more than \(2000\,\text{g}\). 3. If Annie removes the apple, the potatoes and carrots weigh \(1500\,\text{g} + 500\,\text{g} = 2000\,\text{g} = 2\,\text{kg}\).

With the carrots, the bag would weigh \(2\,\text{kg}\ 150\,\text{g}\), so it would be \(150\,\text{g}\) too heavy. Annie must remove the apple; then the potatoes and carrots weigh exactly \(2\,\text{kg}\).

- Express every weight in grams before comparing or adding. - Remember how many grams are in \(1\) kilogram.

1. Convert the flour's weight: \(1\,\text{kg} = 1000\,\text{g}\). 2. Order the weights: \(20\,\text{g} < 100\,\text{g} < 350\,\text{g} < 400\,\text{g} < 1000\,\text{g}\). 3. Add all the weights: \(20\,\text{g} + 100\,\text{g} + 350\,\text{g} + 400\,\text{g} + 1000\,\text{g} = 1870\,\text{g}\). 4. Convert the total: \(1870\,\text{g} = 1\,\text{kg}\ 870\,\text{g}\).

a) Eraser, chocolate bar, bicycle helmet, pencil case, bag of flour b) \(1\,\text{kg}\ 870\,\text{g}\)

- How many meters are in \(1\) kilometer? - Subtract the distance already walked from the whole distance. - A round trip includes the distance there and the same distance back.

1. Since \(1\,\text{km} = 1000\,\text{m}\), Elias takes \(1000\) strides when each stride is \(1\,\text{m}\). 2. After \(400\) strides, he has walked \(400\,\text{m}\). The remaining distance is \(1000\,\text{m} - 400\,\text{m} = 600\,\text{m}\). 3. The trip to the playground and back is \(1000\,\text{m} + 1000\,\text{m} = 2000\,\text{m}\).

a) \(1000\) strides b) \(600\,\text{m}\) c) \(2000\,\text{m}\)

- How many groups of \(100\,\text{m}\) are in \(1000\,\text{m}\)? - Subtract the completed distance from the full distance. - How many times can you add \(250\,\text{m}\) to reach \(1000\,\text{m}\)?

1. Convert the course length: \(1\,\text{km} = 1000\,\text{m}\). Divide by the spacing between cones: \(1000\,\text{m} \div 100\,\text{m} = 10\). Because the starting cone is not counted, the cones are at \(100\,\text{m}\) through \(1000\,\text{m}\), for \(10\) cones. 2. Find the remaining distance: \(1000\,\text{m} - 750\,\text{m} = 250\,\text{m}\). 3. Find the number of laps: \(1000\,\text{m} \div 250\,\text{m} = 4\).

a) \(10\) cones b) \(250\,\text{m}\) c) \(4\) laps

- Remember that \(1\,\text{km} = 1000\,\text{m}\). - Use the number of lengths for \(100\,\text{m}\) to think about the number for \(1000\,\text{m}\). - How many \(50\,\text{m}\) lengths make \(100\,\text{m}\)?

1. Find the number of lengths for \(100\,\text{m}\): \(100\,\text{m} \div 10\,\text{m} = 10\). 2. Convert the full distance and divide: \(1\,\text{km} = 1000\,\text{m}\), and \(1000\,\text{m} \div 10\,\text{m} = 100\). 3. In the competition pool, \(1000\,\text{m} \div 50\,\text{m} = 20\).

a) \(10\) lengths b) \(100\) lengths c) \(20\) lengths

- How many times does Leon walk the route each day? - How many school days are in the week? - Find the distance for one day before finding the weekly distance. - Convert the final distance from meters to kilometers.

1. Leon walks the route twice each day: \(600\,\text{m} \times 2 = 1200\,\text{m}\). 2. For \(5\) days, he walks \(1200\,\text{m} \times 5 = 6000\,\text{m}\). 3. Convert meters to kilometers: \(6000\,\text{m} = 6\,\text{km}\).

Leon walks \(6\,\text{km}\) during one school week.

- How many meters are in \(1\,\text{km}\)? - Add the two lengths already completed. - Subtract that amount from the total length.

1. Add the lengths already completed: \(420\,\text{m} + 390\,\text{m} = 810\,\text{m}\). 2. Convert the total length: \(1\,\text{km} = 1000\,\text{m}\). 3. Subtract: \(1000\,\text{m} - 810\,\text{m} = 190\,\text{m}\).

\(190\,\text{m}\) still need barriers.

- Read carefully to identify what each part asks. - Part a) asks for the distance still remaining. - Part b) asks for the total distance already traveled. - Convert the kilometer to meters before calculating.

1. Convert the total distance: \(1\,\text{km} = 1000\,\text{m}\). 2. For part a), subtract: \(1000\,\text{m} - 480\,\text{m} = 520\,\text{m}\). 3. For part b), add the distances traveled: \(480\,\text{m} + 220\,\text{m} = 700\,\text{m}\).

a) \(520\,\text{m}\) remain. b) The bicyclist is \(700\,\text{m}\) from the starting point.

- How many meters make \(1\) kilometer? - Add all three distances for one day first. - Determine how many times the route is walked in one school week. - Use the weekly distance to find the longer distances.

1. Add the three parts of the daily route: \(450\,\text{m} + 250\,\text{m} + 300\,\text{m} = 1000\,\text{m}\). 2. Convert: \(1000\,\text{m} = 1\,\text{km}\). 3. One school week: \(5 \times 1\,\text{km} = 5\,\text{km}\). 4. Four weeks: \(4 \times 5\,\text{km} = 20\,\text{km}\). 5. Twenty weeks: \(20 \times 5\,\text{km} = 100\,\text{km}\).

One day: \(1\,\text{km}\) One school week: \(5\,\text{km}\) Four weeks: \(20\,\text{km}\) Twenty weeks: \(100\,\text{km}\)

- Convert the daily distance to meters first. - How many times do you need the daily distance for each part? - Multiplying by \(10\) shifts each digit one place to the left.

1. Convert the daily distance to meters: \(1\,\text{km}\ 500\,\text{m} = 1500\,\text{m}\). 2. For \(5\) days: \(5 \times 1500\,\text{m} = 7500\,\text{m} = 7\,\text{km}\ 500\,\text{m}\). 3. For \(10\) days: \(10 \times 1500\,\text{m} = 15{,}000\,\text{m} = 15\,\text{km}\).

a) \(7\,\text{km}\ 500\,\text{m}\) b) \(15\,\text{km}\)

- Remember that she makes two one-way trips each workday. - Find the total miles and total yards separately. - Use \(1760\,\text{yd} = 1\,\text{mi}\) to convert the yards.

1. In \(10\) days, Ms. Miller rides the \(4\)-mile part \(20\) times: \(20 \times 4\,\text{mi} = 80\,\text{mi}\). 2. She also rides the \(440\)-yard part \(20\) times: \(20 \times 440\,\text{yd} = 8800\,\text{yd}\). 3. Since \(1760\,\text{yd} = 1\,\text{mi}\), \(8800\,\text{yd} = 5\,\text{mi}\). 4. Add: \(80\,\text{mi} + 5\,\text{mi} = 85\,\text{mi}\).

Ms. Miller bikes \(85\,\text{mi}\) in \(10\) workdays.

- How many meters are in \(1\) kilometer? - First find how far Jonah and Emma have already walked. - Subtract the distance already walked from the whole route.

1. Convert the total distance: \(1\,\text{km} = 1000\,\text{m}\). 2. Find the distance already walked: \(400\,\text{m} + 250\,\text{m} = 650\,\text{m}\). 3. Find the remaining distance: \(1000\,\text{m} - 650\,\text{m} = 350\,\text{m}\).

They still need to walk \(350\,\text{m}\).

- A full lap follows all four sides of the rectangle. - Multiply the length of one lap by \(4\). - Convert \(1\,\text{km}\) to meters before comparing.

1. A lap follows all four sides of the rectangular garden. Its length is \(20 + 10 + 20 + 10 = 60\,\text{m}\). The student added only one length and one width. 2. Four laps are \(4 \times 60 = 240\,\text{m}\). Since \(1\,\text{km} = 1000\,\text{m}\), \(240\,\text{m}\) is less than \(1\,\text{km}\).

1. One lap is \(60\,\text{m}\). The student forgot to include the opposite length and width. 2. Four laps are \(240\,\text{m}\), which is less than \(1\,\text{km}\).

- How many times does Lisa travel the \(2\,\text{km}\) route each day? - How many meters are in \(1\) kilometer? - Compare \(400\) with the distance you calculate in meters.

1. The trip to school and the trip home total \(2\,\text{km} + 2\,\text{km} = 4\,\text{km}\). 2. Convert kilometers to meters: \(4\,\text{km} = 4000\,\text{m}\). 3. Therefore, Lisa's total of \(400\,\text{m}\) is incorrect. She likely made an error when converting units.

Lisa''s statement is incorrect. The round trip is \(4\,\text{km}\), which is \(4000\,\text{m}\), not \(400\,\text{m}\).

- A frame goes around all four sides. - Find the perimeter in inches first. - Use \(12\,\text{in}=1\,\text{ft}\) to convert.

1. Find the perimeter: \(36+24+36+24=120\,\text{in}\). 2. Convert inches to feet: \(120 \div 12=10\,\text{ft}\).

\(120\,\text{in}\), or \(10\,\text{ft}\)

- Convert each day's distance to meters first. - Pay attention to the place value of the \(75\) meters on the second day. - Then add the two distances.

1. Convert the first day's distance: \(23\,\text{km}\,500\,\text{m} = 23{,}500\,\text{m}\). 2. Convert the second day's distance: \(18\,\text{km}\,75\,\text{m} = 18{,}075\,\text{m}\). 3. Add the distances: \(23{,}500 + 18{,}075 = 41{,}575\,\text{m}\).

The hiker walks \(41{,}575\,\text{m}\) altogether.

- Convert meters to centimeters first. - Find how many times the small side length fits into the quilt side length. - Use the number of rows and the number in each row to find the total. - Keep track of the zeros when dividing the large number.

1. Convert \(1\,\text{m}\) to \(100\,\text{cm}\). One row contains \(100 \div 10=10\) squares. 2. The quilt has \(10\) rows of \(10\) squares, so it contains \(10 \times 10=100\) squares. 3. The number of quilts is \(1{,}000{,}000 \div 100=10{,}000\).

a) \(10\) fabric squares b) \(100\) fabric squares c) \(10{,}000\) quilts

- In the start time, identify the hours, minutes, and seconds. - Add the minutes and seconds separately. - Regroup if the number of seconds reaches \(60\) or more.

1. Add the minutes: \(30\,\text{min} + 3\,\text{min} = 33\,\text{min}\). 2. Add the seconds: \(50\,\text{s} + 20\,\text{s} = 70\,\text{s}\). 3. Regroup \(70\,\text{s}\) as \(1\) minute \(10\) seconds. Add the extra minute: \(33\,\text{min} + 1\,\text{min} = 34\,\text{min}\). 4. The ending time is \(2{:}34{:}10\) p.m.

The song ends at \(2{:}34{:}10\) p.m.

- Express every time in seconds first. - Add the numbers of seconds. - How many full minutes are in the total? Remember that \(60\) seconds make \(1\) minute.

1. Express all the times in seconds: \(25\,\text{s}\), \(70\,\text{s}\), and \(45\,\text{s}\). 2. Add the times: \(25\,\text{s} + 70\,\text{s} + 45\,\text{s} = 140\,\text{s}\). 3. Convert the total: \(140\,\text{s} = 120\,\text{s} + 20\,\text{s} = 2\,\text{min}\ 20\,\text{s}\).

The activities take \(2\) minutes \(20\) seconds altogether.

- A loaded truck's mass includes both the truck and its cargo. - Convert metric tons to kilograms before calculating. - Subtract the cargo or empty mass from the loaded mass, as appropriate.

1. For part a, subtract the empty mass from the loaded mass: \(26{,}450\,\text{kg} - 11{,}820\,\text{kg} = 14{,}630\,\text{kg}\). 2. For part b, convert the masses to kilograms: \(14\) metric tons \(500\,\text{kg} = 14{,}500\,\text{kg}\), and \(8\) metric tons \(900\,\text{kg} = 8900\,\text{kg}\). 3. Start with \(14{,}500\,\text{kg}\). Subtract \(8900\,\text{kg}\). The empty truck has a mass of \(5600\,\text{kg}\).

a) The truck delivered \(14{,}630\,\text{kg}\) of oil. b) The empty truck has a mass of \(5600\,\text{kg}\), or \(5\) metric tons \(600\,\text{kg}\).

- Keep kilograms and grams aligned, or convert everything to grams. - Remember that \(1000\,\text{g} = 1\,\text{kg}\). - Decide whether each part requires subtraction or addition.

1. Find the apples' mass: \(22\,\text{kg}\,450\,\text{g} - 2\,\text{kg}\,150\,\text{g} = 20\,\text{kg}\,300\,\text{g}\). 2. Find the filled pear crate's mass: \(1\,\text{kg}\,850\,\text{g} + 15\,\text{kg}\,400\,\text{g} = 16\,\text{kg}\,1250\,\text{g}\). 3. Regroup \(1000\,\text{g}\) as \(1\,\text{kg}\): \(16\,\text{kg}\,1250\,\text{g} = 17\,\text{kg}\,250\,\text{g}\).

a) The apples have a mass of \(20\,\text{kg}\,300\,\text{g}\). b) The filled pear crate has a mass of \(17\,\text{kg}\,250\,\text{g}\).

- Find the sand load in each car by subtracting the empty mass from the loaded mass. - Compare the two sand loads. - Subtract to find the difference.

1. Find Car 1's sand load: \(78{,}500\,\text{kg} - 22{,}800\,\text{kg} = 55{,}700\,\text{kg}\). 2. Find Car 2's sand load: \(81{,}200\,\text{kg} - 25{,}600\,\text{kg} = 55{,}600\,\text{kg}\). 3. Car 1 carries more because \(55{,}700\,\text{kg} > 55{,}600\,\text{kg}\). 4. Subtract \(55{,}600\,\text{kg}\) from \(55{,}700\,\text{kg}\). The difference is \(100\,\text{kg}\).

Car 1 carries \(100\,\text{kg}\) more sand than Car 2.

- Convert the starting amount to milliliters. - Find the total amount taken by the five children. - Subtract the amount taken from the starting amount.

1. Convert the starting volume: \(1\,\text{L}\,500\,\text{mL} = 1500\,\text{mL}\). 2. Find the amount taken: \(5 \times 120\,\text{mL} = 600\,\text{mL}\). 3. Find the amount remaining: \(1500\,\text{mL} - 600\,\text{mL} = 900\,\text{mL}\).

\(900\,\text{mL}\) of punch remain in the bowl.

- Convert metric tons to kilograms. - The payload is the difference between the maximum total mass and the empty mass. - Put both measurements in the same unit before subtracting.

1. Convert both masses to kilograms: \(3\) metric tons \(500\,\text{kg} = 3500\,\text{kg}\), and \(2\) metric tons \(180\,\text{kg} = 2180\,\text{kg}\). 2. Subtract the empty mass from the maximum total mass: \(3500\,\text{kg} - 2180\,\text{kg} = 1320\,\text{kg}\).

The maximum payload is \(1320\,\text{kg}\).

- Convert the maximum total mass to kilograms. - Subtract the empty trailer's mass to find the maximum payload. - Compare the grain's mass with that payload.

1. Convert the maximum total mass: \(5\) metric tons \(= 5000\,\text{kg}\). 2. Find the maximum payload: \(5000\,\text{kg} - 1650\,\text{kg} = 3350\,\text{kg}\). 3. Since \(3400\,\text{kg} > 3350\,\text{kg}\), the planned load is too heavy. 4. The trailer would exceed its limit by \(3400\,\text{kg} - 3350\,\text{kg} = 50\,\text{kg}\).

No. The trailer would be overloaded by \(50\,\text{kg}\).

- Find the full payload capacity by subtracting the empty mass from the maximum total mass. - Find the combined mass of the three machines. - Subtract the machines' mass from the payload capacity.

1. Convert the maximum total mass to \(12{,}000\,\text{kg}\) and the empty mass to \(5200\,\text{kg}\). Subtract the empty mass from the maximum total mass. The truck's total payload capacity is \(6800\,\text{kg}\). 2. Find the mass of the three machines: \(3 \times 1500\,\text{kg} = 4500\,\text{kg}\). 3. Find the remaining capacity: \(6800\,\text{kg} - 4500\,\text{kg} = 2300\,\text{kg}\).

The truck can carry \(2300\,\text{kg}\) more.

- Convert the metric tons to kilograms first. - For each truck, subtract the empty mass from the maximum total mass. - Compare the payload capacities and find their difference.

1. Find Truck A's payload capacity: \(7500\,\text{kg} - 4230\,\text{kg} = 3270\,\text{kg}\). 2. Convert Truck B's maximum mass: \(8\) metric tons \(= 8000\,\text{kg}\). 3. Find Truck B's payload capacity: \(8000\,\text{kg} - 4850\,\text{kg} = 3150\,\text{kg}\). 4. Compare and subtract: \(3270\,\text{kg} - 3150\,\text{kg} = 120\,\text{kg}\).

Truck A has the greater payload capacity. Truck A can carry \(3270\,\text{kg}\), Truck B can carry \(3150\,\text{kg}\), and the difference is \(120\,\text{kg}\).

- Convert the liter to milliliters. - Divide the total volume equally among \(8\) cups. - Check by multiplying the cup volume by \(8\).

1. Convert the total volume: \(1\,\text{L} = 1000\,\text{mL}\). 2. Divide the volume equally among \(8\) cups: \(1000\,\text{mL} \div 8 = 125\,\text{mL}\).

Each cup holds \(125\,\text{mL}\).

- How many milliliters are in \(1\) liter? - Find the total amount in the five small boxes. - Compare prices only after confirming that the quantities are equal.

1. Find the total amount in the juice boxes: \(5 \times 200\,\text{mL} = 1000\,\text{mL}\). 2. Since \(1000\,\text{mL} = 1\,\text{L}\), both options contain the same amount. 3. Find the difference in price: \(\$2.50 - \$1.90 = \$0.60\). 4. The bottle costs \(\$0.60\) less.

The \(1\,\text{L}\) bottle costs less by \(\$0.60\).

- How many \(250\,\text{mL}\) dispensers equal \(1\,\text{L}\)? - Find the cost of that many small dispensers. - Subtract the refill price from the small-dispenser cost.

1. Convert the refill amount: \(1\,\text{L} = 1000\,\text{mL}\). 2. Since \(4 \times 250\,\text{mL} = 1000\,\text{mL}\), four small dispensers contain the same amount as the refill. 3. Find the cost of four small dispensers: \(4 \times \$1.60 = \$6.40\). 4. Find the savings: \(\$6.40 - \$5.20 = \$1.20\).

Buying the refill saves \(\$1.20\).

- How many \(150\,\text{g}\) servings fit in \(450\,\text{g}\)? - Find the cost of that many small cups. - Compare that cost with the price of the large cup.

1. Since \(3 \times 150\,\text{g} = 450\,\text{g}\), three small cups contain the same amount as the large cup. 2. Find the cost of three small cups: \(3 \times \$0.60 = \$1.80\). 3. Find the price difference: \(\$1.80 - \$1.50 = \$0.30\).

Buying the same amount in small cups would cost \(\$0.30\) more.

- How many pounds are in \(1\) US ton? - How does the lifting capacity change as the reach increases? - Convert the slab’s weight to pounds before comparing it with the table.

1. At \(60\,\text{ft}\), \(9700\,\text{lb} = 4\) tons \(1700\,\text{lb}\). 2. At \(120\,\text{ft}\), \(4300\,\text{lb} = 2\) tons \(300\,\text{lb}\). 3. Convert the slab’s weight: \(3\) tons \(200\,\text{lb} = 6200\,\text{lb}\). 4. At \(90\,\text{ft}\), the crane can lift \(6800\,\text{lb}\), which is enough. At \(120\,\text{ft}\), it can lift only \(4300\,\text{lb}\). Therefore, the greatest possible reach is \(90\,\text{ft}\).

a) At \(60\,\text{ft}\): \(4\) tons \(1700\,\text{lb}\); at \(120\,\text{ft}\): \(2\) tons \(300\,\text{lb}\) b) The greatest safe reach is \(90\,\text{ft}\).

- Convert all masses to kilograms. - Subtract the empty mass from the maximum total mass. - Determine how many \(500\)-kilogram machines fit within the remaining capacity.

1. Convert the maximum mass: \(12\) metric tons \(= 12{,}000\,\text{kg}\). 2. Convert the empty mass: \(7\) metric tons \(500\,\text{kg} = 7500\,\text{kg}\). 3. Start with the maximum mass of \(12{,}000\,\text{kg}\) and subtract the empty mass of \(7500\,\text{kg}\). The remaining payload capacity is \(4500\,\text{kg}\). 4. Nine machines have a mass of \(9 \times 500\,\text{kg} = 4500\,\text{kg}\), which uses the full remaining capacity.

The truck can carry at most \(9\) machines.

- Find the total mass of all four beams. - Convert the crane's lifting limit to kilograms. - Compare the load with the limit.

1. Find the combined mass of the beams: \(4 \times 600\,\text{kg} = 2400\,\text{kg}\). 2. Convert the crane's limit: \(2\) metric tons \(200\,\text{kg} = 2200\,\text{kg}\). 3. Since \(2400\,\text{kg} > 2200\,\text{kg}\), the load exceeds the crane's limit. 4. Find how far the load exceeds the limit: \(2400\,\text{kg} - 2200\,\text{kg} = 200\,\text{kg}\).

No. The four beams have a combined mass of \(2400\,\text{kg}\), which is \(200\,\text{kg}\) above the \(2200\,\text{kg}\) limit.

- Convert all amounts to milliliters before adding. - Convert the total back to liters for part a. - Determine how many \(250\)-milliliter glasses make \(1\) liter, then use the \(2\)-liter total.

1. Convert the apple juice: \(1\,\text{L}\,500\,\text{mL} = 1500\,\text{mL}\). 2. Find the total volume: \(1500\,\text{mL} + 500\,\text{mL} = 2000\,\text{mL}\). 3. Convert for part a: \(2000\,\text{mL} = 2\,\text{L}\). 4. Four \(250\,\text{mL}\) glasses hold \(1000\,\text{mL}\), or \(1\,\text{L}\). Therefore, \(2 \times 4 = 8\) full glasses can be poured.

a) The container holds \(2\,\text{L}\) altogether. b) Jordan can pour \(8\) full glasses.

- Convert the total volume to milliliters. - Divide by the number of milliliters in one bottle. - You can divide both numbers by \(100\) to simplify the calculation.

1. Convert the total volume: \(72\,\text{L} = 72{,}000\,\text{mL}\). 2. Measure both volumes in units of \(100\,\text{mL}\). The total is \(720\) such units, and each bottle holds \(3\) such units. 3. Divide by the one-digit number of units per bottle: \(720 \div 3 = 240\).

The orchard can fill \(240\) bottles completely.

- First find how much water is used by \(10\) watering cans. - Determine how many groups of \(10\) cans are in \(250\) cans. - Subtract the total amount used from the starting volume.

1. Ten watering cans use \(10 \times 400\,\text{mL} = 4000\,\text{mL}\), or \(4\,\text{L}\). 2. There are \(25\) groups of \(10\) cans in \(250\) cans. Find the total used: \(25 \times 4\,\text{L} = 100\,\text{L}\). 3. Subtract from the starting amount: \(150\,\text{L} - 100\,\text{L} = 50\,\text{L}\).

\(50\,\text{L}\) of water remain in the rain barrel.

- Determine how many \(250\)-milliliter cups make \(1\) liter. - Find how many equal groups of that size are in \(80\) cups. - Each group represents \(1\) liter.

1. Four \(250\,\text{mL}\) cups hold \(4 \times 250\,\text{mL} = 1000\,\text{mL}\), or \(1\,\text{L}\). 2. Find how many groups of \(4\) cups are in \(80\) cups: \(80 \div 4 = 20\). Therefore, the total volume is \(20\,\text{L}\).

\(20\,\text{L}\) of punch are prepared altogether.

- How many minutes are in \(1\) hour? Multiply that number by the number of hours in a day. - After you find the number of minutes in a full day, how can you find half as many?

1. a) Multiply \(24\) hours by \(60\) minutes per hour. Using partial products, \(20 \times 60=1200\) and \(4 \times 60=240\). Then \(1200+240=1440\). One full day has \(1440\) minutes. 2. b) Half of \(1440\) is \(720\). Equivalently, half a day is \(12\) hours, and \(12 \times 60=720\).

a) \(1440\,\text{minutes}\) b) \(720\,\text{minutes}\)

- To convert hours to seconds, use two conversion steps: hours to minutes, then minutes to seconds. - Find the number of seconds in \(1\) hour and use that result in the later parts. - For part c), convert the hours and minutes separately, then add the results.

1. a) One hour has \(60\) minutes, and each minute has \(60\) seconds. Calculate \(60 \times 60=3600\). Therefore, \(1\) hour is \(3600\) seconds. 2. b) Multiply the number of seconds in \(1\) hour by \(5\): \(5 \times 3600=18{,}000\). Therefore, \(5\) hours is \(18{,}000\) seconds. 3. c) Convert each part separately. Two hours is \(2 \times 3600=7200\) seconds, and \(15\) minutes is \(15 \times 60=900\) seconds. Then \(7200+900=8100\).

a) \(3600\,\text{seconds}\) b) \(18{,}000\,\text{seconds}\) c) \(8100\,\text{seconds}\)

- How many minutes are in one hour? - Convert the full hiking time to minutes. - Multiply the number of minutes by the steps taken each minute.

1. Convert \(4\) hours to minutes: \(4 \times 60 = 240\) minutes. 2. Multiply the approximate rate by the total number of minutes: \(55 \times 240 \approx 13{,}200\) steps.

The hiker takes about \(13{,}200\) steps.

- How many minutes does the machine run altogether? - Once you know the total number of minutes, how can you use the rate of \(125\) parts per minute?

1. Convert the shift length to minutes: \(8 \times 60 = 480\) minutes. 2. Multiply the production rate by the number of minutes: \(125 \times 480 = 60{,}000\) parts.

The machine makes \(60{,}000\) parts during the shift.

- How many days are in \(1\) week? - First find the elephant's age in days. - How many hours are in each day? - What final operation converts the total days to hours?

1. Convert weeks to days: \(3 \times 7=21\) days. 2. Convert days to hours: \(21 \times 24=504\) hours.

The baby elephant is \(504\,\text{hours}\) old.

- Start by finding the cost of a smaller \(4\)-ounce portion. - Use \(1\,\text{lb} = 16\,\text{oz}\). - For part c), determine how many \(4\)-ounce portions fit the budget.

1. Four ounces is half of \(8\,\text{oz}\), so \(\$1.80 \div 2 = \$0.90\). 2. Since \(1\,\text{lb} = 16\,\text{oz}\), one pound is two \(8\)-ounce bags. The cost is \(2 \times \$1.80 = \$3.60\). 3. Each \(4\,\text{oz}\) costs \(\$0.90\). Since \(5 \times \$0.90 = \$4.50\), the budget buys five \(4\)-ounce portions: \(5 \times 4\,\text{oz} = 20\,\text{oz}\).

a) \(\$0.90\) b) \(\$3.60\) c) \(20\,\text{oz}\)

- Convert half a kilogram to grams. - Compare the masses of \(8\) eggs and \(9\) eggs. - Choose the total that is closer to the target mass.

1. Convert the target mass: \(\frac{1}{2}\,\text{kg} = 500\,\text{g}\). 2. Compare nearby multiples of \(60\,\text{g}\): \(8 \times 60\,\text{g} = 480\,\text{g}\), and \(9 \times 60\,\text{g} = 540\,\text{g}\). 3. Since \(480\,\text{g}\) is closer to \(500\,\text{g}\) than \(540\,\text{g}\), about \(8\) eggs are needed.

About \(8\) eggs.

- Determine how many \(250\)-milliliter glasses make \(1\) liter. - The bottle holds \(2\) liters. - Double the number of glasses in one liter.

1. Four \(250\,\text{mL}\) glasses hold \(4 \times 250\,\text{mL} = 1000\,\text{mL}\), or \(1\,\text{L}\). 2. A \(2\)-liter bottle fills twice as many glasses: \(2 \times 4 = 8\).

\(8\) glasses can be filled completely.

- Determine how many \(200\)-milliliter cups make \(1\) liter. - Find how many groups of that size are in \(15\) cups. - Each group represents \(1\) liter.

1. Five \(200\,\text{mL}\) cups hold \(5 \times 200\,\text{mL} = 1000\,\text{mL}\), or \(1\,\text{L}\). 2. Fifteen cups make \(15 \div 5 = 3\) groups of \(5\) cups, so the students serve \(3\,\text{L}\).

The students serve \(3\,\text{L}\) of tea altogether.

- Convert the carton's volume to milliliters. - Find the total volume poured into the glasses. - Subtract the poured volume from the starting volume.

1. Convert the starting volume: \(1\,\text{L}\,500\,\text{mL} = 1500\,\text{mL}\). 2. Find the volume poured: \(6 \times 200\,\text{mL} = 1200\,\text{mL}\). 3. Find the remaining volume: \(1500\,\text{mL} - 1200\,\text{mL} = 300\,\text{mL}\).

\(300\,\text{mL}\) of juice remain in the carton.

- Check whether the measurements use the same unit before adding or subtracting. - Recall how many smaller units make one larger unit. - What operations do “more than” and “less than” indicate? - Converting everything to the smaller unit can make the arithmetic easier.

1. For parts a) and b), convert \(10\,\text{km}=10{,}000\,\text{m}\). 2. a) \(10{,}000\,\text{m}+400\,\text{m}=10{,}400\,\text{m}=10.4\,\text{km}\). 3. b) \(10{,}000\,\text{m}-400\,\text{m}=9600\,\text{m}=9.6\,\text{km}\). 4. For parts c) and d), convert \(5\,\text{m}=500\,\text{cm}\). 5. c) \(500\,\text{cm}+80\,\text{cm}=580\,\text{cm}=5.8\,\text{m}\). 6. d) \(500\,\text{cm}-80\,\text{cm}=420\,\text{cm}=4.2\,\text{m}\).

a) \(10.4\,\text{km}\), or \(10{,}400\,\text{m}\) b) \(9.6\,\text{km}\), or \(9600\,\text{m}\) c) \(5.8\,\text{m}\), or \(580\,\text{cm}\) d) \(4.2\,\text{m}\), or \(420\,\text{cm}\)

- How many seconds are in \(1\) minute? - How many pounds are in \(1\) ton? - Convert to the smaller unit before calculating. - What operations do “more than” and “less than” indicate? - How many pounds are in half a ton?

1. Use \(1\,\text{min}=60\,\text{s}\) and \(1\,\text{ton}=2000\,\text{lb}\). 2. a) \(3\,\text{min}=180\,\text{s}\). Then \(180\,\text{s}+15\,\text{s}=195\,\text{s}=3\,\text{min}\ 15\,\text{s}\). 3. b) \(180\,\text{s}-15\,\text{s}=165\,\text{s}=2\,\text{min}\ 45\,\text{s}\). 4. c) Half a ton is \(1000\,\text{lb}\). Then \(1000\,\text{lb}+75\,\text{lb}=1075\,\text{lb}\). 5. d) \(1000\,\text{lb}-75\,\text{lb}=925\,\text{lb}\).

a) \(195\,\text{s}\), or \(3\,\text{min}\ 15\,\text{s}\) b) \(165\,\text{s}\), or \(2\,\text{min}\ 45\,\text{s}\) c) \(1075\,\text{lb}\) d) \(925\,\text{lb}\)

- How many meters are in \(1\,\text{km}\)? - Convert the trail length to the smaller unit before dividing. - How many times does each divisor fit into \(1000\)?

1. Convert the trail length: \(1\,\text{km}=1000\,\text{m}\). 2. Divide \(1000\,\text{m}\) by the number of equal sections: a) \(1000 \div 2=500\) b) \(1000 \div 4=250\) c) \(1000 \div 5=200\) d) \(1000 \div 8=125\) e) \(1000 \div 10=100\)

a) \(500\,\text{m}\) b) \(250\,\text{m}\) c) \(200\,\text{m}\) d) \(125\,\text{m}\) e) \(100\,\text{m}\)

- How many minutes are in \(2\) hours? - To divide by \(4\), you can divide by \(2\) twice. - After converting the total time, divide by the number of equal sessions.

1. Convert the total time: \(2\,\text{hr}=120\,\text{min}\), because \(2 \times 60=120\). 2. Divide the total minutes by the number of equal sessions: a) \(120 \div 3=40\) b) \(120 \div 4=30\) c) \(120 \div 5=24\) d) \(120 \div 6=20\) e) \(120 \div 8=15\)

a) \(40\,\text{min}\) b) \(30\,\text{min}\) c) \(24\,\text{min}\) d) \(20\,\text{min}\) e) \(15\,\text{min}\)

- Convert the pitcher's volume to milliliters. - Use multiplication to determine how many \(125\)-milliliter cups make the total. - Use approximation language in your answer.

1. Convert the pitcher's volume: \(1\,\text{L} = 1000\,\text{mL}\). 2. Compare the approximate cup volume with the total: \(8 \times 125\,\text{mL} = 1000\,\text{mL}\). 3. Because the cup volume is approximate, the result is also approximate.

About \(8\) cups can be poured.

- Put both lengths in the same unit before calculating. - How many centimeters are in \(1\) meter? - Determine how many groups of \(3\,\text{cm}\) fit into the total length.

1. Convert the total length to centimeters: \(1.50\,\text{m} = 150\,\text{cm}\). 2. Divide the total length by the length of one paper clip: \(150 \div 3 = 50\).

The line contains \(50\) paper clips.

- How many \(8\)-ounce containers make \(1\) pound? - How many \(4\)-ounce containers make \(1\) pound? - Multiply each container price by the number of containers needed.

1. Since \(1\,\text{lb} = 16\,\text{oz}\), one pound of strawberries requires \(16 \div 8 = 2\) containers. The cost is \(2 \times \$2.90 = \$5.80\). 2. One pound of raspberries requires \(16 \div 4 = 4\) containers. The cost is \(4 \times \$2.10 = \$8.40\).

Strawberries cost \(\$5.80\) per pound. Raspberries cost \(\$8.40\) per pound.

- Find the cost of the grapes and apples separately. - Add the two costs to find the purchase total. - Subtract the total from \(\$10.00\).

1. Eight ounces is half a pound, so the grapes cost \(\$3.60 \div 2 = \$1.80\). 2. The apples cost \(2 \times \$1.90 = \$3.80\). 3. The total cost is \(\$1.80 + \$3.80 = \$5.60\). 4. The change is \(\$10.00 - \$5.60 = \$4.40\).

Ms. Bauer receives \(\$4.40\) in change.

- How many meters are in \(1\,\text{km}\)? - How many grams are in \(1\,\text{kg}\)? - Subtract each given amount from its target amount.

1. Convert the target measurements: \(1\,\text{km}=1000\,\text{m}\) and \(1\,\text{kg}=1000\,\text{g}\). 2. a) \(1000\,\text{m}-420\,\text{m}=580\,\text{m}\), and \(1000\,\text{m}-885\,\text{m}=115\,\text{m}\). 3. b) \(1000\,\text{g}-350\,\text{g}=650\,\text{g}\), and \(1000\,\text{g}-75\,\text{g}=925\,\text{g}\).

a) \(580\,\text{m}\) and \(115\,\text{m}\) b) \(650\,\text{g}\) and \(925\,\text{g}\)

- Remember that \(1\) hour has \(60\) minutes, not \(100\). - How many cups are in \(1\) gallon? - How many pounds are in \(1\) ton?

1. Use \(1\,\text{gallon}=16\,\text{cups}\), \(1\,\text{hour}=60\,\text{min}\), and \(1\,\text{ton}=2000\,\text{lb}\). 2. a) \(16-10=6\), and \(16-15=1\). The missing capacities are \(6\,\text{cups}\) and \(1\,\text{cup}\). 3. b) \(60-12=48\), and \(60-38=22\). The missing times are \(48\,\text{min}\) and \(22\,\text{min}\). 4. c) \(2000-1300=700\), and \(2000-30=1970\). The missing weights are \(700\,\text{lb}\) and \(1970\,\text{lb}\).

a) \(6\,\text{cups}\) and \(1\,\text{cup}\) b) \(48\,\text{min}\) and \(22\,\text{min}\) c) \(700\,\text{lb}\) and \(1970\,\text{lb}\)

- For parts b) and c), convert both amounts to the smaller unit first. - For the money amount, you can count up to the next whole dollar and continue from there. - Include the correct unit with each answer.

1. a) Subtract the money amounts: \(\$20.00-\$12.50=\$7.50\). 2. b) Convert the target distance: \(2\,\text{km}=2000\,\text{m}\). Then \(2000\,\text{m}-850\,\text{m}=1150\,\text{m}\), or \(1\,\text{km}\ 150\,\text{m}\). 3. c) Convert both capacities to milliliters: \(3\,\text{L}=3000\,\text{mL}\), and \(1\,\text{L}\ 200\,\text{mL}=1200\,\text{mL}\). Then \(3000\,\text{mL}-1200\,\text{mL}=1800\,\text{mL}\), or \(1\,\text{L}\ 800\,\text{mL}\).

a) \(\$7.50\) b) \(1150\,\text{m}\), or \(1\,\text{km}\ 150\,\text{m}\) c) \(1800\,\text{mL}\), or \(1\,\text{L}\ 800\,\text{mL}\)

- Find the price of each type of cheese separately. - How many \(4\)-ounce portions are in each amount? - Add the two costs at the end.

1. Eight ounces of Alpine cheese is two \(4\)-ounce portions: \(2 \times \$1.60 = \$3.20\). 2. Sixteen ounces of Edam is four \(4\)-ounce portions: \(4 \times \$1.10 = \$4.40\). 3. Add the two costs: \(\$3.20 + \$4.40 = \$7.60\).

The total cost is \(\$7.60\).

- How many \(4\)-ounce portions are in \(16\,\text{oz}\)? - Compare the total price with \(\$10.00\). - Subtract the price from the amount paid to find the change.

1. Sixteen ounces contains four \(4\)-ounce portions, so the cheese costs \(4 \times \$2.20 = \$8.80\). 2. Since \(\$8.80 < \$10.00\), Tim has enough money. 3. His change is \(\$10.00 - \$8.80 = \$1.20\).

Yes. Tim has enough money and receives \(\$1.20\) in change.

- Start by finding the cost of \(4\,\text{oz}\). - Combine the prices for \(8\,\text{oz}\) and \(4\,\text{oz}\) to find the price of \(12\,\text{oz}\). - How many \(8\)-ounce portions are in \(1\) pound?

1. Four ounces is half of \(8\,\text{oz}\), so \(\$3.60 \div 2 = \$1.80\). 2. Twelve ounces is \(8\,\text{oz} + 4\,\text{oz}\), so \(\$3.60 + \$1.80 = \$5.40\). 3. Since \(1\,\text{lb} = 16\,\text{oz}\), one pound is two \(8\)-ounce portions. The cost is \(2 \times \$3.60 = \$7.20\).

a) \(\$1.80\) b) \(\$5.40\) c) \(\$7.20\)

- Compare each weight with \(4\,\text{oz}\). - When the weight is halved, what happens to the price? - How do you scale the price from \(4\,\text{oz}\) to \(8\,\text{oz}\)? - You can work in cents if that makes the calculations easier.

1. Eight ounces is twice \(4\,\text{oz}\), so \(2 \times \$1.80 = \$3.60\). 2. Two ounces is half of \(4\,\text{oz}\), so \(\$1.80 \div 2 = \$0.90\). 3. One ounce is half of \(2\,\text{oz}\), so \(\$0.90 \div 2 = \$0.45\). 4. Sixteen ounces is four times \(4\,\text{oz}\), so \(4 \times \$1.80 = \$7.20\).

\(8\,\text{oz}\): \(\$3.60\) \(2\,\text{oz}\): \(\$0.90\) \(1\,\text{oz}\): \(\$0.45\) \(16\,\text{oz}\): \(\$7.20\)

- Build each target weight from two of the known weights. - Add the corresponding prices for the two parts.

1. For \(15\,\text{oz}\), combine \(10\,\text{oz}\) and \(5\,\text{oz}\): \(\$4.50 + \$2.25 = \$6.75\). 2. For \(6\,\text{oz}\), combine \(5\,\text{oz}\) and \(1\,\text{oz}\): \(\$2.25 + \$0.45 = \$2.70\). 3. For \(11\,\text{oz}\), combine \(10\,\text{oz}\) and \(1\,\text{oz}\): \(\$4.50 + \$0.45 = \$4.95\).

a) \(\$6.75\) b) \(\$2.70\) c) \(\$4.95\)

- Can each target weight be made by adding or subtracting a known amount? - Choose the simplest sum or difference for each target.

1. For \(15\,\text{oz}\), add the prices for \(10\,\text{oz}\) and \(5\,\text{oz}\): \(\$5.60 + \$2.80 = \$8.40\). 2. For \(11\,\text{oz}\), add the prices for \(10\,\text{oz}\) and \(1\,\text{oz}\): \(\$5.60 + \$0.56 = \$6.16\). 3. For \(9\,\text{oz}\), subtract the price of \(1\,\text{oz}\) from the price of \(10\,\text{oz}\): \(\$5.60 - \$0.56 = \$5.04\).

a) \(\$8.40\) b) \(\$6.16\) c) \(\$5.04\)

- What separate time intervals are given? - Add the hours and minutes. - Remember that \(60\) minutes make one hour. - Make sure the total includes the time before the flight and after landing.

1. Add the preflight time and flight time: \(2\,\text{hours} + 1\,\text{hour}\,45\,\text{minutes} = 3\,\text{hours}\,45\,\text{minutes}\). 2. Add the baggage time: \(3\,\text{hours}\,45\,\text{minutes} + 45\,\text{minutes} = 3\,\text{hours}\,90\,\text{minutes}\). 3. Regroup \(90\) minutes as \(1\) hour \(30\) minutes. The total time is \(4\,\text{hours}\,30\,\text{minutes}\).

The total time is \(4\) hours \(30\) minutes.

- How much time do the two breaks take altogether? - Combine the break time and the construction delay. - Add that total to the driving time. - Regroup the minutes if they total more than \(60\).

1. Find the total break time: \(2 \times 25\,\text{minutes} = 50\,\text{minutes}\). 2. Add the break time and delay to the driving time: \(5\,\text{hours}\,15\,\text{minutes} + 50\,\text{minutes} + 40\,\text{minutes}\). 3. Add the minutes: \(15 + 50 + 40 = 105\) minutes, which is \(1\) hour \(45\) minutes. 4. Add the hours: \(5\,\text{hours} + 1\,\text{hour}\,45\,\text{minutes} = 6\,\text{hours}\,45\,\text{minutes}\).

The trip takes \(6\) hours \(45\) minutes.

- Find the complete trip time for each train. - What time must be added to Train B’s travel time? - Compare the two totals. - Subtract the shorter total from the longer total.

1. Find Train B’s total time: \(2\,\text{hours}\,50\,\text{minutes} + 40\,\text{minutes} = 3\,\text{hours}\,30\,\text{minutes}\). 2. Compare the totals: Train A takes \(3\) hours \(20\) minutes, and Train B takes \(3\) hours \(30\) minutes. Train A is faster. 3. Find the difference: \(3\,\text{hours}\,30\,\text{minutes} - 3\,\text{hours}\,20\,\text{minutes} = 10\,\text{minutes}\).

Train A is faster by \(10\) minutes.

- How many days are in \(3\) months using \(30\) days per month? - First find the calf’s total weight gain. - Express both weights in pounds before comparing them. - How many pounds are in \(1\) US ton?

1. The first \(3\) months contain \(3 \times 30 = 90\) days. 2. The calf gains \(90 \times 5\,\text{lb} = 450\,\text{lb}\). 3. Its new weight is \(100\,\text{lb} + 450\,\text{lb} = 550\,\text{lb}\). 4. Convert the adult weight: \(2\) tons \(= 4000\,\text{lb}\). 5. Find the difference: \(4000\,\text{lb} - 550\,\text{lb} = 3450\,\text{lb}\).

a) The calf weighs \(550\,\text{lb}\). b) It is \(3450\,\text{lb}\) short of the adult weight.

- Convert the starting amount to pounds. - Find the total delivered during the \(12\) days. - Add the delivery to the starting amount. - Use \(2000\,\text{lb} = 1\) ton to express the final amount in tons and pounds.

1. Convert the starting amount: \(8\) tons \(= 16{,}000\,\text{lb}\). 2. Find the total delivered: \(12 \times 1200\,\text{lb} = 14{,}400\,\text{lb}\). 3. Add the amounts: \(16{,}000\,\text{lb} + 14{,}400\,\text{lb} = 30{,}400\,\text{lb}\). 4. Since \(15 \times 2000\,\text{lb} = 30{,}000\,\text{lb}\), there are \(400\,\text{lb}\) left. Therefore, \(30{,}400\,\text{lb} = 15\) tons \(400\,\text{lb}\).

After \(12\) days, the warehouse contains \(30{,}400\,\text{lb}\), or \(15\) tons \(400\,\text{lb}\), of iron.

- Find the total mass of each kind of rhinoceros separately. - Add the two totals. - Convert the final number of kilograms to metric tons and kilograms.

1. Estimate the mass of the white rhinoceroses: \(3 \times 2300\,\text{kg} \approx 6900\,\text{kg}\). 2. Estimate the mass of the black rhinoceroses: \(5 \times 1100\,\text{kg} \approx 5500\,\text{kg}\). 3. Add \(6900\,\text{kg}\) and \(5500\,\text{kg}\). The combined mass is about \(12{,}400\,\text{kg}\). 4. Convert the estimate: \(12{,}400\,\text{kg}\) is \(12\) metric tons \(400\,\text{kg}\).

The animals have a combined mass of about \(12\) metric tons \(400\,\text{kg}\).

- Recall how many quarts equal one gallon. - Multiply to convert the tank's capacity to quarts. - Divide the number of gallons in the tank by the number of gallons in one watering can.

1. Convert gallons to quarts: \(75\,\text{gallons} \times 4 = 300\,\text{quarts}\). 2. Divide the tank's capacity by the watering can's capacity: \(75\,\text{gallons} \div 3\,\text{gallons} = 25\).

a) The full tank contains \(300\,\text{quarts}\) of water. b) The watering can can be filled \(25\) times.

- Recall how many quarts equal one gallon. - Recall how many cups equal one quart. - Convert in two steps: gallons to quarts, then quarts to cups.

1. Convert gallons to quarts: \(50\,\text{gallons} \times 4 = 200\,\text{quarts}\). 2. Each quart contains \(4\) cups, so \(200\,\text{quarts} \times 4 = 800\,\text{cups}\).

a) The event receives \(200\,\text{quarts}\) of cider. b) The event can serve \(800\) one-cup portions.

- Round each amount to the nearest hundred milliliters. - Remember that \(1000\,\text{mL} = 1\,\text{L}\). - Multiply the daily estimate by \(7\) for one week. - Use \(30\) days for the monthly estimate.

1. Round the daily amounts to convenient hundreds: \(800\,\text{mL} + 200\,\text{mL} + 200\,\text{mL} + 200\,\text{mL} + 200\,\text{mL} = 1600\,\text{mL}\). 2. For \(7\) days, \(1600\,\text{mL} \times 7 = 11{,}200\,\text{mL} = 11.2\,\text{L}\), or about \(11\,\text{L}\). 3. For \(30\) days, \(1600\,\text{mL} \times 30 = 48{,}000\,\text{mL} = 48\,\text{L}\).

a) About \(1600\,\text{mL}\) b) About \(11\,\text{L}\) c) About \(48\,\text{L}\)

- Round each daily amount to a nearby hundred for the estimate. - Align place values when adding the exact amounts. - Remember that \(1000\,\text{g} = 1\,\text{kg}\).

1. Estimate by rounding: \(1300\,\text{g} + 800\,\text{g} + 600\,\text{g} + 500\,\text{g} = 3200\,\text{g}\). 2. Add the exact amounts: \(1250\,\text{g} + 810\,\text{g} + 590\,\text{g} + 450\,\text{g} = 3100\,\text{g}\). 3. Find the weekly amount: \(3100\,\text{g} \times 7 = 21{,}700\,\text{g}\). 4. Convert to kilograms: \(21{,}700\,\text{g} = 21.7\,\text{kg}\), which is about \(22\,\text{kg}\).

a) About \(3200\,\text{g}\) b) \(3100\,\text{g}\) c) About \(22\,\text{kg}\)

- Find how many groups of \(10\) are in the target number. - Multiply that scale factor by \(5\,\text{cm}\). - Convert the final distance to meters.

1. a) The distance to \(1000\) is \(1000\div10=100\) times the distance to \(10\). Thus, \(100\times5\,\text{cm}=500\,\text{cm}=5\,\text{m}\). 2. b) The distance to \(20{,}000\) is \(20{,}000\div10=2000\) times the distance to \(10\). Thus, \(2000\times5\,\text{cm}=10{,}000\,\text{cm}=100\,\text{m}\).

a) \(5\,\text{m}\) b) \(100\,\text{m}\)

- First find the total length in millimeters. - Convert millimeters to meters for part a). - For part b), express both lengths in the same unit before dividing.

1. The total length is \(4500\times2\,\text{mm}=9000\,\text{mm}\). 2. Convert to meters: \(9000\,\text{mm}=9\,\text{m}\). 3. Convert the total length to centimeters: \(9\,\text{m}=900\,\text{cm}\). 4. Divide by the width of one notebook: \(900\,\text{cm}\div20\,\text{cm}=45\).

a) \(9\,\text{m}\) b) \(45\) notebooks

- Round each run separately before adding. - Use \(30\) seconds as the halfway point. - Add the rounded minute values.

1. Run 1 rounds to \(2\,\text{min}\) because \(18<30\). 2. Run 2 rounds to \(2\,\text{min}\) because \(42\geq30\). 3. Run 3 rounds to \(3\,\text{min}\) because \(35\geq30\). 4. Add the rounded times: \(2+2+3=7\,\text{min}\).

The total of the rounded times is \(7\,\text{min}\).

- Convert the kilogram amounts to grams first. - Find the total weight in Lucas's backpack. - Find the total weight of Sophie's pasta packages. - Compare the two totals.

1. Find Lucas's total: \(1000\,\text{g} + 1000\,\text{g} + 500\,\text{g} = 2500\,\text{g}\). 2. Find Sophie's total: \(5 \times 500\,\text{g} = 2500\,\text{g}\). 3. Compare: \(2500\,\text{g} = 2500\,\text{g}\), so they carry the same weight.

They carry the same weight: \(2500\,\text{g}\), or \(2\,\text{kg}\ 500\,\text{g}\), each.

- How many ounces are in \(4\) pounds? - Find how much dough was used for all \(12\) rolls. - Subtract the amount used from the starting amount.

1. Convert the total amount of dough: \(4\,\text{lb} = 64\,\text{oz}\). 2. Find the dough used: \(12 \times 4\,\text{oz} = 48\,\text{oz}\). 3. Subtract the amount used: \(64\,\text{oz} - 48\,\text{oz} = 16\,\text{oz}\).

\(16\,\text{oz}\) of dough remain.

- Convert the distance between the houses to meters. - Add the distances Julia and Tom have walked. - Compare their combined distance with the distance between the houses.

1. Convert the distance between the houses: \(1\,\text{km} = 1000\,\text{m}\). 2. Add the distances Julia and Tom have walked: \(460 + 550 = 1010\,\text{m}\). 3. Since \(1010 > 1000\), their combined distance is greater than the distance between the houses. They have already met and passed each other.

Yes. Together they have walked \(1010\,\text{m}\), which is \(10\,\text{m}\) more than the \(1000\,\text{m}\) between their houses, so they have already passed each other.

- Express the cruise ship's entire stay in hours. - How many hours are in \(2\) full days? - Use the same unit before comparing the durations. - Subtract to find the difference.

1. Convert \(2\) days to hours: \(2 \times 24\,\text{hr} = 48\,\text{hr}\). 2. Find the cruise ship's total time: \(48\,\text{hr} + 10\,\text{hr} = 58\,\text{hr}\). 3. Compare: \(60\,\text{hr} > 58\,\text{hr}\), so the cargo ship stays longer. 4. Find the difference: \(60\,\text{hr} - 58\,\text{hr} = 2\,\text{hr}\).

The cargo ship stays longer by \(2\) hours.

- How many minutes are in half an hour? - How many more bottles does Machine B fill in one minute? - Multiply the one-minute difference by the total number of minutes.

1. Convert half an hour to minutes: \(30\) minutes. 2. Find the difference in the machines’ rates: \(55 - 42 = 13\) bottles per minute. 3. Find the total difference over \(30\) minutes: \(13 \times 30 = 390\) bottles.

Machine B fills \(390\) more bottles than Machine A.

- Find how much elastic is needed for each type of item. - Express the needed and available lengths in the same unit. - Add the two amounts needed before comparing the total with the roll.

1. Find the elastic needed for the masks: \(6 \times 12\,\text{in} = 72\,\text{in}\). 2. Find the elastic needed for the hats: \(4 \times 20\,\text{in} = 80\,\text{in}\). 3. Find the total needed: \(72\,\text{in} + 80\,\text{in} = 152\,\text{in}\). 4. Convert the available length: \(15\,\text{ft} = 180\,\text{in}\). 5. Since \(152\,\text{in} < 180\,\text{in}\), there is enough elastic.

Yes, there is enough elastic.

- How many centimeters are in \(1\) meter? - Convert Lucas's rope length to the smaller unit first. - What operation represents “shorter than” in this situation?

1. Convert Lucas's rope length to centimeters: \(2\,\text{m} = 200\,\text{cm}\). 2. Subtract the difference: \(200\,\text{cm} - 45\,\text{cm} = 155\,\text{cm}\).

Tim''s rope is \(155\,\text{cm}\) long.

- Express all the lengths in centimeters first. - Find the combined length of the two pieces that were removed. - Subtract the amount removed from the original length.

1. Convert the full beam length to centimeters: \(3\,\text{m} = 300\,\text{cm}\). 2. Find the total length cut off: \(110\,\text{cm} + 95\,\text{cm} = 205\,\text{cm}\). 3. Subtract the amount removed: \(300\,\text{cm} - 205\,\text{cm} = 95\,\text{cm}\).

\(95\,\text{cm}\) of the beam remain.

- How many fluid ounces are in \(1\) quart? - Find how many fluid ounces are in one glass. - Find the total volume poured into three glasses. - Subtract the poured amount from the full pitcher.

1. Convert the pitcher's volume: \(1\,\text{qt} = 32\,\text{fl oz}\). 2. Find the volume of one glass: \(32\,\text{fl oz} \div 8 = 4\,\text{fl oz}\). 3. Find the volume that was drunk: \(3 \times 4\,\text{fl oz} = 12\,\text{fl oz}\). 4. Find the remaining volume: \(32\,\text{fl oz} - 12\,\text{fl oz} = 20\,\text{fl oz}\).

\(20\,\text{fl oz}\) of juice remain.

- Convert the total weight to ounces. - Subtract the weight of the empty bag. - Find the total flour removed, then subtract it from the starting amount of flour.

1. Convert the total weight: \(2\,\text{lb}\,4\,\text{oz} = 36\,\text{oz}\). 2. Subtract the packaging: \(36\,\text{oz} - 1\,\text{oz} = 35\,\text{oz}\) of flour at the start. 3. Find the amount removed: \(4 \times 8\,\text{oz} = 32\,\text{oz}\). 4. Find the amount remaining: \(35\,\text{oz} - 32\,\text{oz} = 3\,\text{oz}\).

\(3\,\text{oz}\) of flour remain.

- Convert the full path length to meters. - Find how many meters were completed on Tuesday. - Add the first two days and subtract from the full length.

1. Convert the total length: \(1\,\text{km} = 1000\,\text{m}\). 2. Find Tuesday's distance: \(345 + 120 = 465\,\text{m}\). 3. Find the total finished on Monday and Tuesday: \(345 + 465 = 810\,\text{m}\). 4. Find the distance remaining: \(1000 - 810 = 190\,\text{m}\). 5. Since \(190 < 200\), the workers are correct.

Yes. The workers have \(190\,\text{m}\) left, which is less than \(200\,\text{m}\).

- Convert the pound weight to ounces. - Add the known weights on the left side. - Because the scale is balanced, subtract those known weights from the right side.

1. Convert the weight on the right: \(1\,\text{lb} = 16\,\text{oz}\). 2. Add the metal weights on the left: \(2 \times 2\,\text{oz} + 1\,\text{oz} = 5\,\text{oz}\). 3. Subtract to find the nuts' weight: \(16\,\text{oz} - 5\,\text{oz} = 11\,\text{oz}\).

The nuts weigh \(11\,\text{oz}\).

- Convert pounds to ounces. - Subtract the extra weight placed with the jars. - Divide the remaining weight equally among the three jars.

1. Convert the right side: \(4\,\text{lb} = 64\,\text{oz}\). 2. Subtract the added \(1\)-pound weight from the total: \(64\,\text{oz} - 16\,\text{oz} = 48\,\text{oz}\) for the three jars. 3. Divide equally: \(48\,\text{oz} \div 3 = 16\,\text{oz}\) per jar.

One honey jar weighs \(16\,\text{oz}\).

- How many fluid ounces are in \(1\) quart? - How many times does the smaller capacity fit into the larger capacity? - Use division to find the number of equal cupfuls.

1. Convert the bowl's capacity: \(1\,\text{qt} = 32\,\text{fl oz}\). 2. Divide by the capacity of one measuring cup: \(32\,\text{fl oz} \div 2\,\text{fl oz} = 16\). 3. Therefore, \(16\) full measuring cups are needed.

\(16\) full measuring cups are needed.

- Group the \(750\)-milliliter bottles in sets of \(4\) to make \(3\) liters. - Add the apple juice and water volumes in liters. - Determine how many \(300\)-milliliter cups make \(3\) liters, then scale to the total.

1. Four bottles hold \(4 \times 750\,\text{mL} = 3000\,\text{mL}\), or \(3\,\text{L}\). Twelve bottles make \(3\) groups of \(4\) bottles, so the apple juice volume is \(3 \times 3\,\text{L} = 9\,\text{L}\). 2. Add the water: \(9\,\text{L} + 6\,\text{L} = 15\,\text{L}\). 3. Ten \(300\,\text{mL}\) cups hold \(3000\,\text{mL}\), or \(3\,\text{L}\). 4. The punch contains \(5\) groups of \(3\,\text{L}\), so it fills \(5 \times 10 = 50\) cups.

\(50\) cups can be filled completely.

- How far does the scooter move during one wheel rotation? - Multiply that distance by the number of rotations. - Use \(100\,\text{cm} = 1\,\text{m}\) to convert the result.

1. Find the total distance in centimeters: \(9 \times 80\,\text{cm} = 720\,\text{cm}\). 2. Since \(100\,\text{cm} = 1\,\text{m}\), \(720\,\text{cm} = 7\,\text{m}\ 20\,\text{cm}\).

The scooter travels \(720\,\text{cm}\), or \(7\,\text{m}\ 20\,\text{cm}\).

- Express all distances in centimeters before calculating. - Determine how many wheel circumferences fit into the total distance. - Predict which wheel should make more rotations.

1. Convert the total distance to centimeters: \(4\,\text{m} = 400\,\text{cm}\). 2. For the small wheel: \(400\,\text{cm} \div 20\,\text{cm} = 20\) rotations. 3. For the large wheel: \(400\,\text{cm} \div 40\,\text{cm} = 10\) rotations.

The small wheel makes \(20\) rotations, and the large wheel makes \(10\) rotations.

- How many centimeters are in one meter? - Convert the target length to centimeters first. - Use the length of one stone to find the length of several stones. - Use your result from part a) to help with part b).

1. Convert the target length: \(2\,\text{m} = 200\,\text{cm}\). 2. Find the number of stones: \(200\,\text{cm} \div 20\,\text{cm} = 10\). 3. Five stones make a row \(5 \times 20\,\text{cm} = 100\,\text{cm}\) long. 4. Convert: \(100\,\text{cm} = 1\,\text{m}\).

a) \(10\) stones b) \(100\,\text{cm}\), or \(1\,\text{m}\)

- Multiply the pool's length by the number of lengths in each part. - Remember that \(1000\,\text{m} = 1\,\text{km}\). - For part c, find one day's distance before finding the weekly distance.

1. For part a, multiply: \(25 \times 12 = 300\,\text{m}\). 2. For part b, multiply: \(25 \times 60 = 1500\,\text{m}\). Convert: \(1500\,\text{m} = 1.5\,\text{km}\). 3. For part c, find the distance per day: \(25 \times 40 = 1000\,\text{m}\). Then multiply by \(5\): \(1000 \times 5 = 5000\,\text{m}\).

a) \(300\,\text{m}\) b) \(1.5\,\text{km}\) c) \(5000\,\text{m}\)

- How many grams are in \(1\) kilogram? - Convert the total weight to grams first. - What must you subtract to find the weight of only the apples?

1. Convert the total weight: \(1\,\text{kg} = 1000\,\text{g}\). 2. Subtract the empty crate's weight: \(1000\,\text{g} - 245\,\text{g} = 755\,\text{g}\).

The apples weigh \(755\,\text{g}\).

- Find how much the scale reading increased when the flour was added. - Convert the final weight from pounds to ounces. - Find the increase from the flour-only reading to the final reading.

1. Find the flour's weight: \(19\,\text{oz} - 7\,\text{oz} = 12\,\text{oz}\). 2. Convert the final scale reading: \(2\,\text{lb} = 32\,\text{oz}\). 3. Find the sugar's weight: \(32\,\text{oz} - 19\,\text{oz} = 13\,\text{oz}\).

Lucas added \(12\,\text{oz}\) of flour and \(13\,\text{oz}\) of sugar.

- Convert both distances to meters. - Compare the measurements after they use the same unit. - Subtract to find how much longer one route is.

1. Convert Ms. Lee's class's distance: \(1\,\text{km}\ 200\,\text{m} = 1200\,\text{m}\). 2. Compare the distances: \(1500 > 1200\), so Mr. Ortiz's class hikes farther. 3. Find the difference: \(1500 - 1200 = 300\,\text{m}\).

Mr. Ortiz''s class hikes farther. Its route is \(300\,\text{m}\) longer.

- Convert the length in feet to inches first. - Add the ribbon used for all three gifts. - Subtract the amount used from the original amount.

1. Convert the full ribbon length to inches: \(12\,\text{ft} = 144\,\text{in}\). 2. Find the total amount used: \(45\,\text{in} + 33\,\text{in} + 42\,\text{in} = 120\,\text{in}\). 3. Find the amount left: \(144\,\text{in} - 120\,\text{in} = 24\,\text{in}\).

\(24\,\text{in}\) of ribbon remain.

- Convert the maximum mass to kilograms. - Find the combined mass of the pallets. - Add the empty vehicle and cargo, then subtract from the limit.

1. Convert the maximum mass: \(3\) metric tons \(500\,\text{kg} = 3500\,\text{kg}\). 2. Find the pallets' mass: \(4 \times 300\,\text{kg} = 1200\,\text{kg}\). 3. Find the current total mass: \(2100\,\text{kg} + 1200\,\text{kg} = 3300\,\text{kg}\). 4. Find the remaining capacity: \(3500\,\text{kg} - 3300\,\text{kg} = 200\,\text{kg}\).

The van can carry \(200\,\text{kg}\) more.

- Convert both lengths to centimeters. - Compare the two measurements. - Subtract to find the difference.

1. Convert the blue ribbon's length: \(2\,\text{m}\ 40\,\text{cm} = 240\,\text{cm}\). 2. Compare: \(270 > 240\), so the red ribbon is longer. 3. Find the difference: \(270 - 240 = 30\,\text{cm}\).

The red ribbon is \(30\,\text{cm}\) longer.

- How many centimeters make one meter? - Separate \(305\) centimeters into hundreds and leftover centimeters. - Convert \(4\,\text{m}\) to centimeters before finding the difference.

1. Decompose the measurement: \(305\,\text{cm} = 300\,\text{cm} + 5\,\text{cm}\). 2. Since \(300\,\text{cm} = 3\,\text{m}\), the jump is \(3\,\text{m}\ 5\,\text{cm}\). 3. Convert the target: \(4\,\text{m} = 400\,\text{cm}\). 4. Subtract: \(400\,\text{cm} - 305\,\text{cm} = 95\,\text{cm}\).

a) \(3\,\text{m}\ 5\,\text{cm}\) b) \(95\,\text{cm}\)

- Convert both lengths to centimeters. - How many centimeters are in \(1\) meter? In \(1\) decimeter? - Subtract the shorter length from the longer length.

1. Convert the first beam to centimeters: \(1\,\text{m}\,45\,\text{cm} = 100 + 45 = 145\,\text{cm}\). 2. Convert the second beam to centimeters: \(16\,\text{dm} = 160\,\text{cm}\). 3. Since \(160 > 145\), the second beam is longer. 4. Find the difference: \(160 - 145 = 15\,\text{cm}\).

The second beam is longer by \(15\,\text{cm}\).

- Convert every length to centimeters. - Add the three pieces. - Compare the total with \(200\,\text{cm}\) and find the difference.

1. Convert the target length: \(2\,\text{m} = 200\,\text{cm}\). 2. Convert the pieces to centimeters: \(95\,\text{cm}\), \(6\,\text{dm} = 60\,\text{cm}\), and \(300\,\text{mm} = 30\,\text{cm}\). 3. Add the pieces: \(95 + 60 + 30 = 185\,\text{cm}\). 4. Since \(185 < 200\), the pieces are not long enough. 5. Find the missing length: \(200 - 185 = 15\,\text{cm}\).

No. The pieces are \(15\,\text{cm}\) too short.

- Find the combined mass of all four crates. - Convert the forklift's limit to kilograms. - Compare the two amounts.

1. Find the total mass of the crates: \(240\,\text{kg} + 310\,\text{kg} + 180\,\text{kg} + 280\,\text{kg} = 1010\,\text{kg}\). 2. Convert the forklift's limit: \(1\) metric ton \(= 1000\,\text{kg}\). 3. Compare the masses: \(1010\,\text{kg} > 1000\,\text{kg}\). 4. The crates exceed the limit by \(1010\,\text{kg} - 1000\,\text{kg} = 10\,\text{kg}\).

No. The four crates have a total mass of \(1010\,\text{kg}\), which is \(10\,\text{kg}\) over the forklift''s limit.

- Convert all the weights to the same unit before ordering them. - Compare the converted numbers by place value. - Recall the conversion factors between pounds and ounces and between tons and pounds. - Subtract each given amount from its target amount.

1. Convert the weights in part a) to pounds: \(4\,\text{tons}=8000\,\text{lb}\), \(400\,\text{lb}=400\,\text{lb}\), \(400\,\text{oz}=25\,\text{lb}\), and \(40\,\text{lb}=40\,\text{lb}\). 2. Order the values: \(25<40<400<8000\). Therefore, \(400\,\text{oz}<40\,\text{lb}<400\,\text{lb}<4\,\text{tons}\). 3. Since \(1\,\text{lb}=16\,\text{oz}\), \(16-12=4\). The first missing amount is \(4\,\text{oz}\). 4. Since \(1\,\text{ton}=2000\,\text{lb}\), \(2000-1300=700\). The second missing amount is \(700\,\text{lb}\).

a) \(400\,\text{oz}<40\,\text{lb}<400\,\text{lb}<4\,\text{tons}\) b) \(4\,\text{oz}\) and \(700\,\text{lb}\)

- Convert both measurements to kilograms first. - Add the mass of the empty truck and the mass of the sand. - Check that your answer includes the correct unit.

1. Convert the empty truck's mass: \(2\) metric tons \(350\,\text{kg} = 2350\,\text{kg}\). 2. Convert the sand's mass: \(1\) metric ton \(800\,\text{kg} = 1800\,\text{kg}\). 3. Add the masses: \(2350\,\text{kg} + 1800\,\text{kg} = 4150\,\text{kg}\).

The loaded truck has a total mass of \(4150\,\text{kg}\).

- Convert both weights to grams before adding. - Add the backpack's weight and the books' weight. - Convert the total back to kilograms and grams.

1. Convert the books' weight to grams: \(2\,\text{kg}\ 350\,\text{g} = 2350\,\text{g}\). 2. Add the weights: \(1200\,\text{g} + 2350\,\text{g} = 3550\,\text{g}\). 3. Convert the total to mixed-unit form: \(3550\,\text{g} = 3\,\text{kg}\ 550\,\text{g}\).

The filled backpack weighs \(3\,\text{kg}\ 550\,\text{g}\).

- Convert both ages to the same unit. - Change Frieda’s years to months and add the extra months. - Compare the two totals and subtract.

1. Convert Frieda’s age to months: \(6 \times 12 + 10 = 72 + 10 = 82\) months. 2. Compare the ages: \(82 > 80\), so Frieda is older. 3. Find the difference: \(82 - 80 = 2\) months.

Frieda is older by \(2\) months.

- Convert both quantities to years before comparing them. - The phrase “how many are needed” indicates that you should find a difference. - One century is \(100\) years.

1. a) Convert \(8\) centuries to years: \(8 \times 100=800\) years. Then \(800-650=150\), so \(150\) years are needed. 2. b) Convert \(14\) centuries to years: \(14 \times 100=1400\) years. Then \(1400-1050=350\), so \(350\) years are needed.

a) \(150\,\text{years}\) b) \(350\,\text{years}\)

- Convert each time unit to days. - Use the stated value of \(30\) days per month. - Convert the weeks using \(7\) days per week. - Add all three amounts.

1. Convert the months to days: \(2 \times 30 = 60\) days. 2. Convert the weeks to days: \(2 \times 7 = 14\) days. 3. Add all the days: \(60 + 14 + 5 = 79\) days.

The ship is at sea for \(79\) days.

- Convert the weeks to days first. - Add the remaining days. - How many hours are in one day? - Multiply the total number of days by \(24\).

1. Convert the weeks to days: \(3 \times 7 = 21\) days. 2. Add the extra days: \(21 + 4 = 25\) days. 3. Convert days to hours: \(25 \times 24 = 600\) hours.

The expedition lasts \(600\) hours.

- Convert both times to the same unit. - How many minutes are in one hour? - Compare the totals and subtract.

1. Convert Group B’s time to minutes: \(8 \times 60 + 45 = 525\) minutes. 2. Compare the times: \(525 > 510\), so Group B hikes longer. 3. Find the difference: \(525 - 510 = 15\) minutes.

Group B hikes longer by \(15\) minutes.

- How many hours are in one day? - Convert both times to hours. - In a race, the smaller time is faster.

1. Convert Team Blue’s time to hours: \(4 \times 24 + 5 = 101\) hours. 2. Since \(101 < 105\), Team Blue is faster. 3. Find the difference: \(105 - 101 = 4\) hours.

Team Blue is faster by \(4\) hours.

- Divide \(300\) by \(24\) and interpret the remainder. - How many days are in two weeks? - Convert two weeks to hours and compare with \(300\) hours.

1. Divide by the number of hours in a day: \(300 \div 24 = 12\) remainder \(12\). The battery lasts \(12\) days \(12\) hours. 2. Two weeks is \(14\) days, or \(14 \times 24 = 336\) hours. 3. Find the additional time needed: \(336 - 300 = 36\) hours.

a) The battery lasts \(12\) days \(12\) hours. b) It would need to last \(36\) additional hours.

- Add all three weights in grams first. - One kilogram equals \(1000\) grams. - Separate the total into complete thousands and the remainder.

1. Add the three weights: \(12{,}400\,\text{g} + 11{,}950\,\text{g} + 12{,}050\,\text{g} = 36{,}400\,\text{g}\). 2. Since \(1000\,\text{g} = 1\,\text{kg}\), \(36{,}400\,\text{g}\) contains \(36\) kilograms with \(400\) grams remaining. 3. The total weight is \(36\,\text{kg}\ 400\,\text{g}\).

The three crates weigh \(36\,\text{kg}\ 400\,\text{g}\) in all.

- Convert every load to kilograms. - Add the three loads for part a. - Convert \(10\) metric tons to kilograms, then subtract the total load.

1. Convert each load to kilograms: Truck A has \(2450\,\text{kg}\), Truck B has \(1800\,\text{kg}\), and Truck C has \(3050\,\text{kg}\). 2. Add the loads: \(2450\,\text{kg} + 1800\,\text{kg} + 3050\,\text{kg} = 7300\,\text{kg}\). 3. Convert the target mass: \(10\) metric tons \(= 10{,}000\,\text{kg}\). 4. Start with the target of \(10{,}000\,\text{kg}\) and subtract the current total of \(7300\,\text{kg}\). Another \(2700\,\text{kg}\) is needed.

a) The total mass is \(7300\,\text{kg}\), or \(7\) metric tons \(300\,\text{kg}\). b) Another \(2700\,\text{kg}\), or \(2\) metric tons \(700\,\text{kg}\), is needed.

- How many grams are in \(1\) kilogram? - Express both weights in grams before subtracting. - Subtract the empty bag's weight from the packed weight.

1. Convert the total weight to grams: \(2\,\text{kg}\ 100\,\text{g} = 2100\,\text{g}\). 2. Subtract the empty bag's weight: \(2100\,\text{g} - 850\,\text{g} = 1250\,\text{g}\). 3. Convert the result: \(1250\,\text{g} = 1\,\text{kg}\ 250\,\text{g}\).

The books weigh \(1\,\text{kg}\ 250\,\text{g}\).

- Find the weight of only the marbles in each package. - Remember that \(1\,\text{kg} = 1000\,\text{g}\). - Compare the two marble weights. - Subtract to find how much greater one weight is.

1. Find the weight of the marbles in Package A: \(1400\,\text{g} - 250\,\text{g} = 1150\,\text{g}\). 2. Find the weight of the marbles in Package B: \(1200\,\text{g} - 100\,\text{g} = 1100\,\text{g}\). 3. Compare: \(1150\,\text{g} > 1100\,\text{g}\), so Package A contains the greater weight of marbles. 4. Find the difference: \(1150\,\text{g} - 1100\,\text{g} = 50\,\text{g}\).

Package A contains the greater weight of marbles. The difference is \(50\,\text{g}\).

- One pound equals \(16\) ounces. - Find the number of complete groups of \(16\) in \(214\). - Convert \(14\) pounds to ounces before finding the difference.

1. Since \(1\,\text{lb} = 16\,\text{oz}\), \(214\,\text{oz}\) contains \(13\) complete pounds because \(13 \times 16 = 208\). 2. The remainder is \(214 - 208 = 6\), so the dog weighs \(13\,\text{lb}\ 6\,\text{oz}\). 3. Fourteen pounds equals \(14 \times 16 = 224\,\text{oz}\). 4. The dog needs \(224\,\text{oz} - 214\,\text{oz} = 10\,\text{oz}\) more.

The dog weighs \(13\,\text{lb}\ 6\,\text{oz}\), and it needs \(10\,\text{oz}\) more to weigh \(14\,\text{lb}\).

- One meter equals \(100\) centimeters. - Convert both jumps to centimeters before comparing. - Subtract the shorter distance from the longer distance.

1. Convert the first jump: \(305\,\text{cm} = 3\,\text{m}\ 5\,\text{cm}\). 2. Convert the second jump to centimeters: \(3\,\text{m}\ 50\,\text{cm} = 350\,\text{cm}\). 3. Since \(350 > 305\), the second student jumps farther. 4. The difference is \(350\,\text{cm} - 305\,\text{cm} = 45\,\text{cm}\).

The first jump is \(3\,\text{m}\ 5\,\text{cm}\). The second student jumps farther by \(45\,\text{cm}\).

- One year has \(12\) months. - You can often calculate the years and months separately. - If an addition produces at least \(12\) months, regroup \(12\) months as \(1\) year. - If there are not enough months to subtract, regroup \(1\) year as \(12\) months.

1. a) Add years and months separately: \(6+4=10\) years and \(5+3=8\) months. The result is \(10\) years \(8\) months. 2. b) \(8+2=10\) years and \(9+11=20\) months. Since \(20\) months is \(1\) year \(8\) months, the result is \(11\) years \(8\) months. 3. c) Regroup \(1\) year as \(12\) months: \(15\) years \(2\) months becomes \(14\) years \(14\) months. Then \(14-6=8\) years and \(14-7=7\) months, so the result is \(8\) years \(7\) months. 4. d) One year has \(12\) months, and \(12-8=4\). Four months remain.

a) \(10\,\text{years}\ 8\,\text{months}\) b) \(11\,\text{years}\ 8\,\text{months}\) c) \(8\,\text{years}\ 7\,\text{months}\) d) \(4\,\text{months}\)

- How many minutes are in \(1\) hour? - When subtracting, regroup \(1\) hour as \(60\) minutes if needed. - Calculate the hours and minutes separately. - At the end, regroup any set of \(60\) minutes as \(1\) hour.

1. a) Add the hours and minutes: \(4+2=6\) hours and \(35+45=80\) minutes. Since \(80\) minutes is \(1\) hour \(20\) minutes, the result is \(7\) hours \(20\) minutes. 2. b) Regroup \(1\) hour as \(60\) minutes: \(15\) hours \(12\) minutes becomes \(14\) hours \(72\) minutes. Then \(14-6=8\) hours and \(72-45=27\) minutes. The result is \(8\) hours \(27\) minutes. 3. c) Add \(9+4=13\) hours and \(58+7=65\) minutes. Since \(65\) minutes is \(1\) hour \(5\) minutes, the result is \(14\) hours \(5\) minutes.

a) \(7\,\text{hr}\ 20\,\text{min}\) b) \(8\,\text{hr}\ 27\,\text{min}\) c) \(14\,\text{hr}\ 5\,\text{min}\)

- Calculate \(A\) and \(B\) separately. - Remember that \(1\,\text{hr}=60\,\text{min}\). - Subtract the shorter time span from the longer time span.

1. Calculate \(A\): \(5+3=8\) hours, and \(45+25=70\) minutes \(=1\) hour \(10\) minutes. Therefore, \(A=9\,\text{hr}\ 10\,\text{min}\). 2. Calculate \(B\): Regroup \(12\,\text{hr}\ 10\,\text{min}\) as \(11\,\text{hr}\ 70\,\text{min}\). Then \(11-2=9\) hours and \(70-50=20\) minutes, so \(B=9\,\text{hr}\ 20\,\text{min}\). 3. Compare and subtract: \(B\) is longer, and \(9\,\text{hr}\ 20\,\text{min}-9\,\text{hr}\ 10\,\text{min}=10\,\text{min}\).

Time span \(B\) is longer by \(10\,\text{min}\).

- Add like units separately. - How many minutes make \(1\) hour? - How many hours make \(1\) day? - Regroup any total of at least \(60\) minutes or \(24\) hours.

1. a) Add \(4+3=7\) hours and \(40+35=75\) minutes. Since \(75\) minutes is \(1\) hour \(15\) minutes, the result is \(8\,\text{hr}\ 15\,\text{min}\). 2. b) Add \(12+11=23\) hours and \(55+25=80\) minutes. Since \(80\) minutes is \(1\) hour \(20\) minutes, this becomes \(24\) hours \(20\) minutes. Since \(24\) hours is \(1\) day, the result is \(1\,\text{day}\ 20\,\text{min}\). 3. c) Add \(2+1=3\) days and \(15+18=33\) hours. Since \(33\) hours is \(1\) day \(9\) hours, the result is \(4\,\text{days}\ 9\,\text{hr}\).

a) \(8\,\text{hr}\ 15\,\text{min}\) b) \(1\,\text{day}\ 20\,\text{min}\) c) \(4\,\text{days}\ 9\,\text{hr}\)

- Add the two parts of Train A’s trip. - Regroup the minutes if their sum is at least \(60\). - Compare Train A’s total with Train B’s time.

1. Add Train A’s times: \(2\,\text{hours}\,45\,\text{minutes} + 3\,\text{hours}\,35\,\text{minutes}\). 2. Add the minutes: \(45 + 35 = 80\) minutes, which is \(1\) hour \(20\) minutes. 3. Train A’s total time is \(6\) hours \(20\) minutes. 4. Compare with Train B’s \(6\) hours \(10\) minutes. Train B is faster. 5. Find the difference: \(6\,\text{hours}\,20\,\text{minutes} - 6\,\text{hours}\,10\,\text{minutes} = 10\,\text{minutes}\).

Train B is faster by \(10\) minutes.

- How many months are in \(1\) year? - If there are not enough months to subtract, regroup \(1\) year as \(12\) months. - Rewrite each time span in a form that makes subtraction possible.

1. a) Subtract like units: \(9-3=6\) years and \(8-5=3\) months. The result is \(6\) years \(3\) months. 2. b) Regroup \(1\) year as \(12\) months: \(14\) years \(2\) months becomes \(13\) years \(14\) months. Then \(13-6=7\) years and \(14-11=3\) months. The result is \(7\) years \(3\) months. 3. c) Rewrite \(7\) years as \(6\) years \(12\) months. Then \(6-4=2\) years and \(12-9=3\) months. The result is \(2\) years \(3\) months.

a) \(6\,\text{years}\ 3\,\text{months}\) b) \(7\,\text{years}\ 3\,\text{months}\) c) \(2\,\text{years}\ 3\,\text{months}\)

- Find each age difference by subtraction. - Remember that \(1\) year equals \(12\) months. - Regroup one year as \(12\) months when the top number of months is too small.

1. For Lucas and Mia, regroup Lucas’s age as \(10\) years \(15\) months. Then \(10\,\text{years}\,15\,\text{months} - 8\,\text{years}\,7\,\text{months} = 2\,\text{years}\,8\,\text{months}\). 2. For Mia and Ben, regroup Mia’s age as \(7\) years \(19\) months. Then \(7\,\text{years}\,19\,\text{months} - 5\,\text{years}\,10\,\text{months} = 2\,\text{years}\,9\,\text{months}\).

a) Lucas is \(2\) years \(8\) months older than Mia. b) Mia is \(2\) years \(9\) months older than Ben.

- Calculate each side separately. - Regroup any total of at least \(60\) seconds. - Compare the minutes first, then the seconds.

1. Calculate \(A\): \(9+4=13\) minutes, and \(35+40=75\) seconds \(=1\) minute \(15\) seconds. Therefore, \(A=14\,\text{min}\ 15\,\text{s}\). 2. Calculate \(B\): \(6+7=13\) minutes, and \(55+15=70\) seconds \(=1\) minute \(10\) seconds. Therefore, \(B=14\,\text{min}\ 10\,\text{s}\). 3. Since \(14\,\text{min}\ 15\,\text{s}>14\,\text{min}\ 10\,\text{s}\), \(A>B\).

\(A>B\), because \(14\,\text{min}\ 15\,\text{s}>14\,\text{min}\ 10\,\text{s}\).

- What can you regroup when the smaller-unit amount is not large enough to subtract? - How many minutes are in \(1\) hour? - How many hours are in \(1\) day? - How many seconds are in \(1\) minute? - Subtract like units after regrouping.

1. a) Regroup \(14\,\text{hr}\ 20\,\text{min}\) as \(13\,\text{hr}\ 80\,\text{min}\). Then subtract to get \(7\,\text{hr}\ 35\,\text{min}\). 2. b) Regroup \(5\,\text{days}\ 6\,\text{hr}\) as \(4\,\text{days}\ 30\,\text{hr}\). Then subtract to get \(2\,\text{days}\ 15\,\text{hr}\). 3. c) Regroup \(8\,\text{min}\ 12\,\text{s}\) as \(7\,\text{min}\ 72\,\text{s}\). Then subtract to get \(4\,\text{min}\ 32\,\text{s}\).

a) \(7\,\text{hr}\ 35\,\text{min}\) b) \(2\,\text{days}\ 15\,\text{hr}\) c) \(4\,\text{min}\ 32\,\text{s}\)

- Can an inverse operation help you find the missing value? - Decide whether to add or subtract to fill each blank. - Remember that \(1\) day is \(24\) hours and \(1\) minute is \(60\) seconds.

1. a) Find the missing subtrahend by subtracting: \(10\,\text{hr}\ 15\,\text{min}-4\,\text{hr}\ 50\,\text{min}\). Regroup as \(9\,\text{hr}\ 75\,\text{min}-4\,\text{hr}\ 50\,\text{min}=5\,\text{hr}\ 25\,\text{min}\). 2. b) Find the missing minuend by adding: \(3\,\text{days}\ 12\,\text{hr}+1\,\text{day}\ 18\,\text{hr}=4\,\text{days}\ 30\,\text{hr}=5\,\text{days}\ 6\,\text{hr}\). 3. c) Find the missing subtrahend by subtracting: \(20\,\text{min}-12\,\text{min}\ 25\,\text{s}\). Regroup as \(19\,\text{min}\ 60\,\text{s}-12\,\text{min}\ 25\,\text{s}=7\,\text{min}\ 35\,\text{s}\).

a) \(5\,\text{hr}\ 25\,\text{min}\) b) \(5\,\text{days}\ 6\,\text{hr}\) c) \(7\,\text{min}\ 35\,\text{s}\)

- Use an inverse operation to fill each blank. - Count up to the next full hour when that helps. - One hour is \(60\) minutes.

1. a) Subtract to find the missing addend: \(6\,\text{hr}\ 10\,\text{min}-3\,\text{hr}\ 40\,\text{min}=5\,\text{hr}\ 70\,\text{min}-3\,\text{hr}\ 40\,\text{min}=2\,\text{hr}\ 30\,\text{min}\). 2. b) Subtract the difference from the minuend: \(7\,\text{hr}\ 25\,\text{min}-4\,\text{hr}\ 50\,\text{min}=6\,\text{hr}\ 85\,\text{min}-4\,\text{hr}\ 50\,\text{min}=2\,\text{hr}\ 35\,\text{min}\). 3. c) Subtract: \(11\,\text{hr}-8\,\text{hr}\ 12\,\text{min}=10\,\text{hr}\ 60\,\text{min}-8\,\text{hr}\ 12\,\text{min}=2\,\text{hr}\ 48\,\text{min}\). 4. d) Subtract to find the missing addend: \(5\,\text{hr}\ 20\,\text{min}-2\,\text{hr}\ 55\,\text{min}=4\,\text{hr}\ 80\,\text{min}-2\,\text{hr}\ 55\,\text{min}=2\,\text{hr}\ 25\,\text{min}\).

a) \(2\,\text{hr}\ 30\,\text{min}\) b) \(2\,\text{hr}\ 35\,\text{min}\) c) \(2\,\text{hr}\ 48\,\text{min}\) d) \(2\,\text{hr}\ 25\,\text{min}\)

- How many seconds are in \(1\) minute? - If the top number of seconds is smaller, regroup \(1\) minute as \(60\) seconds. - Subtract the minutes and seconds after regrouping.

1. Regroup \(8\,\text{min}\ 15\,\text{s}\) as \(7\,\text{min}\ 75\,\text{s}\). Then \(7\,\text{min}\ 75\,\text{s}-3\,\text{min}\ 45\,\text{s}=4\,\text{min}\ 30\,\text{s}\). 2. Regroup \(20\,\text{min}\ 5\,\text{s}\) as \(19\,\text{min}\ 65\,\text{s}\). Then \(19\,\text{min}\ 65\,\text{s}-12\,\text{min}\ 18\,\text{s}=7\,\text{min}\ 47\,\text{s}\). 3. Regroup \(5\,\text{min}\) as \(4\,\text{min}\ 60\,\text{s}\). Then \(4\,\text{min}\ 60\,\text{s}-3\,\text{min}\ 52\,\text{s}=1\,\text{min}\ 8\,\text{s}\).

1. \(4\,\text{min}\ 30\,\text{s}\) 2. \(7\,\text{min}\ 47\,\text{s}\) 3. \(1\,\text{min}\ 8\,\text{s}\)

- Calculate the subtraction on the left side first. - Convert both time amounts to the same unit before comparing. - How many seconds are in \(5\) or \(6\) minutes?

1. a) Regroup and subtract: \(3\,\text{min}\ 70\,\text{s}-1\,\text{min}\ 50\,\text{s}=2\,\text{min}\ 20\,\text{s}\). Therefore, the two sides are equal. 2. b) The left side is \(9\,\text{min}\ 60\,\text{s}-4\,\text{min}\ 15\,\text{s}=5\,\text{min}\ 45\,\text{s}\). 3. Convert the right side: \(350\,\text{s}=5\,\text{min}\ 50\,\text{s}\), because \(5 \times 60=300\). 4. Since \(5\,\text{min}\ 45\,\text{s}<5\,\text{min}\ 50\,\text{s}\), the left side is less.

a) \(=\) b) \(<\)

- How many hours are in \(1\) day? - Convert to smaller units when that makes subtraction easier. - You can count up from the shorter time to the longer time as a check.

1. a) Add \(18+9=27\) hours. Since \(27\) hours is \(1\) day \(3\) hours, add \(4+2+1=7\) days. The result is \(7\,\text{days}\ 3\,\text{hr}\). 2. b) Convert \(1\) hour \(5\) minutes to \(65\) minutes. Regroup as \(64\,\text{min}\ 60\,\text{s}\). Then subtract \(35\,\text{min}\ 20\,\text{s}\) to get \(29\,\text{min}\ 40\,\text{s}\).

a) \(7\,\text{days}\ 3\,\text{hr}\) b) \(29\,\text{min}\ 40\,\text{s}\)

- For part a, find the difference in grams for one loaf. - Multiply each per-loaf difference by \(100\). - Convert the final amounts from grams to kilograms.

1. Find the flour saved per loaf: \(450\,\text{g} - 380\,\text{g} = 70\,\text{g}\). 2. Find the savings for \(100\) loaves: \(70\,\text{g} \times 100 = 7000\,\text{g}\). 3. Convert the savings: \(7000\,\text{g} = 7\,\text{kg}\). 4. Find the difference per loaf for part b: \(380\,\text{g} - 200\,\text{g} = 180\,\text{g}\). 5. Find the difference for \(100\) loaves: \(180\,\text{g} \times 100 = 18{,}000\,\text{g}\). 6. Convert the difference: \(18{,}000\,\text{g} = 18\,\text{kg}\).

a) The bakery saves \(7\,\text{kg}\) of flour. b) The new recipe uses \(18\,\text{kg}\) more flour than the small-loaf recipe.

- One meter equals \(100\) centimeters. - Find each unit fraction of \(100\) separately. - Compare the three centimeter measurements.

1. One meter equals \(100\) centimeters. 2. Lucas cuts \(100\,\text{cm} \div 2 = 50\,\text{cm}\). 3. Sarah cuts \(100\,\text{cm} \div 4 = 25\,\text{cm}\). 4. Tyler cuts \(100\,\text{cm} \div 5 = 20\,\text{cm}\). 5. Since \(20 < 25 < 50\), Tyler cuts the shortest piece.

Lucas: \(50\,\text{cm}\) Sarah: \(25\,\text{cm}\) Tyler: \(20\,\text{cm}\) Tyler cuts the shortest piece.

- Convert the full route to meters. - How many times does Leonie walk \(75\,\text{m}\) in \(8\) minutes? - Subtract the distance already walked from the full route.

1. Find the approximate distance walked in \(8\) minutes: \(8 \times 75 \approx 600\,\text{m}\). 2. Convert the full route to meters: \(1\,\text{km} = 1000\,\text{m}\). 3. Find the approximate remaining distance: \(1000 - 600 \approx 400\,\text{m}\).

a) Leonie walks about \(600\,\text{m}\). b) She still needs to walk about \(400\,\text{m}\).

- How many seconds are in one minute? - First find the distance in centimeters for one minute. - How many centimeters are in one meter? - Use the one-minute distance to find the five-minute distance.

1. One minute has \(60\) seconds. 2. Find the distance in one minute: \(60 \times 80 = 4800\,\text{cm}\). 3. Convert to meters: \(4800\,\text{cm} = 48\,\text{m}\). 4. Find the distance in \(5\) minutes: \(5 \times 48 = 240\,\text{m}\).

a) The hiker travels \(48\,\text{m}\) in \(1\) minute. b) The hiker travels \(240\,\text{m}\) in \(5\) minutes.

- Find the combined mass of the books. - Find the combined mass of the board games. - Convert the box's limit to grams. - Subtract the current mass from the limit.

1. Find the total mass of the books: \(3 \times 450\,\text{g} = 1350\,\text{g}\). 2. Find the total mass of the board games: \(2 \times 850\,\text{g} = 1700\,\text{g}\). 3. Find the current mass: \(1350\,\text{g} + 1700\,\text{g} = 3050\,\text{g}\). 4. Convert the limit: \(5\,\text{kg} = 5000\,\text{g}\). 5. Find the remaining mass allowed: \(5000\,\text{g} - 3050\,\text{g} = 1950\,\text{g}\).

The remaining contents may have a mass of at most \(1950\,\text{g}\).

- Think about the length of one large step. - Use \(1\,\text{yd}=3\,\text{ft}\). - Compare the measured total with the floor-plan length.

1. A reasonable step length for a child is about \(1.5\) to \(2\,\text{ft}\). At \(2\,\text{ft}\) per step, \(30\div 2=15\) steps. At \(1.5\,\text{ft}\) per step, \(30\div 1.5=20\) steps. An estimate of about \(15\) to \(20\) steps is reasonable. 2. One yard is \(3\,\text{ft}\), so \(6\) yardstick lengths measure \(6\times 3\,\text{ft}=18\,\text{ft}\). 3. Since \(18\,\text{ft}\) is much less than \(30\,\text{ft}\), the student's result is not reasonable.

a) About \(15\) to \(20\) steps, depending on the assumed step length. b) No. Six yardstick lengths equal \(18\,\text{ft}\), not \(30\,\text{ft}\).

- Divide the total volume by the capacity of each larger container. - Convert \(1\,\text{L}\) to milliliters for the cups. - Use multiplication to determine how many \(250\)-milliliter cups make \(1\) liter, then multiply by \(60\).

1. For part a, divide: \(60\,\text{L} \div 2\,\text{L} = 30\) pitchers. 2. For part b, divide: \(60\,\text{L} \div 5\,\text{L} = 12\) containers. 3. For part c, convert \(1\,\text{L} = 1000\,\text{mL}\). Four cups hold \(4 \times 250\,\text{mL} = 1000\,\text{mL}\), so there are \(4\) cups per liter. 4. Scale to \(60\,\text{L}\): \(60 \times 4 = 240\) cups.

a) \(30\) pitchers b) \(12\) containers c) \(4\) cups per liter and \(240\) cups from \(60\,\text{L}\)

- One kilometer equals \(1000\) meters. - Use each denominator to find the corresponding unit fraction of a kilometer in meters. - Subtract the shorter length from the longer length.

1. One kilometer equals \(1000\) meters. 2. Half a kilometer is \(1000\,\text{m} \div 2 = 500\,\text{m}\). 3. One-fifth of a kilometer is \(1000\,\text{m} \div 5 = 200\,\text{m}\). 4. Since \(500 > 200\), the \(\frac{1}{2}\)-kilometer section is longer. The difference is \(500\,\text{m} - 200\,\text{m} = 300\,\text{m}\).

a) \(\frac{1}{2}\,\text{km} = 500\,\text{m}\), and \(\frac{1}{5}\,\text{km} = 200\,\text{m}\). b) The \(\frac{1}{2}\)-kilometer section is longer by \(300\,\text{m}\).

- Write every package mass in the same unit. - Add the packages in each delivery separately. - Compare the two totals.

1. Convert Delivery A's masses to grams and add: \(850\,\text{g} + 1400\,\text{g} + 750\,\text{g} = 3000\,\text{g}\). 2. Convert Delivery B's masses to grams and add: \(1900\,\text{g} + 1100\,\text{g} = 3000\,\text{g}\). 3. Since both totals are \(3000\,\text{g} = 3\,\text{kg}\), the deliveries have the same mass.

The two deliveries have the same mass. Each has a mass of \(3\,\text{kg}\).

- Write the length planted on each day. - Each day after the first is longer than the previous day by the same amount. - You may convert all lengths to centimeters before adding.

1. Find the second day's length: \(12\,\text{m}\,40\,\text{cm} + 1\,\text{m}\,50\,\text{cm} = 13\,\text{m}\,90\,\text{cm}\). 2. Find the third day's length: \(13\,\text{m}\,90\,\text{cm} + 1\,\text{m}\,50\,\text{cm} = 15\,\text{m}\,40\,\text{cm}\). 3. Add all three days: \(12\,\text{m}\,40\,\text{cm} + 13\,\text{m}\,90\,\text{cm} + 15\,\text{m}\,40\,\text{cm} = 41\,\text{m}\,70\,\text{cm}\).

The gardener plants \(41\,\text{m}\,70\,\text{cm}\) of hedge altogether.

- First find the combined length of the North and South sections. - Remember that \(1000\,\text{m} = 1\,\text{km}\). - Then add the amount by which the West section is longer.

1. Add the North and South sections: \(5\,\text{km}\,350\,\text{m} + 4\,\text{km}\,800\,\text{m} = 10\,\text{km}\,150\,\text{m}\). 2. Add the extra length: \(10\,\text{km}\,150\,\text{m} + 1\,\text{km}\,150\,\text{m} = 11\,\text{km}\,300\,\text{m}\).

The West section is \(11\,\text{km}\,300\,\text{m}\) long.

- Add the kilometers and meters separately. - How many meters are in \(1\,\text{km}\)? - Check whether the meter total can be regrouped as kilometers and meters.

1. Add the kilometers: \(26+41+13=80\,\text{km}\). 2. Add the meters: \(380+920+705=2005\,\text{m}\). 3. Regroup \(2005\,\text{m}=2\,\text{km}\ 5\,\text{m}\). 4. Add the results: \(80\,\text{km}+2\,\text{km}\ 5\,\text{m}=82\,\text{km}\ 5\,\text{m}\).

\(82\,\text{km}\ 5\,\text{m}\)

- Add the kilogram parts and gram parts separately. - Regroup every \(1000\,\text{g}\) as \(1\,\text{kg}\). - Combine the regrouped amounts.

1. Add the kilograms: \(3\,\text{kg} + 2\,\text{kg} + 4\,\text{kg} = 9\,\text{kg}\). 2. Add the grams: \(450\,\text{g} + 800\,\text{g} + 50\,\text{g} = 1300\,\text{g}\). 3. Regroup \(1300\,\text{g}\) as \(1\,\text{kg}\,300\,\text{g}\). 4. Combine the amounts: \(9\,\text{kg} + 1\,\text{kg}\,300\,\text{g} = 10\,\text{kg}\,300\,\text{g}\).

The baker measures \(10\,\text{kg}\,300\,\text{g}\) of flour altogether.

- Use \(1\,\text{L}=1000\,\text{mL}\), \(1\,\text{m}=100\,\text{cm}\), and \(1\,\text{cm}=10\,\text{mm}\). - Regroup one larger unit when the smaller-unit amount is not large enough to subtract. - You can check by converting everything to the smallest unit first.

1. a) Regroup \(10\,\text{L}\ 200\,\text{mL}\) as \(9\,\text{L}\ 1200\,\text{mL}\). Then subtract to get \(5\,\text{L}\ 350\,\text{mL}\). 2. b) Regroup \(6\,\text{m}\ 12\,\text{cm}\) as \(5\,\text{m}\ 112\,\text{cm}\). Then subtract to get \(3\,\text{m}\ 67\,\text{cm}\). 3. c) Regroup \(14\,\text{cm}\ 3\,\text{mm}\) as \(13\,\text{cm}\ 13\,\text{mm}\). Then subtract to get \(5\,\text{cm}\ 4\,\text{mm}\).

a) \(5\,\text{L}\ 350\,\text{mL}\) b) \(3\,\text{m}\ 67\,\text{cm}\) c) \(5\,\text{cm}\ 4\,\text{mm}\)

- Calculate the expression on the left before comparing. - Regroup or convert to the smaller unit when needed. - Estimate whether each difference is a little more or less than a whole pound or ton. - Compare both sides using the same unit.

1. a) Convert to ounces: \(197\,\text{oz}-71\,\text{oz}=126\,\text{oz}=7\,\text{lb}\ 14\,\text{oz}\). Since this is less than \(8\,\text{lb}\), use \(<\). 2. b) Convert to pounds: \(10{,}020\,\text{lb}-2030\,\text{lb}=7990\,\text{lb}\). Also, \(3\,\text{tons}\ 1990\,\text{lb}=7990\,\text{lb}\), so use \(=\). 3. c) Convert to ounces: \(132\,\text{oz}-56\,\text{oz}=76\,\text{oz}=4\,\text{lb}\ 12\,\text{oz}\). Since \(4\,\text{lb}\ 12\,\text{oz}>4\,\text{lb}\ 8\,\text{oz}\), use \(>\).

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

- Convert everything to the smallest unit shown. - How many centimeters are in \(1\,\text{m}\)? - How many millimeters are in \(1\,\text{cm}\)? - Convert the final difference back to mixed units.

1. a) \(500\,\text{cm}-72\,\text{cm}=428\,\text{cm}=4\,\text{m}\ 28\,\text{cm}\). 2. b) \(1215\,\text{cm}-840\,\text{cm}=375\,\text{cm}=3\,\text{m}\ 75\,\text{cm}\). 3. c) \(32\,\text{mm}-18\,\text{mm}=14\,\text{mm}=1\,\text{cm}\ 4\,\text{mm}\). 4. d) \(10{,}000\,\text{cm}-4505\,\text{cm}=5495\,\text{cm}=54\,\text{m}\ 95\,\text{cm}\).

a) \(4\,\text{m}\ 28\,\text{cm}\) b) \(3\,\text{m}\ 75\,\text{cm}\) c) \(1\,\text{cm}\ 4\,\text{mm}\) d) \(54\,\text{m}\ 95\,\text{cm}\)

- How many ounces are in \(1\,\text{lb}\)? - How many pounds are in \(1\,\text{ton}\)? - Pay close attention to place value in mixed-unit weights. - Convert both weights to the same smaller unit before subtracting.

1. a) \(16\,\text{oz}-6\,\text{oz}=10\,\text{oz}\). 2. b) \(8000\,\text{lb}-2500\,\text{lb}=5500\,\text{lb}=2\,\text{tons}\ 1500\,\text{lb}\). 3. c) \(165\,\text{oz}-56\,\text{oz}=109\,\text{oz}=6\,\text{lb}\ 13\,\text{oz}\). 4. d) \(5000\,\text{lb}-1600\,\text{lb}=3400\,\text{lb}=1\,\text{ton}\ 1400\,\text{lb}\).

a) \(10\,\text{oz}\) b) \(2\,\text{tons}\ 1500\,\text{lb}\) c) \(6\,\text{lb}\ 13\,\text{oz}\) d) \(1\,\text{ton}\ 1400\,\text{lb}\)

- Recall how many inches are in a foot and how many feet are in a yard. - Regroup when a smaller-unit amount is not large enough to subtract. - Converting each full length to inches can simplify the subtraction.

1. a) Convert to inches: \(7\,\text{yd}\ 1\,\text{ft}\ 2\,\text{in.}=266\,\text{in.}\), and \(3\,\text{yd}\ 2\,\text{ft}\ 5\,\text{in.}=137\,\text{in.}\). Then \(266-137=129\,\text{in.}=3\,\text{yd}\ 1\,\text{ft}\ 9\,\text{in.}\). 2. b) Convert to inches: \(15\,\text{yd}\ 1\,\text{ft}\ 6\,\text{in.}=558\,\text{in.}\), and \(6\,\text{yd}\ 2\,\text{ft}\ 9\,\text{in.}=249\,\text{in.}\). Then \(558-249=309\,\text{in.}=8\,\text{yd}\ 1\,\text{ft}\ 9\,\text{in.}\).

a) \(3\,\text{yd}\ 1\,\text{ft}\ 9\,\text{in.}\) b) \(8\,\text{yd}\ 1\,\text{ft}\ 9\,\text{in.}\)

- Use \(1\,\text{lb}=16\,\text{oz}\) and \(1\,\text{gal}=4\,\text{qt}\). - For part c), convert both lengths to inches before subtracting.

1. a) Convert to ounces: \(197\,\text{oz}-134\,\text{oz}=63\,\text{oz}=3\,\text{lb}\ 15\,\text{oz}\). 2. b) Regroup \(14\,\text{gal}\ 2\,\text{qt}\) as \(13\,\text{gal}\ 6\,\text{qt}\). Then subtract to get \(8\,\text{gal}\ 3\,\text{qt}\). 3. c) Convert to inches: \(20\,\text{yd}=720\,\text{in.}\), and \(12\,\text{yd}\ 1\,\text{ft}\ 8\,\text{in.}=452\,\text{in.}\). Then \(720-452=268\,\text{in.}=7\,\text{yd}\ 1\,\text{ft}\ 4\,\text{in.}\).

a) \(3\,\text{lb}\ 15\,\text{oz}\) b) \(8\,\text{gal}\ 3\,\text{qt}\) c) \(7\,\text{yd}\ 1\,\text{ft}\ 4\,\text{in.}\)

- What must be subtracted from the first measurement to reach the result? - Subtract the result from the starting measurement. - Use the appropriate conversion factor before subtracting. - Think of each equation as an inverse-operation problem.

1. a) Convert to meters and subtract: \(5000-4250=750\). The missing length is \(750\,\text{m}\). 2. b) Convert to ounces: \(197-145=52\,\text{oz}=3\,\text{lb}\ 4\,\text{oz}\). 3. c) Convert to centimeters: \(300-185=115\,\text{cm}=1\,\text{m}\ 15\,\text{cm}\). 4. d) Convert to pounds: \(20{,}000-14{,}800=5200\,\text{lb}=2\,\text{tons}\ 1200\,\text{lb}\).

a) \(750\,\text{m}\) b) \(3\,\text{lb}\ 4\,\text{oz}\) c) \(1\,\text{m}\ 15\,\text{cm}\) d) \(2\,\text{tons}\ 1200\,\text{lb}\)

- One ton is \(2000\,\text{lb}\). - You may convert everything to pounds and convert back at the end. - When subtracting, regroup one ton as \(2000\) pounds if needed. - Calculate tons and pounds separately while tracking regrouping.

1. a) Add \(14+5=19\) tons and \(640+1780=2420\) pounds. Since \(2420\,\text{lb}=1\,\text{ton}\ 420\,\text{lb}\), the result is \(20\,\text{tons}\ 420\,\text{lb}\). 2. b) Regroup \(25\,\text{tons}\ 300\,\text{lb}\) as \(24\,\text{tons}\ 2300\,\text{lb}\). Subtract to get \(12\,\text{tons}\ 1500\,\text{lb}\). 3. c) Regroup \(8\,\text{tons}\ 100\,\text{lb}\) as \(7\,\text{tons}\ 2100\,\text{lb}\). Subtract to get \(4\,\text{tons}\ 600\,\text{lb}\).

a) \(20\,\text{tons}\ 420\,\text{lb}\) b) \(12\,\text{tons}\ 1500\,\text{lb}\) c) \(4\,\text{tons}\ 600\,\text{lb}\)

- How many pounds are in \(1\) ton? - Use inverse operations to solve missing-value equations. - Convert all values to pounds when that makes the calculation easier. - Each full group of \(2000\) pounds is one ton.

1. a) \(3\,\text{tons}=6000\,\text{lb}\), and \(2\,\text{tons}\ 800\,\text{lb}=4800\,\text{lb}\). The difference is \(6000-4800=1200\,\text{lb}\). 2. b) \(6 \times 600\,\text{lb}=3600\,\text{lb}=1\,\text{ton}\ 1600\,\text{lb}\). 3. c) \(10\,\text{tons}=20{,}000\,\text{lb}\), and \(7\,\text{tons}\ 500\,\text{lb}=14{,}500\,\text{lb}\). The difference is \(5500\,\text{lb}=2\,\text{tons}\ 1500\,\text{lb}\).

a) \(1200\,\text{lb}\) b) \(1\,\text{ton}\ 1600\,\text{lb}\) c) \(2\,\text{tons}\ 1500\,\text{lb}\)

- Identify the two parts of the post that are not visible. - Convert the full length to centimeters. - Subtract the hidden parts from the total.

1. Convert the full length: \(4\,\text{m} = 400\,\text{cm}\). 2. Add the hidden parts: \(85\,\text{cm} + 12\,\text{cm} = 97\,\text{cm}\). 3. Subtract the hidden length: \(400\,\text{cm} - 97\,\text{cm} = 303\,\text{cm}\). 4. Convert: \(303\,\text{cm} = 3\,\text{m}\,3\,\text{cm}\).

The visible part is \(3\,\text{m}\,3\,\text{cm}\) long.

- Convert both masses to grams, or regroup kilograms and grams when needed. - “Sum” means add, and “difference” means subtract. - Complete the three operations in the order stated.

1. Find the sum: \(12\,\text{kg}\,450\,\text{g} + 8\,\text{kg}\,700\,\text{g} = 21\,\text{kg}\,150\,\text{g}\). 2. Find the difference: \(12\,\text{kg}\,450\,\text{g} - 8\,\text{kg}\,700\,\text{g} = 3\,\text{kg}\,750\,\text{g}\). 3. Subtract the difference from the sum: \(21\,\text{kg}\,150\,\text{g} - 3\,\text{kg}\,750\,\text{g} = 17\,\text{kg}\,400\,\text{g}\).

The result is \(17\,\text{kg}\,400\,\text{g}\).

- Find the afternoon distance first. - The afternoon distance is shorter, so subtract. - Add the morning and afternoon distances.

1. Find the afternoon distance: \(15\,\text{km}\,650\,\text{m} - 4\,\text{km}\,800\,\text{m} = 10\,\text{km}\,850\,\text{m}\). 2. Add the two distances: \(15\,\text{km}\,650\,\text{m} + 10\,\text{km}\,850\,\text{m} = 26\,\text{km}\,500\,\text{m}\).

The hiker walks \(26\,\text{km}\,500\,\text{m}\) altogether.

- Find the mass of the second load first. - Regroup \(1000\,\text{kg}\) as \(1\) metric ton when needed. - Add the masses of both loads.

1. Find the second load's mass: \(3\) metric tons \(450\,\text{kg} + 1\) metric ton \(600\,\text{kg} = 5\) metric tons \(50\,\text{kg}\). 2. Add both loads: \(3\) metric tons \(450\,\text{kg} + 5\) metric tons \(50\,\text{kg} = 8\) metric tons \(500\,\text{kg}\).

The two loads have a combined mass of \(8\) metric tons \(500\,\text{kg}\).

- Find the dimensions after placing the squares side by side. - Convert all measurements to millimeters first. - Use the rectangle perimeter formula. - Convert the final result back to mixed units.

1. Convert the side length: \(9\,\text{cm}\,4\,\text{mm}=94\,\text{mm}\). 2. The rectangle is \(188\,\text{mm}\) long and \(94\,\text{mm}\) wide. 3. Its perimeter is \(2 \times (188+94)=564\,\text{mm}\). 4. Convert back: \(564\,\text{mm}=56\,\text{cm}\,4\,\text{mm}\).

\(564\,\text{mm}=56\,\text{cm}\,4\,\text{mm}\)

- Convert the total load to kilograms. - Find the mass of the pears. - Add the apples and pears, then subtract their mass from the total.

1. Convert the total load: \(3\) metric tons \(= 3000\,\text{kg}\). 2. Find the mass of the pears: \(1250\,\text{kg} + 420\,\text{kg} = 1670\,\text{kg}\). 3. Find the combined mass of the apples and pears: \(1250\,\text{kg} + 1670\,\text{kg} = 2920\,\text{kg}\). 4. Find the mass of the plums: \(3000\,\text{kg} - 2920\,\text{kg} = 80\,\text{kg}\).

The truck carries \(80\,\text{kg}\) of plums.

- Multiply the number of bags by the weight of each bag. - Convert the trailer capacity to pounds before comparing. - Cargo capacity is the greatest load the trailer may carry.

1. Find the total weight of the potatoes: \(120\times 50\,\text{lb}=6000\,\text{lb}\). 2. Convert the trailer capacity to pounds: \(1.5\times 2000\,\text{lb}=3000\,\text{lb}\). 3. Since \(6000\,\text{lb}>3000\,\text{lb}\), the potatoes weigh twice as much as the trailer can carry. The farmer's statement is not reasonable.

The statement is not reasonable. The potatoes weigh \(6000\,\text{lb}\), or \(3\,\text{tons}\), but the trailer can carry only \(3000\,\text{lb}\), or \(1.5\,\text{tons}\).

- Convert \(1\,\text{km}\) to meters. - Find each student's total distance. - Subtract each distance from \(1000\,\text{m}\) and compare the results.

1. Convert the goal: \(1\,\text{km} = 1000\,\text{m}\). 2. Lucas runs \(4 \times 215\,\text{m} = 860\,\text{m}\), so he needs \(1000\,\text{m} - 860\,\text{m} = 140\,\text{m}\) more. 3. Sarah runs \(6 \times 145\,\text{m} = 870\,\text{m}\), so she needs \(1000\,\text{m} - 870\,\text{m} = 130\,\text{m}\) more. 4. Since \(130\,\text{m} < 140\,\text{m}\), Sarah is closer to \(1\,\text{km}\).

Lucas needs \(140\,\text{m}\) more, and Sarah needs \(130\,\text{m}\) more. Sarah is closer to \(1\,\text{km}\).

- Convert \(1\,\text{kg}\) to grams. - Find the amount in each group of packages. - Add the available flour and compare it with the amount needed. - Subtract to find any shortage.

1. Convert the amount needed: \(1\,\text{kg} = 1000\,\text{g}\). 2. Find the flour in the first group: \(3 \times 185\,\text{g} = 555\,\text{g}\). 3. Find the flour in the second group: \(2 \times 210\,\text{g} = 420\,\text{g}\). 4. Find the total available: \(555\,\text{g} + 420\,\text{g} = 975\,\text{g}\). 5. Since \(975\,\text{g} < 1000\,\text{g}\), there is not enough flour. 6. Find the shortage: \(1000\,\text{g} - 975\,\text{g} = 25\,\text{g}\).

No. The baker needs \(25\,\text{g}\) more flour.

- Which two smaller-unit amounts make exactly \(1000\,\text{g}\) or \(100\,\text{cm}\)? - Add the smaller units first and regroup one whole larger unit when possible. - Sorting or scanning the smaller-unit amounts can help you find useful pairs.

1. a) Pair \(16\,\text{kg}\ 350\,\text{g}\) with \(13\,\text{kg}\ 650\,\text{g}\): \(29\,\text{kg}\ 1000\,\text{g}=30\,\text{kg}\). Pair \(24\,\text{kg}\ 120\,\text{g}\) with \(15\,\text{kg}\ 880\,\text{g}\): \(39\,\text{kg}\ 1000\,\text{g}=40\,\text{kg}\). Then \(30\,\text{kg}+40\,\text{kg}=70\,\text{kg}\). 2. b) Pair \(7\,\text{m}\ 42\,\text{cm}\) with \(12\,\text{m}\ 58\,\text{cm}\): \(19\,\text{m}\ 100\,\text{cm}=20\,\text{m}\). Pair \(12\,\text{m}\ 15\,\text{cm}\) with \(7\,\text{m}\ 85\,\text{cm}\): \(19\,\text{m}\ 100\,\text{cm}=20\,\text{m}\). Then \(20\,\text{m}+20\,\text{m}=40\,\text{m}\).

a) \(70\,\text{kg}\) b) \(40\,\text{m}\)

- How many milliliters make \(1\) liter? - How many minutes make \(1\) hour? - Look for pairs that add to \(1000\) milliliters or \(60\) minutes.

1. a) Pair \(14\,\text{L}\ 250\,\text{mL}\) with \(5\,\text{L}\ 750\,\text{mL}\): \(19\,\text{L}\ 1000\,\text{mL}=20\,\text{L}\). Pair \(22\,\text{L}\ 680\,\text{mL}\) with \(17\,\text{L}\ 320\,\text{mL}\): \(39\,\text{L}\ 1000\,\text{mL}=40\,\text{L}\). Then \(20\,\text{L}+40\,\text{L}=60\,\text{L}\). 2. b) Pair \(1\,\text{hr}\ 15\,\text{min}\) with \(3\,\text{hr}\ 45\,\text{min}\): \(4\,\text{hr}\ 60\,\text{min}=5\,\text{hr}\). Pair \(2\,\text{hr}\ 50\,\text{min}\) with \(1\,\text{hr}\ 10\,\text{min}\): \(3\,\text{hr}\ 60\,\text{min}=4\,\text{hr}\). Then \(5\,\text{hr}+4\,\text{hr}=9\,\text{hr}\).

a) \(60\,\text{L}\) b) \(9\,\text{hr}\)

- Subtract the empty backpack's mass from the total to find the contents' mass. - Compare the contents' mass with the empty backpack's mass. - Convert between kilograms and grams as needed.

1. Convert the total mass: \(4\,\text{kg}\,150\,\text{g} = 4150\,\text{g}\). 2. Find the mass of the contents: \(4150\,\text{g} - 950\,\text{g} = 3200\,\text{g}\). 3. Compare the contents with the empty backpack: \(3200\,\text{g} - 950\,\text{g} = 2250\,\text{g}\). 4. Convert the difference: \(2250\,\text{g} = 2\,\text{kg}\,250\,\text{g}\).

The contents have a mass that is \(2\,\text{kg}\,250\,\text{g}\) greater than the mass of the empty backpack.

- Subtract the empty crate's mass to find the pears' mass. - Divide the pears' mass by \(2\). - Add the empty crate's mass back to the remaining pears.

1. Convert the filled mass to \(14{,}200\,\text{g}\) and the empty crate's mass to \(1350\,\text{g}\). Subtract the empty crate's mass. The pears have a mass of \(12{,}850\,\text{g}\). 2. Divide the pears' mass by \(2\). Half of the pears have a mass of \(6425\,\text{g}\). 3. Add the empty crate's mass: \(6425\,\text{g} + 1350\,\text{g} = 7775\,\text{g}\). 4. Convert the total: \(7775\,\text{g} = 7\,\text{kg}\,775\,\text{g}\).

The crate and the remaining pears have a total mass of \(7\,\text{kg}\,775\,\text{g}\).

- Convert all four times to seconds first. - Add the four values. - Use \(60\) seconds \(= 1\) minute to convert the total back to minutes and seconds.

1. Convert each time to seconds: \(58\), \(62\), \(65\), and \(57\) seconds. 2. Add the times: \(58 + 62 + 65 + 57 = 242\) seconds. 3. Convert \(242\) seconds to minutes and seconds: \(242 \div 60 = 4\) remainder \(2\). 4. The total time was \(4\) minutes \(2\) seconds.

The relay team''s total time was \(4\) minutes \(2\) seconds.

- Find the total time for all class periods. - Add the total break time. - Convert the total number of minutes to hours and minutes before adding it to the start time.

1. Find the total class time: \(5 \times 45 = 225\) minutes. 2. Find the total break time: \(2 \times 15 = 30\) minutes. 3. Add the times: \(225 + 30 = 255\) minutes. 4. Convert \(255\) minutes to hours and minutes: \(255 \div 60 = 4\) remainder \(15\), so the program lasts \(4\) hours \(15\) minutes. 5. Count forward from \(8{:}15\,\text{a.m.}\): \(8{:}15\,\text{a.m.} + 4\,\text{hours}\,15\,\text{minutes} = 12{:}30\,\text{p.m.}\).

The program ends at \(12{:}30\,\text{p.m.}\)

- Use \(1\,\text{kg} = 1000\,\text{g}\). - First evaluate the grouped addition shown in the right branch.

1. Convert the starting mass: \(1\,\text{kg} = 1000\,\text{g}\). 2. Add the masses removed: \(350\,\text{g} + 250\,\text{g} = 600\,\text{g}\). 3. Subtract from the starting mass: \(1000\,\text{g} - 600\,\text{g} = 400\,\text{g}\).

\(400\,\text{g}\) remain.

- Add all five side lengths. - Then convert the total from millimeters to centimeters. - Use \(10\,\text{mm}=1\,\text{cm}\).

1. Add the five side lengths: \(50+40+35+50+35=210\,\text{mm}\). 2. Since \(10\,\text{mm}=1\,\text{cm}\), convert by dividing by \(10\): \(210\div10=21\,\text{cm}\).

The perimeter is \(21\,\text{cm}\).

- Use each row of \(8\) dots as one group of bottles. - Determine the volume in one group before finding the total liters. - Determine how many \(250\)-milliliter cups make \(1\) liter, then scale to the total.

1. Use the groups of \(8\) bottles shown by the diagram. Eight bottles contain \(8 \times 125\,\text{mL} = 1000\,\text{mL}\), or \(1\,\text{L}\). 2. The \(32\) bottles make \(32 \div 8 = 4\) groups, so there are \(4\,\text{L}\) of juice altogether. 3. Four \(250\,\text{mL}\) cups hold \(1\,\text{L}\). Therefore, \(4\,\text{L}\) fills \(4 \times 4 = 16\) cups.

There are \(4\,\text{L}\) of juice, enough to fill \(16\) cups completely.

- First find the total attendance. - Track the units carefully when finding the total waste. - Convert grams to kilograms before finding the number of shifts.

1. a) Multiply the spectators per event by the number of events: \(2500\times4=10{,}000\). 2. b) Find the waste in grams: \(10{,}000\times20\,\text{g}=200{,}000\,\text{g}\). Convert to kilograms: \(200{,}000\,\text{g}=200\,\text{kg}\). 3. c) Divide the total kilograms by the amount removed per shift: \(200\div50=4\).

a) \(10{,}000\) spectators b) \(200\,\text{kg}\) c) \(4\) shifts

- How many grams are in \(3\,\text{kg}\)? - Find the weight of everything already packed. - Find how much weight remains before the bag reaches the limit. - Share the remaining amount equally between the two books.

1. Convert the maximum weight: \(3\,\text{kg} = 3000\,\text{g}\). 2. Find the current weight: \(1150\,\text{g} + 300\,\text{g} + 650\,\text{g} = 2100\,\text{g}\). 3. Find the weight available for both books: \(3000\,\text{g} - 2100\,\text{g} = 900\,\text{g}\). 4. Divide the available weight equally: \(900\,\text{g} \div 2 = 450\,\text{g}\).

Each book can weigh at most \(450\,\text{g}\).

- Convert both trip lengths to minutes before finding the difference. - How many minutes are in one hour? - For the departure time, work backward by the full hours and then the remaining minutes.

1. Convert the bus time to minutes: \(7 \times 60 + 15 = 435\) minutes. 2. Convert the train time to minutes: \(3 \times 60 + 55 = 235\) minutes. 3. Find the difference: \(435 - 235 = 200\) minutes. 4. Work backward \(7\) hours from \(4{:}30\,\text{p.m.}\) to get \(9{:}30\,\text{a.m.}\). 5. Subtract the remaining \(15\) minutes: \(9{:}30\,\text{a.m.} - 15\,\text{minutes} = 9{:}15\,\text{a.m.}\).

a) The train trip is \(200\) minutes shorter. b) The bus left at \(9{:}15\,\text{a.m.}\).

- How many cuts are needed to make \(4\) pieces? - Convert all lengths to millimeters before calculating. - Subtract the material lost in every cut before dividing the remaining length.

1. Cutting one post into \(4\) pieces requires \(3\) cuts. 2. Find the total wood lost: \(3 \times 4 = 12\,\text{mm}\). 3. Convert the original length: \(2\,\text{m} = 2000\,\text{mm}\). 4. Subtract the wood lost: \(2000 - 12 = 1988\,\text{mm}\). 5. Divide the remaining length into \(4\) equal pieces: \(1988 \div 4 = 497\,\text{mm}\).

Each finished piece is \(497\,\text{mm}\) long.

- Find the total length needed for the \(6\) pieces. - Find the total waste from all the cuts and convert it to centimeters. - Subtract both the pieces and the waste from the original ribbon length.

1. Find the total length of the six pieces: \(6 \times 80 = 480\,\text{cm}\). 2. Cutting off \(6\) separate pieces requires \(6\) cuts. 3. Find the total waste: \(6 \times 5 = 30\,\text{mm}\). 4. Convert the waste: \(30\,\text{mm} = 3\,\text{cm}\). 5. Find the total ribbon used: \(480 + 3 = 483\,\text{cm}\). 6. Convert the original length: \(5\,\text{m} = 500\,\text{cm}\). 7. Subtract: \(500 - 483 = 17\,\text{cm}\).

\(17\,\text{cm}\) of ribbon remain.

- First find the number of kernels on one square meter. - Scale that amount to the full \(5\,\text{m}^2\). - How many groups of \(1000\) kernels are in the total? - Convert the final weight from grams to kilograms.

1. Find the approximate number of kernels per square meter: \(400 \times 40 \approx 16{,}000\). 2. Find the approximate number of kernels in \(5\,\text{m}^2\): \(5 \times 16{,}000 \approx 80{,}000\). 3. Find the approximate number of \(1000\)-kernel groups: \(80{,}000 \div 1000 \approx 80\). 4. Find the approximate total weight: \(80 \times 50\,\text{g} \approx 4000\,\text{g}\). 5. Convert to kilograms: \(4000\,\text{g} = 4\,\text{kg}\), so the harvest is about \(4\,\text{kg}\).

The garden can produce about \(4\,\text{kg}\) of wheat.

- Find the number of kernels per square meter, then for the entire field. - Use the weight of \(1000\) kernels to find the total weight. - Divide the total weight by the amount in one bag.

1. Find the approximate number of kernels per square meter: \(400 \times 50 \approx 20{,}000\). 2. Find the approximate total number of kernels: \(10 \times 20{,}000 \approx 200{,}000\). 3. Find the approximate number of \(1000\)-kernel groups: \(200{,}000 \div 1000 \approx 200\). 4. Find the approximate harvest weight: \(200 \times 40\,\text{g} \approx 8000\,\text{g}\). 5. Estimate the number of bags: \(8000\,\text{g} \div 100\,\text{g} \approx 80\).

The harvest could fill about \(80\) bags.

- Convert all weights to ounces. - Subtract the empty crate's weight to find the apples' starting weight. - Subtract the amount already packed and compare the remainder with the fifth bag's target.

1. Convert the filled crate's weight: \(11\,\text{lb} = 176\,\text{oz}\). 2. Convert the empty crate's weight: \(1\,\text{lb}\,8\,\text{oz} = 24\,\text{oz}\). 3. Find the apples' starting weight: \(176\,\text{oz} - 24\,\text{oz} = 152\,\text{oz}\). 4. The first four bags use \(4 \times 32\,\text{oz} = 128\,\text{oz}\). 5. The remaining apples weigh \(152\,\text{oz} - 128\,\text{oz} = 24\,\text{oz}\), or \(1\,\text{lb}\,8\,\text{oz}\). 6. The fifth bag needs \(1\,\text{lb}\,12\,\text{oz} = 28\,\text{oz}\). Since \(24\,\text{oz} < 28\,\text{oz}\), there are not enough apples. 7. The shortage is \(28\,\text{oz} - 24\,\text{oz} = 4\,\text{oz}\).

No. Only \(1\,\text{lb}\,8\,\text{oz}\) remain, so the seller is \(4\,\text{oz}\) short.

- Convert the total length to meters. - Find half of the total length. - Find the second week's amount, then add both weeks before subtracting from the halfway point.

1. Convert the total length: \(2\,\text{km} = 2000\,\text{m}\). 2. Find half of the total length: \(2000 \div 2 = 1000\,\text{m}\). 3. Find the second week's distance: \(320 - 45 = 275\,\text{m}\). 4. Find the total installed after two weeks: \(320 + 275 = 595\,\text{m}\). 5. Find the amount still needed to reach halfway: \(1000 - 595 = 405\,\text{m}\).

The workers must install \(405\,\text{m}\) more to reach half of the total length.

- How many hours are in a full week? - Find the library’s open hours for one weekday and then for all five weekdays. - Add the Saturday hours. - Subtract the total open time from the total time in a week.

1. Find the total hours in a week: \(7 \times 24 = 168\) hours. 2. The library is open \(8\) hours each weekday, so the weekday total is \(5 \times 8 = 40\) hours. 3. On Saturday, the library is open \(5\) hours. 4. Find the total open time: \(40 + 5 = 45\) hours. 5. Subtract from the full week: \(168 - 45 = 123\) hours.

The library is closed for \(123\) hours during a full week.

- For each site, convert all measurements to kilograms. - Add the deliveries for each site. - Subtract each delivered total from that site's target. - Compare the two missing amounts.

1. For Oak Street, find the delivered mass: \(4650\,\text{kg} + 3800\,\text{kg} = 8450\,\text{kg}\). 2. Start with Oak Street's target of \(12{,}000\,\text{kg}\) and subtract the delivered mass of \(8450\,\text{kg}\). Oak Street still needs \(3550\,\text{kg}\). 3. For Robin Street, find the delivered mass: \(5200\,\text{kg} + 2950\,\text{kg} = 8150\,\text{kg}\). 4. Start with Robin Street's target of \(10{,}000\,\text{kg}\) and subtract the delivered mass of \(8150\,\text{kg}\). Robin Street still needs \(1850\,\text{kg}\). 5. Compare: \(3550\,\text{kg} > 1850\,\text{kg}\), so Oak Street still needs more sand.

The Oak Street site still needs the greater amount. Oak Street needs \(3550\,\text{kg}\), while Robin Street needs \(1850\,\text{kg}\).

- Subtract the age difference to find Bertha’s age. - Regroup one year as \(12\) months when subtracting. - Add the two ages for the second question. - Regroup any total of \(12\) or more months.

1. Convert one year to \(12\) months before subtracting: \(14\,\text{years}\,8\,\text{months} = 13\,\text{years}\,20\,\text{months}\). 2. Subtract the age difference: \(13\,\text{years}\,20\,\text{months} - 5\,\text{years}\,9\,\text{months} = 8\,\text{years}\,11\,\text{months}\). 3. Add the two ages: \(14\,\text{years}\,8\,\text{months} + 8\,\text{years}\,11\,\text{months} = 22\,\text{years}\,19\,\text{months}\). 4. Regroup \(19\) months as \(1\) year \(7\) months. The total is \(23\) years \(7\) months.

1. Bertha is \(8\) years \(11\) months old. 2. The sum of their ages is \(23\) years \(7\) months.

- Find the total mass of the seed potatoes used in both beds. - Find the harvest from the second bed by subtracting. - Add the harvests from both beds. - Subtract the seed-potato mass from the total harvested mass.

1. Find the total mass of seed potatoes: \(2 \times 12\,\text{kg}\,500\,\text{g} = 25\,\text{kg}\). 2. Find the harvest from the second bed: \(110\,\text{kg}\,400\,\text{g} - 15\,\text{kg}\,250\,\text{g} = 95\,\text{kg}\,150\,\text{g}\). 3. Find the total harvest: \(110\,\text{kg}\,400\,\text{g} + 95\,\text{kg}\,150\,\text{g} = 205\,\text{kg}\,550\,\text{g}\). 4. Compare the harvest with the seed potatoes: \(205\,\text{kg}\,550\,\text{g} - 25\,\text{kg} = 180\,\text{kg}\,550\,\text{g}\).

The total harvest is \(180\,\text{kg}\,550\,\text{g}\) greater than the mass of the seed potatoes.

- Find the total mass of the two wax blocks. - Find the mass of the candles made from the second block. - Add the masses of all the candles. - Subtract the candle mass from the original wax mass.

1. Find the total mass of the wax blocks: \(2 \times 5\,\text{kg}\,200\,\text{g} = 10\,\text{kg}\,400\,\text{g}\). 2. Find the mass of the candles from the second block: \(4\,\text{kg}\,850\,\text{g} - 420\,\text{g} = 4\,\text{kg}\,430\,\text{g}\). 3. Find the total mass of the candles: \(4\,\text{kg}\,850\,\text{g} + 4\,\text{kg}\,430\,\text{g} = 9\,\text{kg}\,280\,\text{g}\). 4. Find the waste: \(10\,\text{kg}\,400\,\text{g} - 9\,\text{kg}\,280\,\text{g} = 1\,\text{kg}\,120\,\text{g}\).

The total mass of the wax waste is \(1\,\text{kg}\,120\,\text{g}\).

- Convert weeks to days first. - One day equals \(24\) hours. - Regroup one day as \(24\) hours when subtracting. - Compare Beta with each longer mission separately.

1. Convert Mission Gamma to days: \(2 \times 7 + 3 = 17\) days. 2. Compare Alpha and Beta. Regroup Alpha as \(13\) days \(29\) hours: \(13\,\text{days}\,29\,\text{hours} - 9\,\text{days}\,18\,\text{hours} = 4\,\text{days}\,11\,\text{hours}\). 3. Compare Gamma and Beta. Regroup Gamma as \(16\) days \(24\) hours: \(16\,\text{days}\,24\,\text{hours} - 9\,\text{days}\,18\,\text{hours} = 7\,\text{days}\,6\,\text{hours}\).

Mission Beta is \(4\) days \(11\) hours shorter than Mission Alpha and \(7\) days \(6\) hours shorter than Mission Gamma.

- First find the length of the daylight period. - A full day has \(24\) hours. - Subtract the daylight period from \(24\) hours to find the night length. - Compare the night length with the daylight length.

1. Find the daylight period: From \(8{:}15\,\text{a.m.}\) to \(4{:}05\,\text{p.m.}\) is \(7\) hours \(50\) minutes. 2. Find the night length: \(24\,\text{hours} - 7\,\text{hours}\,50\,\text{minutes} = 16\,\text{hours}\,10\,\text{minutes}\). 3. Find the difference: \(16\,\text{hours}\,10\,\text{minutes} - 7\,\text{hours}\,50\,\text{minutes} = 8\,\text{hours}\,20\,\text{minutes}\).

The night is \(8\) hours \(20\) minutes longer than the daylight period.

- Find each day's distance in order. - Pay attention to whether each comparison says farther or less. - Convert all distances to meters before calculating, then convert the total to kilometers.

1. Convert the first day's distance: \(14\,\text{km}\,600\,\text{m} = 14{,}600\,\text{m}\). 2. Add the second-day increase. Adding \(3850\,\text{m}\) to \(14{,}600\,\text{m}\) gives \(18{,}450\,\text{m}\). 3. Subtract the third-day decrease. Subtracting \(2500\,\text{m}\) from \(18{,}450\,\text{m}\) gives \(15{,}950\,\text{m}\). 4. Add the three days: \(14{,}600 + 18{,}450 + 15{,}950 = 49{,}000\,\text{m}\). 5. Convert the total: \(49{,}000\,\text{m} = 49\,\text{km}\).

The hiker walks \(49\,\text{km}\) altogether.

- Find Trail C first. - Calculate twice the length of Trail B. - Compare the two lengths.

1. Add Trails A and B: \(12\,\text{km}\,250\,\text{m} + 9\,\text{km}\,900\,\text{m} = 22\,\text{km}\,150\,\text{m}\). 2. Find Trail C: \(22\,\text{km}\,150\,\text{m} - 4\,\text{km}\,500\,\text{m} = 17\,\text{km}\,650\,\text{m}\). 3. Find twice the length of Trail B: \(2 \times 9\,\text{km}\,900\,\text{m} = 19\,\text{km}\,800\,\text{m}\). 4. Since \(17\,\text{km}\,650\,\text{m} < 19\,\text{km}\,800\,\text{m}\), Trail C is not more than twice as long as Trail B.

No. Trail C is \(17\,\text{km}\,650\,\text{m}\), while twice the length of Trail B is \(19\,\text{km}\,800\,\text{m}\).

- Find the second stage first. - Add the lengths of the first two stages. - Subtract that sum from the total trail length.

1. Find the second stage: \(4\,\text{km}\,250\,\text{m} - 2\,\text{km}\,100\,\text{m} = 2\,\text{km}\,150\,\text{m}\). 2. Add the first two stages: \(4\,\text{km}\,250\,\text{m} + 2\,\text{km}\,150\,\text{m} = 6\,\text{km}\,400\,\text{m}\). 3. Subtract from the full trail: \(15\,\text{km} - 6\,\text{km}\,400\,\text{m} = 8\,\text{km}\,600\,\text{m}\).

The third stage is \(8\,\text{km}\,600\,\text{m}\) long.

- Think about what remains when the difference between two quantities is removed from their sum. - The remaining amount represents two equal copies of the lighter mass. - Divide that amount by \(2\).

1. When the difference between two masses is subtracted from their sum, the result is twice the lighter mass. 2. Therefore, twice the lighter suitcase's mass is \(18\,\text{kg}\,600\,\text{g}\). 3. Divide by \(2\): \(18\,\text{kg}\,600\,\text{g} \div 2 = 9\,\text{kg}\,300\,\text{g}\).

The lighter suitcase has a mass of \(9\,\text{kg}\,300\,\text{g}\).

- Find the red roll's original length first. - For each roll, subtract the amount remaining from the original amount. - Compare the amounts used and find their difference.

1. Find the red roll's original length: \(12\,\text{m}\,40\,\text{cm} + 1\,\text{m}\,50\,\text{cm} = 13\,\text{m}\,90\,\text{cm}\). 2. Find the amount used from the blue roll: \(12\,\text{m}\,40\,\text{cm} - 3\,\text{m}\,80\,\text{cm} = 8\,\text{m}\,60\,\text{cm}\). 3. Find the amount used from the red roll: \(13\,\text{m}\,90\,\text{cm} - 4\,\text{m}\,25\,\text{cm} = 9\,\text{m}\,65\,\text{cm}\). 4. Compare and subtract: \(9\,\text{m}\,65\,\text{cm} - 8\,\text{m}\,60\,\text{cm} = 1\,\text{m}\,5\,\text{cm}\).

More ribbon was used from the red roll, by \(1\,\text{m}\,5\,\text{cm}\).

- Represent the two trail lengths as a longer part and a shorter part. - Adding the difference to the combined length gives two copies of the longer trail. - Convert the measurements to meters before calculating.

1. Convert the combined length and difference to meters: \(12\,\text{km}\,400\,\text{m} = 12{,}400\,\text{m}\) and \(2\,\text{km}\,200\,\text{m} = 2200\,\text{m}\). 2. Add the difference to the combined length. Adding \(2200\,\text{m}\) to \(12{,}400\,\text{m}\) gives \(14{,}600\,\text{m}\). This equals twice the longer trail. 3. Divide by \(2\). Half of \(14{,}600\,\text{m}\) is \(7300\,\text{m}\). 4. Convert: \(7300\,\text{m} = 7\,\text{km}\,300\,\text{m}\).

The longer trail is \(7\,\text{km}\,300\,\text{m}\) long.

- Convert all masses to grams. - Find the mass of the second package. - Add the first two packages, then subtract their mass from the total. - Convert the result back to kilograms and grams.

1. Convert the total and first package to grams: \(12\,\text{kg} = 12{,}000\,\text{g}\) and \(4\,\text{kg}\,250\,\text{g} = 4250\,\text{g}\). 2. Find the second package's mass: \(4250\,\text{g} - 850\,\text{g} = 3400\,\text{g}\). 3. Find the combined mass of the first two packages: \(4250\,\text{g} + 3400\,\text{g} = 7650\,\text{g}\). 4. Start with the total of \(12{,}000\,\text{g}\) and subtract \(7650\,\text{g}\). The third package has a mass of \(4350\,\text{g}\). 5. Convert: \(4350\,\text{g} = 4\,\text{kg}\,350\,\text{g}\).

The third package has a mass of \(4\,\text{kg}\,350\,\text{g}\).

- Convert both length measurements to inches first. - What fraction is \(24\,\text{in}\) of the total length? - Convert the total weight to ounces before finding the fraction. - Convert the remaining ounces back to pounds and ounces.

1. Convert the length: \(16\,\text{ft} = 192\,\text{in}\). 2. The cut piece is \(\frac{24}{192} = \frac{1}{8}\) of the total length. 3. Convert the total weight: \(46\,\text{lb} = 736\,\text{oz}\). 4. Find the cut piece's weight: \(\frac{1}{8} \times 736\,\text{oz} = 92\,\text{oz}\). 5. Find the remaining weight: \(736\,\text{oz} - 92\,\text{oz} = 644\,\text{oz}\). 6. Convert back: \(644\,\text{oz} = 40\,\text{lb}\,4\,\text{oz}\).

a) \(\frac{1}{8}\) b) \(40\,\text{lb}\,4\,\text{oz}\)

- Convert the full length of the break from weeks and days to days. - Remember that July has 31 days. - Find the final day of summer break before counting the days that remain.

1. Convert the length of the break to days: \(6 \times 7 + 2 = 44\) days. 2. From July 12 through July 31, there are \(20\) days. Therefore, \(44 - 20 = 24\) days of the break are in August, so August 24 is the last day of break. 3. After August 20, the remaining days are August 21, 22, 23, and 24. That is \(4\) days.

There are \(4\) days of summer break remaining after the family returns.