- What changes before the books go on the shelves? - What must be known before sharing books equally?

1. Before sharing the books equally, find how many are still available. 2. Subtract the books on loan: \(14 - 6 = 8\).

She must first find how many books are left: \(8\) books.

- Which event happened before any seedlings wilted? - How could the rows help you find the starting number of seedlings?

1. To find how many remain growing, the club first needs the number planted. 2. The club planted \(4 \times 5 = 20\) seedlings.

First find the number planted: \(20\) seedlings.

- Will the bags hold only the original balls, or also the borrowed ones? - What do all four bags have to share?

1. Before finding the number in each bag, the coach needs the full number of balls. 2. Add the original and borrowed balls: \(9 + 7 = 16\).

First find the total available: \(16\) balls.

- List all the items Leon plans to buy. - Can you find the cost of the pasta? - Is the price known for every item? - What information is needed to add the complete total?

1. Find the price of one box of pasta: \(\$6.00 \div 3 = \$2.00\). 2. Find the price of five boxes: \(5 \times \$2.00 = \$10.00\). 3. The price of the jar of tomato sauce is not given. 4. Therefore, the total cost cannot be determined.

Leon cannot find the total because the price of the tomato sauce is missing.

- Identify the operation at each branch of the tree. - Work from the starting numbers toward the final result. - Match each number in the word problem with its position in the tree.

1. The juice costs \(4\times\$14=\$56\). 2. The pretzels cost \(3\times\$8=\$24\). 3. The pretzels and decorations cost \(\$24+\$6=\$30\). 4. The total cost is \(\$56+\$30=\$86\).

a) The intermediate values are \(56\), \(24\), and \(30\). b) The total cost is \(\$86\).

- Read the three different edge lengths from the prism. - Use the stated fact about how often each dimension occurs among the edges. - Check that your final quantity is a length, not an area or volume.

1. Read the three dimensions from the diagram: \(5\,\text{cm}\), \(4\,\text{cm}\), and \(3\,\text{cm}\). 2. Their sum is \(5\,\text{cm}+4\,\text{cm}+3\,\text{cm}=12\,\text{cm}\). 3. There are four edges of each dimension, so the total edge length is \(4\times12\,\text{cm}=48\,\text{cm}\).

a) \(12\,\text{cm}\) b) \(48\,\text{cm}\)

- How many weekend days are there in five full weeks? - Read both extra kinds of days off in the chart. - Count each day off once before finding the days at school.

1. Five full weeks contain \(5 \times 7 = 35\) days, including \(5 \times 2 = 10\) weekend days. 2. The chart shows \(3\) vacation days and \(2\) snow days. 3. Students attend for \(35 - 10 - 3 - 2 = 20\) days.

Students attend school on \(20\) days.

- Identify the quantity before any batteries are used. - Your equation should include both actions from the story and the unknown \(x\). - Check that the equation's final value has the meaning “batteries left.”

1. The starting number of batteries is \(7\times18=126\). 2. The situation can be represented by \(7\times18-25=x\). 3. Therefore, \(x=126-25=101\).

\(7\times18-25=x\), so \(x=101\).

- Separate the pencils into the boxed part and the loose part. - The equation should combine both parts into one total \(p\). - Check that the unknown represents the whole amount, not one box.

1. The boxed pencils total \(8\times24=192\). 2. The situation is represented by \(p=8\times24+35\). 3. Therefore, \(p=192+35=227\).

\(p=8\times24+35\), so \(p=227\).

- Picture how one tray looks before and after the baker removes rolls. - Can every tray have the same number of rolls at both times?

1. Each tray starts with \(36 \div 6 = 6\) rolls. 2. Removing \(2\) leaves \(6 - 2 = 4\) rolls on each tray.

\(4\) rolls on each tray: \(36 \div 6 = 6\), then \(6 - 2 = 4\).

- Think of the people who arrived in the vehicles and the people who joined later. - Can one count describe everyone from the beginning?

1. The minibuses carry \(2 \times 7 = 14\) campers. 2. After the others join, there are \(14 + 3 = 17\) campers.

\(17\) campers: \(2 \times 7 = 14\), then \(14 + 3 = 17\).

- Are the boxes separate amounts or equal groups? - Will giving pencils away increase or decrease the count?

1. First find the total in the boxes: \(3 \times 12 = 36\). 2. Subtract the pencils given away: \(36 - 7 = 29\).

\(3 \times 12 = 36\) pencils at first; \(36 - 7 = 29\) pencils left.

- Which rolls are set aside and not packed? - Would a bag's share depend on the rolls already set aside?

1. The rolls available for bags are \(42 - 6 = 36\). 2. Share the remaining rolls: \(36 \div 4 = 9\) per bag.

\(42 - 6 = 36\) rolls to pack; \(36 \div 4 = 9\) rolls per bag.

- Are the campers sharing one box or the complete supply? - Would the answer change if the boxes were emptied into a single pile first?

1. Find the total: \(4 \times 9 = 36\) pencils. 2. Divide among the campers: \(36 \div 6 = 6\) pencils each.

\(4 \times 9 = 36\) pencils altogether; \(36 \div 6 = 6\) pencils per camper.

- Look at how many pencils are removed from every box. - Compare what the two trees subtract before or after combining boxes. - Check your choice against one box first.

1. Each box has \(8 - 3 = 5\) pencils left. 2. Tree b) represents \(6 \times (8 - 3)\), giving \(6 \times 5 = 30\). 3. Tree a) subtracts only \(3\) pencils after counting all boxes, so it does not fit the story.

Tree b); \(30\) pencils.

- Would five tickets cost the same as one ticket? - Think about what the museum keeps and what the students get back.

1. The tickets cost \(5 \times \$7 = \$35\) altogether. 2. Subtract the cost from the payment: \(\$50 - \$35 = \$15\).

The tickets cost \(5 \times \$7 = \$35\). The change is \(\$50 - \$35 = \$15\).

- Think about what changed in one basket. - Would looking at just one basket tell you the total sold?

1. Each basket has \(7 - 5 = 2\) fewer pears. 2. Across \(4\) baskets, \(4 \times 2 = 8\) pears were sold.

\(8\) pears were sold: \(7 - 5 = 2\) from each basket, and \(4 \times 2 = 8\).

- Was the bus carrying the same number of passengers at both times? - Does the three-times statement compare the van with the bus before or after the stop?

1. Before the new people boarded, the bus carried \(45 - 9 = 36\) passengers. 2. The van carried one-third as many: \(36 \div 3 = 12\) passengers.

\(12\) passengers.

- Which distance is repeated for every lap? - Is the cool-down distance repeated after every lap or added only at the end? - Use the story to decide where the multiplication stops and the addition begins.

1. The runner covers \(4\,\text{km}\) three times for the laps, but walks the extra \(2\,\text{km}\) only once. 2. The correct calculation is \(3 \times 4 + 2 = 14\) kilometers. 3. Nico's calculation repeats the cool-down walk three times.

\(14\,\text{km}\). The extra \(2\,\text{km}\) happens only once.

- How does the value of one token relate to the total value? - How can the number of tokens help with the box count and the total mass? - Which unit is requested for the final mass?

1. The park buys \(6000\div3=2000\) tokens. 2. Packing \(5\) tokens per box needs \(2000\div5=400\) boxes. 3. The tokens have a total mass of \(2000\times4=8000\,\text{g}\). Since \(1000\,\text{g}=1\,\text{kg}\), that is \(8\,\text{kg}\).

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

- Find the cost of each group of supplies. - Then add the two costs. - Break the problem into smaller steps.

1. Find the cost of the colored pencils: \(15 \times 4 = 60\), so they cost \(\$60\). 2. Find the cost of the sketch pads: \(20 \times 3 = 60\), so they cost \(\$60\). 3. Add the two costs: \(\$60 + \$60 = \$120\).

Mr. Weber spends \(\$120\) altogether.

- Find the cost for each group separately. - Use the different admission price for each group. - Add the two group costs.

1. Find the admission cost for the students: \(25 \times 6 = 150\), so the students cost \(\$150\). 2. Find the admission cost for the chaperones: \(4 \times 9 = 36\), so the chaperones cost \(\$36\). 3. Add the two costs: \(\$150 + \$36 = \$186\).

The total admission cost is \(\$186\).

- First find the total number of bulbs. - Subtract the bulbs that cannot be used. - Equal sharing indicates division.

1. Find the total number delivered: \(8 \times 25 = 200\). 2. Remove the damaged bulbs: \(200 - 20 = 180\). 3. Divide the remaining bulbs equally: \(180 \div 6 = 30\).

\(30\) flower bulbs are planted in each bed.

- First find how many marbles the brother receives. - Giving marbles away decreases his amount. - Solve the problem in two steps.

1. Find half of the marbles: \(160 \div 2 = 80\). 2. Subtract the marbles given to the friend: \(80 - 15 = 65\).

Paul’s brother has \(65\) marbles left.

- First find the count of reserved seats. - Decide whether the newly sold seats overlap the reserved seats. - Combine the nonoverlapping counts.

1. Half the seats are reserved: \(720 \div 2 = 360\). 2. The \(85\) new sales are for other seats, so add them: \(360 + 85 = 445\).

\(445\) seats are reserved or sold.

- Find the number left after Monday. - Planting more bulbs decreases the amount remaining. - The question asks for the bulbs not yet planted.

1. After half are planted, \(840 \div 2 = 420\) bulbs remain. 2. Subtract the bulbs planted Tuesday: \(420 - 155 = 265\).

\(265\) flower bulbs remain unplanted.

- First find how many books all three bookcases can hold. - Then subtract the books already on the shelves. - Use the capacity of one bookcase to find the total capacity.

1. Find the total capacity of the three bookcases: \(3 \times 85 = 255\). 2. Subtract the books already placed on the shelves: \(255 - 140 = 115\).

The bookcases can hold \(115\) more books.

- Find the number of juice bottles and water bottles separately. - Add to find the total number purchased. - Subtract the number that guests drank.

1. Find the number of juice bottles: \(5 \times 12 = 60\). 2. Find the number of water bottles: \(4 \times 20 = 80\). 3. Find the total number of bottles: \(60 + 80 = 140\). 4. Subtract the bottles that were consumed: \(140 - 85 = 55\).

There are \(55\) bottles left.

- First find the number planted in the small beds. - Subtract to find the number left. - Divide the remaining bulbs equally between two planters.

1. Find the number planted in the beds: \(6 \times 45 = 270\). 2. Find the number remaining: \(400 - 270 = 130\). 3. Divide equally between the planters: \(130 \div 2 = 65\).

Each planter receives \(65\) tulip bulbs.

- Find the total number of muffins on the four trays. - Compare that total with \(55\). - Subtract to find the difference.

1. Find the number baked by the first class: \(4 \times 12 = 48\). 2. Compare the amounts: \(55 > 48\), so the second class bakes more. 3. Find the difference: \(55 - 48 = 7\).

The second class bakes more muffins. It bakes \(7\) more muffins.

- Find the total number of red tulips. - Find the total number of yellow tulips. - Add the two totals.

1. Find the number of red tulips: \(6 \times 15 = 90\). 2. Find the number of yellow tulips: \(4 \times 20 = 80\). 3. Add the two amounts: \(90 + 80 = 170\).

The gardener plants \(170\) tulips altogether.

- First add the numbers in the table. - The school receives the same amount for each toy. - Multiply the total number of toys by \(3\).

1. Add the numbers of toys sold: \(34 + 41 + 38 + 45 + 42 + 50 = 250\). 2. Multiply by the amount received for each toy: \(250 \times 3 = 750\).

a) The classes sell \(250\) toys altogether. b) The school receives \(\$750\).

- First add the numbers of boxes collected by all four teams. - Each box is worth \(\$2\). - Double the total number of boxes to find the credit.

1. Add the numbers of boxes: \(115 + 122 + 108 + 135 = 480\). 2. Multiply by the credit for each box: \(480 \times 2 = 960\).

a) The teams collect \(480\) boxes altogether. b) The school receives a \(\$960\) recycling credit.

- Combine the package sizes to make exactly the required number of passes. - Try more than one combination. - Add the prices and compare the totals.

1. For \(4\) passes, one \(3\)-pass package plus one single pass costs \(10 + 4 = 14\) dollars. This costs less than four single passes. 2. For \(8\) passes, one \(5\)-pass package plus one \(3\)-pass package costs \(16 + 10 = 26\) dollars. 3. Another way is two \(3\)-pass packages plus two single passes, which costs \(10 + 10 + 4 + 4 = 28\) dollars.

a) One \(3\)-pass package and one single pass cost \(\$14\). b) One \(5\)-pass package plus one \(3\)-pass package costs \(\$26\). Two \(3\)-pass packages plus two single passes cost \(\$28\).

- In each situation, identify what must be found before the final change can be applied. - Use the result of the first step as information for the second step. - Check that your final answer matches what is being counted after both changes.

1. The two shelves hold \(300 \times 2=600\) notebooks. After the delivery, there are \(600+150=750\) notebooks. 2. Half of \(800\) is \(400\), so \(400\) flyers remain after the morning. After \(60\) more are handed out, \(400-60=340\) flyers remain.

a) \(750\) notebooks b) \(340\) flyers

- Use \(100\) cents in one dollar and \(10\) cents in one dime. - Divide the total number of cents by the amount saved each day. - Compare your number of days with \(182\).

1. Since \(100\) cents equals \(1\) dollar, \(1000 \div 100 = 10\). Therefore, \(1000\) cents is \(\$10\). 2. One dime is \(10\) cents. Divide the goal by the daily amount: \(1000 \div 10 = 100\) days. 3. Compare \(100\) and \(182\). Since \(100 < 182\), the saving time is less than half a year.

a) \(\$10\) b) \(100\) days c) Less than half a year

- Find the cost of the first two types of supplies. - Subtract that total from the budget. - Determine how many \(\$2\) glue sticks fit within the remaining amount.

1. Find the cost of the construction paper: \(7 \times \$12 = \$84\). 2. Find the cost of the watercolor sets: \(6 \times \$9 = \$54\). 3. Find the total spent: \(\$84 + \$54 = \$138\). 4. Find the money remaining: \(\$150 - \$138 = \$12\). 5. Find the number of glue sticks: \(\$12 \div \$2 = 6\).

Ms. Miller can buy \(6\) glue sticks.

- Find a subtotal for each group of ordered items. - Combine the subtotals before deciding whether the family has enough money. - Use the budget comparison to justify your conclusion.

1. The two kids’ meals cost \(2 \times \$5=\$10\). 2. The adults’ entrées cost \(\$9+\$7=\$16\). 3. Four soups cost \(4 \times \$4=\$16\). 4. The total bill is \(\$10+\$16+\$16=\$42\). 5. Since \(\$42<\$50\), \(\$50\) is enough, with \(\$50-\$42=\$8\) remaining.

Yes. The bill is \(\$42\), so \(\$50\) is enough and \(\$8\) remains.

- Find each person's total spending separately. - Remember that Anna buys two identical items. - Subtract each total from \(\$25\) to find the money left. - Compare the two spending totals.

1. Tim spends \(\$18 + \$4 = \$22\) and has \(\$25 - \$22 = \$3\) left. 2. Anna spends \(2 \times \$12 = \$24\) and has \(\$25 - \$24 = \$1\) left. 3. Since \(\$24 > \$22\), Anna spends more.

Anna spends more. Tim has \(\$3\) left, and Anna has \(\$1\) left.

- Determine how many times Lucas walks the route each school day. - Find his total walking distance for one day. - Use the daily distance to find the distance for all five days.

1. The distance Lucas walks each day is \(70\,\text{m} \times 2=140\,\text{m}\). 2. The distance for five days is \(140\,\text{m} \times 5=700\,\text{m}\).

Lucas walks \(700\,\text{m}\) during the school week.

- Find the walking distance for one round trip. - Account for both lessons in one week. - Then extend the weekly distance across two weeks.

1. One round trip is \(110\,\text{m} \times 2=220\,\text{m}\). 2. Two lessons in one week require \(220\,\text{m} \times 2=440\,\text{m}\) of walking. 3. Over two weeks, Sophie walks \(440\,\text{m} \times 2=880\,\text{m}\).

Sophie walks \(880\,\text{m}\) over two weeks.

- First find the bus cost for the entire class without the contribution. - Subtract the amount paid by the parent organization.

1. Find the full cost for \(24\) students: \(24 \times \$13 = \$312\). 2. Subtract the contribution: \(\$312 - \$100 = \$212\).

\(\$212\)

- How much do all four children save in one month? - How many monthly deposits of that amount are needed to reach \(\$320\)?

1. Find the amount the four children save each month: \(4 \times \$8 = \$32\). 2. Let \(m\) be the number of months. The total saved is \(\$32 \times m\). Since \(\$32 \times 10 = \$320\), \(m=10\). 3. They will reach the goal after \(10\) months.

They will have enough money after \(10\) months.

- Find the total number of seats from the rows and seats per row. - What does it mean that all seats are occupied? - Subtract the number of adults from the total number of moviegoers. - Decide which details are not needed.

1. Find the total number of seats: \(15 \times 20 = 300\). 2. Because all seats are occupied, there are \(300\) moviegoers. 3. Subtract the number of adults: \(300-180=120\). 4. The ticket prices, start time, and movie duration are not needed.

There are \(120\) children in the theater.

- Find the adult and child costs separately. - Multiply each price by the number of people. - Add the two subtotals.

1. The adult tickets cost \(2 \times \$14 = \$28\). 2. The child tickets cost \(3 \times \$9 = \$27\). 3. The total is \(\$28 + \$27 = \$55\).

The family pays \(\$55\) altogether.

- Find the total train cost for all \(4\) people. - Compare the train cost with the total car cost. - Subtract the travel times to find the time savings.

1. The train costs \(4 \times \$21 = \$84\) for the family. 2. Driving saves \(\$84 - \$60 = \$24\). 3. Taking the train saves \(4 - 2 = 2\) hours.

Driving saves \(\$24\). Taking the train saves \(2\) hours of travel time.

- First determine how many small cups contain the same mass as one large cup. - Compare the cost of that many small cups with the large-cup price. - Make sure the two options contain equal amounts of yogurt before comparing prices.

1. Three small cups contain \(3 \times 150\,\text{g}=450\,\text{g}\). 2. Three small cups cost \(3 \times \$2=\$6\). 3. The difference is \(\$6-\$5=\$1\).

Buying \(450\,\text{g}\) in small cups would cost \(\$1\) more.

- Which cost occurs once, and which cost occurs for every ball? - First find the total cost of all the balls. - Then add the net’s price.

1. Find the cost of the soccer balls: \(6 \times \$24 = \$144\). 2. Add the cost of the goal net: \(\$144 + \$155 = \$299\).

\(\$299\)

- First find the cost of the \(4\) trees. - The delivery and planting fee is charged only once. - Use multiplication, then addition.

1. Find the cost of the trees: \(4 \times \$142 = \$568\). 2. Add the delivery and planting fee: \(\$568 + \$75 = \$643\).

\(\$643\)

- Find the admission cost for all the students. - Add the one-time bus fee. - Written multiplication may help with the first step.

1. Find the total admission cost: \(26 \times \$14 = \$364\). 2. Add the bus fee: \(\$364 + \$85 = \$449\).

\(\$449\)

- Combine the two one-minute observations before extending them across time. - The problem asks what would happen if the observed rates stayed constant. - Check that the final quantity counts breaths for both animals over all \(10\) minutes.

1. The combined observed rate is \(163+90=253\) breaths per minute. 2. Over \(10\) minutes, the total is \(253 \times 10=2530\) breaths.

At those observed rates, the two animals would take \(2530\) breaths altogether in \(10\) minutes.

- Find the combined heartbeats of the two animals in one minute. - Convert the rates to totals over the same hour. - The question asks for a difference, not a multiplication factor.

1. The combined animal rate is \(24+36=60\) beats per minute. 2. Over one hour, the two animals have \(60\times60=3600\) heartbeats together. 3. The person has \(65\times60=3900\) heartbeats. 4. The difference is \(3900-3600=300\) heartbeats.

The person would have \(300\) more heartbeats in one hour.

- First compare the two one-minute rates. - Then extend that one-minute difference across the full time interval. - Check that the final answer is a difference in heartbeat counts, not a rate.

1. The difference in the two rates is \(98-18=80\) beats per minute. 2. Over \(20\) minutes, the difference is \(80 \times 20=1600\) beats.

The hibernation rate produces \(1600\) fewer heartbeats over \(20\) minutes.

- Find the total distance for one week first. - Count how many days she drives the longer route and how many days she drives the shorter route. - Then multiply the weekly distance by \(4\).

1. Find the distance from Monday through Friday: \(5 \times 36 = 180\,\text{miles}\). 2. Add Saturday's route: \(180 + 15 = 195\,\text{miles}\) per week. 3. Find the distance for four weeks: \(195 \times 4 = 780\,\text{miles}\).

The mail carrier drives \(780\,\text{miles}\) in \(4\) weeks.

- Find the distance of one complete round trip first. - Use the number of daily round trips to determine one day’s distance. - Extend the daily distance across the \(30\)-day month.

1. One round trip is \(6 \times 2=12\,\text{miles}\). 2. The daily distance is \(12 \times 8=96\,\text{miles}\). 3. The distance for \(30\) days is \(96 \times 30=2880\,\text{miles}\).

The shuttle travels \(2880\,\text{miles}\) in the month.

- Work from the smallest container up to one shipping crate. - Multiply the watches in one shipping crate by the number of shipping crates. - Count each type of container separately before adding them.

1. One shipping crate contains \(4\times 4\times 4=64\) watches. 2. With \(4\) shipping crates, the warehouse contains \(64\times 4=256\) watches. 3. There are \(4\) shipping crates, \(4\times 4=16\) wooden boxes, and \(16\times 4=64\) metal cases. 4. The total number of containers is \(4+16+64=84\).

a) \(64\) watches b) \(256\) watches c) \(84\) containers

- Break the problem into smaller steps. - First find the total number of haystacks. - After finding the number of families, multiply by the number of baby mice in each family. - Work from the largest group, barns, down to the smallest group, baby mice.

1. Find the total number of haystacks: \(5\times 8=40\). 2. Find the total number of mouse families: \(40\times 12=480\). 3. Find the total number of baby mice: \(480\times 6=2880\).

There are \(2880\) baby mice altogether.

- Find the total amount paid over all \(12\) months. - Use multiplication to combine the monthly payments. - Subtract the full-payment price from the payment-plan total.

1. Find the total cost of the payment plan: \(12 \times \$52 = \$624\). 2. Find the difference from the full-payment price: \(\$624 - \$565 = \$59\).

The payment plan costs \(\$59\) more.

- Find how much each group of visitors pays in total. - Multiplication gives the revenue when the same price is paid many times. - Add the two revenue amounts.

1. Find the revenue from child tickets: \(312 \times \$6 = \$1872\). 2. Find the revenue from adult tickets: \(145 \times \$9 = \$1305\). 3. Add the two amounts: \(\$1872 + \$1305 = \$3177\).

The museum’s total ticket revenue is \(\$3177\).

- Find the total amount paid under each plan separately. - Compare the two totals rather than the monthly payment amounts alone. - Use the difference between the totals to quantify how much cheaper one plan is.

1. Plan A totals \(12 \times \$78=\$936\). 2. Plan B totals \(24 \times \$40=\$960\). 3. Plan A costs less because \(\$936<\$960\). 4. The difference is \(\$960-\$936=\$24\).

Plan A totals \(\$936\), and Plan B totals \(\$960\). Plan A costs \(\$24\) less.

- Find the unpaid part of the tablet price before considering the monthly plan. - Relate the monthly payment amount to the unpaid balance to determine the number of months. - For part c), compare the amount left from one month’s allowance with Lucas’s stated minimum.

1. Lucas still needs to repay \(\$240-\$60=\$180\). 2. Since \(\$20 \times 9=\$180\), the repayment takes \(9\) months. 3. After each payment, Lucas keeps \(\$25-\$20=\$5\), so the plan meets his stated goal exactly.

a) \(\$180\) b) \(9\) months c) Yes. He keeps \(\$5\) each month, which meets his goal.

- Multiply the monthly savings by the number of months. - Compare the result with the violin’s price. - Find the monthly amount that, when multiplied by \(20\), equals the goal.

1. Find the amount saved in \(24\) months: \(24 \times \$15 = \$360\). 2. Compare it with the goal: \(\$400 - \$360 = \$40\), so she is short by \(\$40\). 3. Use an inverse multiplication fact for \(20\) months: \(20 \times \$20 = \$400\), so she needs to save \(\$20\) per month.

a) Mia will save \(\$360\). b) No. She will need \(\$40\) more. c) She would need to save \(\$20\) per month.

- Work through the nested groups one level at a time. - Use each intermediate total as the starting amount for the next grouping level. - Check that the final count refers to individual markers, not boxes or cartons.

1. The pallets hold \(8 \times 25=200\) cartons. 2. The cartons hold \(200 \times 10=2000\) boxes. 3. The boxes contain \(2000 \times 5=10{,}000\) markers.

The warehouse contains \(10{,}000\) markers.

- Determine how many \(4\)-ounce portions are in each purchased amount. - Find each cheese cost separately before combining them. - Check that both purchased amounts are included in the final total.

1. Eight ounces of Alpine cheese is two \(4\)-ounce portions, so it costs \(2 \times \$2=\$4\). 2. Sixteen ounces of Edam is four \(4\)-ounce portions, so it costs \(4 \times \$1=\$4\). 3. The total cost is \(\$4+\$4=\$8\).

The total cost is \(\$8\).

- Relate \(16\,\text{oz}\) to the \(4\)-ounce price unit. - Find the total purchase price before comparing it with the money Tim has. - If the purchase is affordable, the remaining amount is the change.

1. Sixteen ounces contains four \(4\)-ounce portions, so the cheese costs \(4 \times \$2=\$8\). 2. Since \(\$8<\$10\), Tim has enough money. 3. His change is \(\$10-\$8=\$2\).

Yes. Tim has enough money and receives \(\$2\) in change.

- Find the adult and child costs under the family special separately. - To find the savings, compare the special price with the total using regular fares. - Keep track of which child fare applies in each calculation.

1. With the special, the two adult fares cost \(2 \times \$22 = \$44\), and the three child fares cost \(3 \times \$8 = \$24\). 2. The family’s total with the special is \(\$44 + \$24 = \$68\). 3. At the regular fares, the adults cost \(\$44\), and the children cost \(3 \times \$11 = \$33\), for a total of \(\$77\). 4. The savings are \(\$77 - \$68 = \$9\).

a) The total fare with the family special is \(\$68\). b) The family saves \(\$9\).

- Decide which travelers pay full price, half price, or no fare. - “Half price” means divide the full fare by \(2\). - A round trip includes two one-way trips.

1. The first adult pays \(\$84\). 2. Each of the other two adults pays half of \(\$84\): \(\$84 \div 2 = \$42\). 3. The children ride free, so the one-way total is \(\$84 + \$42 + \$42 = \$168\). 4. The round trip costs twice the one-way total: \(2 \times \$168 = \$336\).

a) The one-way trip costs \(\$168\). b) The round trip costs \(\$336\).

- Calculate the cost of buying all individual tickets. - Then calculate the cost of one family pass plus the extra child ticket. - Compare the two totals.

1. Buying individual tickets would cost \(2 \times \$45 = \$90\) for the adults and \(4 \times \$30 = \$120\) for the children. 2. The individual-ticket total is \(\$90 + \$120 = \$210\). 3. A family pass covers the two adults and three children. One additional child ticket is needed, so this option costs \(\$130 + \$30 = \$160\). 4. Since \(\$160 < \$210\), the family pass plus one child ticket is less expensive.

The least expensive total admission cost is \(\$160\).

- How many equal juice portions fit into each order? - Find the price of each order using the price per portion. - Compare the two total prices, not their amounts of juice.

1. The \(36\,\text{fl oz}\) order contains \(36\div4=9\) portions and costs \(\$9\). 2. The \(28\,\text{fl oz}\) order contains \(28\div4=7\) portions and costs \(\$7\). 3. The difference is \(\$9-\$7=\$2\).

The larger order costs \(\$2\) more.

- Which listed price applies to more than one pack? - Find the total cost before thinking about the change. - Check your result against the money Maya pays.

1. Two \(5\,\text{oz}\) packs cost \(2 \times \$2.25 = \$4.50\). 2. All the packs cost \(\$4.20 + \$4.50 + \$0.50 = \$9.20\). 3. The change is \(\$10.00 - \$9.20 = \$0.80\).

Maya gets \(\$0.80\) in change.

- Find the round-trip cost for the first traveler. - Find half of the one-way fare for the second traveler. - Remember that each traveler needs a ticket in both directions.

1. The first traveler’s round-trip cost is \(2 \times \$50 = \$100\). 2. The second traveler’s one-way fare is half of \(\$50\): \(\$50 \div 2 = \$25\). 3. The second traveler’s round-trip cost is \(2 \times \$25 = \$50\). 4. The total cost is \(\$100 + \$50 = \$150\).

The tickets cost \(\$150\) in all.

- Add half of the coach fare to the coach fare. - First find the one-way first-class fare. - A round trip includes two one-way tickets.

1. Half of the coach fare is \(\$48 \div 2 = \$24\). 2. A one-way first-class ticket costs \(\$48 + \$24 = \$72\). 3. A round-trip ticket costs \(2 \times \$72 = \$144\).

One round-trip first-class ticket costs \(\$144\).

- Compare only the ticket types named in the question. - What extra amount is added to the Boston coach price? - Find the two full ticket prices before comparing them.

1. Half the Boston coach price is \(\$58 \div 2 = \$29\). 2. The Boston first-class ticket costs \(\$58 + \$29 = \$87\). 3. It costs \(\$87 - \$79 = \$8\) more than the Philadelphia coach ticket.

The first-class ticket to Boston costs \(\$8\) more.

- One person pays the full price; the other people all pay the same reduced price. - Find the group’s total admission cost before checking the budget. - If the budget is sufficient, compare it with the total cost to find what remains.

1. Each additional person pays \(\$8 \div 2=\$4\). 2. Four additional people cost \(4 \times \$4=\$16\). 3. The group’s total admission cost is \(\$8+\$16=\$24\). 4. Since \(\$24<\$25\), the group has enough money, with \(\$25-\$24=\$1\) remaining.

Yes. Admission costs \(\$24\), so \(\$1\) remains.

- Separate the first ticket from the additional tickets because they have different prices. - Use the total bill to determine how much was spent on the additional tickets. - Check your proposed family size by recomputing the complete ticket cost.

1. Each additional ticket costs \(\$14 \div 2=\$7\). 2. After the first ticket, \(\$42-\$14=\$28\) remains for additional tickets. 3. Since \(\$7 \times 4=\$28\), the family buys four additional tickets. 4. Including the first ticket, the family has \(1+4=5\) people.

The family has \(5\) people.

- Work out the complete group cost for each offer before comparing them. - Offer B uses one price for the first traveler and a different repeated price for the others. - The final answer needs both the cheaper offer and the amount of the savings.

1. Offer A costs \(4 \times \$12=\$48\). 2. Under Offer B, each additional person pays \(\$18 \div 2=\$9\). 3. Offer B costs \(\$18+3 \times \$9=\$45\). 4. Offer B is cheaper, and the difference is \(\$48-\$45=\$3\).

Offer B costs \(\$3\) less.

- Determine each train’s total capacity from its cars and seats per car. - Compare the two capacities only after both totals are known. - The final question also asks how far apart the two capacities are.

1. Mountain Express has \(7 \times 64=448\) seats. 2. Coastal Limited has \(5 \times 92=460\) seats. 3. Coastal Limited has more seats because \(460>448\). 4. The difference is \(460-448=12\) seats.

Coastal Limited has more seats. It has \(12\) more seats than Mountain Express.

- All three trailer types count toward the total length, but only two provide sleeping spaces. - Read the capacity of each trailer type carefully. - Find each subtotal before adding the lengths and sleeping spaces.

1. Find the length of each type: \(15 \times 20\,\text{ft} = 300\,\text{ft}\), \(8 \times 30\,\text{ft} = 240\,\text{ft}\), and \(4 \times 40\,\text{ft} = 160\,\text{ft}\). 2. Add the lengths: \(300\,\text{ft} + 240\,\text{ft} + 160\,\text{ft} = 700\,\text{ft}\). 3. Find the sleeping spaces: \(15 \times 4 = 60\) and \(8 \times 7 = 56\). 4. Add the sleeping spaces: \(60 + 56 = 116\). The equipment trailers do not add sleeping spaces.

The convoy is \(700\,\text{ft}\) long, and \(116\) people can sleep in the trailers.

- Add the sugar from the food and drink. - Determine how many groups of \(3\,\text{g}\) are in the total amount.

1. Find the total amount of sugar: \(12\,\text{g} + 18\,\text{g} = 30\,\text{g}\). 2. Divide by the mass of one sugar cube: \(30\,\text{g} \div 3\,\text{g} = 10\).

The total sugar is equal to \(10\) sugar cubes.

- Find the equivalent number of sugar cubes for each drink. - Then subtract the two results. - Another method is to find the difference in grams first.

1. Super Sweet contains the equivalent of \(48\,\text{g} \div 3\,\text{g} = 16\) sugar cubes. 2. Fruit Mix contains the equivalent of \(36\,\text{g} \div 3\,\text{g} = 12\) sugar cubes. 3. Find the difference: \(16 - 12 = 4\) sugar cubes.

The drinks differ by the equivalent of \(4\) sugar cubes.

- Determine what part of the project cost is not covered by the grant. - The towns share only that remaining amount, not the full project cost. - Check that three equal shares reproduce the uncovered cost.

1. The amount not covered by the grant is \(\$450{,}000-\$360{,}000=\$90{,}000\). 2. Sharing that amount equally among three towns gives \(\$90{,}000 \div 3=\$30{,}000\) per town.

Each town must pay \(\$30{,}000\).

- Use all four expense rows to determine the amount the families actually need to cover. - Compare one family’s equal share with what that family deposited over the year. - Part b) depends on the share found in part a).

1. The yearly costs total \(\$456+\$192+\$74+\$142=\$864\). 2. Each family’s share is \(\$864 \div 4=\$216\). 3. Each family deposits \(12 \times \$19=\$228\) during the year. 4. Each family should receive \(\$228-\$216=\$12\) back.

a) The total cost is \(\$864\), and each family’s share is \(\$216\). b) Each family receives \(\$12\) back.

- Determine the complete cost before considering how it is shared. - Every class pays an equal share of the same total. - Check that five copies of your proposed share make the full festival cost.

1. The festival costs total \(\$115+\$240+\$35+\$60=\$450\). 2. Sharing the total equally among five classes gives \(\$450 \div 5=\$90\) per class.

Each class must pay \(\$90\).

- Work from the quantities that left the warehouse and the quantity that remained. - Determine the combined number sent to the three schools before including the donation. - Check that subtracting all shipments from your starting amount leaves \(4000\).

1. The three school shipments used \(350 \times 3=1050\) notebooks. 2. Altogether, the warehouse sent out \(1050+2000=3050\) notebooks. 3. The starting amount was \(4000+3050=7050\) notebooks.

The warehouse started with \(7050\) notebooks.

- Use the two equal booth amounts to reconstruct how many tickets remained after the sale. - Then account for the tickets that had already been sold. - Use the interval and multiple-of-\(25\) conditions as a final check.

1. The two booths receive \(100 \times 2=200\) tickets altogether after the sale. 2. The starting number was \(200+150=350\) tickets. 3. The check passes because \(350\) lies between \(300\) and \(400\), and \(25 \times 14=350\).

The school had \(350\) raffle tickets at first.

- Determine how much the family paid over the entire year. - Compare the amount paid with the actual yearly cost. - A refund is the amount paid beyond what was actually owed.

1. The family pays \(12 \times \$52=\$624\) during the year. 2. The refund is \(\$624-\$595=\$29\).

The family should receive a refund of \(\$29\).

- Put both payment options on the same one-year time period before comparing them. - Determine the complete cost of the monthly option. - The savings are the difference between the two yearly totals.

1. Twelve monthly payments cost \(12 \times \$26=\$312\). 2. The yearly option saves \(\$312-\$285=\$27\).

The yearly payment saves \(\$27\).

- Use the sample plant count and kernels-per-plant estimate to find a kernel total. - Relate that total to the given \(1000\)-kernel mass sample. - Keep approximation language in the final answer because the inputs are estimates.

1. The estimated number of kernels is \(8 \times 500=4000\). 2. Four groups of \(1000\) kernels make \(4000\) kernels. 3. The estimated mass is \(4 \times 380\,\text{g}=1520\,\text{g}\).

Based on the sample data, the estimated mass is about \(1520\,\text{g}\) per square meter.

- Determine the amount actually eaten in one day from the amount offered and the amount left. - Use that daily result across all \(7\) days. - Check that the final amount counts food eaten, not food offered or left over.

1. The elephant eats \(150-75=75\,\text{kg}\) each day. 2. Over \(7\) days, it eats \(75 \times 7=525\,\text{kg}\).

The elephant eats \(525\,\text{kg}\) of plant food during the week.

- First find how many conifer trees there are altogether. - If the conifer count is one-tenth of the deciduous count, how many times as large is the deciduous count? - Find the number of deciduous trees. - What final operation combines the two groups?

1. Find the total number of conifer trees: \(34{,}000 + 16{,}000 = 50{,}000\). 2. The conifer count is one-tenth of the deciduous count, so the deciduous count is \(10\) times as great: \(10 \times 50{,}000 = 500{,}000\). 3. Add both groups: \(50{,}000 + 500{,}000 = 550{,}000\).

There are \(550{,}000\) trees in the forest altogether.

- Use the edge-count facts given in the problem rather than relying on memory about the solids. - Find the complete wire length for each model separately. - Compare the two totals only after both have been calculated.

1. Lucas needs \(12 \times 8\,\text{cm}=96\,\text{cm}\) of wire. 2. Maya needs \(4 \times 10\,\text{cm}+4 \times 6\,\text{cm}+4 \times 8\,\text{cm}=40\,\text{cm}+24\,\text{cm}+32\,\text{cm}=96\,\text{cm}\). 3. The two models require the same total length of wire.

a) \(96\,\text{cm}\) b) \(96\,\text{cm}\) c) The two models require the same amount of wire.

- Start with the horse’s daily amount. - Determine how many bale weights fit in the total amount. - Multiply the number of bales by the price per bale.

1. The horse eats \(12\,\text{lb} \times 100 = 1200\,\text{lb}\). 2. Since \(4 \times 300\,\text{lb} = 1200\,\text{lb}\), \(4\) bales are needed. 3. The cost is \(4 \times \$45 = \$180\).

a) \(1200\,\text{lb}\) b) \(4\) bales c) \(\$180\)

- First find the paper used by the entire class in one day. - Find the total number of school days in \(20\) weeks. - Find how many groups of \(500\) sheets make the total by using multiplication.

1. The class uses \(25 \times 2 = 50\) sheets per day. 2. Twenty school weeks contain \(20 \times 5 = 100\) school days. 3. The class uses \(50 \times 100 = 5000\) sheets. 4. Since \(10 \times 500 = 5000\), the class needs \(10\) packages. 5. The total cost is \(10 \times \$6 = \$60\).

a) The class uses \(50\) sheets per day. b) The class uses \(5000\) sheets in \(20\) school weeks. c) The class needs \(10\) packages, costing \(\$60\) in all.

- Add the three uses for one student. - Multiply by \(22\) students. - Multiply the daily class amount by \(5\) days. - Use approximation language because the per-student amounts are averages.

1. Estimate one student's daily use: \(2\,\text{L} + 4\,\text{L} + 18\,\text{L} \approx 24\,\text{L}\). 2. Estimate the class's daily use: \(22 \times 24\,\text{L} \approx 528\,\text{L}\). 3. Estimate the class's weekly use: \(5 \times 528\,\text{L} \approx 2640\,\text{L}\).

a) About \(24\,\text{L}\) b) About \(528\,\text{L}\) c) About \(2640\,\text{L}\)

- Account for both the number of people and the number of days in each listed category. - Compare the combined known categories with the complete weekend total. - Keep approximation language because the category amounts are averages.

1. Personal-care use is about \(4 \times 2 \times 45\,\text{L}=360\,\text{L}\). 2. Toilet-flushing use is about \(4 \times 2 \times 30\,\text{L}=240\,\text{L}\). 3. Those known uses total about \(360\,\text{L}+240\,\text{L}=600\,\text{L}\). 4. The remaining use is about \(1100\,\text{L}-600\,\text{L}=500\,\text{L}\).

The family uses about \(500\,\text{L}\) of water for other purposes during the weekend.

- Subtract the shower amount from the bath amount. - Multiply by \(4\) people for one day. - Multiply the daily family savings by \(7\) days. - Use approximation language because the starting amounts are estimates.

1. Estimate the savings for one person: \(150\,\text{L} - 40\,\text{L} \approx 110\,\text{L}\). 2. Estimate the family's daily savings: \(4 \times 110\,\text{L} \approx 440\,\text{L}\). 3. Estimate the weekly savings: \(440\,\text{L} \times 7 \approx 3080\,\text{L}\).

1. About \(110\,\text{L}\) 2. About \(440\,\text{L}\) 3. About \(3080\,\text{L}\)

- Add all the entries in the table for one day. - Multiply the daily amount by \(30\) for June. - For the last part, find the sister's daily amount before multiplying by \(7\).

1. Add Liam's daily amounts: \(3\,\text{L} + 45\,\text{L} + 32\,\text{L} + 6\,\text{L} + 4\,\text{L} = 90\,\text{L}\). 2. Find Liam's June use: \(90\,\text{L} \times 30 = 2700\,\text{L}\). 3. Find his sister's daily use: \(90\,\text{L} + 15\,\text{L} = 105\,\text{L}\). Then find her weekly use: \(105\,\text{L} \times 7 = 735\,\text{L}\).

1. Liam uses \(90\,\text{L}\) per day. 2. Liam uses \(2700\,\text{L}\) in June. 3. His sister uses \(735\,\text{L}\) in one week.

- First find the total number of people. - Compare the total cost of the two ordering plans. - Find the savings by subtracting the lesser total from the greater total. - Use the cost and number served by one family-size pizza to find the per-person cost.

1. There are \(22+3=25\) people. 2. a) Individual pizzas cost \(25\times\$8=\$200\). 3. b) The team needs \(25\div5=5\) family-size pizzas. They cost \(5\times\$35=\$175\). 4. c) Family-size pizzas cost less. The savings are \(\$200-\$175=\$25\). 5. d) Each family-size pizza costs \(\$35\) for \(5\) people, so the cost per person is \(\$35\div5=\$7\).

a) \(\$200\) b) \(5\) family-size pizzas for \(\$175\) c) Family-size pizzas cost less and save \(\$25\). d) \(\$7\) per person

- Find each plan's total cost separately. - For part b, determine how many people are not covered by the two group tickets. - Compare the three totals after calculating them.

1. a) Thirty individual tickets cost \(30\times\$6=\$180\). 2. b) Two group tickets cover \(20\) people and cost \(2\times\$52=\$104\). The remaining \(10\) individual tickets cost \(10\times\$6=\$60\). The total is \(\$104+\$60=\$164\). 3. c) Three group tickets cost \(3\times\$52=\$156\). 4. d) Compare \(\$180, \$164\), and \(\$156\). Three group tickets are least expensive.

a) \(\$180\) b) \(\$164\) c) \(\$156\) d) Three group tickets are least expensive.

- The seven-day plan applies only to the flour left after Monday morning. - Determine the amount available for those seven equal daily portions. - Check that seven copies of your daily amount use exactly the remaining flour.

1. The bakery has \(250\,\text{kg}-40\,\text{kg}=210\,\text{kg}\) remaining. 2. Sharing the remaining flour equally across seven days gives \(210\,\text{kg} \div 7=30\,\text{kg}\) per day.

The bakery will use \(30\,\text{kg}\) each day.

- First find the number of jars left for sale. - Then determine how many groups of \(8\) can be made. - Break apart \(168\) into convenient multiples of \(8\) if helpful.

1. Find the number of jars available for sale: \(180 - 12 = 168\). 2. Divide by the number of jars per box: \(168 \div 8 = 21\).

The beekeeper can pack \(21\) full boxes.

- First think about what one-fourth of a number means. - What operation can you use to find a fraction of a set? - Find the number of books in each category one at a time. - Read carefully to see how the number of space books is related to the number of animal books.

1. Find the number of animal books: \(\frac{1}{4} \times 120 = 30\). 2. Find the number of space books: \(30 + 15 = 45\). 3. Add the two categories: \(30 + 45 = 75\).

There are \(75\) animal and space books altogether.

- How many pieces did Lucas place if he completed one-fifth of the puzzle? - How many pieces did Anna place compared with Lucas? - How many pieces have they placed altogether? - What operation finds the number still missing from the whole puzzle?

1. Find the number of pieces Lucas placed: \(\frac{1}{5} \times 500 = 100\). 2. Find the number of pieces Anna placed: \(100 + 35 = 135\). 3. Find the total number already placed: \(100 + 135 = 235\). 4. Subtract from the total number of pieces: \(500 - 235 = 265\).

There are \(265\) puzzle pieces left to place.

- The bouquets can use flowers from both types, so determine how many flowers are available altogether. - Relate the complete flower total to the fixed bouquet size. - Check that your number of bouquets accounts for all the flowers.

1. There are \(160+240=400\) flowers altogether. 2. Grouping \(400\) flowers into bouquets of \(8\) gives \(400 \div 8=50\) bouquets.

The nursery can make \(50\) bouquets.

- First find one-half of \(\$60\). - Then find one-fourth of the original amount. - How much was spent on the bus and museum altogether? - What remains after subtracting those expenses from \(\$60\)?

1. Find the cost of the bus ride: \(\frac{1}{2} \times \$60 = \$30\). 2. Find the cost of the museum tickets: \(\frac{1}{4} \times \$60 = \$15\). 3. Subtract both expenses from the total: \(\$60 - \$30 - \$15 = \$15\).

The class has \(\$15\) available for ice cream.

- First find how many rolls were made in the morning. - Was the afternoon amount greater or less than the morning amount? - How many rolls were made altogether? - What operation finds how many more are needed to reach the goal?

1. Find the number of rolls made in the morning: \(\frac{1}{5} \times 150 = 30\). 2. Find the number made in the afternoon: \(30 + 12 = 42\). 3. Find the total number made: \(30 + 42 = 72\). 4. Subtract from the goal: \(150 - 72 = 78\).

The baker must make \(78\) more rolls.

- What amount stays the same when the type of ball changes? - First find the total amount the club plans to spend. - Divide that amount equally among the six basketballs.

1. Find the total planned budget: \(8 \times \$30 = \$240\). 2. Divide the budget equally among six basketballs: \(\$240 \div 6 = \$40\).

One basketball costs \(\$40\).

- First find how many apples and pears remain separately. - What does it mean mathematically to use one-half of a group? - What fraction remains after one-third is used? - How can you combine the remaining amounts from both crates?

1. Find the number of apples used: \(\frac{1}{2} \times 40 = 20\). 2. Find the number of pears used: \(\frac{1}{3} \times 30 = 10\). 3. Find the amounts remaining: \(40 - 20 = 20\) apples and \(30 - 10 = 20\) pears. 4. Add the remaining amounts: \(20 + 20 = 40\).

There are \(40\) pieces of fruit remaining altogether.

- How much money does each person keep after contributing to the gift? - Find each contribution separately. - What operation can you use to find one-fourth or one-eighth of an amount? - Remember to add the two remaining amounts at the end.

1. Find Lucas's contribution: \(\frac{1}{4} \times \$120 = \$30\). 2. Find Lucas's remaining money: \(\$120 - \$30 = \$90\). 3. Find Sofia's contribution: \(\frac{1}{8} \times \$160 = \$20\). 4. Find Sofia's remaining money: \(\$160 - \$20 = \$140\). 5. Add the remaining amounts: \(\$90 + \$140 = \$230\).

Together, they have \(\$230\) left.

- First find how many students from each class enter the contest. - How many students remain in each class after subtracting the contestants? - Could you also begin by finding the total number of students in both classes? - What operation finds one-fourth of a group?

1. In Class 4A, \(\frac{1}{4} \times 24 = 6\) students enter the contest, so \(24 - 6 = 18\) students go to the gym. 2. In Class 4B, \(\frac{1}{4} \times 28 = 7\) students enter the contest, so \(28 - 7 = 21\) students go to the gym. 3. Add the numbers going to the gym: \(18 + 21 = 39\).

A total of \(39\) students go to the gym.

- How many pads does one class receive? - First divide the original total equally among the classes. - Once you know the amount for one class, how can you find the amount for fifteen classes?

1. Find the number of drawing pads for one class: \(120 \div 6 = 20\). 2. Find the number needed for fifteen classes: \(20 \times 15 = 300\).

The school needs \(300\) drawing pads.

- Use the comparison to find the red marble count. - Which marble colors are included in the total? - Split the complete set into five equal parts to find one-fifth.

1. There are \(4\times15=60\) red marbles. 2. The bag has \(15+60=75\) marbles altogether. 3. One-fifth of that total is \(75\div5=15\) glass marbles.

There are \(15\) glass marbles.

- First find the amount of sparkling water. - How much apple juice and sparkling water are available altogether? - How can you find one-third of the total amount? - Compare your result with the original amount of apple juice.

1. Find the amount of sparkling water: \(4 \times 6\,\text{gal} = 24\,\text{gal}\). 2. Find the total amount of liquid: \(6\,\text{gal} + 24\,\text{gal} = 30\,\text{gal}\). 3. Find one-third of the total: \(\frac{1}{3} \times 30\,\text{gal} = 10\,\text{gal}\). 4. Compare the amounts: \(10\,\text{gal} > 6\,\text{gal}\), so the fruit drink amount is greater.

There are \(10\,\text{gal}\) of fruit drink in the dispenser. This is greater than the \(6\,\text{gal}\) of apple juice.

- First find the number of green notebooks. - Multiply that number by the cost of one notebook. - Express the total in dollars and cents.

1. Find the number of green notebooks: \(18 - 12 = 6\). 2. Find the total cost in cents: \(6 \times 60 = 360\) cents. 3. Convert to dollars: \(360\) cents is \(\$3.60\).

All the green notebooks cost \(\$3.60\).

- First find the number of nonfiction shelves. - Then multiply the number of shelves by the number of books on each shelf. - Pay attention to which operation must come first.

1. Find the number of nonfiction shelves: \(18 - 7 = 11\). 2. Find the number of nonfiction books: \(11 \times 40 = 440\).

The library has \(440\) nonfiction books.

- First find Lisa's number of rows. - Find each person's total number of carrots. - Compare the two totals.

1. Find Lisa's number of rows: \(15 - 6 = 9\). 2. Find Lisa's number of carrots: \(9 \times 40 = 360\). 3. Find Tim's number of carrots: \(15 \times 20 = 300\). 4. Compare: \(360 > 300\), so Lisa has more carrots.

Lisa has more carrots. She has \(360\) carrots, while Tim has \(300\).

- First find how many bags can be filled from the harvest. - Once you know the number of bags, how can you find the total revenue?

1. Find the number of full bags: \(120\,\text{kg} \div 10\,\text{kg} = 12\). 2. Find the total revenue: \(12 \times \$7 = \$84\).

The farmer earns \(\$84\) altogether.

- How many sets are made when every set contains five cards? - First find the money earned from selling all the sets. - Which operation finds the amount left after the donation?

1. Find the number of card sets: \(200 \div 5 = 40\). 2. Find the total sales: \(40 \times \$4 = \$160\). 3. Subtract the donation: \(\$160 - \$65 = \$95\).

The class has \(\$95\) left for the class fund.

- First determine how many seedlings remain after the teachers plant some. - The remaining seedlings are split into \(6\) equal shares. - Use division after finding the remainder.

1. Find the number of seedlings remaining: \(120 - 30 = 90\). 2. Divide the remaining seedlings equally among the classes: \(90 \div 6 = 15\).

Each class receives \(15\) seedlings.

- Find how many balls remain after the missing balls are removed from the total. - Each can holds the same number of balls. - Divide the remaining number of balls by the number that fits in one can.

1. Find the number of tennis balls remaining: \(100 - 28 = 72\). 2. Divide by the number of balls in each can: \(72 \div 4 = 18\).

The club needs \(18\) cans.

- How many hours do Mia and Ben work altogether? - Use the total bulbs and total hours to find the shared hourly rate. - Multiply the hourly rate by each student’s work time.

1. Find the total work time: \(4 + 5 = 9\) hours. 2. Find the number of bulbs planted per hour: \(135 \div 9 = 15\) bulbs per hour. 3. Find Mia’s share: \(4 \times 15 = 60\) bulbs. 4. Find Ben’s share: \(5 \times 15 = 75\) bulbs.

Mia plants \(60\) bulbs, and Ben plants \(75\) bulbs.

- Add the class sizes to find the total number of students. - Find the amount per student by using a multiplication fact with the total number of students. - Multiply the per-student cost by each class size.

1. The total number of students is \(20 + 22 + 18 = 60\). 2. Since \(60 \times \$8 = \$480\), the cost per student is \(\$8\). 3. Class 4A pays \(20 \times \$8 = \$160\). 4. Class 4B pays \(22 \times \$8 = \$176\). 5. Class 4C pays \(18 \times \$8 = \$144\). 6. The class totals check because \(\$160 + \$176 + \$144 = \$480\).

a) Class 4A pays \(\$160\), Class 4B pays \(\$176\), and Class 4C pays \(\$144\). b) Each student pays \(\$8\).

- First find how many shells Lucas collects in the afternoon. - How many shells does he collect during the whole day? - What does “equally” tell you about the final operation? - Which operation separates a total into equal groups?

1. Find the number of shells collected in the afternoon: \(125 + 30 = 155\). 2. Find the total number of shells: \(125 + 155 = 280\). 3. Divide the shells equally among four bags: \(280 \div 4 = 70\).

Lucas puts \(70\) shells in each bag.

- Find the number of apples in each crate one at a time. - Pay attention to which crate is used for each comparison. - How many apples are there altogether? - To divide \(630\) by \(3\), consider the hundreds and tens separately.

1. Find the number of apples in the second crate: \(240 - 60 = 180\). 2. Find the number of apples in the third crate: \(180 + 30 = 210\). 3. Find the total number of apples: \(240 + 180 + 210 = 630\). 4. Divide equally among the three shelves: \(630 \div 3 = 210\).

Each shelf holds \(210\) apples.

- First find how many books are already in the first four bins. - Subtract to find how many books remain. - Determine how many empty bins are left. - Use division to share the remaining books equally.

1. Find the number of books placed in the first four bins: \(4\times 12=48\). 2. Find the number of books left: \(150-48=102\). 3. Find the number of remaining bins: \(10-4=6\). 4. Divide the remaining books equally: \(102\div 6=17\).

Each remaining bin will hold \(17\) books.

- Find how many pretzels remain after filling the first six sheets. - Find how many pretzels the last four sheets require. - Compare the amount available with the amount needed. - If the amount needed is greater, the difference is the shortage.

1. The first six sheets use \(6\times 15=90\) pretzels. 2. The number left is \(160-90=70\). 3. The remaining four sheets would require \(4\times 18=72\) pretzels. 4. Since \(72-70=2\), the baker is short by \(2\) pretzels.

No. The baker needs \(2\) more pretzels.

- First find how much flour was in each bin. - What happens to the amount in one bin when some flour is used? - Remember that flour is taken from only one bin.

1. Find the amount of flour in each bin: \(120\,\text{kg} \div 3 = 40\,\text{kg}\). 2. Subtract the amount taken from the first bin: \(40\,\text{kg} - 15\,\text{kg} = 25\,\text{kg}\).

The first bin has \(25\,\text{kg}\) of flour remaining.

- How much water is placed in each trough at the beginning? - How much water do the two horses drink altogether? - Subtract the total amount they drink from the amount in the first trough.

1. Find the amount of water in each trough: \(450\,\text{L} \div 5 = 90\,\text{L}\). 2. Find the total amount the horses drink: \(22\,\text{L} + 18\,\text{L} = 40\,\text{L}\). 3. Subtract the amount they drink: \(90\,\text{L} - 40\,\text{L} = 50\,\text{L}\).

The first trough has \(50\,\text{L}\) of water remaining.

- How many books were on each shelf at the beginning? - For each shelf, decide whether books are added or removed. - The two shelves will have different answers.

1. Find the original number of books on each shelf: \(150 \div 5 = 30\). 2. Find the number remaining on the top shelf: \(30 - 12 = 18\). 3. Find the new number on the bottom shelf: \(30 + 7 = 37\).

The top shelf has \(18\) nonfiction books, and the bottom shelf has \(37\) nonfiction books.

- Use multiplication facts to find each class’s price per student. - Then subtract the smaller per-student price from the larger one. - Check each unit price by multiplying by the class size.

1. Since \(20 \times \$6 = \$120\), Class 4A pays \(\$6\) per student. 2. Since \(25 \times \$4 = \$100\), Class 4B pays \(\$4\) per student. 3. Each Class 4B student saves \(\$6 - \$4 = \$2\).

Class 4A pays \(\$6\) per student, and Class 4B pays \(\$4\) per student. Each Class 4B student saves \(\$2\).

- Find how much each person spends. - Subtract each spending amount from \(\$25\). - Compare the amounts left and find their difference.

1. Leon spends \(3 \times \$4 = \$12\), so he has \(\$25 - \$12 = \$13\) left. 2. Sophie spends \(5 \times \$2 = \$10\), so she has \(\$25 - \$10 = \$15\) left. 3. Since \(\$15 > \$13\), Sophie has more money left. 4. The difference is \(\$15 - \$13 = \$2\).

Sophie has more money left. She has \(\$2\) more than Leon.

- First find the total cost of the special feed. - How many times does the cost of one bale of hay fit into the total amount? - A related basic multiplication fact may help with \(360 \div 9\).

1. Find the total cost of the special feed: \(8 \times \$45 = \$360\). 2. Divide by the cost of one bale of hay: \(\$360 \div \$9 = 40\).

The zookeeper could buy \(40\) bales of hay.

- First find the cost of one jump rope. - Then multiply the unit price by \(9\). - Check that your unit price gives the original cost for \(6\) ropes.

1. One jump rope costs \(\$24 \div 6 = \$4\). 2. Nine jump ropes cost \(9 \times \$4 = \$36\).

\(9\) jump ropes cost \(\$36\).

- How many seedlings went into one original bed? - The new beds will use the same number per bed. - You can find the total for all beds or add the new seedlings to those already planted.

1. Each original bed has \(48\div4=12\) seedlings. 2. All seven beds need \(7\times12=84\) seedlings. 3. As a check, the three new beds need \(3\times12=36\) seedlings, and \(48+36=84\).

The gardener needs \(84\) seedlings altogether.

- Find the amount drunk during each break separately. - Both fractions refer to the original \(8\,\text{gal}\). - How much lemonade was drunk altogether? - Subtract the total amount drunk from the original amount.

1. Find the amount drunk during the first break: \(\frac{1}{4} \times 8\,\text{gal} = 2\,\text{gal}\). 2. Find the amount drunk during the second break: \(\frac{1}{8} \times 8\,\text{gal} = 1\,\text{gal}\). 3. Find the total amount drunk: \(2\,\text{gal} + 1\,\text{gal} = 3\,\text{gal}\). 4. Subtract from the original amount: \(8\,\text{gal} - 3\,\text{gal} = 5\,\text{gal}\).

There are \(5\,\text{gal}\) of lemonade remaining.

- Find how many bags they pack together in \(1\) hour. - Use that rate to find how many hours \(100\) bags take. - Add the needed hours to the start time and compare with noon.

1. Find their combined rate: \(8 + 12 = 20\) bags per hour. 2. Find the time needed: \(100 \div 20 = 5\) hours. 3. Add \(5\) hours to the start time: \(8{:}00\) a.m. plus \(5\) hours is \(1{:}00\) p.m. 4. Since \(1{:}00\) p.m. is later than \(12{:}00\) p.m., Sarah is not correct.

No. Together they pack \(20\) bags per hour, so \(100\) bags take \(5\) hours. They finish at \(1{:}00\) p.m.

- How much does one shrub cost? - Find how many shrubs can be bought for \(\$117\). - Remember that the question asks for the total from both orders.

1. Find the cost of one shrub: \(\$72 \div 8 = \$9\). 2. Find the number of shrubs in the second order: \(\$117 \div \$9 = 13\). 3. Add the shrubs from both orders: \(8 + 13 = 21\).

The gardener orders \(21\) shrubs altogether.

- Does the second fraction refer to the original rope or only to the amount that remains? - Work one step at a time and record how much rope remains after each cut. - A drawing of a line divided into parts may help.

1. Find the length of the first piece: \(\frac{1}{2} \times 32\,\text{ft} = 16\,\text{ft}\). 2. After the first cut, \(32\,\text{ft} - 16\,\text{ft} = 16\,\text{ft}\) remain. 3. Find one-fourth of the remaining rope: \(\frac{1}{4} \times 16\,\text{ft} = 4\,\text{ft}\). 4. Subtract the second piece: \(16\,\text{ft} - 4\,\text{ft} = 12\,\text{ft}\).

a) The second piece is \(4\,\text{ft}\) long. b) \(12\,\text{ft}\) of rope remain.

- First find the cost of one basketball. - Once you know the unit price, find the cost of nine basketballs. - Subtract the total cost from the amount paid to find the change.

1. Find the cost of one basketball: \(\$32 \div 4 = \$8\). 2. Find the cost of nine basketballs: \(9 \times \$8 = \$72\). 3. Subtract the cost from the amount paid: \(\$100 - \$72 = \$28\).

The teacher receives \(\$28\) in change.

- First find how many muffins remain after removing the chocolate-decorated muffins. - Which operation divides a quantity equally among several trays? - Break the problem into two steps.

1. Find the number of muffins without chocolate decorations: \(250 - 70 = 180\). 2. Divide the remaining muffins equally among the trays: \(180 \div 6 = 30\).

Each serving tray holds \(30\) muffins.

- How many blocks remain after the broken ones are removed? - Next subtract the blocks used for the castle. - How many blocks are left for the small bins? - Divide that amount equally among the bins.

1. Find the number of unbroken blocks: \(640 - 40 = 600\). 2. Subtract the blocks used for the castle: \(600 - 120 = 480\). 3. Divide the remaining blocks equally among eight bins: \(480 \div 8 = 60\).

Each small bin holds \(60\) building blocks.

- First find one-half of \(60\). - How can you find one-fifth of a number? - After finding the numbers of apples and bananas, how can you find the rest?

1. Find the number of apples: \(\frac{1}{2} \times 60 = 30\). 2. Find the number of bananas: \(\frac{1}{5} \times 60 = 12\). 3. Subtract the apples and bananas from the total: \(60 - 30 - 12 = 18\).

There are \(18\) pears in the crate.

- How many students are one-fourth of the whole group? - Find the size of each group whose fraction is given. - What remains after subtracting those groups from the total? - The question asks for the number of students, not the distance.

1. Find the number who run \(1\,\text{mi}\): \(\frac{1}{4} \times 120 = 30\). 2. Find the number who run \(2\,\text{mi}\): \(\frac{1}{3} \times 120 = 40\). 3. Subtract the first two groups from the total: \(120 - 30 - 40 = 50\).

\(50\) students run the \(3\,\text{mi}\) distance.

- First find one-fourth of \(120\). - How can you find one-tenth of a number? - After finding the sunflower and rose seedlings, how can you find what remains?

1. Find the number of sunflower seedlings: \(\frac{1}{4} \times 120 = 30\). 2. Find the number of rose seedlings: \(\frac{1}{10} \times 120 = 12\). 3. Subtract both groups from the total: \(120 - 30 - 12 = 78\).

There are \(78\) tulip seedlings.

- Find the exact number in each of the three groups whose fractions are given. - What operation helps you find a fraction such as one-sixth of a set? - Compare all four group sizes to identify the design that appears most often.

1. Find the number of space stickers: \(\frac{1}{8} \times 240 = 30\). 2. Find the number of animal stickers: \(\frac{1}{3} \times 240 = 80\). 3. Find the number of sports stickers: \(\frac{1}{6} \times 240 = 40\). 4. Find the number of smiley-face stickers: \(240 - 30 - 80 - 40 = 90\). 5. Compare the groups: \(90 > 80 > 40 > 30\), so smiley faces appear most often.

Julia has \(90\) smiley-face stickers. Smiley faces appear most often.

- Find the number of packages of each flavor. - Add to find how many packages are in the freezer. - Subtract from the freezer's capacity.

1. Find the number of vanilla packages: \(4 \times 12 = 48\). 2. Find the number of chocolate packages: \(3 \times 15 = 45\). 3. Find the total number added: \(48 + 45 = 93\). 4. Find the remaining space: \(100 - 93 = 7\).

The freezer can hold \(7\) more packages.

- First find the cost of one notebook. - How much do all seven notebooks cost? - Subtract the total cost from the amount paid. - Compare the change with \(\$5\).

1. Find the cost of one notebook: \(\$12 \div 4 = \$3\). 2. Find the cost of seven notebooks: \(7 \times \$3 = \$21\). 3. Find the change: \(\$25 - \$21 = \$4\). 4. Compare the change with five dollars: \(\$4 < \$5\), so the change is less than \(\$5\).

Ms. Weber receives less than \(\$5\) in change because she receives exactly \(\$4\).

- First determine the cost of one soccer ball. - How much more does one basketball cost? - Find the costs of the soccer balls and basketballs separately. - Add the two costs to find the total.

1. Find the cost of one soccer ball: \(\$40 \div 4 = \$10\). 2. Find the cost of one basketball: \(\$10 + \$5 = \$15\). 3. Find the cost of twelve soccer balls: \(12 \times \$10 = \$120\). 4. Find the cost of five basketballs: \(5 \times \$15 = \$75\). 5. Add the two costs: \(\$120 + \$75 = \$195\).

The total cost of the order is \(\$195\).

- First find how many bags the new machine packs in one hour. - Then compare the two hourly rates. - The question asks for a difference, not the new machine’s full rate.

1. Find the new machine’s hourly rate: \(1648 \div 8 = 206\) bags per hour. 2. Find the difference between the hourly rates: \(206 - 198 = 8\) bags per hour.

The new machine packs \(8\) more bags per hour.

- Find the total number of blue chairs. - Find the total number of red chairs. - Add the two totals.

1. Find the number of blue chairs: \(8 \times 15 = 120\). 2. Find the number of red chairs: \(6 \times 14 = 84\). 3. Add the two amounts: \(120 + 84 = 204\).

There are \(204\) chairs altogether.

- Find the cost of \(100\) sheets. - Determine how many sheets remain for Class 4C. - Express each class’s share as groups of \(100\) sheets.

1. The paper costs \(\$24 \div 6 = \$4\) for each group of \(100\) sheets. 2. Class 4A takes two groups of \(100\), so it pays \(2 \times \$4 = \$8\). 3. Class 4B takes three groups of \(100\), so it pays \(3 \times \$4 = \$12\). 4. Class 4C receives \(600 - 200 - 300 = 100\) sheets, so it pays \(\$4\).

Class 4A pays \(\$8\), Class 4B pays \(\$12\), and Class 4C pays \(\$4\).

- How many laptops are placed in the computer labs altogether? - Find how many laptops remain after the labs receive theirs. - Which operation divides the remaining laptops equally among the classrooms?

1. Find the number of laptops placed in the computer labs: \(4 \times 16 = 64\). 2. Find the number of laptops remaining: \(144 - 64 = 80\). 3. Divide the remaining laptops equally among eight classrooms: \(80 \div 8 = 10\).

Each classroom receives \(10\) laptops.

- How can you find the price of one kilogram in each bag? - Does a higher total price always mean a higher unit price? - Find each unit price separately and compare them.

1. Find the price per kilogram for the small bag: \(\$12 \div 4 = \$3\) per kilogram. 2. Find the price per kilogram for the large bag: \(\$20 \div 10 = \$2\) per kilogram. 3. Since \(\$2 < \$3\), the large bag has the lower price per kilogram. Lucas is not correct.

Lucas is not correct. Apples cost \(\$3\) per kilogram in the small bag and \(\$2\) per kilogram in the large bag.

- Compare the teams’ practice times for one practice first. - Determine how that difference accumulates over all practices in one week. - Extend the weekly difference across four weeks.

1. The difference for one practice is \(60-45=15\) minutes. 2. The weekly difference is \(3 \times 15=45\) minutes. 3. Over four weeks, the difference is \(4 \times 45=180\) minutes.

The Sharks practice \(180\) more minutes over four weeks.

- Identify what the customer is trying to compare. - First find the number of seedlings in one package. - Find twice the original number of packages. - Calculate the actual packages needed for \(126\) seedlings and compare.

1. Find the number of seedlings per package: \(56\div 4=14\). 2. Find the packages needed for \(126\) seedlings: \(126\div 14=9\). 3. Twice the original \(4\) packages would be \(4\times 2=8\) packages. 4. Since \(9\neq 8\), the customer is incorrect.

No. The customer needs \(9\) packages, but twice \(4\) packages is only \(8\) packages.

- How many kilograms of apples are harvested on Tuesday? - What is the total harvest for both days? - If each crate holds more apples, are more or fewer crates needed?

1. Find Tuesday's harvest: \(56\,\text{kg} - 16\,\text{kg} = 40\,\text{kg}\). 2. Find the total harvest: \(56\,\text{kg} + 40\,\text{kg} = 96\,\text{kg}\). 3. Find the number of eight-kilogram crates: \(96\,\text{kg} \div 8\,\text{kg} = 12\). 4. Fewer crates would be needed if each crate held \(12\,\text{kg}\), because each crate would hold more apples.

a) \(12\) crates are filled altogether. b) Fewer crates would be needed because each crate would hold more apples.

- First determine the cost of one package. - Then determine how many equal \(\$3\) amounts fit in the budget. - Check that the total cost does not exceed \(\$45\).

1. Find the cost of one package: \(\$12 \div 4 = \$3\). 2. Divide the budget by the cost per package: \(\$45 \div \$3 = 15\).

Ms. Weber can buy at most \(15\) packages of juice.

- First find the cost of one bush. - How much do all fifteen bushes cost? - Find the total amount the gardener pays. - Subtract the cost from the amount paid to find the change.

1. Find the cost of one rose bush: \(\$36 \div 4 = \$9\). 2. Find the cost of fifteen bushes: \(15 \times \$9 = \$135\). 3. Find the amount paid: \(\$100 + \$50 = \$150\). 4. Find the change: \(\$150 - \$135 = \$15\).

The gardener receives \(\$15\) in change.

- Think about how the total weight is shared by six apples. - What single-apple weight would fit the total? - Use it to find the weight of four apples.

1. Each apple weighs \(900\div6=150\,\text{g}\). 2. Four apples weigh \(4\times150=600\,\text{g}\).

Four apples weigh \(600\,\text{g}\).

- First determine the total number packaged during the full run. - Compare \(20\) minutes with the given \(5\)-minute interval. - Then account for the boxes that cannot be used.

1. Twenty minutes contains \(4\) groups of \(5\) minutes, so the machine packages \(45 \times 4=180\) boxes. 2. Remove the \(35\) boxes that fail inspection: \(180-35=145\).

There are \(145\) usable boxes.

- First find the number of blocks in each given color group. - How many blocks remain after subtracting the red and blue blocks? - Compare the two numbers named in the question.

1. Find the number of red blocks: \(\frac{1}{3} \times 120 = 40\). 2. Find the number of blue blocks: \(\frac{1}{8} \times 120 = 15\). 3. Add the red and blue blocks: \(40 + 15 = 55\). 4. Subtract from the total to find the yellow blocks: \(120 - 55 = 65\). 5. Since \(65 > 40\), there are more yellow blocks than red blocks.

There are \(65\) yellow blocks. Because \(65 > 40\), there are more yellow blocks than red blocks.

- First find how many flowers were planted in all the boxes. - Include the flowers that remained on the cart. - Calculate the two box sizes separately before combining them.

1. The large boxes use \(22\times 12=264\) pansies. 2. The small boxes use \(18\times 8=144\) pansies. 3. The boxes use \(264+144=408\) pansies altogether. 4. Add the remaining plants: \(408+25=433\).

There were \(433\) pansies at the beginning.

- Find the number of bottles for each type of drink. - Find the total number of bottles in the proposed order. - Compare the proposed total with the answer from part a.

1. The water cases contain \(16\times 12=192\) bottles. 2. The juice cases contain \(14\times 15=210\) bottles. 3. The actual delivery contains \(192+210=402\) bottles. 4. The proposed order would contain \(20\times 20=400\) bottles. 5. Since \(402>400\), the volunteer is incorrect. The actual delivery has \(2\) more bottles.

a) \(402\) bottles b) No. The proposed order would have \(400\) bottles, which is \(2\) fewer than the actual delivery.

- Find the total cost of each kind of drink. - Add those costs to find the amount spent. - Combine the amount spent with the money remaining.

1. The apple juice costs \(15 \times \$2 = \$30\). 2. The sparkling water costs \(24 \times \$1 = \$24\). 3. The class spends \(\$30 + \$24 = \$54\). 4. The starting amount was \(\$54 + \$13 = \$67\).

The class fund originally contained \(\$67\).

- Find the combined length of all the \(50\)-centimeter pieces. - Find the combined length of all the \(30\)-centimeter pieces. - The original strip included both groups of pieces and the leftover piece.

1. Find the total length of the \(50\)-centimeter pieces: \(18 \times 50 = 900\,\text{cm}\). 2. Find the total length of the \(30\)-centimeter pieces: \(25 \times 30 = 750\,\text{cm}\). 3. Add the lengths of all the cut pieces: \(900 + 750 = 1650\,\text{cm}\). 4. Add the remaining piece: \(1650 + 85 = 1735\,\text{cm}\).

The original wooden strip was \(1735\,\text{cm}\) long.

- First find how many seedlings have already been planted. - If the same number were planted each hour, divide the total planted among the four hours. - Break the problem into two operations.

1. Find the number of seedlings already planted: \(120 - 32 = 88\). 2. Divide by the number of hours: \(88 \div 4 = 22\).

The gardener planted \(22\) seedlings per hour.

- How many cars have already been packaged? - If that amount was completed in three hours, how many were packaged per hour? - Think about hundreds and tens when dividing.

1. Find the number of cars already packaged: \(400 - 160 = 240\). 2. Divide by the number of hours: \(240 \div 3 = 80\).

An average of \(80\) cars were packaged per hour.

- First find how many jars remain after the market sale. - Divide the remaining jars equally among the boxes. - Compare your result with the helper's claim.

1. Find the number of jars remaining after the sale: \(380 - 80 = 300\). 2. Divide the remaining jars equally among five boxes: \(300 \div 5 = 60\). 3. Compare the result with the helper's claim: \(60 \ne 50\). 4. The helper is not correct. Each box should contain \(60\) jars.

No. The helper is not correct; each box should contain \(60\) jars.

- How many books are already on the shelf if there is room for \(120\) more? - Those books came from eight equal boxes. How can you find the number in one box? - First find the total number already unpacked.

1. Find the number of books already on the shelf: \(600 - 120 = 480\). 2. Divide by the number of boxes: \(480 \div 8 = 60\).

Each box contained \(60\) books.

- How many bulbs remain after the rotten ones are removed? - The remaining bulbs are equally distributed among all six boxes. Which operation finds the number in each box?

1. Find the total number of bulbs remaining: \(450 - 30 = 420\). 2. Divide the remaining bulbs equally among the six boxes: \(420 \div 6 = 70\).

Each box now contains \(70\) flower bulbs.

- Add the numbers of marbles to find the total. - Convert the total price to cents and use multiplication to find the cost per marble. - Multiply the cost per marble by each person’s number of marbles.

1. The bag contains \(15 + 25 + 20 = 60\) marbles. 2. The total cost is \(1200\) cents. Since \(60 \times 20 = 1200\), each marble costs \(20\) cents. 3. Lucas pays \(15 \times 20 = 300\) cents, or \(\$3\). 4. Tim pays \(25 \times 20 = 500\) cents, or \(\$5\). 5. Sarah pays \(20 \times 20 = 400\) cents, or \(\$4\).

Lucas should pay \(\$3\), Tim should pay \(\$5\), and Sarah should pay \(\$4\).

- Find how many packages are needed for \(30\) stickers. - Multiply the number of packages by the price per package. - Compare the total cost with \(\$3\).

1. Thirty stickers require \(30 \div 5 = 6\) packages. 2. The six packages cost \(6 \times 60 = 360\) cents, or \(\$3.60\). 3. Since \(\$3.60 > \$3.00\), Lucas does not have enough money.

No. The \(30\) stickers cost \(\$3.60\), so Lucas is \(60\) cents short.

- The room amounts are linked, so determine them in an order that uses the comparison statements. - Part 2 requires all three room amounts, including the one given at the start. - Check that the amounts satisfy both comparison statements.

1. Room 10 collects \(14 \times 4=56\,\text{kg}\). 2. Room 12 collects \(56\,\text{kg}-20\,\text{kg}=36\,\text{kg}\). 3. The three rooms collect \(14\,\text{kg}+56\,\text{kg}+36\,\text{kg}=106\,\text{kg}\) altogether.

1) Room 12 collects \(36\,\text{kg}\). 2) The three rooms collect \(106\,\text{kg}\) altogether.

- Determine the Friday amounts in order because each later amount depends on an earlier one. - For part 2, compare the third-Friday result with the combined amount from the first two Fridays. - Check that the comparison uses the correct totals.

1. On the second Friday, they plant \(15 \times 3=45\) flowers. 2. On the third Friday, they plant \(45+12=57\) flowers. 3. The first two Fridays total \(15+45=60\) flowers. 4. Since \(57<60\), the third-Friday amount is smaller.

1) They plant \(57\) flowers on the third Friday. 2) No. They plant \(60\) flowers on the first two Fridays combined, and \(57<60\).

- First find the total amount Julia spends during the first \(3\) days. - How many days remain in the week? - Interpret “\(\$12\) more than the total she has spent” carefully. - Organizing the amounts and days in a small table may help.

1. Find how much Julia spends during the first \(3\) days: \(3 \times \$8 = \$24\). 2. Find how much money she has left: \(\$24 + \$12 = \$36\). 3. Find the number of days remaining: \(7 - 3 = 4\) days. 4. Divide the remaining money equally: \(\$36 \div 4 = \$9\) per day.

Julia can spend \(\$9\) each day for the remaining \(4\) days.

- How much water is used during the first \(4\) days? - The amount left is \(90\,\text{gal}\) less than that total. What amount remains? - Divide the remaining amount among the next \(5\) days.

1. Find the total amount used during the first \(4\) days: \(4 \times 60 = 240\,\text{gal}\). 2. Find the amount left in the tank: \(240 - 90 = 150\,\text{gal}\). 3. Divide the remaining water equally among \(5\) days: \(150 \div 5 = 30\,\text{gal}\) per day.

They can use \(30\,\text{gal}\) each day for the next \(5\) days.

- Find the money earned from books first. - Subtract that amount from the total earnings. - Use the remaining earnings to find the number of stuffed animals, then add both item counts.

1. The books bring in \(4 \times \$6 = \$24\). 2. The stuffed animals bring in \(\$84 - \$24 = \$60\). 3. The booth sells \(\$60 \div \$5 = 12\) stuffed animals. 4. The total number of items is \(4 + 12 = 16\).

The booth sold \(16\) items altogether.

- Find the total weight in the large crates. - Subtract that amount from the full delivery. - Determine how many \(11\)-pound crates make the remaining weight, then add both kinds of crates.

1. Find the apples in the large crates: \(6 \times 18\,\text{lb} = 108\,\text{lb}\). 2. Find the apples in the small crates: \(152\,\text{lb} - 108\,\text{lb} = 44\,\text{lb}\). 3. Four small crates contain \(4 \times 11\,\text{lb} = 44\,\text{lb}\), so there are \(4\) small crates. 4. Add all the crates: \(6 + 4 = 10\).

The delivery contains \(10\) crates altogether.

- How many spaces are there between \(7\) posts? - Draw a shorter row of posts and count the spaces. - Divide the total length by the number of equal spaces.

1. Seven posts create \(7 - 1 = 6\) equal spaces between neighboring posts. 2. Divide the total length by the number of spaces: \(24 \div 6 = 4\,\text{m}\).

Neighboring posts are \(4\,\text{m}\) apart.

- First find how many marbles Sophie removes. - Add the amounts removed by Max and Sophie. - The starting number includes the marbles removed and the marbles left.

1. Find the number Sophie removes: \(24 \times 3 = 72\). 2. Find the total number removed: \(24 + 72 = 96\). 3. Add the marbles left: \(96 + 15 = 111\).

There were \(111\) marbles in the box at the beginning.

- Find the amount saved in each month in order. - Add the three monthly amounts. - Subtract the total savings from the bicycle's price.

1. Find the February savings: \(15 \times 3 = 45\), so he saves \(\$45\). 2. Find the March savings: \(\$45 - \$20 = \$25\). 3. Find the total saved: \(\$15 + \$45 + \$25 = \$85\). 4. Find the amount still needed: \(\$240 - \$85 = \$155\).

Lucas still needs \(\$155\).

- Think about the spaces between the rosebushes. - How many spaces are there between \(3\) plants? Between \(4\) plants? - Multiply the number of spaces, not the number of plants, by \(4\,\text{m}\).

1. There is one fewer space than rosebushes: \(21 - 1 = 20\) spaces. 2. Multiply the number of spaces by the distance between neighboring rosebushes: \(20 \times 4 = 80\,\text{m}\).

The row is \(80\,\text{m}\) long.

- First find the number of equal spaces along the string. - Remember that flags are placed at both the beginning and the end. - Repeat the process for the second spacing, then subtract.

1. With \(3\)-meter spacing, there are \(30 \div 3 = 10\) spaces. Including both endpoints requires \(10 + 1 = 11\) flags. 2. With \(5\)-meter spacing, there are \(30 \div 5 = 6\) spaces. Including both endpoints requires \(6 + 1 = 7\) flags. 3. The number saved is \(11 - 7 = 4\) flags.

a) \(11\) flags are needed. b) The larger spacing uses \(4\) fewer flags.

- First find how far the small robot travels in one minute. - Which operation does the word “farther” suggest? - Use place value and a related basic division fact.

1. Find the distance the first robot travels in one minute: \(320\,\text{m} \div 4 = 80\,\text{m}\). 2. Add the additional distance traveled by the newer robot: \(80\,\text{m} + 15\,\text{m} = 95\,\text{m}\).

The newer robot travels \(95\,\text{m}\) in one minute.

- How many bottles are in one case? - Decompose a factor by place value if helpful. - Remember that every case contains the same number of bottles.

1. Find the number of bottles in one case: \(72 \div 6 = 12\). 2. Find the number of bottles in fifteen cases: \(15 \times 12 = 180\).

\(180\) bottles are delivered for the celebration.

- What operation finds half of an amount? - First find the number of daylight hours in December. - Subtract the daylight hours from the full \(24\)-hour day.

1. Find the December daylight period: \(12\,\text{hr} \div 2 = 6\,\text{hr}\). 2. Find the nighttime period: \(24\,\text{hr} - 6\,\text{hr} = 18\,\text{hr}\).

The night lasts \(18\) hours.

- First find the one-way distance to the mountains. - A round trip includes the trip there and the trip back. - How many times is the one-way distance traveled?

1. Find the one-way distance to the mountains: \(534 + 218 = 752\,\text{miles}\). 2. Double the one-way distance for the round trip: \(752 \times 2 = 1504\,\text{miles}\).

The family would travel \(1504\,\text{miles}\) on the round trip.

- First find how many sheets all nine groups receive. - Use the number of groups and the amount per group. - Subtract the distributed amount from \(750\).

1. Find the total number distributed: \(9 \times 65 = 585\). 2. Subtract from the amount purchased: \(750 - 585 = 165\).

There are \(165\) sheets left in storage.

- Find each person's total number of pencils. - Break apart the larger factors if helpful. - Compare the products and subtract to find the difference.

1. Find Ms. Miller's total: \(4 \times 125 = 500\). 2. Find Mr. Smith's total: \(3 \times 160 = 480\). 3. Compare: \(500 > 480\), so Ms. Miller buys more. 4. Find the difference: \(500 - 480 = 20\).

Ms. Miller buys \(20\) more colored pencils than Mr. Smith.

- Find the total amount needed for all five bags. - Break apart \(145\) to make the multiplication easier. - Compare the amount needed with \(750\,\text{g}\).

1. Find the amount needed for five bags: \(5 \times 145\,\text{g} = 725\,\text{g}\). 2. Since \(725\,\text{g} < 750\,\text{g}\), there are enough grapes. 3. Find the amount left: \(750\,\text{g} - 725\,\text{g} = 25\,\text{g}\).

Yes. Lucas can fill the five bags and will have \(25\,\text{g}\) of grapes left.

- Find each child's total number of stickers. - Use multiplication for each set of equal groups. - Subtract the smaller total from the larger total.

1. Find Lucas's total: \(8 \times 15 = 120\). 2. Find Mia's total: \(6 \times 22 = 132\). 3. Since \(132 > 120\), Mia has more stickers. 4. Find the difference: \(132 - 120 = 12\).

Mia has \(12\) more stickers than Lucas.

- How many tulips are used in the large bouquets altogether? - How many tulips remain for the small bouquets? - Which operation divides a quantity into equal groups? - Check that you have used every important number in the problem.

1. Find the number of tulips used in the large bouquets: \(8 \times 15 = 120\). 2. Find the number of tulips remaining: \(400 - 120 = 280\). 3. Divide the remaining tulips into groups of seven: \(280 \div 7 = 40\).

The workers can make \(40\) small bouquets.

- First find the amount spent on rose bushes. - Subtract that amount from the total to find the amount spent on lavender plants. - How many times does the price of one lavender plant fit into that amount? - Work through the problem one step at a time.

1. Find the total cost of the rose bushes: \(8 \times \$12 = \$96\). 2. Find the amount spent on lavender plants: \(\$151 - \$96 = \$55\). 3. Divide by the cost of one lavender plant: \(\$55 \div \$5 = 11\).

The gardener buys \(11\) lavender plants.

- How much does Sarah spend after receiving change? - How much do all the pens cost? - Subtract the pen cost to find the amount spent on drawing pads. - Which operation finds the number of pads from their total cost?

1. Find the total amount spent: \(\$100 - \$7 = \$93\). 2. Find the cost of the pens: \(15 \times \$4 = \$60\). 3. Find the amount spent on drawing pads: \(\$93 - \$60 = \$33\). 4. Divide by the cost of one drawing pad: \(\$33 \div \$3 = 11\).

Sarah buys \(11\) drawing pads.

- Find each family’s plant cost first. - Add the fertilizer or soil cost to each subtotal. - Compare the totals and subtract to find the difference.

1. The Johnson family spends \(4 \times \$125 + \$38 = \$500 + \$38 = \$538\). 2. The Patel family spends \(6 \times \$82 + \$54 = \$492 + \$54 = \$546\). 3. Since \(\$546 > \$538\), the Patel family spends more. 4. The difference is \(\$546 - \$538 = \$8\).

The Patel family spends more by \(\$8\).

- Find the total admission cost under Offer A. - Add the one-time bus fee to that amount. - Compare the result with the flat price and find the difference.

1. Offer A costs \(24 \times \$12 + \$55 = \$288 + \$55 = \$343\). 2. Offer B costs \(\$310\). 3. Since \(\$310 < \$343\), Offer B costs less. 4. The class saves \(\$343 - \$310 = \$33\).

Offer B costs less. The class saves \(\$33\).

- How can you find the weight in one bag for each class? - Which operation divides a total equally among several bags? - Compare the two per-bag weights at the end.

1. Find the weight in each Class 3A bag: \(72\,\text{kg} \div 8 = 9\,\text{kg}\). 2. Find the weight in each Class 3B bag: \(36\,\text{kg} \div 6 = 6\,\text{kg}\). 3. Compare the results: \(9\,\text{kg} > 6\,\text{kg}\), so Class 3A collects more per bag.

a) Each Class 3A bag contains \(9\,\text{kg}\). b) Each Class 3B bag contains \(6\,\text{kg}\). c) Class 3A collects more per bag.

- How many plants are on one table in the first section? - How many plants are on one table in the second section? - Once you know both values, find their difference. - Related basic division facts may help.

1. Find the number of plants on each table in the first section: \(450 \div 9 = 50\). 2. Find the number of plants on each table in the second section: \(490 \div 7 = 70\). 3. Find the difference: \(70 - 50 = 20\).

Each table in the second section has \(20\) more plants.

- First find the number of notebooks one student receives. - Multiply that amount by the number of students. - Another method is to find the total for each notebook type separately and then add.

1. One student receives \(4+3=7\) notebooks. 2. For \(24\) students, the total is \(24\times 7=168\) notebooks. 3. Another method is to calculate \(24\times 4=96\) lined notebooks and \(24\times 3=72\) graph-paper notebooks, then add \(96+72=168\).

\(168\) notebooks must be ordered.

- How much do all the ponies eat together in one day? - How many days must the hay last? - Compare the amount needed with the hay available.

1. The ponies eat \(6\times10=60\,\text{lb}\) of hay in one day. 2. Over \(14\) days, they eat \(60\times14=840\,\text{lb}\). 3. The supply is enough because \(840<900\), and \(900-840=60\,\text{lb}\) remains.

Yes. The ponies need \(840\,\text{lb}\) of hay, and \(60\,\text{lb}\) will be left.

- First find the mother's number of seashells. - Use that amount to find the cousin's number. - Then compare the cousin's amount with Leonie's.

1. Find the mother's number of seashells: \(12 \times 6 = 72\). 2. Find the cousin's number: \(72 - 50 = 22\). 3. Find the difference between the cousin and Leonie: \(22 - 12 = 10\).

The cousin has \(10\) more seashells than Leonie.

- First find the total amount the coach spends. - Find the cost of all the basketballs. - The remaining part of the spending is the cost of the sports bags. - Divide that amount equally among the four bags.

1. Find the total amount spent: \(\$450 - \$194 = \$256\). 2. Find the cost of the basketballs: \(6 \times \$24 = \$144\). 3. Find the amount spent on sports bags: \(\$256 - \$144 = \$112\). 4. Divide by the number of sports bags: \(\$112 \div 4 = \$28\).

One sports bag costs \(\$28\).

- Find the starting total on all eight shelves. - Subtract the notebooks sold. - Then add the new delivery.

1. Find the starting number of notebooks: \(8 \times 42 = 336\). 2. Subtract the notebooks sold: \(336 - 175 = 161\). 3. Add the delivery: \(161 + 60 = 221\).

There are \(221\) notebooks on the shelves Saturday morning.

- Find the amount paid to all workers for one day before extending it across the work period. - Keep both the number of workers and the number of days in your reasoning. - Check that the final total covers every worker-day in the situation.

1. All \(6\) workers earn \(\$125 \times 6=\$750\) in one day. 2. Over \(4\) days, the owner pays \(\$750 \times 4=\$3000\).

The owner will pay \(\$3000\) in total.

- Find the total number of cases first. - Multiply the number of cases by the cost per case. - Subtract the purchase cost from the budget.

1. The committee buys \(6 \times 3 = 18\) cases. 2. The cases cost \(18 \times \$14 = \$252\). 3. The amount remaining is \(\$300 - \$252 = \$48\).

\(\$48\) will remain.

- Find the total weight of the small sacks. - Find the total weight of the large sacks. - Add the two results.

1. Find the weight of the small sacks: \(24 \times 15\,\text{lb} = 360\,\text{lb}\). 2. Find the weight of the large sacks: \(16 \times 45\,\text{lb} = 720\,\text{lb}\). 3. Add the two amounts: \(360\,\text{lb} + 720\,\text{lb} = 1080\,\text{lb}\).

The potato harvest weighs \(1080\,\text{lb}\).

- Find the number of seats in each auditorium separately. - Multiply the rows by the seats in each row. - Add the two auditorium totals.

1. The first auditorium has \(14\times 18=252\) seats. 2. The second auditorium has \(12\times 22=264\) seats. 3. Together, they have \(252+264=516\) seats.

The two auditoriums have \(516\) seats altogether.

- Find how many bulbs would be planted if none were damaged. - Find the total number of damaged bulbs. - You may subtract the damaged bulbs by type or from the combined total. - Organize the tulip and daffodil calculations separately.

1. The tulip beds were planned for \(8\times 55=440\) bulbs. 2. The daffodil beds were planned for \(12\times 45=540\) bulbs. 3. The total planned number is \(440+540=980\) bulbs. 4. The number of damaged bulbs is \(15+20=35\). 5. The number of healthy bulbs is \(980-35=945\). 6. Alternatively, \(440-15=425\) healthy tulip bulbs and \(540-20=520\) healthy daffodil bulbs, and \(425+520=945\).

\(945\) healthy bulbs will be planted.

- First find the total cost of the type of rose whose quantity is known. - How much of the total cost remains for the other roses? - How can you find the number of roses bought with that amount? - Finally, compare the two total costs.

1. Find the total cost of the red rose plants: \(12 \times \$5 = \$60\). 2. Find the amount spent on white rose plants: \(\$95 - \$60 = \$35\). 3. Find the number of white rose plants: \(\$35 \div \$7 = 5\). 4. Find the difference between the two total costs: \(\$60 - \$35 = \$25\).

The gardener buys \(5\) white rose plants. The red rose plants cost \(\$25\) more altogether than the white rose plants.

- First find how many children attend. - How many adults attend if their tickets cost \(\$54\) altogether? - To divide \(78\) by \(6\), you can break \(78\) into easier parts. - Add the numbers of children and adults at the end.

1. Find the number of children: \(\$78 \div \$6 = 13\). 2. Find the number of adults: \(\$54 \div \$9 = 6\). 3. Find the total number of people: \(13 + 6 = 19\).

\(19\) people attend the theater performance altogether.

- Find the price of one lavender plant. - Use the equal quantities to find the total cost efficiently. - Compare the total with the budget and find the difference.

1. One lavender plant costs \(\$16 - \$9 = \$7\). 2. One rose bush and one lavender plant cost \(\$16 + \$7 = \$23\). 3. Twenty pairs cost \(20 \times \$23 = \$460\). 4. Since \(\$460 < \$500\), the budget is enough. 5. The amount left is \(\$500 - \$460 = \$40\).

Yes. The total cost is \(\$460\), and \(\$40\) remains.

- How many children's fiction books are there? - Find the total in the two groups that are not comics. - Add all three groups only after finding the number of comics.

1. There are \(64\times5=320\) children's fiction books. 2. The first two groups contain \(64+320=384\) books. 3. There are \(384\times2=768\) comic books. 4. The library has \(384+768=1152\) books altogether.

The library has \(1152\) books altogether.

- Find Tuesday’s amount first. - Combine Monday and Tuesday before finding Wednesday’s amount. - Use that intermediate sum to calculate Wednesday’s production. - Include all three days in the final total.

1. Tuesday’s amount is \(132\times 3=396\) rolls. 2. Monday and Tuesday together total \(132+396=528\) rolls. 3. Wednesday’s amount is \(528\times 4=2112\) rolls. 4. The three-day total is \(528+2112=2640\) rolls.

The bakery makes \(2640\) rolls over the three days.

- Find the total weight of all the apple crates. - Find the total weight of all the pear crates. - Subtract to find the difference.

1. Find the total weight of the apples: \(14 \times 12\,\text{lb} = 168\,\text{lb}\). 2. Find the total weight of the pears: \(16 \times 9\,\text{lb} = 144\,\text{lb}\). 3. Find the difference: \(168\,\text{lb} - 144\,\text{lb} = 24\,\text{lb}\).

The apples weigh \(24\,\text{lb}\) more than the pears.

- Find the weight of each kind of material. - Add the two cargo weights. - Subtract the current cargo from the maximum capacity.

1. Find the sand's weight: \(25 \times 20\,\text{lb} = 500\,\text{lb}\). 2. Find the gravel's weight: \(18 \times 15\,\text{lb} = 270\,\text{lb}\). 3. Find the current cargo weight: \(500\,\text{lb} + 270\,\text{lb} = 770\,\text{lb}\). 4. Find the remaining capacity: \(1000\,\text{lb} - 770\,\text{lb} = 230\,\text{lb}\).

The truck can carry \(230\,\text{lb}\) more.

- Use each multiplicative comparison to find the amount raised by each class. - Add the three amounts. - Compare the total with the equipment cost.

1. Class 4B raises \(3 \times \$135 = \$405\). 2. Class 4C raises \(2 \times \$405 = \$810\). 3. Together, the classes raise \(\$135 + \$405 + \$810 = \$1350\). 4. Since \(\$1350 < \$1500\), they did not raise enough. 5. They are short by \(\$1500 - \$1350 = \$150\).

No. The classes raised \(\$1350\), so they are \(\$150\) short.

- Find the knee-pad price first. - Use that price to find the helmet price. - Add all three costs, then subtract from the amount saved.

1. The knee pads cost \(\$84 \div 2 = \$42\). 2. The helmet costs \(\$42 + \$15 = \$57\). 3. The three items cost \(\$84 + \$42 + \$57 = \$183\). 4. Lucas has \(\$220 - \$183 = \$37\) left.

Lucas will have \(\$37\) left.

- Find one-third of the rental fee. - Use the drink cost to find the food cost. - Add all planned expenses, then subtract from the amount raised.

1. The drinks cost \(\$120 \div 3 = \$40\). 2. The food costs \(3 \times \$40 = \$120\). 3. The rental, drinks, and food cost \(\$120 + \$40 + \$120 = \$280\). 4. The amount available for decorations is \(\$350 - \$280 = \$70\).

\(\$70\) is available for decorations.

- A week has \(7\) days. - Find each family’s total water use separately. - Compare the two totals. - Subtract to find the difference.

1. Find the Miller family’s use in \(7\) days: \(145 \times 7 = 1015\,\text{gal}\). 2. Find the Schmidt family’s use in \(9\) days: \(112 \times 9 = 1008\,\text{gal}\). 3. Since \(1015 > 1008\), the Miller family uses more water. 4. Find the difference: \(1015 - 1008 = 7\,\text{gal}\).

The Miller family uses \(1015\,\text{gal}\), which is \(7\,\text{gal}\) more than the Schmidt family’s \(1008\,\text{gal}\).

- Find the total number of rolls for each part of the day. - The number of sheets alone does not determine the total. - Compare the two products.

1. In the morning, the baker makes \(14\times 18=252\) rolls. 2. In the afternoon, the baker makes \(16\times 15=240\) rolls. 3. Since \(252>240\), the baker is incorrect. 4. The morning total is \(252-240=12\) rolls greater.

No. The baker made \(252\) rolls in the morning and \(240\) in the afternoon, so the morning total was \(12\) greater.

- First find the weight of one package. - Once you know one package's weight, how can you find the weight of four packages? - Break the problem into two steps.

1. Find the weight of one package: \(240\,\text{kg} \div 6 = 40\,\text{kg}\). 2. Find the weight of four packages: \(4 \times 40\,\text{kg} = 160\,\text{kg}\).

a) One package weighs \(40\,\text{kg}\). b) Four packages weigh \(160\,\text{kg}\) altogether.

- First find each bird's distance for one day. - Break the larger numbers into parts that are easier to divide by three or four. - Compare the two daily distances.

1. Find the stork's daily distance: \(450\,\text{km} \div 3 = 150\,\text{km}\). 2. Find the crane's daily distance: \(560\,\text{km} \div 4 = 140\,\text{km}\). 3. Compare the daily distances: \(150\,\text{km} > 140\,\text{km}\). 4. The stork travels farther in one day.

The stork travels farther in one day.

- First find the total after the number of apples becomes five times as large. - The question asks only for the added apples. - Subtract the original amount from the new total.

1. Find the new total: \(24 \times 5 = 120\). 2. Subtract the original apples: \(120 - 24 = 96\).

The vendor adds \(96\) apples.

- “Triple” means multiply by \(3\). - Follow the events at the stops in order. - Subtract people who leave and add people who board.

1. Find the number after the first stop: \(14 \times 3 = 42\). 2. Subtract the passengers who leave: \(42 - 8 = 34\). 3. Add the passengers who board: \(34 + 5 = 39\).

There are \(39\) passengers on the bus.

- First find the current number of streetlights. - The question asks for the increase, not the new total. - Subtract the old number from the new number.

1. Find the current number: \(8 \times 6 = 48\). 2. Find the increase: \(48 - 8 = 40\).

The number of streetlights increased by \(40\).

- First find the amount of orange juice. - “Four times as many” indicates multiplication. - Then add both amounts.

1. Find the amount of orange juice: \(15\,\text{L} \times 4 = 60\,\text{L}\). 2. Add both kinds of juice: \(15\,\text{L} + 60\,\text{L} = 75\,\text{L}\).

There are \(75\,\text{L}\) of juice altogether.

- First find the cost of one case. - Once you know the unit price, how can you find the cost of twelve cases?

1. Find the cost of one case: \(\$72 \div 8 = \$9\). 2. Find the cost of twelve cases: \(12 \times \$9 = \$108\).

Twelve cases of lemonade cost \(\$108\) altogether.

- Find the number of pencils in each container one at a time. - What operation finds one-half of a quantity? - How many pencils are there altogether? - Divide the total equally among the new boxes.

1. Find the number of pencils in the second container: \(120 \div 2 = 60\). 2. Find the number in the third container: \(60 - 20 = 40\). 3. Find the total number of pencils: \(120 + 60 + 40 = 220\). 4. Divide equally among four boxes: \(220 \div 4 = 55\).

Each box contains \(55\) pencils.

- For each class, find how much has been raised and how much is still needed. - How can you compare the two remaining amounts? - A difference can be found by subtraction.

1. Class 4A has raised \(\frac{1}{8} \times \$720 = \$90\), so it still needs \(\$720 - \$90 = \$630\). 2. Class 4B has raised \(\frac{1}{5} \times \$600 = \$120\), so it still needs \(\$600 - \$120 = \$480\). 3. Since \(\$630 > \$480\), Class 4A still needs to raise more money. 4. Find the difference: \(\$630 - \$480 = \$150\).

Class 4A still needs to raise more money. The difference is \(\$150\).

- Use the money collected and the price per package to determine how many packages were sold. - Then relate the number of packages to the number of cards in each package. - Check that your final answer counts individual cards rather than packages.

1. The child sold \(\$48 \div \$2=24\) packages. 2. The number of cards sold is \(24 \times 10=240\).

The child sold \(240\) trading cards altogether.

- How much do all three books cost? - What is the total cost including the board game? - Which operation divides the total cost equally among four children? - You can break the total into parts that are easy to divide by four.

1. Find the cost of the three books: \(3 \times \$12 = \$36\). 2. Find the total cost: \(\$36 + \$32 = \$68\). 3. Divide the cost equally among four friends: \(\$68 \div 4 = \$17\).

Each child pays \(\$17\).

- What do the tickets and food cost altogether? - Does the coupon increase or decrease the amount owed? - How much does the group still need to pay after using the coupon? - Divide that amount equally among nine children.

1. Add the ticket and food costs: \(\$108 + \$45 = \$153\). 2. Subtract the coupon: \(\$153 - \$9 = \$144\). 3. Divide the remaining cost equally among nine children: \(\$144 \div 9 = \$16\).

Each child pays \(\$16\).

- Subtract the damaged apples first. - Divide the remaining amount equally among \(6\) crates. - Check by multiplication.

1. Find the amount remaining: \(600\,\text{lb} - 48\,\text{lb} = 552\,\text{lb}\). 2. Divide equally among the crates: \(552\,\text{lb} \div 6 = 92\,\text{lb}\).

Each crate contains \(92\,\text{lb}\) of apples.

- Add the amounts raised by both classes. - Subtract the expenses. - Divide the remaining amount equally between the two shelters.

1. The classes raise \(\$245 + \$275 = \$520\). 2. After paying expenses, \(\$520 - \$50 = \$470\) remains. 3. Each shelter receives \(\$470 \div 2 = \$235\).

Each animal shelter receives \(\$235\).

- First find how much all the machines fill in one hour. - Then divide that amount equally among the \(5\) machines. - Keep track of the unit in each step.

1. Find how much juice all \(5\) machines fill in one hour: \(900 \div 6 = 150\). They fill \(150\,\text{L}\) per hour together. 2. Divide that hourly amount equally among the \(5\) machines: \(150 \div 5 = 30\). 3. One machine fills \(30\,\text{L}\) in one hour.

One machine fills \(30\,\text{L}\) in one hour.

- Find the total notebook supply first. - Then look for the number of \(20\)-notebook bundles that exactly makes that total. - Check your bundle count with a multiplication equation.

1. The school receives \(8 \times 50=400\) notebooks. 2. Since \(20 \times 20=400\), there are enough notebooks for \(20\) classes.

There are enough notebooks for \(20\) classes.

- Find the total weight of all the apples. - Divide by each bag size separately. - Subtract the two numbers of bags.

1. Find the total weight of the apples: \(5 \times 24\,\text{lb} = 120\,\text{lb}\). 2. With \(3\)-pound bags: \(120\,\text{lb} \div 3\,\text{lb} = 40\) bags. 3. With \(4\)-pound bags: \(120\,\text{lb} \div 4\,\text{lb} = 30\) bags. 4. Find the difference: \(40 - 30 = 10\) bags.

a) The seller can fill \(40\) three-pound bags. b) The seller needs \(10\) fewer bags when using four-pound bags.

- First find the cost of one package of paving stones. - How much more does one package of edging stones cost? - Which operation finds the cost of eight packages?

1. Find the cost of one package of paving stones: \(\$120 \div 6 = \$20\). 2. Find the cost of one package of edging stones: \(\$20 + \$15 = \$35\). 3. Find the cost of eight packages of edging stones: \(8 \times \$35 = \$280\).

Eight packages of edging stones cost \(\$280\).

- Find the price of one package. - Use the unit price to find the cost of \(12\) packages. - Compare the cost with \(\$50\).

1. One package costs \(\$32 \div 8 = \$4\). 2. Twelve packages cost \(12 \times \$4 = \$48\). 3. Since \(\$48 < \$50\), the money is enough.

Yes. The \(12\) packages cost \(\$48\), so \(\$50\) is enough.

- Find the cost of one bookcase together with its bookend set. - Multiply that cost by \(6\). - Compare the total cost with the budget.

1. Find the cost of one bookcase with its bookend set: \(\$120 + \$15 = \$135\). 2. Find the cost of six sets: \(6 \times \$135 = \$810\). 3. Compare with the budget: \(\$810 < \$850\), so the budget is enough.

Yes. The order costs \(\$810\), which is less than the \(\$850\) budget.

- First find the total number of bulbs delivered. - Calculate each type of bag separately before adding. - Include the single bag of \(110\) bulbs. - Use division to share the total equally.

1. The four larger bags contain \(4\times 145=580\) bulbs. 2. The three other bags contain \(3\times 120=360\) bulbs. 3. The total delivery is \(580+360+110=1050\) bulbs. 4. Divide equally among six beds: \(1050\div 6=175\).

Each flower bed will receive \(175\) bulbs.

- How much hay is available per goat if it is shared evenly? - How much does each goat eat in one day? - You can check your answer using the amount the whole herd eats each day.

1. Divide the hay equally among the goats: \(960\div8=120\,\text{lb}\) per goat. 2. Each goat's share lasts \(120\div4=30\) days. 3. Check with the whole herd: \(8\times4=32\,\text{lb}\) per day and \(32\times30=960\,\text{lb}\).

The hay will last \(30\) days.

- First find the number of herb plants. - Find the cost of each type of plant. - Add the costs, then subtract from \(\$100\).

1. Find the number of herb plants: \(30 - 12 = 18\). 2. Find the cost of the tomato plants: \(12 \times \$2 = \$24\). 3. Find the cost of the herb plants: \(18 \times \$3 = \$54\). 4. Find the total cost: \(\$24 + \$54 = \$78\). 5. Find the change: \(\$100 - \$78 = \$22\).

The class receives \(\$22\) in change.

- Find the total cost for all student tickets. - Include the teacher tickets. - Compare the individual-ticket total with the group-ticket price.

1. Find the cost of the student tickets: \(24 \times \$5 = \$120\). 2. Find the cost of the teacher tickets: \(2 \times \$8 = \$16\). 3. Find the total for individual tickets: \(\$120 + \$16 = \$136\). 4. Compare with the group ticket: \(\$130 < \$136\). 5. Find the difference: \(\$136 - \$130 = \$6\).

The group ticket costs less by \(\$6\).

- First find the total cost of the apples. - Determine both the number of pounds of pears and their price per pound. - Add the two costs and compare the total with the available money.

1. Find the cost of the apples: \(20 \times \$2 = \$40\). 2. Find the amount of pears: \(20 - 4 = 16\,\text{lb}\). 3. Find the price per pound of pears: \(\$2 + \$1 = \$3\). 4. Find the cost of the pears: \(16 \times \$3 = \$48\). 5. Find the total cost: \(\$40 + \$48 = \$88\). 6. Compare the total with the budget: \(\$100 - \$88 = \$12\).

Yes, \(\$100\) is enough. Mr. Weber will have \(\$12\) left.

- Find the price of one oil-paint set. - Find the two class totals separately. - Add the class totals.

1. One oil-paint set costs \(\$14 + \$6 = \$20\). 2. The watercolor sets cost \(7 \times \$14 = \$98\). 3. The oil-paint sets cost \(9 \times \$20 = \$180\). 4. The total cost is \(\$98 + \$180 = \$278\).

The school pays \(\$278\) altogether.

- Find the apple-crate revenue. - Use “twice as much” to find the price of one pear crate. - Find the pear-crate revenue, then subtract the two totals.

1. The apple-crate revenue is \(15 \times \$18 = \$270\). 2. One pear crate costs \(2 \times \$18 = \$36\). 3. The pear-crate revenue is \(12 \times \$36 = \$432\). 4. The difference is \(\$432 - \$270 = \$162\).

The pear-crate revenue is \(\$162\) greater.

- You could draw a bar for the whole route and mark the known parts. - What do one-fourth and one-third mean as operations? - Find the lengths of the known sections first. - What remains after subtracting the known sections from the whole route?

1. Find the forest distance: \(\frac{1}{4} \times 12\,\text{mi} = 3\,\text{mi}\). 2. Find the meadow distance: \(\frac{1}{3} \times 12\,\text{mi} = 4\,\text{mi}\). 3. Add the known sections: \(3\,\text{mi} + 4\,\text{mi} = 7\,\text{mi}\). 4. Subtract from the total: \(12\,\text{mi} - 7\,\text{mi} = 5\,\text{mi}\).

The climbing section is \(5\,\text{mi}\) long.

- Find the price per basketball set. - Use the difference between the two total bills to find the volleyball total. - Find the price per volleyball set and compare the unit prices.

1. One basketball set costs \(\$432 \div 6 = \$72\). 2. The nine volleyball sets cost \(\$432 + \$54 = \$486\). 3. One volleyball set costs \(\$486 \div 9 = \$54\). 4. Since \(\$72 > \$54\), a basketball set costs more. 5. The difference is \(\$72 - \$54 = \$18\).

A basketball set costs \(\$18\) more than a volleyball set.

- Find the total number of rabbits in each area. - Use multiplication for each set of equal groups. - Subtract the smaller total from the larger total.

1. Find the number in the first area: \(14 \times 4 = 56\). 2. Find the number in the second area: \(11 \times 6 = 66\). 3. Find the difference: \(66 - 56 = 10\).

The second area has \(10\) more rabbits.

- Find the number of animals in each species one at a time. - Pay close attention to “four times as many,” “one-sixth as many,” and “120 fewer.” - Which operation matches each phrase? - Add all four groups at the end.

1. The number of zebras is \(144\). 2. Find the number of antelopes: \(4 \times 144 = 576\). 3. Find the number of lions: \(\frac{1}{6} \times 144 = 24\). 4. Find the number of giraffes: \(144 - 120 = 24\). 5. Add all four groups: \(144 + 576 + 24 + 24 = 768\).

There are \(768\) animals of these four species altogether.

- Find how many of each product were sold. - Find the revenue from each product separately. - Compare the three revenues, then add them.

1. The market sells \(30 - 12 = 18\) jars of honey, \(45 - 18 = 27\) jars of jam, and \(60 - 15 = 45\) cartons of eggs. 2. Honey earns \(18 \times \$7 = \$126\). 3. Jam earns \(27 \times \$4 = \$108\). 4. Eggs earn \(45 \times \$3 = \$135\). 5. The egg cartons earn the most money. 6. The total revenue is \(\$126 + \$108 + \$135 = \$369\).

The egg cartons earn the most, with \(\$135\). The total revenue is \(\$369\).

- Find the total number of students and the number of items each student receives. - Identify what each \(6 \times 12\) represents in Lucas's method. - Find the number of items received by one group.

1. There are \(6 \times 12 = 72\) students, and each receives \(2\) items, so \(72 \times 2 = 144\) items. 2. Lucas counts \(72\) water bottles and \(72\) apples: \(6 \times 12 + 6 \times 12 = 144\). 3. One group receives \(12 + 12 = 24\) items, so Julia counts \(6\) groups of \(24\): \(6 \times 24 = 144\). Both methods are correct.

a) \(144\) items are distributed. b) Yes. Julia's \(24\) is the \(12\) water bottles and \(12\) apples received by one group.

- Find the total number in each bed. - Compare the two products. - Subtract to find how many more.

1. Find the number in the first bed: \(4 \times 18 = 72\). 2. Find the number in the second bed: \(5 \times 14 = 70\). 3. Since \(72 > 70\), the first bed has more tulips. 4. Find the difference: \(72 - 70 = 2\).

The first bed has \(2\) more tulips than the second bed.

- Reconstruct the gardener’s original amount of money from what was spent and what remained. - Then compare that original amount with the price of one yellow flower. - Check the number of yellow flowers with a multiplication equation.

1. The red flowers cost \(15 \times \$4=\$60\). 2. The gardener originally had \(\$60+\$20=\$80\). 3. Since \(\$8 \times 10=\$80\), the gardener could have bought \(10\) yellow flowers.

The gardener could have bought \(10\) yellow flowers.

- Find the cost of the six small pretzels. - Use the money left to find Felix's starting amount. - Find the cost of five large pretzels and compare.

1. Find the cost of the small pretzels: \(6 \times 40 = 240\) cents, or \(\$2.40\). 2. Find Felix's starting amount: \(\$2.40 + \$0.60 = \$3.00\). 3. Find the cost of five large pretzels: \(5 \times 70 = 350\) cents, or \(\$3.50\). 4. Since \(\$3.00 < \$3.50\), he would be \(\$0.50\) short.

No. Felix had \(\$3.00\), but five large pretzels cost \(\$3.50\), so he would be \(\$0.50\) short.

- Put the information in the order needed for the calculations. - How can you find one-half of a quantity? - What operation combines all the fruit amounts at the end? - Notice which fruit amount depends on another amount.

1. Find the pear harvest: \(\frac{1}{2} \times 456\,\text{lb} = 228\,\text{lb}\). 2. Find the plum harvest: \(228\,\text{lb} - 85\,\text{lb} = 143\,\text{lb}\). 3. Add all three harvests: \(456\,\text{lb} + 228\,\text{lb} + 143\,\text{lb} = 827\,\text{lb}\).

The farm harvests \(827\,\text{lb}\) of fruit altogether.

- Find the amount raised in each week. - Add the three weekly amounts. - Subtract the total raised from the goal.

1. The second week raises \(2 \times \$420 = \$840\). 2. The third week raises \(\$840 - \$150 = \$690\). 3. The class has raised \(\$420 + \$840 + \$690 = \$1950\). 4. The amount still needed is \(\$2000 - \$1950 = \$50\).

The class still needs \(\$50\).

- Find each child's total number of marbles. - Use multiplication for each set of equal groups. - Compare the totals and subtract.

1. Find Lucas's total: \(14 \times 6 = 84\). 2. Find Marie's total: \(8 \times 11 = 88\). 3. Since \(88 > 84\), Marie has more. 4. Find the difference: \(88 - 84 = 4\).

Marie has \(4\) more marbles than Lucas.

- Find the total number of flowers in each bed. - Use a multiplication equation for each bed. - Compare the totals and find their difference.

1. Find the number of tulips: \(28 \times 7 = 196\). 2. Find the number of daffodils: \(35 \times 5 = 175\). 3. Since \(196 > 175\), the first bed has more flowers. 4. Find the difference: \(196 - 175 = 21\).

The first bed has \(21\) more flowers than the second bed.

- How many tulip bulbs are there altogether? - First find the number used in twelve rows of nine. - Add the bulbs that were left over. - Divide the total by the new number in each row.

1. Find the number of bulbs in the original twelve rows: \(12 \times 9 = 108\). 2. Find the total number of bulbs: \(108 + 12 = 120\). 3. Divide by the new number of bulbs per row: \(120 \div 8 = 15\).

The gardener can plant \(15\) rows.

- Divide the final total equally among the four children. - Compare each current amount with that target. - Add the four missing amounts.

1. Each child’s target amount is \(\$100 \div 4 = \$25\). 2. Lucas must add \(\$25 - \$12 = \$13\). 3. Mia must add \(\$25 - \$17 = \$8\). 4. Noah must add \(\$25 - \$9 = \$16\). 5. Sara must add \(\$25 - \$22 = \$3\). 6. Altogether, they add \(\$13 + \$8 + \$16 + \$3 = \$40\).

Lucas must add \(\$13\), Mia \(\$8\), Noah \(\$16\), and Sara \(\$3\). Altogether, they add \(\$40\).

- First find the number of yellow roses. - The number of white roses is compared with the yellow roses. - “Fewer than” indicates subtraction.

1. Find the number of yellow roses: \(45 \times 3 = 135\). 2. Find the number of white roses: \(135 - 56 = 79\).

There are \(79\) white roses.

- First find the second-year harvest. - The question asks for the difference, not the new total. - Subtract the first-year harvest from the second-year harvest.

1. Find the second-year harvest: \(15\,\text{kg} \times 6 = 90\,\text{kg}\). 2. Find the increase: \(90\,\text{kg} - 15\,\text{kg} = 75\,\text{kg}\).

The second-year harvest is \(75\,\text{kg}\) greater.

- First find the total weight of all the apple crates. - How much of the total weight remains for the pear crates? - Once you know the pear crates' total weight, how can you find their number? - Add the two numbers of crates at the end.

1. Find the total weight of the apple crates: \(15 \times 4\,\text{kg} = 60\,\text{kg}\). 2. Find the total weight of the pear crates: \(180\,\text{kg} - 60\,\text{kg} = 120\,\text{kg}\). 3. Find the number of pear crates: \(120\,\text{kg} \div 8\,\text{kg} = 15\). 4. Find the total number of crates: \(15 + 15 = 30\).

The seller receives \(30\) crates altogether.

- Find the total cost if every student pays separately. - Compare that total with the class rate. - Subtract the lower cost from the higher cost to find the savings.

1. Find the cost of individual admission: \(32 \times 6 = 192\) dollars. 2. Compare the prices: \(\$180 < \$192\), so the class rate costs less. 3. Find the savings: \(192 - 180 = 12\) dollars.

The class rate costs less, and the class saves \(\$12\).

- Find each truck's total for one day. - Compare the two totals. - Subtract to find the difference. - Be sure to use the correct number of trips for each truck.

1. The small truck carries \(4 \times 250 = 1000\,\text{kg}\) per day. 2. The large truck carries \(2 \times 450 = 900\,\text{kg}\) per day. 3. Since \(1000 > 900\), the small truck carries more. 4. The difference is \(1000 - 900 = 100\,\text{kg}\).

The small truck carries more sand, by \(100\,\text{kg}\).

- First find the total value of all the coins. - Calculate the value of each kind of coin separately. - Once you know the price of the whole bag, divide by the number of rolls.

1. Find the value of each group of coins: \(7 \times 25 = 175\) cents, \(5 \times 10 = 50\) cents, and \(5 \times 5 = 25\) cents. 2. Find the total cost of the bag: \(175 + 50 + 25 = 250\) cents. 3. Divide by the number of rolls: \(250 \div 5 = 50\) cents.

One roll costs \(50\) cents.

- How many pinecones do the two classes collect altogether? - How many figures can be made from the total? - Divide the total by the number needed for one figure. - Does your result match the teacher's number?

1. Find the total number of pinecones: \(135 + 145 = 280\). 2. Find the number of figures that can be made: \(280 \div 7 = 40\). 3. The result matches the teacher's claim, so the claim is correct.

Yes. The classes have \(280\) pinecones altogether, and \(280 \div 7 = 40\).

- Find the cost of buying all \(24\) sets individually. - Find how many packs of \(4\) are needed. - Compare the two total costs.

1. Buying \(24\) individual sets costs \(24 \times 5 = 120\) dollars. 2. The teacher needs \(24 \div 4 = 6\) packs. 3. Six packs cost \(6 \times 18 = 108\) dollars. 4. The savings are \(120 - 108 = 12\) dollars.

The teacher saves \(\$12\).

- Use the case count to determine the cost of one case first. - Then use the number of bottles in one case to determine the cost of one bottle. - Check that the bottle price rebuilds the full cost of all five cases.

1. One case costs \(\$40 \div 5=\$8\). 2. One bottle costs \(\$8 \div 4=\$2\).

One bottle of lemonade costs \(\$2\).

- Use the number of crates to find the cost of one crate before working with kilograms. - The mass in one crate then lets you find the price for one kilogram. - Use the unit price only after both earlier relationships are resolved.

1. One crate costs \(\$120 \div 8=\$15\). 2. Since one crate contains \(5\,\text{kg}\), one kilogram costs \(\$15 \div 5=\$3\). 3. Three kilograms cost \(3 \times \$3=\$9\).

Three kilograms of apples cost \(\$9\).

- How much does one box weigh if six boxes weigh \(48\,\text{kg}\)? - Which operation divides the total weight equally among the boxes? - Once you know one box's weight, how can you find the weight of fifteen boxes?

1. Find the weight of one box: \(48\,\text{kg} \div 6 = 8\,\text{kg}\). 2. Find the weight of fifteen boxes: \(15 \times 8\,\text{kg} = 120\,\text{kg}\).

a) One box weighs \(8\,\text{kg}\). b) Fifteen boxes weigh \(120\,\text{kg}\) altogether.

- First find the number of marbles in one small package. - Pay attention to how many small packages Paul buys. - Add the marbles from all the packages at the end.

1. Find the number of marbles in one small package: \(240 \div 4 = 60\). 2. Find the number in two small packages: \(2 \times 60 = 120\). 3. Add the marbles in all three packages: \(240 + 120 = 360\).

Paul buys \(360\) marbles altogether.

- First find Julia's age. - Use Julia's age to find the father's current age. - Go back the same number of years as Tim's age.

1. Find Julia's age: \(9 + 3 = 12\). 2. Find the father's current age: \(12 \times 3 = 36\). 3. Tim was born \(9\) years ago, so the father was \(36 - 9 = 27\) years old.

Their father was \(27\) years old when Tim was born.

- First find the elephant's current age. - Once you know both ages, find their difference. - Does the age difference between two living things change over time?

1. Find the elephant's current age: \(60 \div 5 = 12\) years. 2. Find the difference in their ages: \(60 - 12 = 48\) years. 3. Their age difference stays constant, so the turtle was \(48\) years old when the elephant was born.

The turtle was \(48\) years old when the elephant was born.

- Determine Mia’s current age from the comparison with Lucas. - A parent’s age at a child’s birth can be found from their current age difference. - Compare the two ages you obtain for Ms. Berger.

1. Mia is \(2 \times 7=14\) years old. 2. At Mia’s birth, Ms. Berger was \(42-14=28\) years old. 3. At Lucas’s birth, Ms. Berger was \(42-7=35\) years old. 4. Since \(35>28\), she was older when Lucas was born.

a) Ms. Berger was \(28\) years old when Mia was born. b) She was \(35\) years old when Lucas was born. c) She was older when Lucas was born.

- Find the total amount on all four pallets. - Compare the available amount with the amount needed. - Subtract to find the shortage or excess.

1. Find the amount available: \(4 \times 1245\,\text{lb} = 4980\,\text{lb}\). 2. Since \(4980\,\text{lb} < 5500\,\text{lb}\), there is not enough flour. 3. Find the shortage: \(5500\,\text{lb} - 4980\,\text{lb} = 520\,\text{lb}\).

No. The bakery is short by \(520\,\text{lb}\).

- Find how many marbles they have altogether. - If they share the total equally, how many marbles will each person have? - Find how many marbles Lena has above that equal-share amount.

1. Find the total number of marbles: \(134 + 158 = 292\). 2. Divide the total equally: \(292 \div 2 = 146\). 3. Lena must give away the amount she has above \(146\): \(158 - 146 = 12\).

Lena must give Max \(12\) marbles.

- First find the total number of pinecones. - How many should each friend have after an equal redistribution? - How many more do Marie and Jonas each need? - Who supplies the pinecones they need?

1. Find the total number of pinecones: \(150 + 90 + 60 = 300\). 2. Find the equal share for each person: \(300 \div 3 = 100\). 3. Find how many Marie receives: \(100 - 90 = 10\). 4. Find how many Jonas receives: \(100 - 60 = 40\). 5. Find how many Paul gives away altogether: \(10 + 40 = 50\).

Marie receives \(10\) pinecones, Jonas receives \(40\), and Paul gives away \(50\) pinecones altogether.

- First determine the whole morning batch from the information about one half. - Keep track of which quantity the phrase “twice the original morning batch” refers to. - Check that your final amount is larger than the original batch.

1. If half the morning batch is \(40\) rolls, the whole morning batch is \(40 \times 2=80\) rolls. 2. Twice the original morning batch is \(80 \times 2=160\) rolls.

\(160\) rolls

- Multiply the weight of one crate by the number of crates. - For part b, compare the total with the trailer's maximum load. - Find the amount over or under the limit.

1. For part a, multiply: \(18\,\text{lb} \times 30 = 540\,\text{lb}\). 2. For part b, find the weight of \(60\) crates: \(18\,\text{lb} \times 60 = 1080\,\text{lb}\). 3. Since \(1080\,\text{lb} > 1000\,\text{lb}\), the load exceeds the trailer's limit. 4. Find the amount over the limit: \(1080\,\text{lb} - 1000\,\text{lb} = 80\,\text{lb}\).

a) \(30\) crates weigh \(540\,\text{lb}\). b) No. \(60\) crates weigh \(1080\,\text{lb}\), which is \(80\,\text{lb}\) over the limit.

- Find the student and adult costs separately. - Use the correct ticket price for each group. - Add the two costs.

1. Student admission costs \(24\times\$6=\$144\). 2. Teacher admission costs \(2\times\$14=\$28\). 3. The total cost is \(\$144+\$28=\$172\).

\(\$172\)

- Find the total ticket cost first. - Compare that cost with the amount in the fund. - Subtract to find the amount remaining.

1. The student tickets cost \(28\times\$4=\$112\). 2. Add the teacher's ticket: \(\$112+\$9=\$121\). 3. Since \(\$121<\$150\), there is enough money. 4. The amount remaining is \(\$150-\$121=\$29\).

Yes. \(\$29\) remains.

- Find the individual-ticket total for the teenagers and adults. - Find the total number of people before using the group price. - Compare the two totals and subtract to find the savings.

1. Individual tickets cost \(25\times\$7+2\times\$12=\$175+\$24=\$199\). 2. The group has \(25+2=27\) people. 3. The group-price total is \(27\times\$6=\$162\). 4. The group price costs less and saves \(\$199-\$162=\$37\).

The group price costs less and saves \(\$37\).

- One method can compare the numbers of pallets before using the number of pots per pallet. - For a different method, use the \(9\) pallets that the two colors can be matched in common, then account for the unmatched pallets. - Both methods should give the same difference in numbers of geraniums.

1. Method 1: There are \(14-9=5\) extra red pallets. The number of extra geraniums is \(5 \times 120=600\). 2. Method 2: The \(9\) white pallets contain \(9 \times 120=1080\) geraniums. The red pallets contain those same \(1080\) geraniums plus the \(5\) extra pallets, or \(1080+600=1680\). The difference is \(1680-1080=600\).

The gardener bought \(600\) more red geraniums than white geraniums.

- First compare the two bag counts. - Use the mass of one bag to convert that difference to pounds. - For part b, interpret the center’s stated daily portion as equal groups of the extra amount.

1. The difference in bag counts is \(23-18=5\) bags. 2. The extra mass is \(5 \times 15=75\,\text{lb}\). 3. The number of daily portions is \(75 \div 3=25\), so the extra amount lasts about \(25\) days.

a) Mr. Green’s class collects \(75\,\text{lb}\) more. b) The extra amount will last about \(25\) days.

- One method can combine the two daily bag counts before extending the total to the month. - For a second method, keep the two crops separate until the monthly amounts are known. - For the crate count, use place value to reinterpret groups of \(50\) as groups of five tens.

1. Method 1 for part a: The farm packs \(125+75=200\) bags per day. In \(3\) days it packs \(3 \times 200=600\) bags, so in \(30\) days it packs \(10 \times 600=6000\) bags. 2. Method 2 for part a: In \(3\) days the farm packs \(3 \times 125=375\) potato bags and \(3 \times 75=225\) onion bags. In \(30\) days those amounts are \(3750\) and \(2250\), for a total of \(3750+2250=6000\) bags. 3. For part b, think of each crate as \(5\) groups of \(10\) bags. The \(6000\) bags make \(600\) groups of \(10\), and \(600 \div 5=120\). Therefore \(120\) crates are needed.

a) \(6000\) bags b) \(120\) crates

- Divide the total cost by \(8\) to find the unit price. - Multiply the unit price by \(15\). - You can split \(15\) into \(10 + 5\) if that makes the multiplication easier.

1. One ball costs \(\$144 \div 8 = \$18\). 2. Fifteen balls cost \(15 \times \$18 = \$270\).

a) One ball costs \(\$18\). b) Fifteen balls cost \(\$270\).

- Divide the prize equally among the four friends. - Add Leon’s share to his existing savings. - Check the equal share by multiplying it by \(4\).

1. Each friend receives \(\$1348 \div 4 = \$337\). 2. Leon then has \(\$337 + \$57 = \$394\).

Leon has \(\$394\).

- First determine the total number of balloons purchased. - The balloons used and the balloons remaining make the starting total. - Compare the starting total with the amount left after decorating.

1. The school bought \(2 \times 250=500\) balloons. 2. The number used is \(500-112=388\).

The school used \(388\) balloons.

- Find the total weight of all the apples. - Subtract the amount sold. - Determine how many \(3\,\text{lb}\) groups are in the remaining amount.

1. Find the total weight of the apples: \(15 \times 28 = 420\,\text{lb}\). 2. Find the weight remaining after the sales: \(420 - 135 = 285\,\text{lb}\). 3. Find the amount of cider: \(285 \div 3 = 95\,\text{qt}\).

The remaining apples can make \(95\,\text{qt}\) of cider.

- First find the total number of markers delivered. - Subtract the markers given to the classes. - Compare the number of cups you find with the teacher’s estimate.

1. The school receives \(14\times 24=336\) markers. 2. The classes receive \(8\times 12=96\) markers. 3. The number left is \(336-96=240\) markers. 4. The number of cups is \(240\div 6=40\). 5. Since \(40<50\), the teacher is incorrect.

a) \(40\) cups b) No. Only \(40\) cups can be filled.

- First find Ben's savings. - Use Ben's amount to find Noah's savings. - Add all three amounts and compare the total with the tent's price.

1. Find Ben's savings: \(2 \times \$150 = \$300\). 2. Find Noah's savings: \(\$300 - \$80 = \$220\). 3. Add all three amounts: \(\$150 + \$300 + \$220 = \$670\). 4. Since \(\$670 > \$600\), they have enough money.

The children have saved \(\$670\) altogether, so they have enough for the tent.

- Count both the starting year and ending year in the full period. - Find how many \(3\)-year intervals fit between \(2002\) and \(2023\). - Remember that counting intervals is not the same as counting events. - Subtract the festival years from the total number of years.

1. Find the number of years in the inclusive period: \(2023 - 2002 + 1 = 22\) years. 2. Find the number of \(3\)-year intervals: \((2023 - 2002) \div 3 = 7\). 3. Include the festival at the start of the period: \(7 + 1 = 8\) festivals. 4. Subtract the festival years from all years in the period: \(22 - 8 = 14\) years without a festival.

a) There were \(8\) festivals. b) There were \(14\) years without a festival.

- Find the amount of each type of fruit in the order needed. - Read carefully to identify which combined amount the plum fraction refers to. - A table may help you organize the intermediate results. - What operation finds one-fourth of a quantity?

1. Find the pear harvest: \(4250\,\text{lb} - 1340\,\text{lb} = 2910\,\text{lb}\). 2. Find the combined apple and pear harvest: \(4250\,\text{lb} + 2910\,\text{lb} = 7160\,\text{lb}\). 3. Find the plum harvest: \(\frac{1}{4} \times 7160\,\text{lb} = 1790\,\text{lb}\). 4. Add all three fruit harvests: \(7160\,\text{lb} + 1790\,\text{lb} = 8950\,\text{lb}\).

The farm harvests \(8950\,\text{lb}\) of fruit altogether.

- What must you calculate before you can find the amount of plastic? - Which operations match “three times as much” and “one-sixth of”? - Keep the place values aligned in your calculations. - Make sure your final answer gives the total of all three materials.

1. Find the weight of the glass: \(3 \times 1284\,\text{lb} = 3852\,\text{lb}\). 2. Find the combined weight of the paper and glass: \(1284\,\text{lb} + 3852\,\text{lb} = 5136\,\text{lb}\). 3. Find the weight of the plastic: \(\frac{1}{6} \times 5136\,\text{lb} = 856\,\text{lb}\). 4. Find the total weight: \(5136\,\text{lb} + 856\,\text{lb} = 5992\,\text{lb}\).

The students collect \(5992\,\text{lb}\) of recyclable material altogether.

- Find the hay amount for each barn in order. - Which operation matches “five times as much,” and which matches “one-third as much”? - Did you add all three amounts at the end?

1. Find the hay used by the cattle barn: \(5 \times 1356\,\text{lb} = 6780\,\text{lb}\). 2. Find the hay used by the goat barn: \(\frac{1}{3} \times 6780\,\text{lb} = 2260\,\text{lb}\). 3. Add the three amounts: \(1356\,\text{lb} + 6780\,\text{lb} + 2260\,\text{lb} = 10{,}396\,\text{lb}\).

The three barns use \(10{,}396\,\text{lb}\) of hay altogether in one week.

- Use the multiplicative comparison to determine the train’s one-hour distance. - Keep the one-hour comparison separate from the repeated five-hour distance. - Check that the five-hour result is five times the train’s one-hour distance.

1. The train travels \(2 \times 50=100\) miles in one hour. 2. In five hours, the train travels \(5 \times 100=500\) miles.

The train travels \(500\) miles in five hours.

- Count the shelves and the spaces separately. - Find the combined thickness of the shelves. - Find the combined height of the spaces, then add the two amounts.

1. The shelves have a combined thickness of \(6\times3\,\text{cm}=18\,\text{cm}\). 2. The spaces have a combined height of \(5\times32\,\text{cm}=160\,\text{cm}\). 3. The total height is \(18\,\text{cm}+160\,\text{cm}=178\,\text{cm}\).

\(178\,\text{cm}\)

- First find the number of daffodils. - Use the daffodil count to find the number of crocuses. - Add the three flower counts.

1. Find the number of daffodils: \(124 \times 2 = 248\). 2. Find the number of crocuses: \(248 - 50 = 198\). 3. Add all three amounts: \(124 + 248 + 198 = 570\).

There are \(570\) flowers altogether.

- Find the number of laps for each class one at a time. - What operation does “more” suggest? - How do you find one-half of a number? - Add the results for all three classes at the end.

1. Find the number of laps completed by Class 3B: \(210 + 40 = 250\). 2. Find the number completed by Class 3C: \(250 \div 2 = 125\). 3. Add the laps from all three classes: \(210 + 250 + 125 = 585\).

The three classes complete \(585\) laps altogether.

- Find the sales for each part of the day in order. - Use the afternoon amount to find the evening amount. - Add the three amounts.

1. Find the afternoon sales: \(185 + 40 = 225\). 2. Find the evening sales: \(225 \times 2 = 450\). 3. Add all three amounts: \(185 + 225 + 450 = 860\).

Students sell \(860\) raffle tickets during the day.

- Compare the delivery cost with the cost of \(10\) apples. - Use equal groups of \(\$4\) to find the total number of apples. - Divide the apples equally among the four crates.

1. The delivery price of \(\$64\) is \(64 \div 4 = 16\) groups of \(\$4\). 2. Each \(\$4\) group represents \(10\) apples, so the delivery contains \(16 \times 10 = 160\) apples. 3. Each crate contains \(160 \div 4 = 40\) apples.

Each crate contains \(40\) apples.

- First find the total amount delivered to all four schools. - Use multiplication for the equal deliveries. - Subtract the delivered amount from the starting amount.

1. Find the total amount delivered: \(4 \times 185\,\text{kg} = 740\,\text{kg}\). 2. Subtract from the starting amount: \(850\,\text{kg} - 740\,\text{kg} = 110\,\text{kg}\).

\(110\,\text{kg}\) of apples remain.

- Focus first on how the afternoon sales relate to the morning sales. - Keep track of the total number sold before deciding how many tickets remain. - Use the remaining-ticket count to judge the claim about \(300\).

1. Find the afternoon sales: \(215 \times 2 = 430\). 2. Find the total sold: \(215 + 430 = 645\). 3. Find the number left: \(950 - 645 = 305\). 4. Since \(305 > 300\), Linnea's statement is correct.

Yes. There are \(305\) tickets left, and \(305 > 300\).

- The red-package count must be determined before the overall total can be found. - Use the stated relationship between the red and blue package counts. - After finding both color counts, combine them to answer the question asked.

1. The number of red packages is \(135 \times 4=540\). 2. The total number of packages is \(135+540=675\).

The store has \(675\) packages altogether.

- First find the number of relay-race awards. - Add the awards for both events. - Compare the total needed with \(1000\).

1. Find the number of relay-race awards: \(124 \times 7 = 868\). 2. Find the total needed: \(124 + 868 = 992\). 3. Since \(992 < 1000\), there are enough awards.

Yes. The school needs \(992\) awards, so the \(1000\) printed awards are enough.

- First find the total number baked. - Add the morning and afternoon sales. - Subtract the total sold from the total baked.

1. Find the total number baked: \(6 \times 45 = 270\). 2. Find the total number sold: \(142 + 78 = 220\). 3. Find the number left: \(270 - 220 = 50\).

There are \(50\) rolls left.

- First find the capacity of all four bookcases. - Then subtract the number of books placed on them. - The difference is the number of open spaces.

1. Find the total capacity: \(4 \times 115 = 460\). 2. Subtract the books placed on the shelves: \(460 - 325 = 135\).

There are \(135\) spaces left.

- Find the amount actually spent after the change is returned. - Find the total cost of the construction paper. - Divide the remaining cost equally among the eight boxes.

1. The full purchase costs \(6 \times \$100 - \$50 = \$550\). 2. The construction paper costs \(25 \times \$14 = \$350\). 3. The eight boxes of glitter gems cost \(\$550 - \$350 = \$200\). 4. One box costs \(\$200 \div 8 = \$25\).

One box of glitter gems costs \(\$25\).

- First find how far each train travels in one hour. - How can you use the total distance and total time to find a one-hour distance? - Compare the two one-hour distances.

1. Find Train A's distance in one hour: \(270 \div 3 = 90\) miles. 2. Find Train B's distance in one hour: \(320 \div 4 = 80\) miles. 3. Since \(90 > 80\), Train A travels farther in one hour.

Train A travels farther in one hour: \(90\) miles compared with Train B's \(80\) miles.

- First find half of \(960\). - Subtract that amount from the total. - Divide the remaining tickets equally between the two classes. - Break large numbers into place-value parts if that makes the division easier.

1. Class 4A receives half of \(960\): \(960 \div 2 = 480\). 2. The number of tickets left is \(960 - 480 = 480\). 3. Divide the remaining tickets equally between two classes: \(480 \div 2 = 240\).

Class 4C receives \(240\) tickets.

- Find how much money remains after the purchase. - Divide the remaining amount by \(5\). - Break the remaining amount into \(500\) and another convenient part if helpful. - You can also compare the original number of bills with the number of bills spent.

1. One method is to find the remaining amount first: \(\$750 - \$200 = \$550\). 2. Divide the remaining amount by the value of each bill: \(550 \div 5 = (500 \div 5) + (50 \div 5) = 100 + 10 = 110\). 3. A second method is to find \(750 \div 5 = 150\) bills at first and \(200 \div 5 = 40\) bills spent. Then \(150 - 40 = 110\).

There are \(110\) five-dollar bills remaining.

- How many pencils remain after the reserve is set aside? - Which operation divides a quantity equally among groups? - Use place value and a related basic division fact.

1. Find the number of pencils remaining after the reserve is removed: \(845 - 125 = 720\). 2. Divide equally among eight classes: \(720 \div 8 = 90\).

Each class receives \(90\) pencils.

- How much do the food and drinks cost altogether? - How much money remains after those costs are paid? - Once you know the remaining amount, how do you find one class's share? - A related basic division fact may help with the larger number.

1. Find the total cost of food and drinks: \(\$345 + \$285 = \$630\). 2. Find the money remaining: \(\$950 - \$630 = \$320\). 3. Divide equally among eight classes: \(\$320 \div 8 = \$40\).

Each class receives \(\$40\).

- First determine the total number of baseballs in all six bins. - The question asks for what remains after a known amount leaves the collection. - Check your result against both the equal-bin information and the amount moved.

1. The starting total is \(6 \times 80=480\) baseballs. 2. After \(150\) are moved, \(480-150=330\) baseballs remain.

\(330\) baseballs remain.

- First find the combined cost. - Divide that total by \(10\). - For part b, decide whether adding an amount to one cost and subtracting the same amount from the other changes their sum.

1. Find the total cost: \(\$4200 + \$4200 = \$8400\). 2. Divide the total into \(10\) equal payments: \(\$8400 \div 10 = \$840\). 3. For part b, increasing one cost by \(\$500\) and decreasing the other by \(\$500\) leaves the total unchanged. The number of payments also stays the same, so the monthly payment remains \(\$840\).

a) Each monthly payment is \(\$840\). b) No. The total remains \(\$8400\), since \(\$4700 + \$3700 = \$8400\), so the monthly payment does not change.

- Find the total amount raised by all the grade levels. - Divide the total equally among the months. - Begin by combining all three amounts.

1. Find the total amount raised: \(\$1420 + \$1250 + \$2530 = \$5200\). 2. Divide the total equally among \(5\) months: \(\$5200 \div 5 = \$1040\).

The school can spend \(\$1040\) each month.

- Add the two donations first. - Divide the total equally among \(8\) elephants. - Check the quotient by multiplication.

1. Find the total amount of hay: \(4125\,\text{kg} + 3755\,\text{kg} = 7880\,\text{kg}\). 2. Divide equally among the elephants: \(7880\,\text{kg} \div 8 = 985\,\text{kg}\).

Each elephant receives \(985\,\text{kg}\) of hay.

- First find how much the two classes raise together each day. - Use that daily amount to find the total after \(10\) days. - Find the number of daily amounts that make the goal by using multiplication.

1. Find the combined daily amount: \(\$36 + \$44 = \$80\). 2. Find the amount after \(10\) days: \(10 \times \$80 = \$800\). 3. Use an inverse multiplication fact to reach the goal: \(15 \times \$80 = \$1200\), so the classes need \(15\) school days.

a) After \(10\) school days, they have raised \(\$800\). b) They need \(15\) school days to reach the goal.

- The pencil count is given in relation to the marker count, so determine the marker count first. - The eraser comparison uses that same marker count as its reference quantity. - Check that the eraser count is greater than the marker count and consistent with “three times as many.”

1. Since \(600\) pencils is twice the marker count, the school orders \(600\div2=300\) markers. 2. The eraser count is three times the marker count: \(3 \times 300=900\).

The school orders \(900\) erasers.

- Pair one comic book with one adventure book. - Find how many equal pairs make the total revenue. - For part b, find the total number of items and use the new unit price.

1. One comic book and one adventure book bring in \(\$3 + \$5 = \$8\). 2. Jan sells \(\$112 \div \$8 = 14\) pairs, so he sells \(14\) comic books and \(14\) adventure books. 3. He sells \(14 + 14 = 28\) items altogether. 4. At \(\$3\) per item, he would earn \(28 \times \$3 = \$84\).

a) Jan sells \(14\) comic books. b) He would earn \(\$84\).

- First determine the total number of boxes. - For part a, compare finding each color separately with combining the box counts first. - For part b, determine how many boxes or blocks remain after \(2\) boxes are removed.

1. Part a, Method 1: The red boxes contain \(6 \times 80 = 480\) blocks, and the blue boxes contain \(4 \times 80 = 320\) blocks. Altogether, \(480 + 320 = 800\). 2. Part a, Method 2: There are \(6 + 4 = 10\) boxes, so \(10 \times 80 = 800\) blocks. 3. Part b: The preschool buys \(2 \times 80 = 160\) blocks, so \(800 - 160 = 640\) blocks remain. Equivalently, \(10 - 2 = 8\) boxes remain, and \(8 \times 80 = 640\).

a) \(800\) blocks b) \(640\) blocks

- How many days are in one week? - Find the total amount eaten during the week. - Compare that total with \(900\,\text{kg}\). - Decide whether your total is greater than or less than \(900\).

1. Multiply the daily amount by the number of days: \(124 \times 7 = 868\,\text{kg}\). 2. Compare the weekly amount with the supply: \(868 < 900\). 3. The supply is enough because the elephants need less than \(900\,\text{kg}\).

The elephants eat \(868\,\text{kg}\) of hay in one week. The \(900\,\text{kg}\) supply is enough.

- Multiply the number of marbles in one bag by \(4\). - Think about how the starting supply changes after the bags are filled. - Subtract the packed marbles from the starting amount.

1. Find the number packed: \(235 \times 4 = 940\) marbles. 2. Subtract from the starting amount: \(1000 - 940 = 60\) marbles.

a) \(940\) marbles b) \(60\) marbles

- First find the number placed on Sunday. - Add the Saturday and Sunday amounts. - Subtract the completed pieces from \(1000\).

1. Find the number placed on Sunday: \(165 \times 3 = 495\). 2. Find the total placed during the weekend: \(165 + 495 = 660\). 3. Find the number missing: \(1000 - 660 = 340\).

The puzzle is missing \(340\) pieces.

- Find the total weight of each load. - Compare the two products. - Subtract the smaller weight from the larger weight.

1. Find the weight of the potatoes: \(7 \times 45\,\text{kg} = 315\,\text{kg}\). 2. Find the weight of the onions: \(6 \times 55\,\text{kg} = 330\,\text{kg}\). 3. The onion load is heavier because \(330 > 315\). 4. Find the difference: \(330\,\text{kg} - 315\,\text{kg} = 15\,\text{kg}\).

The onion load is heavier by \(15\,\text{kg}\).

- Identify the total number of markers before thinking about the equal shares. - Let the letter stand for one classroom’s share, and relate three equal shares to the total. - Check your equation by substituting your value for the letter.

1. The total number of markers is \(4 \times 60=240\). 2. An equation for the equal sharing is \(3m=240\). 3. Solving the unknown-factor equation gives \(m=240\div3=80\). 4. Check: \(3 \times 80=240\).

\(3m=240\), so \(m=80\). Each classroom receives \(80\) markers.

- How many five-minute intervals are in one hour? - Once you know the number of intervals, how can you find the total number of posters? - Will the machine print more or fewer posters in one hour than in five minutes?

1. Find how many five-minute intervals are in one hour: \(60 \div 5 = 12\). 2. Multiply by the number of posters printed in each interval: \(12 \times 50 = 600\).

The machine prints \(600\) posters in one hour.

- How many trays can the farmer buy for \(\$125\) if each tray costs \(\$5\)? - Each tray has \(6\) plants. How can you find the total number of plants once you know the number of trays? - Could you break \(125\) into smaller parts to make the division easier?

1. Find the number of trays the farmer buys: \(\$125 \div \$5 = 25\). 2. Find the total number of plants: \(25 \times 6 = 150\).

The farmer buys \(150\) tomato plants altogether.

- First find how many rows are needed altogether. - Identify how many rows are already planted. - Subtract to find the number of rows remaining.

1. Find the total number of rows: \(360 \div 9 = 40\) rows. 2. Subtract the rows already planted: \(40 - 15 = 25\) rows.

The gardener must plant \(25\) more rows.

- Use the blue-pen row to find the number of boxes. - The red pens fill the same number of boxes. - How many red pens fit in each box?

1. There are \(480\div8=60\) boxes of blue pens. 2. There are also \(60\) boxes of red pens. 3. The total number of red pens is \(60\times6=360\).

There are \(360\) red pens altogether.

- First find how much money Lukas has after adding the gift. - Subtract that amount from the skateboard’s price. - Determine how many weekly savings amounts fit into the amount still needed.

1. Find how much money Lukas has after the gift: \(\$85 + \$20 = \$105\). 2. Find how much more he needs: \(\$140 - \$105 = \$35\). 3. Find the number of weeks: \(\$35 \div \$5 = 7\).

Lukas must save for \(7\) more weeks.

- Find the total mass of the barrels. - Subtract that mass from the truck's limit. - Find the total mass of the \(8\) boxes. - Compare the box mass with the remaining capacity.

1. The barrels weigh \(4 \times 80 = 320\,\text{kg}\). 2. The remaining capacity is \(600 - 320 = 280\,\text{kg}\). 3. The boxes weigh \(8 \times 40 = 320\,\text{kg}\). 4. Since \(320 > 280\), all \(8\) boxes would exceed the remaining capacity.

No. The boxes weigh \(320\,\text{kg}\), but only \(280\,\text{kg}\) of capacity remains.

- How much juice does Lucas drink from the full bottle? - How much remains after Lucas drinks? - Marie drinks one-half of the amount that remains, not one-half of the original amount. - After finding Marie's amount, subtract it from the amount that was left.

1. Find the amount Lucas drinks: \(\frac{1}{4} \times 32\,\text{fl oz} = 8\,\text{fl oz}\). 2. Find the amount remaining after Lucas drinks: \(32\,\text{fl oz} - 8\,\text{fl oz} = 24\,\text{fl oz}\). 3. Marie drinks one-half of the remaining juice: \(\frac{1}{2} \times 24\,\text{fl oz} = 12\,\text{fl oz}\). 4. Find the final amount remaining: \(24\,\text{fl oz} - 12\,\text{fl oz} = 12\,\text{fl oz}\).

\(12\,\text{fl oz}\) of juice remain in the bottle.

- How many minutes are in one hour? - First find the minutes spent warming up and running. - Add those times to find how much of the hour is already planned. - How many minutes remain in the full hour?

1. Convert the total time to minutes: \(1\,\text{hour} = 60\,\text{minutes}\). 2. Find the warm-up time: \(\frac{1}{4} \times 60\,\text{minutes} = 15\,\text{minutes}\). 3. Find the running-event time: \(\frac{1}{2} \times 60\,\text{minutes} = 30\,\text{minutes}\). 4. Find the time already used: \(15\,\text{minutes} + 30\,\text{minutes} = 45\,\text{minutes}\). 5. Find the remaining time: \(60\,\text{minutes} - 45\,\text{minutes} = 15\,\text{minutes}\).

There are \(15\,\text{minutes}\) left for the awards ceremony.

- Find the number of pieces in each given fruit group separately. - How can you find a fraction of a known total? - After finding the pears and apples, how can you find the rest?

1. Find the number of pears: \(\frac{1}{5} \times 30 = 6\). 2. Find the number of apples: \(\frac{1}{3} \times 30 = 10\). 3. Since \(10 > 6\), there are more apples than pears. 4. Find the number of bananas: \(30 - 6 - 10 = 14\).

There are \(6\) pears, \(10\) apples, and \(14\) bananas. There are more apples than pears.

- First determine the cracked fraction of the full collection. - The cartons are filled only with the eggs that remain sellable. - Interpret the carton size as an equal-group question after finding the uncracked total.

1. The number of cracked eggs is \(\frac{1}{6}\times720=120\). 2. The number of uncracked eggs is \(720-120=600\). 3. The farmer can fill \(600\div6=100\) small cartons.

The farmer can pack \(100\) full cartons.

- Find the bicycle and walking groups separately. - How many students are in those two groups altogether? - How can you find the number remaining for the bus? - A drawing with \(24\) dots may help you organize the groups.

1. Find the number who ride bicycles: \(\frac{1}{3} \times 24 = 8\). 2. Find the number who walk: \(\frac{1}{4} \times 24 = 6\). 3. Find the number who do not ride the bus: \(8 + 6 = 14\). 4. Subtract from the class total: \(24 - 14 = 10\).

\(10\) students ride the bus.

- Which part of the route does Lucas walk twice because he forgot the pass? - Find the total extra distance caused by returning home. - Subtract that extra distance from the amount shown on the fitness tracker.

1. The forgotten pass adds a trip of \(300\,\text{m}\) away from home and \(300\,\text{m}\) back: \(2 \times 300 = 600\,\text{m}\). 2. Subtract the extra distance from the total: \(1400 - 600 = 800\,\text{m}\). 3. The direct distance from home to the bus stop is \(800\,\text{m}\).

The bus stop is \(800\,\text{m}\) from Lucas’s home.

- First find the number of bulbs already planted. - Use multiplication for the equal rows. - Include the bulbs still in the basket.

1. Find the number planted: \(9 \times 35 = 315\). 2. Add the bulbs left in the basket: \(315 + 78 = 393\).

There were \(393\) tulip bulbs in the basket at the beginning.

- Find each class's total separately. - Compare the two products. - Subtract the smaller total from the larger total.

1. Find Class A's total: \(8 \times 45 = 360\). 2. Find Class B's total: \(6 \times 55 = 330\). 3. Since \(360 > 330\), Class A collects more. 4. Find the difference: \(360 - 330 = 30\).

Class A collects \(30\) more pinecones than Class B.

- Find each group's total separately. - Compare the two totals. - Add them to find the combined number of jump ropes.

1. Find Group A's total: \(7 \times 12 = 84\). 2. Find Group B's total: \(5 \times 16 = 80\). 3. Since \(84 > 80\), Group A receives more. 4. Find the combined total: \(84 + 80 = 164\).

Group A receives more jump ropes. The two groups receive \(164\) jump ropes altogether.

- First find the exact number of red blocks. - Dividing by \(3\) finds one-third of a quantity. - Once you know both quantities, find their difference.

1. Find the number of red blocks: \(150 \div 3 = 50\). 2. Find the difference: \(150 - 50 = 100\).

The bin contains \(100\) more blue blocks than red blocks.

- The question asks for a difference, but one of the two card counts must be found first. - Use the relationship between Finn’s amount and Lucas’s amount to determine both counts. - Check that your final result answers “how many more,” not Finn’s total.

1. Finn has \(48 \times 4=192\) cards. 2. The difference is \(192-48=144\).

Finn has \(144\) more cards than Lucas.

- Determine both drink amounts before answering the comparison question. - Use the relationship between the water amount and the apple-juice amount. - Make sure the final number represents “how many more,” not the total amount of water.

1. The fair prepares \(115 \times 3=345\) cups of water. 2. The difference is \(345-115=230\) cups.

The fair prepares \(230\) more cups of water than apple juice.

- What operation finds half as many books? - Find the number of books on each shelf one step at a time. - The question asks for a difference, not the combined total.

1. Find the number of books on Shelf B: \(124 \div 2 = 62\). 2. Find the number of books on Shelf C: \(62 \times 3 = 186\). 3. Find the difference between Shelves C and A: \(186 - 124 = 62\).

Shelf C has \(62\) more books than Shelf A.

- Find both shelf counts before determining the requested comparison. - Use the stated relationship between the mystery and nonfiction counts. - Check that the final value is a difference rather than the larger shelf total.

1. The mystery shelf has \(45 \times 3=135\) books. 2. The difference is \(135-45=90\) books.

There are \(90\) more mystery books than nonfiction books.

- The second and third beds are each described in relation to the first bed, but by different kinds of comparison. - Determine both missing harvest amounts before comparing them with each other. - Check which comparison phrase describes a factor and which describes a difference.

1. The second-bed harvest is \(120 \times 2=240\,\text{kg}\). 2. The third-bed harvest is \(120-50=70\,\text{kg}\). 3. The difference is \(240-70=170\,\text{kg}\).

The second bed produces \(170\,\text{kg}\) more potatoes than the third bed.

- First find Marie's total savings. - The question asks for the difference, not Marie's total. - Subtract Lucas's amount from Marie's amount.

1. Find Marie's savings: \(\$160 \times 4 = \$640\). 2. Find the difference: \(\$640 - \$160 = \$480\).

Marie has saved \(\$480\) more than Lucas.

- The two players use the same starting value, bonus, and multiplier, but the order is different. - Keep each player’s sequence of actions separate. - Compare only after both final scores have been found.

1. Leon has \(80+20=100\) points before the multiplier, so his final score is \(100 \times 4=400\). 2. Sophie’s score is first \(80 \times 4=320\), then \(320+20=340\). 3. Since \(400>340\), Leon has the greater final score.

Leon has \(400\) points. Sophie has \(340\) points. Leon has the greater score.

- Look for two equivalent ways to represent the same change over an eight-hour day. - Your two methods should organize the given rates differently rather than merely rewriting the same arithmetic. - Use the agreement of the two results as a check.

1. Method 1: The hourly increase is \(165-140=25\) rolls. Across \(8\) hours, the increase is \(25 \times 8=200\) rolls. 2. Method 2: The new daily total is \(165 \times 8=1320\) rolls and the old daily total is \(140 \times 8=1120\) rolls. Their difference is \(1320-1120=200\) rolls.

The bakery produces \(200\) more rolls during an \(8\)-hour workday.

- Find the cost of each kind of sticker. - Convert the amount paid to cents. - Subtract the total cost from the amount paid.

1. The glitter stickers cost \(4 \times 35 = 140\) cents. 2. The animal stickers cost \(6 \times 45 = 270\) cents. 3. The total cost is \(140 + 270 = 410\) cents. 4. A \(\$10\) bill is \(1000\) cents, so the change is \(1000 - 410 = 590\) cents, or \(\$5.90\).

Paul receives \(\$5.90\) in change.

- What calculation represents “one out of every \(3\) passengers”? - How many passengers pay the regular price? - Find the revenue from each group before adding.

1. The number of discount tickets is \(54\div3=18\). 2. The number of regular tickets is \(54-18=36\). 3. Revenue from discount tickets is \(18\times\$24=\$432\). 4. Revenue from regular tickets is \(36\times\$30=\$1080\). 5. The total revenue is \(\$432+\$1080=\$1512\).

The trip generated \(\$1512\) in ticket revenue.

- Find the nonfiction book count from the children's fiction count. - The two shelving groups include every book today. - Compare today's total with the earlier count.

1. The library now has \(140\times3=420\) adult nonfiction books. 2. There are \(140+420=560\) books today. 3. That is \(560-90=470\) more books than before.

The library has \(470\) more books today.

- Find the number of adult members first. - Add to find the club’s current membership. - Determine the target membership that is \(600\) above the founding amount. - Compare the current membership with the target.

1. The number of adult members is \(110\times 4=440\). 2. The club currently has \(110+440=550\) members. 3. A membership that is \(600\) greater than the founding membership would be \(55+600=655\). 4. The club needs \(655-550=105\) more members. 5. Equivalently, the current increase is \(550-55=495\), and \(600-495=105\).

The club needs \(105\) more members.

- Find how much Class 4B collects. - Find how much Class 4C collects. - Add all three classes' amounts.

1. Find the amount collected by Class 4B: \(24\,\text{kg} \times 3 = 72\,\text{kg}\). 2. Find the amount collected by Class 4C: \(24\,\text{kg} + 15\,\text{kg} = 39\,\text{kg}\). 3. Add the three amounts: \(24\,\text{kg} + 72\,\text{kg} + 39\,\text{kg} = 135\,\text{kg}\).

The three classes collect \(135\,\text{kg}\) of paper altogether.

- Divide the total cost by \(5\) to find the unit price. - Then divide the budget by the price of one bag. - Check using multiplication.

1. One bag costs \(\$25 \div 5 = \$5\). 2. With \(\$125\), the gardener can buy \(\$125 \div \$5 = 25\) bags.

a) One bag costs \(\$5\). b) The gardener can buy \(25\) bags.

- Find the unit price first. - Divide the teacher’s total cost by the unit price. - For the final part, use the number of notebooks per package.

1. One package costs \(\$48 \div 6 = \$8\). 2. The teacher buys \(\$96 \div \$8 = 12\) packages. 3. The packages contain \(12 \times 10 = 120\) notebooks.

a) One package costs \(\$8\). b) The teacher buys \(12\) packages. c) The teacher buys \(120\) notebooks altogether.

- Find the total price of five individual tickets. - Find how much is missing for each option. - Compare the two missing amounts.

1. Five individual tickets cost \(5 \times \$8 = \$40\). 2. The group needs \(\$40 - \$33 = \$7\) more for individual tickets. 3. The group needs \(\$38 - \$33 = \$5\) more for the small-group pass. 4. Since \(\$5 < \$7\), the child is correct.

a) The group needs \(\$7\) more for individual tickets. b) Yes. The group needs only \(\$5\) more for the small-group pass, compared with \(\$7\) for individual tickets.

- Determine the fruit count in one basket before scaling to all baskets. - For part b, first determine the total number of apples. - To avoid a difficult direct product, group the apples into tens and find the cost of one ten-apple group.

1. Each basket contains \(6+4+2=12\) pieces of fruit. 2. All \(25\) baskets contain \(25 \times 12=300\) pieces of fruit. 3. The baskets contain \(25 \times 6=150\) apples. 4. Ten apples cost \(10 \times 30=300\) cents, or \(\$3.00\). There are \(15\) groups of ten apples, so the total apple cost is \(15 \times \$3.00=\$45.00\).

a) The seller uses \(300\) pieces of fruit. b) All the apples cost \(\$45.00\).

- Try a shorter example and compare the number of spaces with the number of posts. - Divide the total length by the distance between posts. - Remember to include the post at the starting point.

1. Find the number of equal spaces between posts: \(180 \div 10 = 18\). 2. A row of \(18\) spaces has one more post than spaces, so \(18 + 1 = 19\) posts are needed.

\(19\) posts are needed.

- Find the total mass of each type of fruit crate. - Add the two partial masses and compare the result with the limit. - For part c, combine the two multiplication expressions with addition.

1. Find the mass of the apple crates: \(12 \times 25\,\text{kg} = 300\,\text{kg}\). 2. Find the mass of the pear crates: \(8 \times 20\,\text{kg} = 160\,\text{kg}\). 3. Find the total mass: \(300\,\text{kg} + 160\,\text{kg} = 460\,\text{kg}\). 4. Since \(460\,\text{kg} \le 500\,\text{kg}\), the load is within the limit. 5. One expression for the total is \(12 \times 25 + 8 \times 20\).

a) The load has a total mass of \(460\,\text{kg}\). b) Yes. The load is within the \(500\)-kg limit. c) \(12 \times 25 + 8 \times 20\)

- Find the total mass of each material. - Subtract the smaller total from the greater total. - Combine the multiplication and subtraction into one expression.

1. Find the total mass of the cement: \(15 \times 25\,\text{kg} = 375\,\text{kg}\). 2. Find the total mass of the plaster mix: \(12 \times 30\,\text{kg} = 360\,\text{kg}\). 3. Since \(375\,\text{kg} > 360\,\text{kg}\), the cement has the greater total mass. 4. Find the difference: \(375\,\text{kg} - 360\,\text{kg} = 15\,\text{kg}\). 5. One expression for the difference is \(15 \times 25 - 12 \times 30\).

The cement has the greater total mass by \(15\,\text{kg}\). One expression for the difference is \(15 \times 25 - 12 \times 30\).

- Count the days in each part of the week. - Find the number of rolls for each part separately. - Add the three subtotals.

1. Find the Monday-through-Wednesday total: \(3 \times 125 = 375\) rolls. 2. Find the Thursday-and-Friday total: \(2 \times 150 = 300\) rolls. 3. Find the weekend total: \(2 \times 210 = 420\) rolls. 4. Add the three totals: \(375 + 300 + 420 = 1095\) rolls.

The bakery makes \(1095\) rolls during the week.

- Find the number of books received in each week. - Add the three weekly totals. - Compare the total with the goal of \(500\). - Subtract to find the shortage or extra amount.

1. The first week brings \(5\times 24=120\) books. 2. The second week brings \(8\times 18=144\) books. 3. The third week brings \(12\times 15=180\) books. 4. The school collects \(120+144+180=444\) books. 5. Since \(444<500\), the goal is not reached. The school still needs \(500-444=56\) books.

No. The school collects \(444\) books and still needs \(56\) more.

- Use the row and column labels to identify each entry. - For part a), first find the total attendance for each of the two days. - What operation represents “three times” a number? - For part b), compare only the entries in the Children column.

1. Find Friday’s total attendance: \(124 + 88 = 212\). 2. Find Sunday’s total attendance: \(582 + 295 = 877\). 3. Find the increase: \(877 - 212 = 665\) people. 4. Find three times Friday’s number of children: \(124 \times 3 = 372\). 5. Since \(456 > 372\), Saturday’s number of children was more than three times Friday’s number.

a) \(665\) people b) Yes. Since \(124 \times 3 = 372\) and \(456 > 372\), the number was more than three times as great.

- Think of covering the floor with long strips placed side by side. - Find how many strips are needed across the room. - Find the total strip length and compare it with one roll's length.

1. Find the number of strips needed across the room: \(12\div6=2\) strips. 2. Each strip must be \(15\,\text{ft}\) long, so the total length needed is \(2\times15\,\text{ft}=30\,\text{ft}\). 3. One roll is \(36\,\text{ft}\) long. Since \(30\,\text{ft}<36\,\text{ft}\), one roll is enough.

\(1\) roll is needed.

- Use the stated prism facts to connect vertices with clay balls and edges with straws. - Keep the three edge lengths separate before finding the total straw length. - Check that your straw count accounts for all \(12\) edges.

1. There are \(8\) vertices, so Ethan needs \(8\) clay balls. 2. He needs four \(12\,\text{cm}\) straws, four \(8\,\text{cm}\) straws, and four \(5\,\text{cm}\) straws. 3. The total straw length is \(4 \times 12+4 \times 8+4 \times 5=48+32+20=100\,\text{cm}\).

a) \(8\) clay balls b) Four \(12\,\text{cm}\) straws, four \(8\,\text{cm}\) straws, and four \(5\,\text{cm}\) straws c) \(100\,\text{cm}\)

- First find the total numbers of adults and children. - How do you find the admission revenue for each group? - Add the adult and child revenue.

1. The total number of adults is \(24+38=62\). 2. Adult admission revenue is \(62 \times \$9=\$558\). 3. The total number of children is \(36+22=58\). 4. Child admission revenue is \(58 \times \$5=\$290\). 5. The total revenue is \(\$558+\$290=\$848\).

The museum's total admission revenue was \(\$848\).

- Find the total weight of the fuel. - Add the truck, driver, and fuel weights. - Subtract the current weight from the maximum allowed weight.

1. Find the fuel weight: \(20 \times 7\,\text{lb} = 140\,\text{lb}\). 2. Find the current total weight: \(5460 + 190 + 140 = 5790\), so the truck, driver, and fuel weigh \(5790\,\text{lb}\). 3. Subtract from the maximum: \(7700\,\text{lb} - 5790\,\text{lb} = 1910\,\text{lb}\).

The truck can carry \(1910\,\text{lb}\) of goods.

- How many bottles do both machines fill together in one minute? - Once you know the combined rate, how many groups of that amount make \(400\)?

1. Find the combined number of bottles filled per minute: \(45 + 55 = 100\). 2. Divide the target number of bottles by the combined rate: \(400 \div 100 = 4\) minutes.

The machines must work together for \(4\) minutes.

- Determine one equal shelf section before combining several sections. - The same section size applies to all seven sections. - Check that your per-section amount would reproduce the original shipment across seven sections.

1. Each shelf section receives \(84\div7=12\) card games. 2. Three sections contain \(12 \times 3=36\) card games.

There are \(36\) card games in \(3\) shelf sections altogether.

- How many liters of juice are in one barrel? - Use place value and a related basic division fact. - For part b), find the amount in \(10\) barrels and compare it with \(300\,\text{L}\).

1. Find the amount of juice in each barrel: \(240\,\text{L} \div 8 = 30\,\text{L}\). 2. Find the amount in five barrels: \(5 \times 30\,\text{L} = 150\,\text{L}\). 3. Check the employee's statement: \(10 \times 30\,\text{L} = 300\,\text{L}\). 4. The result agrees with the statement, so the employee is correct.

a) The \(5\) barrels hold \(150\,\text{L}\) of juice. b) Yes. The employee is correct because \(10 \times 30\,\text{L} = 300\,\text{L}\).

- Find the total cost of each type of item. - Express \(\$5.00\) in cents. - Subtract the total cost from the starting amount.

1. Find the cost of the pencils: \(4 \times 45 = 180\) cents. 2. Find the cost of the erasers: \(5 \times 60 = 300\) cents. 3. Find the total cost: \(180 + 300 = 480\) cents. 4. Convert the starting amount: \(\$5.00 = 500\,\text{cents}\). 5. Find the amount left: \(500 - 480 = 20\) cents.

Lucas has \(20\) cents left.

- Find the cost of each type of book. - Add the two costs. - Compare the total with \(\$100\) and find the difference.

1. Find the cost of the nonfiction books: \(5 \times \$12 = \$60\). 2. Find the cost of the fiction books: \(4 \times \$9 = \$36\). 3. Find the total cost: \(\$60 + \$36 = \$96\). 4. Since \(\$96 < \$100\), there is enough money. 5. Find the amount left: \(\$100 - \$96 = \$4\).

Yes. The class has enough money and will have \(\$4\) left.

- First find the number sold during the five weekdays. - Add the weekend sales. - Subtract the total sold from the starting amount.

1. Find the weekday sales: \(5 \times 12 = 60\). 2. Find the total sales: \(60 + 24 = 84\). 3. Find the amount remaining: \(95 - 84 = 11\).

There are \(11\) cans left.

- Find the flour used in each week separately. - Subtract the total used from the starting amount. - Compare the amount left with \(45\,\text{kg}\).

1. Find the first-week use: \(6 \times 35\,\text{kg} = 210\,\text{kg}\). 2. Find the second-week use: \(2 \times 110\,\text{kg} = 220\,\text{kg}\). 3. Find the total used: \(210\,\text{kg} + 220\,\text{kg} = 430\,\text{kg}\). 4. Find the flour left: \(500\,\text{kg} - 430\,\text{kg} = 70\,\text{kg}\). 5. Since \(70\,\text{kg} > 45\,\text{kg}\), there is enough flour. After the order, \(70 - 45 = 25\,\text{kg}\) would remain.

Yes. The bakery has \(70\,\text{kg}\) left, which is enough for the \(45\,\text{kg}\) order.

- First find how much Ms. Miller spends on colored pencils. - Subtract that amount from the total to find the cost of all the drawing pads. - How can you find the price of one pad from the price of four pads?

1. Find the total cost of the colored pencils: \(6 \times \$5 = \$30\). 2. Find the total cost of the four drawing pads: \(\$54 - \$30 = \$24\). 3. Divide by the number of drawing pads: \(\$24 \div 4 = \$6\).

One drawing pad costs \(\$6\).

- Match each comparison with the correct balloon color. - Find the number of each color separately. - Add the two results.

1. Find the number of blue balloons: \(35 \times 3 = 105\). 2. Find the number of red balloons: \(18 + 124 = 142\). 3. Add both colors: \(105 + 142 = 247\).

The school prepares \(247\) balloons altogether.

- Find the number of books on each bookcase. - Add the two totals. - Compare the two bookcase totals.

1. Find the number on the first bookcase: \(6 \times 8 = 48\). 2. Find the number on the second bookcase: \(4 \times 15 = 60\). 3. Find the combined total: \(48 + 60 = 108\). 4. Since \(60 > 48\), the second bookcase has more books.

There are \(108\) books altogether, and the second bookcase has more books.

- Find the number of chairs in each group of rows. - Add the two amounts. - Compare the total with \(130\).

1. Find the chairs in the first group: \(8 \times 9 = 72\). 2. Find the chairs in the added rows: \(4 \times 12 = 48\). 3. Find the total: \(72 + 48 = 120\). 4. Since \(120 < 130\), there are not enough seats. The school needs \(130 - 120 = 10\) more chairs.

No. There are \(120\) chairs, so \(10\) more chairs are needed.

- First find the total number of trees. - What operation finds one-third of a group? - After finding the number already harvested, how can you find the rest?

1. Find the total number of trees: \(13 + 11 = 24\). 2. Find one-third of the trees: \(\frac{1}{3} \times 24 = 8\). 3. Find the number still needing harvest: \(24 - 8 = 16\).

The fruit has been harvested from \(8\) trees, and \(16\) trees still need to be harvested.

- Find the total cost of the paintbrushes. - Use the full bill to find the cost of the three paint tubes. - Find the price of one tube, then compare the cost of four tubes with \(\$10\).

1. The five paintbrushes cost \(5 \times \$3 = \$15\). 2. The three paint tubes cost \(\$24 - \$15 = \$9\). 3. One tube costs \(\$9 \div 3 = \$3\). 4. Four tubes cost \(4 \times \$3 = \$12\). 5. Since \(\$12 > \$10\), Lucas is not correct.

No. Four tubes cost \(\$12\), but Lucas has only \(\$10\).

- One of the two geranium counts must be determined from the comparison statement. - Keep the park amount and the flower-bed amount separate until both are known. - The question asks for the combined amount, not only the larger amount.

1. The park needs \(24 \times 8=192\) geraniums. 2. The flower bed and park need \(192+24=216\) geraniums altogether.

The nursery needs \(216\) geraniums altogether.

- Find how much water enters during the entire storm. - Add that amount to the water already in the tank. - Subtract the new amount from the tank’s capacity.

1. Find the amount added during the storm: \(30 \times 35 = 1050\,\text{gal}\). 2. Find the new amount in the tank: \(1250 + 1050 = 2300\,\text{gal}\). 3. Find the amount still needed: \(3000 - 2300 = 700\,\text{gal}\).

The tank needs \(700\,\text{gal}\) more to be full.

- First determine the amount in one original bed. - The new amount per bed is different from the original amount per bed. - Build the new seven-bed total before comparing it with the bulbs already available.

1. Each original bed contains \(240\div4=60\) bulbs. 2. Each new bed should contain \(60+15=75\) bulbs. 3. Seven new beds require \(7\times75=525\) bulbs. 4. The school already has \(240\) bulbs, so it must obtain \(525-240=285\) additional bulbs. 5. One expression for the whole calculation is \(7\times(240\div4+15)-240\).

The school must obtain \(285\) additional bulbs. One expression is \(7\times(240\div4+15)-240=285\). Here, \(240\div4\) finds the original bulbs per bed, \(+15\) gives the new amount per bed, multiplying by \(7\) gives the new total needed, and subtracting \(240\) finds the additional bulbs required.

- Find the number of bags under each possible bag size. - Compare each bag count with the \(65\) available labels. - State both the workable plan and the size of the surplus or shortage.

1. With \(6\) rolls per bag, \(420 \div 6=70\) bags are needed, so \(70-65=5\) more labels would be needed. 2. With \(7\) rolls per bag, \(420 \div 7=60\) bags are needed, leaving \(65-60=5\) labels. 3. Therefore, only the \(7\)-roll plan can be completed with the available labels.

The \(6\)-roll plan needs \(70\) bags and is short \(5\) labels. The \(7\)-roll plan needs \(60\) bags and leaves \(5\) labels. Use the \(7\)-roll plan.

- Determine the case count for each school. - Use those results both for the comparison and for the warehouse total. - Compare the combined case count with the warehouse capacity.

1. School A receives \(560 \div 8=70\) cases. 2. School B receives \(540 \div 6=90\) cases, so School B receives more. 3. Together they have \(70+90=160\) cases. 4. Since the warehouse holds \(155\), it needs \(160-155=5\) more case spaces.

School B receives more cases. The two schools receive \(160\) cases altogether, so the warehouse is short by \(5\) case spaces.

- First recover each per-bicycle price from the given totals. - Compare the unit prices before working on part b. - Use the per-bicycle difference to find the difference for four bicycles.

1. Ms. Miller’s price per bicycle is \(\$450 \div 3=\$150\). 2. Mr. Schmidt’s price per bicycle is \(\$320 \div 2=\$160\), so he pays more per bicycle. 3. The per-bicycle difference is \(\$160-\$150=\$10\). For \(4\) bicycles, the difference is \(4 \times \$10=\$40\).

a) Mr. Schmidt pays more per bicycle: \(\$160\) versus \(\$150\). b) Four bicycles at Mr. Schmidt’s price cost \(\$40\) more than four at Ms. Miller’s price.

- Find the number of spaces between \(16\) attachment points. - Find the distance from the first point to the last point. - Include the extension at both ends.

1. Sixteen attachment points create \(16 - 1 = 15\) spaces. 2. The distance from the first attachment point to the last is \(15 \times 6\,\text{in} = 90\,\text{in}\). 3. The two end extensions total \(2 \times 5\,\text{in} = 10\,\text{in}\). 4. The rail length is \(90\,\text{in} + 10\,\text{in} = 100\,\text{in}\).

The rail is \(100\,\text{in}\) long.

- Find the distance ridden on each day separately. - How can you find the total distance ridden during both days? - What operation finds the distance still remaining?

1. Find Monday's distance: \(\frac{1}{8} \times 24\,\text{mi} = 3\,\text{mi}\). 2. Find Tuesday's distance: \(\frac{1}{4} \times 24\,\text{mi} = 6\,\text{mi}\). 3. Add the two distances: \(3\,\text{mi} + 6\,\text{mi} = 9\,\text{mi}\). 4. Find the remaining distance: \(24\,\text{mi} - 9\,\text{mi} = 15\,\text{mi}\).

The class rides \(9\,\text{mi}\) altogether, and \(15\,\text{mi}\) remain.

- Use the employee's comparison to find the number of tickets still available. - Once you know the sold and unsold amounts, how can you find the total? - Breaking \(480\) into hundreds and tens may help with the division.

1. Find the number of tickets still available: \(480 \div 4 = 120\). 2. Add the tickets already sold and the tickets still available: \(480 + 120 = 600\).

There were \(600\) tickets available for the concert altogether.

- First find the money collected from all the cups. - Multiply the number sold by the price per cup. - Profit is revenue minus cost.

1. Find the total revenue: \(450 \times \$2 = \$900\). 2. Subtract the cost: \(\$900 - \$300 = \$600\).

The class collects \(\$900\) and makes a profit of \(\$600\).

- First find how many feet are repaired on the first day. - How does the second-day amount compare with the first-day amount? - How much is repaired during both days altogether? - What operation finds the part still needing repair?

1. Find the distance repaired on the first day: \(\frac{1}{3} \times 945\,\text{ft} = 315\,\text{ft}\). 2. Find the distance repaired on the second day: \(315\,\text{ft} - 42\,\text{ft} = 273\,\text{ft}\). 3. Find the total distance repaired: \(315\,\text{ft} + 273\,\text{ft} = 588\,\text{ft}\). 4. Subtract from the full trail: \(945\,\text{ft} - 588\,\text{ft} = 357\,\text{ft}\).

\(357\,\text{ft}\) of the trail still need to be repaired.

- Use the Saturday revenue and price per bag to find the number sold. - Add \(15\) to find the number sold on Sunday. - Multiply the Sunday number of bags by the price per bag.

1. Saturday sales: \(\$320 \div \$4 = 80\) bags. 2. Sunday sales: \(80 + 15 = 95\) bags. 3. Sunday revenue: \(95 \times \$4 = \$380\).

a) \(80\) bags b) \(\$380\)

- Find the cost of four pots bought separately. - Compare that amount with the box price. - Subtract to find the savings.

1. Four individual pots cost \(4 \times 3 = 12\) dollars. 2. Compare \(\$12\) with the box price of \(\$10\). 3. The savings are \(12 - 10 = 2\) dollars.

a) \(\$12\) b) \(\$2\)

- Find the number of each color separately. - Compare the two products. - Add the two amounts.

1. Find the number of red flowers: \(15 \times 4 = 60\). 2. Find the number of blue flowers: \(12 \times 5 = 60\). 3. Compare: \(60 = 60\), so the numbers are equal. 4. Find the total: \(60 + 60 = 120\).

There are \(60\) red flowers and \(60\) blue flowers, so the amounts are equal. There are \(120\) flowers altogether.

- Find the distance traveled by each type of transportation. - Use distance per hour and number of hours. - Add the two distances.

1. Find the bicycle distance: \(2 \times 9 = 18\) miles. 2. Find the bus distance: \(3 \times 30 = 90\) miles. 3. Add the distances: \(18 + 90 = 108\) miles.

The entire trip is \(108\) miles long.

- Read each row separately when finding yearly changes for children and adults. - Part b) asks for both calculations and an observation about the results. - For part c), compare the total membership at the beginning and the end. - It may help to find the combined total for each of those two years first.

1. Find each yearly increase in child members: \(188 - 156 = 32\) for 2022, \(245 - 188 = 57\) for 2023, and \(290 - 245 = 45\) for 2024. The greatest increase was \(57\), in 2023. 2. Find the adult increases: \(255 - 235 = 20\) for 2023 and \(295 - 255 = 40\) for 2024. The 2024 increase is twice the 2023 increase. 3. Find the total membership in 2021: \(156 + 210 = 366\). 4. Find the total membership in 2024: \(290 + 295 = 585\). 5. Find the overall increase: \(585 - 366 = 219\) members.

a) 2023 b) The increases were \(20\) members for 2023 and \(40\) members for 2024. The 2024 increase was twice the 2023 increase. c) \(219\) members

- Use the lengths and scale of the bars to read each count. - Add all the counts for part a). - Compare the combined Reading-and-Music count with one-fourth of the total.

1. Read the counts: Reading \(14\), Sports \(24\), Music \(8\), Gaming \(18\), Friends \(16\). 2. The total is \(14+24+8+18+16=80\) students. 3. Sports has the greatest count, \(24\). 4. One-fourth of \(80\) is \(80\div4=20\); Reading and Music total \(14+8=22\), which is greater than \(20\).

a) \(80\) students b) Sports, \(24\) students c) Yes. Reading and Music total \(22\), more than one-fourth of \(80\), which is \(20\).

- Read each starting amount from the vertical scale. - Complete the changes in the order given and record each new amount. - When two amounts are combined and shared equally, add first and then divide by \(2\). - Check that the total amount of paper remains the same.

1. Read the starting amounts: Class A, \(35\,\text{lb}\); Class B, \(50\,\text{lb}\); Class C, \(45\,\text{lb}\); Class D, \(20\,\text{lb}\). 2. After Class C transfers \(10\,\text{lb}\), Class A has \(35 + 10 = 45\,\text{lb}\), and Class C has \(45 - 10 = 35\,\text{lb}\). 3. Classes B and D have \(50 + 20 = 70\,\text{lb}\) together. Dividing equally gives \(70 \div 2 = 35\,\text{lb}\) for each class. 4. The final amounts are Class A, \(45\,\text{lb}\), and Classes B, C, and D, \(35\,\text{lb}\) each. Class A has the greatest amount.

a) Class A: \(35\,\text{lb}\); Class B: \(50\,\text{lb}\); Class C: \(45\,\text{lb}\); Class D: \(20\,\text{lb}\) b) Class A: \(45\,\text{lb}\); Class B: \(35\,\text{lb}\); Class C: \(35\,\text{lb}\); Class D: \(35\,\text{lb}\) c) Class A

- Add all the bar values to verify the total. - Add \(15\) to the basketball value for part b). - Find one-fourth of the total and compare it with the combined swimming and gymnastics votes.

1. Add all five values: \(40 + 25 + 15 + 10 + 10 = 100\) students. 2. Basketball received \(25\) votes. A value \(15\) greater is \(25 + 15 = 40\), which is the soccer value. 3. Swimming and gymnastics received \(15 + 10 = 25\) votes. One-fourth of \(100\) is \(100 \div 4 = 25\). Therefore, the statement is true.

a) \(100\) students b) Soccer c) **True**, because swimming and gymnastics received \(25\) votes altogether, and \(100 \div 4 = 25\).

- Read both stamp counts from the labeled bars. - Find the value of each kind using its price per stamp. - Add the two values and convert cents to dollars.

1. Read the counts: \(30\) blue stamps and \(24\) green stamps. 2. The blue stamps are worth \(30\times2=60\) cents. 3. The green stamps are worth \(24\times3=72\) cents. 4. Together they are worth \(60+72=132\) cents, or \(\$1.32\).

The stamps are worth \(\$1.32\) altogether.

- Find the total spaces from the number of rows and spaces per row. - Read the two unavailable-space counts from the chart. - Subtract both nonoverlapping groups from the overall total.

1. The lot has \(6\times11=66\) spaces. 2. There are \(8\) reserved spaces and \(17\) occupied spaces, a total of \(8+17=25\) unavailable spaces. 3. There are \(66-25=41\) spaces that are neither reserved nor occupied.

\(41\) spaces are neither reserved nor occupied.

- Read the charged-tablet total from the chart rather than adding both categories. - Find how many charged tablets the classes need altogether. - Check whether there are enough and then find any leftover.

1. The chart shows \(46\) charged tablets. 2. The classes need \(3\times15=45\) tablets. 3. Since \(46>45\), there are enough, with \(46-45=1\) charged tablet left.

Yes. There is enough for all three classes, and \(1\) charged tablet remains.

- Use the array to count the dots efficiently. - Each dot stands for the same number of seats. - Compare the total capacity with the occupied-seat count.

1. The diagram has \(6\) rows of \(8\) dots, so it contains \(6\times8=48\) dots. 2. The total capacity is \(48\times10=480\) seats. 3. The number available is \(480-85=395\) seats.

\(395\) seats are still available.

- Use the diagram to determine the number of booklets. - Find the total number of pages before sharing them. - The final share must be equal for all six students.

1. Count the dots: there are \(35\) booklets. 2. The total number of pages is \(35\times24=35\times(20+4)=700+140=840\). 3. Dividing equally gives \(840\div6=140\) pages per student.

Each student reads \(140\) pages.

- Use the diagram to determine how many pallets there are. - Each pallet contains the same number of cartons. - Account for the shipment only after finding the full inventory.

1. Count the dots: there are \(56\) pallets. 2. The total number of cartons is \(56\times12=56\times(10+2)=560+112=672\). 3. After shipping \(95\) cartons, \(672-95=577\) cartons remain.

\(577\) cartons remain.

- Use the rows and columns to count the dots efficiently. - Four groups of \(\$25\) make \(\$100\). - Subtract the expenses from the amount raised.

1. The array has \(6\) rows of \(10\) dots, so it shows \(6 \times 10 = 60\) dots. 2. The total amount raised is \(60 \times \$25 = \$1500\). 3. After expenses, \(\$1500 - \$420 = \$1080\) remains.

\(\$1080\) remains after expenses.

- Use the diagram to determine the number of panels. - Find the total number of cells before accounting for damage. - Check that the final count is less than the original cell total.

1. Count the dots: there are \(45\) panels. 2. The total number of cells is \(45\times18=45\times(20-2)=900-90=810\). 3. Subtract the damaged cells: \(810-27=783\).

\(783\) cells still work.

- Use the diagram to determine how many equal tablet groups are represented. - Keep track of what one dot represents. - Your two-step calculation should connect the diagram total to one class's share.

1. The diagram contains \(90\) dots, so it represents \(90\times3=270\) tablets. 2. Dividing equally among \(6\) classes gives \(270\div6=45\) tablets per class.

\(90\times3=270\), then \(270\div6=45\). Each class receives \(45\) tablets.

- Each row belongs to one bus. Count the planned trips in all rows. - Use the unfilled dots to work out how many trips will run. - Compare that number with the trips the school needs.

1. The picture has \(5\) rows of \(8\) trips, so \(5\times8=40\) trips were planned. 2. Seven dots are unfilled, so \(40-7=33\) trips will run. 3. The school needs \(30\) trips. There are enough, with \(33-30=3\) extra trips.

Yes. There are \(3\) extra trips available.

- Read the four tile counts from their labeled bars. - Multiply each count by that kind of tile's price. - Add the four costs after you have found them separately.

1. Read the chart: \(24\) blue, \(18\) red, \(12\) green and \(6\) yellow tiles. 2. Blue tiles cost \(24\times\$2=\$48\). 3. Red tiles cost \(18\times\$3=\$54\). 4. Green tiles cost \(12\times\$4=\$48\). 5. Yellow tiles cost \(6\times\$5=\$30\). 6. Add the costs: \(\$48+\$54+\$48+\$30=\$180\).

The mosaic costs \(\$180\).

- Interpret each intermediate quantity in words before calculating it. - The \(11\) empty seats refer to the whole seating arrangement, not to one table or to a difference between two givens. - A plausible numerical result is not enough; every operation must match an action or relationship in the story.

1. The total number of seats is \(6 \times 8=48\). 2. The \(11\) empty seats are removed from that total, so equation A models the story. 3. Equation B removes only \(11-6=5\) seats, which has no meaning in the situation. Equation C treats \(11-8\) as seats per table, which also has no meaning in the story. 4. Thus \(s=48-11=37\).

Equation A is correct: \(s=6\times8-11\), so \(s=37\). Equation B subtracts \(11-6=5\) from the total seats, but that \(5\) has no meaning in the situation. Equation C treats \(11-8=3\) as a number of occupied seats per table, which also does not match the situation.

- Round the quantities to nearby friendly numbers before doing exact arithmetic. - Use the estimate only to judge the size of the claim. - Then compute the exact bus total and include the students arriving by car.

1. Estimate \(36\) as \(40\) and \(47\) as \(50\): \(9\times40+50=410\). A total near \(410\) makes \(520\) look too large. 2. The exact bus total is \(9\times36=324\). 3. The exact overall total is \(324+47=371\). 4. The claim of \(520\) is not reasonable and is not correct.

A reasonable estimate is about \(410\) students. The exact total is \(371\), so the claim of \(520\) is not reasonable.

- Decide whether the \(5\) empty seats belong to each row or to the theater as a whole. - Both equations have positive intermediate values, so judge them by meaning rather than by whether a calculation looks possible. - Interpret the parentheses in words.

1. The theater has \(12 \times 24=288\) seats altogether. 2. The \(5\) empty seats are removed once from the theater total, so \(o=12 \times 24-5\) is correct. 3. The other equation subtracts \(5\) seats from every row, as if each of the \(12\) rows had \(5\) empty seats. 4. The occupied-seat count is \(288-5=283\).

The correct equation is \(o=12 \times 24-5\). The other equation incorrectly removes \(5\) seats from every row. Thus \(o=283\).

- Round the two factors to nearby tens before multiplying. - Round the loose-notebook count to a nearby hundred for the estimate. - After finding the exact total, compare it with your estimate.

1. Round \(28\) to \(30\), \(42\) to \(40\), and \(115\) to about \(100\). This gives \(30\times40+100=1300\), so \(1300\) is the best estimate. 2. The exact case total is \(28\times42=1176\). 3. Add the loose notebooks: \(1176+115=1291\). 4. The exact total \(1291\) is only \(9\) away from \(1300\), so the estimate is reasonable.

The most reasonable estimate is \(1300\), because \(30\times40+100=1300\). The exact total is \(28\times42+115=1291\), which is only \(9\) away from the estimate, so \(1300\) is reasonable.

- Build the requested number of orders from the choices in the table. - There may be more than one way to make the same number of orders. - Find and compare the total cost of each combination. - Subtract the deal price from the cost of four single orders to find the savings.

1. For \(3\) orders, three single orders cost \(3 \times 3 = 9\) dollars. One \(2\)-order deal plus one single order costs \(5 + 3 = 8\) dollars. 2. For \(5\) orders, a \(4\)-order deal plus one single order costs \(9 + 3 = 12\) dollars. Two \(2\)-order deals plus one single order cost \(5 + 5 + 3 = 13\) dollars. One \(2\)-order deal plus three single orders costs \(5 + 3 + 3 + 3 = 14\) dollars. Five single orders cost \(5 \times 3 = 15\) dollars. Therefore, the least expensive price is \(\$12\). 3. Four single orders cost \(4 \times 3 = 12\) dollars. The \(4\)-order deal costs \(9\) dollars, so the savings are \(12 - 9 = 3\) dollars.

a) Three single orders cost \(\$9\). One \(2\)-order deal plus one single order costs \(\$8\). b) The least expensive price is \(\$12\), using the \(4\)-order deal plus one single order. c) You save \(\$3\).

- Start with the larger packages, but make sure the total number is exact. - Compare the cost of buying tickets separately with the package prices. - Test different combinations and add their prices. - For part c), calculate each total separately before comparing.

1. Fourteen single tickets would cost \(14 \times 2 = 28\) dollars. Compared with single-ticket prices, a \(4\)-ticket package saves \(\$1\), a \(6\)-ticket package saves \(\$2\), and a \(10\)-ticket package saves \(\$5\). For exactly \(14\) tickets, the greatest possible savings are \(\$6\), using \(10 + 4\) tickets. The cost is \(28 - 6 = 22\) dollars. 2. Eleven single tickets would cost \(11 \times 2 = 22\) dollars. The greatest possible savings are \(\$5\), using one \(10\)-ticket package and one single ticket. The cost is \(22 - 5 = 17\) dollars. 3. Two \(6\)-ticket packages cost \(2 \times 10 = 20\) dollars. Three \(4\)-ticket packages cost \(3 \times 7 = 21\) dollars. The first way costs \(1\) dollar less.

a) \(\$22\), using \(10 + 4\) tickets b) \(\$17\), using \(10 + 1\) tickets c) Two \(6\)-ticket packages cost less by \(\$1\).

- Find the number of objects at one level before moving to the next level. - Determine the total number of cabinets before finding the drawers. - Multiply the number of drawers by the folders in each drawer. - Check that each step uses the total from the previous level.

1. Find the total number of aisles: \(4\times 12=48\). 2. Find the total number of filing cabinets: \(48\times 15=720\). 3. Find the total number of drawers: \(720\times 6=4320\). 4. Find the total number of file folders: \(4320\times 8=34{,}560\).

The archive contains \(34{,}560\) file folders.

- Find the number of seats already provided by the first-class cars. - Subtract to find how many seats are still needed. - Find how many groups of \(70\) seats make the remaining number of seats. - Include every passenger car and the locomotive in the total length.

1. The first-class cars provide \(3 \times 40 = 120\) seats. 2. The number of seats still needed is \(400 - 120 = 280\). 3. Since \(4 \times 70 = 280\), the train needs \(4\) coach cars. 4. The train has \(3 + 4 = 7\) passenger cars. 5. The passenger cars are \(7 \times 72\,\text{ft} = 504\,\text{ft}\) long in all. 6. Including the locomotive, the train is \(504\,\text{ft} + 50\,\text{ft} = 554\,\text{ft}\) long.

a) The train needs \(4\) coach cars. b) The train is \(554\,\text{ft}\) long.

- Find the number of tickets sold during each hour in order. - What operation finds one-fourth of the starting number? - After finding all three hourly amounts, how can you find the total number sold? - How can you find what remains from the starting number?

1. Find the number sold during the first hour: \(\frac{1}{4} \times 400 = 100\). 2. Find the number sold during the second hour: \(100 - 35 = 65\). 3. Find the number sold during the third hour: \(2 \times 65 = 130\). 4. Find the total number sold: \(100 + 65 + 130 = 295\). 5. Subtract from the starting number: \(400 - 295 = 105\).

There are \(105\) raffle tickets remaining.

- Try finding the number of sheets remaining in each package separately. - Could you instead begin by finding the total number of sheets in both packages? - What happens if you combine the removed sheets before subtracting? - How many sheets are one-eighth of \(80\)? How many are one-eighth of \(48\)?

Method 1: 1. In the first package, \(80 - (\frac{1}{8} \times 80) = 80 - 10 = 70\) sheets remain. 2. In the second package, \(48 - (\frac{1}{8} \times 48) = 48 - 6 = 42\) sheets remain. 3. Add the remaining sheets: \(70 + 42 = 112\). Method 2: 1. Find the total number of sheets: \(80 + 48 = 128\). 2. One-eighth of all the sheets is removed: \(\frac{1}{8} \times 128 = 16\). 3. Subtract the removed sheets: \(128 - 16 = 112\).

There are \(112\) sheets of construction paper remaining altogether.

- Find the total number of bags and use multiplication to find the reward per bag. - Use each friend’s number of bags to find that person’s share. - For part b, compare Mia’s share with one-third of the total reward.

1. The friends collected \(4 + 5 + 3 = 12\) bags. 2. Since \(12 \times \$5 = \$60\), the reward is \(\$5\) per bag. 3. Leo receives \(4 \times \$5 = \$20\), Mia receives \(5 \times \$5 = \$25\), and Tom receives \(3 \times \$5 = \$15\). 4. An equal three-way split gives each friend \(\$60 \div 3 = \$20\). 5. Mia would receive \(\$25 - \$20 = \$5\) less under an equal split, so she is correct.

a) Leo receives \(\$20\), Mia receives \(\$25\), and Tom receives \(\$15\). b) Yes. An equal split would give Mia \(\$20\), which is \(\$5\) less than \(\$25\).

- Find the number of bottles of each kind first. - Use multiplication to find the price per bottle that makes the total bill. - Multiply the price per bottle by each kind’s bottle count.

1. The apple juice order contains \(3 \times 6 = 18\) bottles. 2. The orange juice order contains \(2 \times 10 = 20\) bottles. 3. The total number of bottles is \(18 + 20 = 38\). 4. Since \(38 \times \$2 = \$76\), one bottle costs \(\$2\). 5. The apple juice costs \(18 \times \$2 = \$36\), and the orange juice costs \(20 \times \$2 = \$40\).

a) \(38\) bottles were ordered. b) One bottle costs \(\$2\). c) The apple juice costs \(\$36\), and the orange juice costs \(\$40\).

- How many books are shelved during the first phase? - How many books remain after the first \(5\) days? - Use the faster daily rate to find the remaining number of days. - Add the days from both phases.

1. Find the number shelved during the first \(5\) days: \(5 \times 24 = 120\) books. 2. Find the number remaining: \(240 - 120 = 120\) books. 3. Find the additional days needed: \(120 \div 30 = 4\) days. 4. Add the two time periods: \(5 + 4 = 9\) days.

It takes \(9\) days to shelve all the books.

- Find how many pieces the rope will make. - Does the final piece require another cut? - Sketch the rope and mark each place where a cut is needed.

1. Find the number of pieces: \(18 \div 6 = 3\) pieces. 2. To make \(3\) pieces from one rope, Lukas needs \(3 - 1 = 2\) cuts. 3. Find the total cutting time: \(2 \times 15 = 30\) seconds.

It will take Lukas \(30\) seconds.

- Find the amount spent on each item in order. - Both fractions refer to the original \(\$600\). - Which two costs must you compare to find the difference?

1. Find the bicycle cost: \(\frac{1}{2} \times \$600 = \$300\). 2. Find the helmet cost: \(\frac{1}{5} \times \$600 = \$120\). 3. Find the total spent on the bicycle and helmet: \(\$300 + \$120 = \$420\). 4. Find the rack and lock set cost: \(\$600 - \$420 = \$180\). 5. Find the difference: \(\$300 - \$180 = \$120\).

The bicycle costs \(\$120\) more than the rack and lock set.

- Find the total flour used in the first week. - Use each multiplier to find the second week's amounts. - Add the second week's amounts and subtract the first week's total.

1. Find the first week's total: \(156\,\text{lb} + 98\,\text{lb} + 45\,\text{lb} = 299\,\text{lb}\). 2. Find the second week's amounts: \(156\,\text{lb} \times 2 = 312\,\text{lb}\), \(98\,\text{lb} \times 3 = 294\,\text{lb}\), and \(45\,\text{lb} \times 4 = 180\,\text{lb}\). 3. Find the second week's total: \(312\,\text{lb} + 294\,\text{lb} + 180\,\text{lb} = 786\,\text{lb}\). 4. Find the increase: \(786\,\text{lb} - 299\,\text{lb} = 487\,\text{lb}\).

The bakery uses \(487\,\text{lb}\) more flour in the second week.

- Find the number needed for each section in order. - How many stones are needed for the entire project? - Compare the total need with the number ordered.

1. Find the stones needed for the second section: \(4 \times 4320 = 17{,}280\). 2. Find the stones needed for the third section: \(\frac{1}{2} \times 17{,}280 = 8640\). 3. Find the total number needed: \(4320 + 17{,}280 + 8640 = 30{,}240\). 4. Since \(30{,}240 > 25{,}000\), the order is not enough. 5. Find the shortage: \(30{,}240 - 25{,}000 = 5240\).

No. The order is short by \(5240\) paving stones.

- First find the number of pounds used for the first three purposes. - How much remains from the total of \(4800\,\text{lb}\)? - How many eighths are equal to one-half and one-fourth? - What do the three given fractions add to?

1. Find the amount sent to grocery stores: \(\frac{1}{2} \times 4800\,\text{lb} = 2400\,\text{lb}\). 2. Find the amount used for juice: \(\frac{1}{4} \times 4800\,\text{lb} = 1200\,\text{lb}\). 3. Find the amount used for applesauce: \(\frac{1}{8} \times 4800\,\text{lb} = 600\,\text{lb}\). 4. Add the assigned amounts: \(2400\,\text{lb} + 1200\,\text{lb} + 600\,\text{lb} = 4200\,\text{lb}\). 5. Find the amount donated: \(4800\,\text{lb} - 4200\,\text{lb} = 600\,\text{lb}\). 6. The donated amount equals the applesauce amount. This is because \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{7}{8}\), leaving \(\frac{1}{8}\).

The farm receives \(600\,\text{lb}\) of apples. This is the same amount used for applesauce because the assigned fractions total \(\frac{7}{8}\), leaving another \(\frac{1}{8}\).

- Use \(12\) months for the monthly deposits. - Remember that both children contribute each month. - Add all deposits before subtracting the withdrawals.

1. The parents contribute \(12\times\$35=\$420\). 2. The two children contribute \(12\times2\times\$5=\$120\). 3. The grandparents contribute \(2\times\$50=\$100\). 4. Total deposits are \(\$420+\$120+\$100=\$640\). 5. The family removes \(3\times\$20=\$60\). 6. The final amount is \(\$640-\$60=\$580\).

\(\$580\)

- Include the starting savings, monthly income, and gifts. - Subtract the total monthly spending for the year. - Compare the final savings with the bicycle price.

1. Monthly income for the year is \(12\times\$15=\$180\). 2. The two gifts total \(\$60+\$30=\$90\). 3. Yearly spending is \(12\times\$4=\$48\). 4. The amount saved after one year is \(\$45+\$180+\$90-\$48=\$267\). 5. Since \(\$267<\$320\), more money is needed. The shortage is \(\$320-\$267=\$53\).

No. Jordan needs \(\$53\) more.

- First find how much hay each animal group receives. - What operations represent one-half and one-fifth of a quantity? - The final question asks for a difference, not only the remaining amount.

1. Find the hay used by the elephant habitat: \(\frac{1}{2} \times 4800\,\text{lb} = 2400\,\text{lb}\). 2. Find the hay used by the zebra habitat: \(\frac{1}{5} \times 4800\,\text{lb} = 960\,\text{lb}\). 3. Find the hay used by the first two habitats: \(2400\,\text{lb} + 960\,\text{lb} = 3360\,\text{lb}\). 4. Find the hay for the ponies: \(4800\,\text{lb} - 3360\,\text{lb} = 1440\,\text{lb}\). 5. Find the difference: \(2400\,\text{lb} - 1440\,\text{lb} = 960\,\text{lb}\).

The elephant habitat uses \(960\,\text{lb}\) more hay than the ponies.

- Break the problem into one day at a time. - After each step, record how many pounds remain. - Check whether each fraction refers to the original amount or to the amount remaining. - A table or diagram may help you track the changing amount.

1. On Monday, \(\frac{1}{4} \times 80\,\text{lb} = 20\,\text{lb}\) is used, leaving \(80\,\text{lb} - 20\,\text{lb} = 60\,\text{lb}\). 2. On Tuesday, \(\frac{1}{5} \times 60\,\text{lb} = 12\,\text{lb}\) is used, leaving \(60\,\text{lb} - 12\,\text{lb} = 48\,\text{lb}\). 3. On Wednesday, \(\frac{1}{2} \times 48\,\text{lb} = 24\,\text{lb}\) is donated, leaving \(48\,\text{lb} - 24\,\text{lb} = 24\,\text{lb}\).

\(24\,\text{lb}\) of apples remain.

- Compare the actual distance with the direct trail distance first. - The same backtracked section contributes to the trip twice: once backward and once when retraced forward. - Use the turnaround location only after determining the one-way backtrack distance.

1. The extra distance is \(22-18=4\,\text{miles}\). 2. The search section is walked once backward and once forward, so the one-way backtrack is \(4\div2=2\,\text{miles}\). 3. The hiker turns around at mile \(8\) and walks back \(2\) miles, so the bottle is \(8-2=6\,\text{miles}\) from Maple Point.

a) The hiker walks \(4\,\text{miles}\) extra. b) The bottle is found \(6\,\text{miles}\) from Maple Point.

- Find the number printed during each hour in order. - How can you find the fraction printed during the first hour? - After finding the first two hourly amounts, how can you find the third? - Compare all three results.

1. Find the first-hour amount: \(\frac{1}{6} \times 720 = 120\). 2. Find the second-hour amount: \(120 + 55 = 175\). 3. Find the total for the first two hours: \(120 + 175 = 295\). 4. Find the third-hour amount: \(720 - 295 = 425\). 5. Since \(425 > 175 > 120\), the most posters are printed during the third hour.

The first hour has \(120\) posters, the second hour has \(175\), and the third hour has \(425\). The most posters are printed during the third hour.

- First group the soccer balls in sets of \(6\). - Convert those groups to the equivalent number of basketballs. - Add that amount to the basketballs already produced. - Compare the result with \(14{,}000\).

1. The \(9150\) soccer balls form \(9150\div 6=1525\) groups of \(6\). 2. That material could make \(1525\times 5=7625\) basketballs. 3. The total possible number of basketballs is \(6420+7625=14{,}045\). 4. Since \(14{,}045>14{,}000\), the employee is correct.

Yes. The available material could have produced \(14{,}045\) basketballs.

- Represent both purchase costs in the same equation before finding the remainder. - Keep the meaning of \(m\) fixed as “money left after the purchases.” - After solving for \(m\), compare it with the separate event fee.

1. The shirt cost is \(8\times\$27=\$216\). 2. The hat cost is \(5\times\$18=\$90\). 3. One equation is \(m=500-8\times27-5\times18\). 4. Thus \(m=500-216-90=194\), so the club cannot pay the \(\$200\) fee. 5. The shortfall is \(\$200-\$194=\$6\).

\(m=500-8\times27-5\times18\), so \(m=194\). The club is \(\$6\) short of the \(\$200\) event fee.

- Determine the amount actually spent and express it in cents. - Find the cost of one notebook-and-cover pair. - Use convenient groups such as ten pairs and then refine the remaining amount rather than multiplying a three-digit number by a two-digit number directly.

1. The teacher spends \(\$30.00-\$2.40=\$27.60\), or \(2760\) cents. 2. One notebook-and-cover pair costs \(80+35=115\) cents. 3. Ten pairs cost \(1150\) cents, so twenty pairs cost \(2 \times 1150=2300\) cents. 4. The remaining amount is \(2760-2300=460\) cents. Since \(4 \times 115=460\), four more pairs use the remainder. 5. The teacher buys \(20+4=24\) pairs, so she buys \(24\) notebooks.

The teacher buys \(24\) notebooks.