- Read the dividend and divisor from the written setup. - Find the one-digit multiple of the divisor that equals the displayed partial dividend.

1. The displayed division is \(84\div12\). 2. Since \(12\times7=84\), the missing quotient digit is \(7\), and the missing subtraction row is \(84\). 3. The remainder is \(0\).

Quotient digit: \(7\); subtraction row: \(84\).

- Use a related multiplication fact. - Ask how many groups of \(12\) make \(120\).

1. Since \(12\times10=120\), \(120\div12=10\). 2. Check: \(10\times12=120\).

\(10\); check: \(10\times12=120\).

- Use the package size to complete the rows with known package counts. - In the last row, the total number of blocks is known and the number of packages is unknown. - Check the package count by multiplying it by \(20\).

1. Two packages contain \(2 \times 20=40\) blocks. 2. Ten packages contain \(10 \times 20=200\) blocks. 3. The missing package count is \(1000 \div 20=50\).

<table><tr><th>Packages</th><th>Blocks</th></tr><tr><td>\(1\)</td><td>\(20\)</td></tr><tr><td>\(2\)</td><td>\(40\)</td></tr><tr><td>\(10\)</td><td>\(200\)</td></tr><tr><td>\(50\)</td><td>\(1000\)</td></tr></table>

- Divide the total cost equally among all the stands. - Which operation finds the price of one item from a total cost? - Use an estimate to check whether your answer is reasonable.

1. Divide the total cost by the number of stands: \(\$525 \div 15\). 2. Calculate: \(525 \div 15 = 35\). 3. One stand costs \(\$35\).

One music stand costs \(\$35\).

- Divide the total cost equally among the \(14\) people. - Break the total into parts that are easy to divide by \(14\). - Estimate how many times \(14\) fits into about \(400\).

1. Divide the total cost by the number of people: \(\$406 \div 14\). 2. Calculate: \(406 \div 14 = 29\). 3. One ticket costs \(\$29\).

One admission ticket costs \(\$29\).

- Convert the total volume to milliliters. - Divide the total equally among \(12\) bottles. - Check by multiplying one bottle's volume by \(12\).

1. \(6\,\text{L}=6000\,\text{mL}\). 2. \(6000\,\text{mL}\div12=500\,\text{mL}\).

Each bottle contains \(500\,\text{mL}\).

- Use related multiplication facts for each two-digit divisor. - Estimate before calculating if a quotient is not immediate. - Compare the four quotient values after evaluating them.

1. A: \(270\div30=9\). 2. B: \(720\div24=30\). 3. C: \(960\div12=80\). 4. D: \(180\div20=9\). 5. The greatest quotient is \(80\), so C is greatest.

C has the greatest quotient, \(80\).

- Determine how many daily portions fit in the current supply. - Use the same daily amount for the record-year supply. - Divide each total amount by \(50\,\text{lb}\) per day.

1. Divide the current supply by the daily amount: \(3000 \div 50 = 60\). The current supply lasts \(60\) days. 2. Divide the record-year supply by the daily amount: \(22{,}500 \div 50 = 450\). The record-year supply lasted \(450\) days.

The current supply will last \(60\) days. The record-year supply lasted \(450\) days.

- Divide the total amount into equal groups of \(15\,\text{g}\). - Think about what number multiplied by \(15\) gives the total amount. - Check your answer by multiplication.

1. Divide the total amount needed by the amount in one packet: \(450\,\text{g} \div 15\,\text{g} = 30\). 2. Therefore, the bakery must open \(30\) packets.

The bakery must open \(30\) packets.

- Use multiplication to connect each division equation to a fact with a two-digit divisor. - When the divisor is missing, ask what number times the quotient gives the dividend. - Check every completed equation with multiplication.

1. \(360\div12=30\) because \(12\times30=360\). 2. For b), the missing divisor is \(30\) because \(30\times14=420\). 3. For c), the missing dividend is \(360\) because \(24\times15=360\). 4. \(624\div26=24\) because \(26\times24=624\).

a) \(30\) b) \(30\) c) \(360\) d) \(24\)

- Use multiplication as the inverse of division. - Identify whether the missing value is the dividend, divisor, or quotient. - Check each completed equation.

1. a) \(320 \div 40 = 8\). 2. b) Since \(6 \times 80 = 480\), the divisor is \(80\). 3. c) \(3 \times 70 = 210\), so the dividend is \(210\). 4. d) \(900 \div 90 = 10\). 5. e) \(10 \times 50 = 500\), so the dividend is \(500\).

a) \(8\) b) \(80\) c) \(210\) d) \(10\) e) \(500\)

- Notice how the dividend changes from one expression to the next. - Think about how many additional groups of \(13\) are added each step. - Look for a pattern in the quotients.

1. The quotients are \(39 \div 13 = 3\), \(52 \div 13 = 4\), \(65 \div 13 = 5\), \(78 \div 13 = 6\), and \(91 \div 13 = 7\). 2. Each dividend increases by one more group of \(13\), so the quotient increases by \(1\).

a) \(3\) b) \(4\) c) \(5\) d) \(6\) e) \(7\) The quotient increases by \(1\) each time.

- Use multiplication to find a missing dividend. - Determine how many groups of \(15\) make each known dividend. - Use completed columns to recognize the pattern.

1. \(45 \div 15 = 3\). 2. For quotient \(4\), use the inverse operation: \(4 \times 15 = 60\). 3. \(75 \div 15 = 5\). 4. For quotient \(6\), use the inverse operation: \(6 \times 15 = 90\).

The missing quotients are \(3\) and \(5\). The missing dividends are \(60\) and \(90\).

- Rewrite each division as an unknown-factor multiplication equation. - Test small whole-number factors. - For a larger divisor, look for a nearby multiple that matches the dividend. - Check each quotient by multiplication.

1. a) \(36 \div 12 = 3\). 2. b) \(48 \div 16 = 3\). 3. c) \(52 \div 13 = 4\). 4. d) \(75 \div 15 = 5\). 5. e) \(44 \div 22 = 2\). 6. f) \(66 \div 33 = 2\). 7. g) \(84 \div 21 = 4\). 8. h) \(96 \div 24 = 4\).

a) \(3\) b) \(3\) c) \(4\) d) \(5\) e) \(2\) f) \(2\) g) \(4\) h) \(4\)

- Equal sharing uses division. - Determine how many groups of \(25\) are in \(175\).

1. Divide the total bill by the number of students: \(\$175\div25=\$7\).

\(\$7\) per student

- Use division to make equal groups of \(15\). - Interpret the quotient as full bags. - Interpret the remainder as marbles that do not fill another bag.

1. Divide the total number of marbles by the number in each bag: \(1240\div15\). 2. The result is \(1240\div15=82\) remainder \(10\), because \(15\times82=1230\) and \(1240-1230=10\). 3. The quotient gives the number of full bags, and the remainder gives the number of unpacked marbles.

\(82\) full bags with \(10\) marbles remaining

- Focus on the displayed final partial dividend. - Find the one-digit multiple of \(24\) that equals \(144\). - That multiplier determines both missing entries.

1. The displayed final partial dividend is \(144\). 2. Since \(144=24\times6\), the missing quotient digit is \(6\) and the missing subtraction row is \(144\). 3. The quotient is \(36\).

Since \(144=24\times6\), the missing quotient digit is \(6\) and the subtraction row is \(144\). The quotient is \(36\).

- Use the displayed final partial dividend \(43\). - Choose the greatest one-digit multiple of \(31\) that does not exceed it. - Compare the amount left with the divisor.

1. The displayed final partial dividend is \(43\). 2. The greatest multiple of \(31\) not exceeding \(43\) is \(31\times1=31\). The missing quotient digit is \(1\) and the subtraction row is \(31\). 3. The remainder is \(43-31=12\). Since \(12<31\), it is valid. The result is \(31\text{ R }12\).

The missing quotient digit is \(1\), the subtraction row is \(31\), and the remainder is \(12\). The remainder is valid because \(12<31\). Thus \(973\div31=31\text{ R }12\).

- Look specifically at the last displayed partial dividend. - Decide how many groups of \(34\) fit into that amount. - Consider what would happen to the place value of the quotient if the final digit were omitted.

1. The first quotient digit is \(1\), and the next is \(2\); after the second subtraction the remainder is \(0\). 2. Bringing down the final dividend digit gives a partial dividend of \(0\), as shown. 3. Since \(34\) goes into \(0\) zero times, the missing final quotient digit is \(0\). It must be written to preserve the ones place, giving quotient \(120\).

The missing quotient digit is \(0\). The displayed final partial dividend is \(0\), so \(34\) fits zero times; the \(0\) must be written in the ones place to give quotient \(120\).

- Use the displayed second partial dividend. - Find the one-digit multiple of \(28\) that matches it. - Interpret the completed quotient in the ticket context.

1. The displayed final partial dividend is \(224\). 2. Since \(224=28\times8\), the missing quotient digit is \(8\) and the subtraction row is \(224\). 3. The quotient is \(48\), so each class receives \(48\) tickets.

\(28\times8=224\). The hidden quotient digit is \(8\), the subtraction row is \(224\), and each class receives \(48\) tickets.

- Focus on the second partial dividend and subtraction row. - Ask what one-digit multiple of \(28\) equals \(56\). - Match that step to the place of the dividend digit being used.

1. The second partial dividend is \(56\), and the subtraction row is \(56\). 2. Since \(56=28\times2\), the missing quotient digit is \(2\). 3. This step uses the final dividend digit, so \(2\) is in the ones place. The quotient is \(32\).

The missing digit is \(2\), and the quotient is \(32\). The \(2\) belongs in the ones place because this step uses the final digit of the dividend.

- Use the given starting partial quotient of \(30\). - Divide the remainder by \(21\). - Add the partial quotients and check with multiplication.

1. \(21\times30=630\), leaving \(672-630=42\). 2. \(42=21\times2\), so there are \(2\) more groups. 3. The quotient is \(30+2=32\). 4. Check: \(32\times21=672\).

\(30\) groups use \(630\), leaving \(42\). Since \(42=21\times2\), there are \(2\) more groups, so the quotient is \(32\). Check: \(32\times21=672\).

- Find the greatest multiple of \(24\) that does not exceed \(155\). - The number of full groups is the quotient. - Subtract that multiple from \(155\) to find the remainder.

1. \(24\times6=144\), and \(24\times7=168\), which is too large. 2. Six full boxes can be filled. 3. The remainder is \(155-144=11\), so \(11\) books are left over.

\(6\) full boxes, with \(11\) books left over

- Find how many full groups of \(48\) fit in \(250\). - Decide whether a nonzero remainder can be ignored in this situation. - The question asks for enough buses for every student.

1. \(48\times5=240\), so five full buses carry \(240\) students. 2. There are \(250-240=10\) students left. 3. Those \(10\) students need another bus, so \(6\) buses are needed.

\(6\) buses. The remainder of \(10\) means \(10\) students still need seats after five full buses.

- Divide the total number of flyers by each box capacity. - For part c, first find how many groups of \(60\) flyers there are. - Find the height in centimeters, then convert to meters.

1. a) Divide by the number of flyers in each small box: \(72{,}000\div24=3000\). Therefore, \(3000\) small boxes are needed. 2. b) Divide by the number of flyers in each large box: \(72{,}000\div48=1500\). Therefore, \(1500\) large boxes are needed. 3. c) First find the number of groups of \(60\) flyers: \(72{,}000\div60=1200\). The total height is \(1200\times3\,\text{cm}=3600\,\text{cm}\). Since \(100\,\text{cm}=1\,\text{m}\), the stack is \(36\,\text{m}\) high.

a) \(3000\) small boxes b) \(1500\) large boxes c) \(36\,\text{m}\)

- Estimate how many times each two-digit divisor fits into its dividend. - Multiply the divisor by your quotient and subtract to find the remainder. - Match both the quotient and the remainder, not just one of them.

1. \(126\div20=6\text{ R }6\) and \(186\div30=6\text{ R }6\). 2. \(213\div34=6\text{ R }9\) and \(279\div45=6\text{ R }9\). 3. \(257\div41=6\text{ R }11\) and \(371\div60=6\text{ R }11\).

\(126\div20\) pairs with \(186\div30\): \(6\text{ R }6\). \(213\div34\) pairs with \(279\div45\): \(6\text{ R }9\). \(257\div41\) pairs with \(371\div60\): \(6\text{ R }11\).

- First find how much remains after the class fund pays its share. - Divide the remaining amount equally among the students. - The problem requires two operations.

1. Find the remaining cost: \(\$840 - \$240 = \$600\). 2. Divide the remaining cost equally: \(\$600 \div 24 = \$25\).

Each student pays \(\$25\).

- Use related multiplication equations. - Compare how changing the dividend by a factor of \(10\) changes the quotient when the divisor stays fixed. - Check each completed division with multiplication.

1. a) \(630\div70=9\), and making the dividend ten times as large gives \(6300\div70=90\). 2. b) \(3500\div50=70\), while \(350\div50=7\). 3. c) Since \(30\times8=240\), the missing divisor is \(30\). Also, \(900\div30=30\).

a) \(9\), \(90\) b) \(70\), \(7\) c) \(30\), \(30\)

- First find how many yogurt cups are filled in one hour. - Then determine how many groups of \(12\) cups are in that hourly amount.

1. Find the number of yogurt cups filled each hour: \(48{,}000 \div 8 = 6000\) cups per hour. 2. Find the number of cases packed each hour: \(6000 \div 12 = 500\) cases per hour.

The dairy packs \(500\) cases each hour.

- How many bags are filled in one hour? - How many cartons can be made from those bags in one hour? - Scale the hourly number of cartons to the full shift.

1. Find the number of bags packed each hour: \(3600 \div 15 = 240\) bags per hour. 2. Find the number of cartons packed each hour: \(240 \div 20 = 12\) cartons per hour. 3. Find the number of cartons packed in \(7\) hours: \(12 \times 7 = 84\) cartons.

The company packs \(84\) shipping cartons during the shift.

- How many minutes are in one hour? - First find how many chocolate bars are made in one hour. - Divide that total by the number of bars in each case.

1. One hour has \(60\) minutes. 2. Find the number of chocolate bars made in one hour: \(120 \times 60 = 7200\) bars. 3. Find the number of cases: \(7200 \div 24 = 300\) cases.

The factory fills \(300\) cases in one hour.

- How many car lengths fit in the train’s total length? - After finding the number of cars, use the load carried by each car. - Which operation finds the number of equal-size groups?

1. Divide the total length by the length of one car: \(3000\,\text{ft} \div 50\,\text{ft} = 60\). The train has \(60\) freight cars. 2. Multiply the number of cars by the load per car: \(60 \times 25\,\text{tons} = 1500\,\text{tons}\).

The train has \(60\) freight cars and carries \(1500\) tons of gravel.

- Convert the total length to meters first. - Divide by the amount of line used by each truck. - Multiply the number of trucks by \(8\) pallets per truck.

1. \(3\,\text{km}=3000\,\text{m}\). 2. \(3000\div25=120\), so there are about \(120\) trucks. 3. \(120\times8=960\), so the trucks carry about \(960\) pallets.

About \(120\) trucks and \(960\) pallets.

- Divide the total repayment equally among \(12\) months. - Compare the amount borrowed with the amount repaid. - You may break the division into smaller steps.

1. Find the monthly payment: \(\$1656 \div 12 = \$138\). 2. Find the additional amount repaid: \(\$1656 - \$1500 = \$156\).

The monthly payment is \(\$138\), and Mr. Alt repays \(\$156\) more than he borrowed.

- First find how many bags are used in one week. - Multiply the weekly number of bags by \(52\). - A year has \(52\) weeks.

1. Find the number of bags used each week: \(4200\,\text{lb} \div 60\,\text{lb} = 70\). 2. Multiply by the number of weeks: \(52 \times 70 = 3640\). 3. The bakery uses \(3640\) bags in one year.

The bakery uses \(3640\) bags of flour in one year.

- Divide the total number of tickets by the number in one bundle. - For part c), count how many groups of \(50\) tickets are in the full stack. - Use the given centimeter-to-meter conversion at the end.

1. a) \(600{,}000\div50=12{,}000\) bundles. 2. b) \(600{,}000\div20=30{,}000\) bundles. 3. c) \(600{,}000\div50=12{,}000\) groups of \(50\) tickets. 4. The stack is about \(12{,}000\times0.5\,\text{cm}=6000\,\text{cm}=60\,\text{m}\).

a) \(12{,}000\) bundles b) \(30{,}000\) bundles c) About \(60\,\text{m}\)

- First find the total number of cents the first group spends on cookies. - Both groups have the same amount of money. - How many times does the price of one juice pack fit into the total amount? - Removing a factor of ten from both numbers may make the division easier.

1. Find the amount of money collected by the first group: \(12 \times 60 = 720\) cents. 2. The second group also has \(720\) cents. 3. Divide by the price of one juice pack: \(720 \div 80 = 9\).

The second group buys \(9\) juice packs.

- Add the numbers of pizzas to find the total order. - Divide the total bill by the total number of pizzas. - Multiply the price per pizza by each class’s order.

1. The classes ordered \(12 + 15 = 27\) pizzas. 2. One pizza costs \(\$216 \div 27 = \$8\). 3. Class 4A pays \(12 \times \$8 = \$96\). 4. Class 4B pays \(15 \times \$8 = \$120\).

Class 4A should pay \(\$96\), and Class 4B should pay \(\$120\).

- How much money remains after Jonas buys the stickers? - How many times does the price of one bouncy ball fit into the remaining amount? - Check whether the remainder is less than the cost of another bouncy ball.

1. Find the money remaining after the sticker purchase: \(700 - 180 = 520\) cents. 2. Divide by the cost of one bouncy ball: \(520 \div 60 = 8\) remainder \(40\). 3. Jonas can buy \(8\) bouncy balls because \(8 \times 60 = 480\) cents, and the remaining \(40\) cents is not enough for another one.

Jonas can buy \(8\) bouncy balls, and \(40\) cents will remain.

- One method begins by finding how many books there are altogether. - Another method begins by finding the cost of one box. - State what each intermediate quotient represents.

1. Method 1: First find the total number of books: \(4 \times 5 = 20\). Then divide the total cost by the number of books: \(\$160 \div 20 = \$8\). 2. Method 2: First find the cost of one box: \(\$160 \div 4 = \$40\). Then divide by the \(5\) books in the box: \(\$40 \div 5 = \$8\).

One book costs \(\$8\).

- Convert pounds to ounces using the stated conversion. - One method can find the amount per crate and then per container. - Another method can find the total number of containers before dividing.

1. \(18\,\text{lb}=288\,\text{oz}\). 2. Method 1: \(288\div3=96\,\text{oz}\) per crate, then \(96\div12=8\,\text{oz}\) per container. 3. Method 2: There are \(3\times12=36\) containers, and \(288\div36=8\,\text{oz}\) per container.

Method 1: \(18\,\text{lb}=288\,\text{oz}\), then \(288\div3=96\) and \(96\div12=8\,\text{oz}\) per container. Method 2: \(3\times12=36\) containers, and \(288\div36=8\,\text{oz}\) per container.

- First find each person's share of the group fare. - Some costs are given for the whole group, while another cost is already per person. - Add the two per-person costs.

1. Divide the group gondola fare equally: \(\$432\div18=\$24\) per person. 2. Add the guided-walk fee: \(\$24+\$5=\$29\).

\(\$29\) per person

- Remove the shipping charge before finding the unit price. - Then divide the cost of the balls by \(15\). - Perform the steps in the order described by the situation.

1. Subtract the shipping charge to find the cost of all the soccer balls: \(\$342-\$12=\$330\). 2. Divide by the number of soccer balls: \(\$330\div15=\$22\).

\(\$22\)

- Follow each displayed partial dividend to the next quotient digit. - In a), use the displayed \(93\) to choose the final quotient digit. - In b), compare the middle partial dividend with \(37\) before deciding what quotient digit belongs there.

1. In a), the displayed final partial dividend is \(93\). Since \(25\times3=75\), the missing quotient digit is \(3\) and the final subtraction row is \(75\). The remainder is \(18\), so the result is \(313\text{ R }18\). 2. In b), bringing down the next \(0\) gives \(20\), which is smaller than \(37\), so the tens quotient digit is \(0\). Bringing down the last \(0\) gives \(200\); \(37\times5=185\), leaving \(15\). The result is \(405\text{ R }15\). 3. Checks: \(313\times25+18=7843\) and \(405\times37+15=15{,}000\).

a) Missing quotient digit: \(3\); subtraction row: \(75\); \(7843\div25=313\text{ R }18\). b) Missing middle quotient digit: \(0\); \(15{,}000\div37=405\text{ R }15\). The \(0\) preserves the tens place because the partial dividend \(20<37\). Checks: \(313\times25+18=7843\) and \(405\times37+15=15{,}000\).

- Divide the original seats by the seats in each row. - Adding chairs does not change the number of rows. - You may calculate only the added chairs and add them to \(240\).

1. The number of rows is \(240\div 12=20\). 2. Each row then has \(12+3=15\) seats. 3. The new total is \(20\times 15=300\) seats. 4. Another method is to find the added chairs: \(20\times 3=60\), then calculate \(240+60=300\).

a) \(20\) rows b) \(300\) seats. One method is \(20\times 15=300\); another is \(240+20\times 3=300\).

- Focus on the displayed partial dividend \(201\). - Choose the greatest one-digit multiple of \(24\) that does not exceed \(201\). - Use the remainder to form the next partial dividend.

1. The first quotient digit is \(7\) because \(24\times7=168\), leaving \(20\); bringing down \(1\) gives \(201\). 2. The greatest multiple of \(24\) not exceeding \(201\) is \(24\times8=192\). The missing quotient digit is \(8\), and the missing subtraction row is \(192\). 3. \(201-192=9\); bringing down \(6\) gives \(96\), and \(96=24\times4\). The quotient is \(784\).

Since \(24\times8=192\) is the greatest multiple of \(24\) that does not exceed \(201\), the missing quotient digit is \(8\) and the missing subtraction row is \(192\). The quotient is \(784\).

- Find how many full groups of \(18\) fit into \(400\). - Decide what a nonzero remainder means in this context. - Every muffin must be included in a batch.

1. \(400\div18=22\) remainder \(4\). 2. Twenty-two full batches make \(22\times18=396\) muffins, leaving \(4\). 3. The remaining muffins need one more batch, so \(23\) batches are needed.

\(23\) batches

- Interpret both the quotient and remainder. - Compare the remainder with the capacity of one tray. - Find the amount needed to complete the next group of \(24\).

1. Divide: \(1500\div24=62\) remainder \(12\). 2. The quotient means \(62\) trays can be filled completely, with \(12\) seedlings left. 3. One tray holds \(24\) seedlings, so \(24-12=12\) more seedlings are needed to fill another tray.

\(62\) full trays; \(12\) additional seedlings needed

- Divide the wall-paint total by \(12\) to find its unit price. - Add the price difference to find the enamel-paint unit price. - Multiply that unit price by \(15\).

1. One bucket of wall paint costs \(\$216 \div 12 = \$18\). 2. One bucket of enamel paint costs \(\$18 + \$7 = \$25\). 3. Fifteen buckets of enamel paint cost \(15 \times \$25 = \$375\).

The \(15\) buckets of enamel paint cost \(\$375\).

- For a), divide the total by the number of pavers in each row. - For b), divide the total by the desired number of rows. - Check both quotients with the same multiplication equation.

1. Divide the total by the number in each row: \(480 \div 8 = 60\), so there are \(60\) rows. 2. Divide the total by the number of rows: \(480 \div 60 = 8\), so each row has \(8\) pavers. 3. The divisor and quotient switch roles: if \(480 \div 8 = 60\), then \(480 \div 60 = 8\). Both equations come from \(8 \times 60 = 480\).

a) \(60\) rows b) \(8\) pavers per row c) The divisor and quotient switch places; both equations are related to \(8 \times 60 = 480\).

- First subtract the trade-in credit from the bicycle price. - Then subtract the amount already paid. - Divide the remaining balance equally among the \(12\) months.

1. Subtract the trade-in credit: \(\$2450 - \$350 = \$2100\). 2. Subtract the immediate payment: \(\$2100 - \$540 = \$1560\). 3. Divide the remaining balance equally among \(12\) months: \(1560 \div 12 = 130\).

Each monthly payment is \(\$130\).

- Start with the division that has the two-digit divisor. - Compare how both numbers in part b relate to the numbers in part a). - Multiply the quotient by \(1800\) to check part b).

1. \(5220\div18=290\). 2. Dividing both \(522{,}000\) and \(1800\) by \(100\) gives the equivalent division \(5220\div18\), so the second quotient is also \(290\). 3. \(290\times1800=522{,}000\), so the quotient checks.

a) \(290\) b) \(290\) c) \(290\times1800=522{,}000\)

- Find the total cost of the basketballs. - Subtract that amount from the full bill. - Divide the remaining cost by the price of one soccer ball.

1. The basketballs cost \(25 \times \$15 = \$375\). 2. The amount spent on soccer balls is \(\$711 - \$375 = \$336\). 3. The number of soccer balls is \(\$336 \div \$12 = 28\).

The school orders \(28\) soccer balls.

- Divide the total weight by the capacity of one crate. - Compare \(30\) with \(60\). - When each crate holds less, decide how the number of crates changes.

1. For the larger crates: \(480 \div 60 = 8\) crates. 2. A \(30\)-pound crate holds half as much as a \(60\)-pound crate, so twice as many crates are needed. 3. For the smaller crates: \(480 \div 30 = 16\) crates.

a) \(8\) crates b) More crates are needed because each crate holds less. The farm needs \(16\) smaller crates.

- In part a), divide the total cost by the number of people. - In part b), first find the total amount paid by the children. - Divide the remaining cost among the \(60\) adults.

1. a) \(3000\div75=40\), so each person would pay \(\$40\). 2. The children pay \(15\times\$20=\$300\). 3. The adults must cover \(\$3000-\$300=\$2700\). 4. \(2700\div60=45\), so each adult pays \(\$45\).

a) \(\$40\) b) \(\$45\)

- Determine how many times the length of one stone fits into the path. - If the path becomes three times as long while the stones stay the same size, how does the number of stones change? - Compare the lengths of the \(40\)-centimeter and \(20\)-centimeter stones.

1. For part a, divide the path length by the length of one stone: \(240 \div 40 = 6\). 2. A path that is three times as long needs three times as many stones: \(6 \times 3 = 18\). 3. For the smaller stones, divide: \(240 \div 20 = 12\).

a) \(6\) stones b) \(18\) stones c) \(12\) stones

- Divide the total mass by the mass each crate holds. - For part b, compare \(10\,\text{kg}\) with \(20\,\text{kg}\). - For part c, compare \(1200\,\text{kg}\) with \(600\,\text{kg}\).

1. For part a, divide the total mass by the mass in each crate: \(600\,\text{kg} \div 20\,\text{kg} = 30\) crates. 2. A \(10\)-kg crate holds half as much as a \(20\)-kg crate, so twice as many crates are needed: \(30 \times 2 = 60\). 3. The \(1200\,\text{kg}\) harvest is twice the original harvest, so the number of \(20\)-kg crates also doubles: \(30 \times 2 = 60\).

a) \(30\) crates are needed. b) \(60\) crates are needed because each crate holds half as much, so the number of crates doubles. c) \(60\) crates are needed.

- Find the total yarn used for one bracelet. - Use the given conversion to express the total yarn in centimeters. - Find the number of bracelets, then find each color's total.

1. One bracelet uses \(20+30=50\,\text{cm}\) of yarn. 2. \(10\,\text{m}=1000\,\text{cm}\). 3. \(1000\div50=20\) bracelets. 4. Green yarn: \(20\times20\,\text{cm}=400\,\text{cm}=4\,\text{m}\). 5. Yellow yarn: \(20\times30\,\text{cm}=600\,\text{cm}=6\,\text{m}\).

\(4\,\text{m}\) of green yarn and \(6\,\text{m}\) of yellow yarn

- First find the total number of screws in all \(48\) bags. - Then determine how many groups of \(24\) fit into that total. - Check the carton count with multiplication.

1. Find the total number of screws: \(48 \times 15 = 48 \times (10 + 5) = 480 + 240 = 720\). 2. Divide the screws into cartons: \(720 \div 24 = 30\). 3. Check: \(24 \times 30 = 720\).

There are \(720\) screws, which fill \(30\) cartons.

- Start from the displayed middle partial dividend \(82\). - Choose the largest one-digit multiple of \(33\) that does not exceed it. - After subtracting, bring down the next dividend digit.

1. The displayed middle partial dividend is \(82\). 2. The greatest multiple of \(33\) not exceeding \(82\) is \(66=33\times2\). The missing quotient digit is \(2\), and the subtraction row is \(66\). 3. \(82-66=16\); bringing down \(5\) gives \(165\). 4. The quotient is \(225\).

Missing quotient digit: \(2\); subtraction row: \(66\). Since \(82-66=16\), bringing down \(5\) gives \(165\). The quotient is \(225\).

- Compare the displayed remainder directly with the divisor. - Ask whether the remainder still contains another complete group of the divisor. - If it does, determine how the quotient must change.

1. A final remainder must be less than the divisor. 2. The displayed remainder is \(22\), equal to the divisor \(22\), so another full group remains. 3. Add one more group to the quotient: \(56+1=57\), leaving remainder \(0\). 4. The correct result is \(57\).

The displayed remainder is not valid because it equals the divisor. The correct quotient is \(57\) with remainder \(0\).

- Use the displayed partial dividend \(273\). - Select the greatest one-digit multiple of \(48\) that fits in it. - Check that the remainder followed by the final digit produces \(336\).

1. The greatest multiple of \(48\) not exceeding \(273\) is \(48\times5=240\). The missing quotient digit is \(5\), and the subtraction row is \(240\). 2. \(273-240=33\). Bringing down the final \(6\) gives \(336\). 3. Since \(336=48\times7\), the quotient is \(357\).

The missing quotient digit is \(5\), the subtraction row is \(240\), and the quotient is \(357\). Since \(273-240=33\), bringing down the final \(6\) gives the next partial dividend \(336\).

- Divide each remnant length by \(20\) separately. - Record the full strips and leftover length from each remnant. - Decide whether two separate leftovers can form one continuous strip.

1. From the \(50\,\text{in.}\) remnant, she can cut \(50 \div 20 = 2\) strips with \(10\,\text{in.}\) left. 2. From the \(75\,\text{in.}\) remnant, she can cut \(75 \div 20 = 3\) strips with \(15\,\text{in.}\) left. 3. She can cut \(2 + 3 = 5\) full strips. Adding the two lengths first incorrectly treats the separate leftover pieces as one continuous piece of fabric.

She can cut \(5\) full strips. The two separate leftover pieces cannot be combined into one unbroken \(20\,\text{in.}\) strip.