Free math worksheets

- 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 subtrahend is \(84\). 3. The remainder is \(0\).

Quotient digit: \(7\); subtrahend: \(84\), because \(12\times7=84\).

- Use the package size to complete the rows with known package counts. - For the last row, the total number of blocks is known and the number of packages is unknown. - Check your missing 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. To find the missing package count, divide \(1000\) by the two-digit divisor \(20\): \(1000 \div 20=50\).

<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 liters to milliliters. - Divide the total volume equally among \(12\) bottles. - Check by multiplying the volume of one bottle by \(12\).

1. Convert the total volume: \(6\,\text{L} = 6000\,\text{mL}\). 2. Divide equally among the \(12\) bottles: \(6000\,\text{mL} \div 12 = 500\,\text{mL}\).

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

- Find the missing factor in a multiplication equation. - Use a related basic fact and place value. - Check by multiplying the quotient and divisor.

1. a) \(6 \times 40 = 240\), so \(240 \div 40 = 6\). 2. b) \(70 \times 6 = 420\), so \(420 \div 6 = 70\). 3. c) \(8 \times 80 = 640\), so \(640 \div 80 = 8\). 4. d) \(7 \times 50 = 350\), so \(350 \div 50 = 7\).

a) \(6 \times 40 = 240\); quotient \(6\) b) \(70 \times 6 = 420\); quotient \(70\) c) \(8 \times 80 = 640\); quotient \(8\) d) \(7 \times 50 = 350\); quotient \(7\)

- Evaluate each quotient separately. - Use related multiplication facts. - Compare the four results.

1. A: \(270 \div 30 = 9\). 2. B: \(450 \div 9 = 50\). 3. C: \(360 \div 4 = 90\). 4. D: \(180 \div 20 = 9\). 5. Since \(90\) is greatest, expression C has the greatest quotient.

C has the greatest quotient, \(90\). The other quotients are A: \(9\), B: \(50\), and D: \(9\).

- Use a multiplication equation to find each quotient. - Compare division by a one-digit number with division by a related multiple of \(10\). - Check each quotient by multiplication.

1. \(150 \div 30 = 5\), because \(5 \times 30 = 150\). 2. \(150 \div 3 = 50\), because \(50 \times 3 = 150\). 3. \(320 \div 80 = 4\), because \(4 \times 80 = 320\). 4. \(320 \div 8 = 40\), because \(40 \times 8 = 320\). 5. \(450 \div 90 = 5\), because \(5 \times 90 = 450\). 6. \(450 \div 9 = 50\), because \(50 \times 9 = 450\).

a) \(5\) b) \(50\) c) \(4\) d) \(40\) e) \(5\) f) \(50\)

- Think about how many times each known factor fits into \(180\). - Use a related fact with \(18\) for the equations involving \(30\) or \(60\). - For those equations, divide both \(180\) and the known factor by \(10\) to make an equivalent simpler quotient. - Check each answer by multiplying the two factors.

1. In a), \(180 \div 2 = 90\). 2. In b), \(180 \div 30 = 6\). 3. In c), \(180 \div 60 = 3\).

a) \(90\) b) \(6\) c) \(3\)

- Which number multiplied by \(4\) gives \(32\)? Use that fact to reason about \(320\). - For division by \(80\) or \(40\), divide both numbers by \(10\) to use a related basic fact. - Break each problem into smaller steps. - Check with the inverse operation.

1. In a), \(320 \div 4 = 80\). 2. In b), \(320 \div 80 = 4\). 3. In c), \(320 \div 40 = 8\).

a) \(80\) b) \(4\) c) \(8\)

- Look for a related multiplication fact with product \(54\). - Decide whether the missing factor should be a one-digit number or a multiple of \(10\). - Think about how many tens are in \(540\). - Use the inverse operation and check by multiplication.

1. In a), \(540 \div 9 = 60\). 2. In b), \(540 \div 60 = 9\). 3. In c), \(540 \div 90 = 6\).

a) \(60\) b) \(9\) c) \(6\)

- Use a related basic multiplication or division fact. - Compare \(48\) with \(480\) and \(8\) with \(80\). - Check quotients with multiplication.

1. a) \(48 \div 6 = 8\) and \(48 \div 8 = 6\). Since \(480\) is ten times \(48\), \(480 \div 6 = 80\). Also, \(480 \div 80 = 6\) because \(80 \times 6 = 480\). 2. b) \(72 \div 8 = 9\) and \(72 \div 9 = 8\). Therefore \(720 \div 80 = 9\) and \(720 \div 9 = 80\).

a) \(8\), \(6\), \(80\), \(6\) b) \(9\), \(8\), \(9\), \(80\)

- Use related multiplication facts. - Compare each one-digit divisor with the related two-digit divisor. - Explain how a tenfold change in the divisor affects the quotient.

1. a) \(240 \div 3 = 80\) and \(240 \div 30 = 8\). 2. b) \(350 \div 5 = 70\) and \(350 \div 50 = 7\). 3. c) \(480 \div 6 = 80\) and \(480 \div 60 = 8\). 4. d) \(720 \div 8 = 90\) and \(720 \div 80 = 9\). 5. In each pair, making the divisor ten times as great makes the quotient one-tenth as great.

a) \(80\), \(8\) b) \(70\), \(7\) c) \(80\), \(8\) d) \(90\), \(9\)

- Pair divisors that differ by a factor of \(10\). - Use related multiplication facts. - Compare the quotients within each pair.

1. \(600 \div 6 = 100\) and \(600 \div 60 = 10\). 2. \(600 \div 3 = 200\) and \(600 \div 30 = 20\). 3. \(600 \div 2 = 300\) and \(600 \div 20 = 30\). 4. \(600 \div 100 = 6\) and \(600 \div 10 = 60\). 5. In each related pair, making the divisor ten times as great makes the quotient one-tenth as great.

\(600 \div 6 = 100\) \(600 \div 60 = 10\) \(600 \div 3 = 200\) \(600 \div 30 = 20\) \(600 \div 2 = 300\) \(600 \div 20 = 30\) \(600 \div 100 = 6\) \(600 \div 10 = 60\)

- Evaluate both quotients in each comparison. - For the same dividend, a larger divisor gives a smaller quotient. - Check each quotient with multiplication.

1. a) \(400 \div 4 = 100\) and \(400 \div 40 = 10\), so \(100 > 10\). 2. b) \(800 \div 20 = 40\) and \(800 \div 40 = 20\), so \(40 > 20\). 3. c) \(200 \div 5 = 40\) and \(200 \div 50 = 4\), so \(40 > 4\). 4. d) \(900 \div 3 = 300\) and \(900 \div 30 = 30\), so \(300 > 30\).

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

- 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 as the inverse of division. - Identify whether the missing value is a divisor, dividend, or quotient. - Use place-value relationships to connect each equation to a basic fact.

1. a) \(150 \div 3 = 50\). 2. b) Since \(3 \times 50 = 150\), the divisor is \(50\). 3. c) \(40 \times 5 = 200\), so the dividend is \(200\). 4. d) \(320 \div 4 = 80\). 5. e) Since \(4 \times 80 = 320\), the divisor is \(80\).

a) \(50\) b) \(50\) c) \(200\) d) \(80\) e) \(80\)

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

1. a) Since \(8 \times 10 = 80\), the divisor is \(10\). 2. b) \(120 \div 20 = 6\). 3. c) \(4 \times 30 = 120\), so the dividend is \(120\). 4. d) Since \(4 \times 60 = 240\), the divisor is \(60\). 5. e) \(300 \div 50 = 6\).

a) \(10\) b) \(6\) c) \(120\) d) \(60\) e) \(6\)

- 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

- Rewrite each division as an unknown-factor multiplication equation. - Use related basic facts and place value. - Check by multiplication.

1. a) \(8 \times 40 = 320\), so the quotient is \(8\). 2. b) \(6 \times 80 = 480\), so the quotient is \(6\). 3. c) \(9 \times 70 = 630\), so the quotient is \(9\). 4. d) \(7 \times 30 = 210\), so the quotient is \(7\). 5. e) \(8 \times 50 = 400\), so the quotient is \(8\). 6. f) \(6 \times 90 = 540\), so the quotient is \(6\).

a) \(8\) b) \(6\) c) \(9\) d) \(7\) e) \(8\) f) \(6\)

- Rewrite each division equation as multiplication. - When the dividend is missing, multiply the quotient by the divisor. - When the divisor is missing, divide the dividend by the quotient. - If both dividend and divisor end in \(0\), divide both by \(10\) to check an equivalent simpler quotient.

1. a) \(5 \times 60 = 300\), so the dividend is \(300\). 2. b) Since \(8 \times 90 = 720\), the divisor is \(90\). 3. c) \(350 \div 70 = 5\). 4. d) Since \(4 \times 60 = 240\), the divisor is \(60\). 5. e) \(9 \times 30 = 270\), so the dividend is \(270\).

a) \(300\) b) \(90\) c) \(5\) d) \(60\) e) \(270\)

- Use place value to divide by \(10\). - For division by \(30\), use related multiplication facts or divide by \(10\) and then by \(3\). - Check with multiplication.

1. Dividing by \(10\) gives \(15\), \(30\), \(45\), \(60\), and \(75\). 2. Dividing by \(30\) gives \(5\), \(10\), \(15\), \(20\), and \(25\).

Divided by \(10\): \(15\), \(30\), \(45\), \(60\), \(75\) Divided by \(30\): \(5\), \(10\), \(15\), \(20\), \(25\)

- Find each quotient after division by \(10\) first. - Compare the two quotients in each pair. - Think about how many times \(20\) fits compared with \(10\). - Formulate a rule using halving.

1. a) \(340 \div 10 = 34\). Since \(20\) is twice \(10\), \(340 \div 20 = 34 \div 2 = 17\). 2. b) \(520 \div 10 = 52\), so \(520 \div 20 = 52 \div 2 = 26\). 3. c) \(780 \div 10 = 78\), so \(780 \div 20 = 78 \div 2 = 39\). 4. Dividing by \(20\) gives half the quotient obtained by dividing the same number by \(10\).

a) \(34\) and \(17\) b) \(52\) and \(26\) c) \(78\) and \(39\) Halve the quotient from division by \(10\).

- Think about how many equal groups of the second number make the first number. - You may count backward by the amount being subtracted. - Divide both numbers by a common factor of \(10\) to find a related basic fact. - Use the inverse multiplication equation to check your result.

1. Repeatedly subtracting \(90\) from \(720\) is equivalent to finding \(720 \div 90\). Since \(90 \times 8 = 720\), the number of subtractions is \(8\). 2. Repeatedly subtracting \(60\) from \(240\) is equivalent to finding \(240 \div 60\). Since \(60 \times 4 = 240\), the number of subtractions is \(4\).

a) \(8\) times b) \(4\) times

- Ask which whole number multiplied by the divisor gives the dividend. - Test likely multiplication facts, such as a product with \(5\), if helpful. - Use multiplication to check each quotient. - Break a two-digit factor into tens and ones if needed.

1. \(84 \div 12 = 7\). The related multiplication equation is \(7 \times 12 = 84\). 2. \(75 \div 15 = 5\). The related multiplication equation is \(5 \times 15 = 75\). 3. \(98 \div 14 = 7\). The related multiplication equation is \(7 \times 14 = 98\).

a) \(7\); \(7 \times 12 = 84\) b) \(5\); \(5 \times 15 = 75\) c) \(7\); \(7 \times 14 = 98\)

- Find each quotient. - Write the quotient next to each letter. - Compare the quotients and match equal results. - Check by multiplying the quotient by the divisor.

1. The quotients are A: \(90 \div 18 = 5\), B: \(72 \div 12 = 6\), C: \(65 \div 13 = 5\), D: \(96 \div 16 = 6\), E: \(52 \div 13 = 4\), and F: \(64 \div 16 = 4\). 2. Match expressions with equal quotients: A with C, B with D, and E with F.

A and C both have quotient \(5\). B and D both have quotient \(6\). E and F both have quotient \(4\).

- Focus on the displayed final partial dividend rather than restarting the whole division. - Find the one-digit multiple of the divisor that matches that partial dividend. - That multiplier determines both the missing quotient digit and the subtraction row.

1. The first quotient digit is \(3\), since \(24\times3=72\). After subtracting and bringing down the final digit, the displayed partial dividend is \(144\). 2. Since \(144=24\times6\), the missing quotient digit is \(6\) and the missing subtrahend is \(144\). 3. The quotient is \(36\).

The displayed partial dividend is \(144\), and \(24\times6=144\). Therefore the missing quotient digit is \(6\), the missing subtrahend is \(144\), and the quotient is \(36\).

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

1. The first subtraction is \(97-93=4\); bringing down \(3\) gives the displayed partial dividend \(43\). 2. The greatest multiple of \(31\) not exceeding \(43\) is \(31\times1=31\). The missing quotient digit is \(1\) and the missing subtrahend 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 \(1\) and the missing subtrahend is \(31\), because \(31\times1\le43<31\times2\). The remainder is \(43-31=12\), and it is valid because \(12<31\). Thus the result is \(31\text{ R }12\).

- 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 first quotient digit is \(4\), because \(28\times4=112\). After subtracting from \(134\) and bringing down \(4\), the displayed partial dividend is \(224\). 2. Since \(224=28\times8\), the missing quotient digit is \(8\) and the missing subtrahend is \(224\). 3. Each class receives \(48\) tickets.

The displayed final partial dividend is \(224\), and \(28\times8=224\). Therefore the missing quotient digit is \(8\), the missing subtrahend is \(224\), and each class receives \(48\) tickets.

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

1. The first quotient digit is \(3\), since \(28\times3=84\). After subtracting and bringing down \(6\), the second partial dividend is \(56\). 2. The displayed subtrahend is also \(56\), and \(56=28\times2\). Therefore the missing quotient digit is \(2\). 3. This step uses the final dividend digit, so \(2\) belongs in the ones place. The quotient is \(32\).

The missing digit is \(2\), because the second displayed step shows \(56=28\times2\). That step uses the final dividend digit, so \(2\) is the ones digit. The quotient is \(32\).

- Find how many whole times the divisor fits into the dividend. - Subtract the corresponding product to find the remainder. - For larger divisors, estimate the quotient first, but calculate the remainder from the original numbers.

1. \(26\div 4=6\text{ R }2\) and \(44\div 7=6\text{ R }2\). 2. \(310\div 50=6\text{ R }10\) and \(490\div 80=6\text{ R }10\). 3. \(19\div 3=6\text{ R }1\) and \(55\div 9=6\text{ R }1\). 4. The expressions in each pair have the same quotient and remainder.

\(26\div 4\) pairs with \(44\div 7\): \(6\text{ R }2\). \(310\div 50\) pairs with \(490\div 80\): \(6\text{ R }10\). \(19\div 3\) pairs with \(55\div 9\): \(6\text{ R }1\).

- 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 \(\$25\).

- Rewrite each division equation as multiplication. - Decide whether the missing value is the dividend, divisor, or quotient. - Check with multiplication.

1. a) \(6 \times 70 = 420\). 2. b) \(9 \times 60 = 540\), so the divisor is \(60\). 3. c) \(4 \times 80 = 320\), so the quotient is \(4\). 4. d) \(40 \times 5 = 200\), so the divisor is \(5\).

a) \(420\) b) \(60\) c) \(4\) d) \(5\)

- Rewrite each division equation as multiplication. - Identify whether the missing value is the divisor or dividend. - Check by substituting the value.

1. a) \(280 \div 7 = 40\), so the divisor is \(40\). 2. b) \(280 \div 70 = 4\), so the divisor is \(4\). 3. c) \(4 \times 60 = 240\), so the dividend is \(240\). 4. d) \(40 \times 6 = 240\), so the dividend is \(240\).

a) \(40\) b) \(4\) c) \(240\) d) \(240\)

- Evaluate both sides of each comparison. - Use related multiplication facts. - Compare the quotients.

1. a) \(240 \div 30 = 8\) and \(240 \div 3 = 80\), so \(8 < 80\). 2. b) \(400 \div 50 = 8\) and \(400 \div 80 = 5\), so \(8 > 5\). 3. c) \(630 \div 70 = 9\) and \(810 \div 90 = 9\), so the quotients are equal. 4. d) \(560 \div 8 = 70\) and \(560 \div 80 = 7\), so \(70 > 7\).

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

- Determine how the dividend changes each time. - Compare the divisors \(80\) and \(40\). - Calculate the given quotients before predicting the relationship. - Think about what happens to a quotient when the divisor is smaller.

1. The dividends decrease by \(80\), so a) continues with \(560 \div 80 = 7\) and \(480 \div 80 = 6\). 2. Sequence b) uses the same dividends and continues with \(560 \div 40 = 14\) and \(480 \div 40 = 12\). 3. Each quotient in b) is twice the matching quotient in a) because dividing by \(40\) uses a divisor half as large as \(80\).

a) \(560 \div 80 = 7\), \(480 \div 80 = 6\) b) \(560 \div 40 = 14\), \(480 \div 40 = 12\) Each quotient in b) is twice the matching quotient in a).

- Imagine sharing \(600\) objects among more equal groups. - Compare \(20\) with \(10\). - Compare \(30\) with \(60\).

1. The dividend stays \(600\). As the divisor increases, the quotient decreases. 2. The divisor \(20\) is twice \(10\). 3. The corresponding quotient, \(30\), is half of \(60\). Thus, for a fixed dividend, doubling the divisor halves the quotient.

As the divisor increases, the quotient decreases. When the divisor doubles from \(10\) to \(20\), the quotient is halved from \(60\) to \(30\).

- Use related multiplication equations. - Compare how multiplying the dividend or divisor by \(10\) changes the quotient. - Check each completed division.

1. a) \(63 \div 7 = 9\). Since \(630\) is ten times \(63\), \(630 \div 7 = 90\). 2. b) \(350 \div 5 = 70\). Dividing the same dividend by a divisor ten times as great gives \(350 \div 50 = 7\). 3. c) Since \(30 \times 8 = 240\), \(240 \div 30 = 8\) and \(240 \div 8 = 30\).

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

- Rewrite each division equation as multiplication. - Find the factor that pairs with the quotient to make \(600\). - Check each divisor in the original equation.

1. Rewrite each equation as multiplication: the quotient times the divisor equals \(600\). 2. a) \(60 \times 10 = 600\), so the divisor is \(10\). 3. b) \(6 \times 100 = 600\), so the divisor is \(100\). 4. c) \(10 \times 60 = 600\), so the divisor is \(60\). 5. d) \(100 \times 6 = 600\), so the divisor is \(6\). 6. e) \(20 \times 30 = 600\), so the divisor is \(30\).

a) \(10\) b) \(100\) c) \(60\) d) \(6\) e) \(30\)

- Evaluate the quotient on each side of every comparison. - When both the dividend and divisor end in \(0\), divide both by \(10\) to make an equivalent simpler quotient. - Write down the intermediate quotients before comparing them. - Without calculating, predict how a larger divisor affects the quotient when the dividend stays the same.

1. a) \(560 \div 7 = 80\) and \(560 \div 70 = 8\), so \(80 > 8\). 2. b) \(810 \div 90 = 9\) and \(810 \div 9 = 90\), so \(9 < 90\). 3. c) \(400 \div 5 = 80\) and \(800 \div 10 = 80\), so the quotients are equal. 4. d) \(270 \div 30 = 9\) and \(270 \div 90 = 3\), so \(9 > 3\).

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

- 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.

- 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 space used by each truck. - Once you know the number of trucks, multiply by the pallets on each truck.

1. Convert the total length to meters: \(3\,\text{km} = 3000\,\text{m}\). 2. Estimate the number of trucks: \(3000 \div 25 \approx 120\). 3. Estimate the total number of pallets: \(120 \times 8 \approx 960\).

There are about \(120\) trucks carrying about \(960\) pallets altogether.

- Rewrite each division equation as multiplication. - Find the factor that pairs with the quotient to make \(500\). - Check each divisor in the original equation.

1. Rewrite each equation as quotient times divisor equals \(500\). 2. a) \(50 \times 10 = 500\), so the divisor is \(10\). 3. b) \(5 \times 100 = 500\), so the divisor is \(100\). 4. c) \(100 \times 5 = 500\), so the divisor is \(5\). 5. d) \(10 \times 50 = 500\), so the divisor is \(50\). 6. e) \(250 \times 2 = 500\), so the divisor is \(2\). 7. f) \(25 \times 20 = 500\), so the divisor is \(20\).

a) \(10\) b) \(100\) c) \(5\) d) \(50\) e) \(2\) f) \(20\)

- 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.

- Try dividing both the dividend and divisor by the same number. - Try multiplying both numbers by the same number. - Verify that each new quotient is \(30\).

1. Multiplying or dividing both the dividend and divisor by the same nonzero number keeps the quotient unchanged. 2. Dividing both numbers by \(2\) gives \(180\div 6=30\). 3. Multiplying both numbers by \(2\) gives \(720\div 24=30\). 4. Dividing both numbers by \(6\) gives \(60\div 2=30\).

One possible answer is: \(180\div 6=30\) \(720\div 24=30\) \(60\div 2=30\)

- 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 prize by the value of one bill. - For part c, first find the number of groups of \(100\) bills. - Convert centimeters to meters at the end.

1. a) Divide the prize by the value of each bill: \(6{,}000{,}000\div100=60{,}000\). 2. b) Divide by \(20\): \(6{,}000{,}000\div20=300{,}000\). 3. c) The number of 100-bill groups is \(300{,}000\div100=3000\). Using the approximate thickness, the stack is about \(3000\,\text{cm}\) tall. 4. Convert the estimated height to meters: \(3000\,\text{cm}=30\,\text{m}\). The stack is about \(30\,\text{m}\) tall.

a) \(60{,}000\) bills b) \(300{,}000\) bills c) About \(30\,\text{m}\)

- 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.

- Evaluate both quotients in each comparison. - Use related multiplication equations. - Compare the results.

1. a) \(240 \div 30 = 8\) and \(280 \div 40 = 7\), so \(8 > 7\). 2. b) \(180 \div 20 = 9\) and \(270 \div 30 = 9\), so the quotients are equal. 3. c) \(450 \div 50 = 9\) and \(540 \div 60 = 9\), so the quotients are equal. 4. d) \(1000 \div 10 = 100\) and \(100 \div 1 = 100\), so the quotients are equal.

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

- Rewrite each division as an unknown-factor multiplication equation. - Repeated addition can help you test a factor. - Check by multiplying the quotient and divisor.

1. a) \(12 \times 5 = 60\), so \(60 \div 12 = 5\). 2. b) \(16 \times 3 = 48\), so \(48 \div 16 = 3\). 3. c) \(25 \times 3 = 75\), so \(75 \div 25 = 3\). 4. d) \(15 \times 6 = 90\), so \(90 \div 15 = 6\).

a) \(5\) b) \(3\) c) \(3\) d) \(6\)

- 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 first. - One method can find the weight per crate and then per container. - Another method can find the total number of containers before dividing.

1. Convert the total weight: \(18\,\text{lb} = 288\,\text{oz}\). 2. Method 1: Find the weight per crate: \(288\,\text{oz} \div 3 = 96\,\text{oz}\). Then find the weight per container: \(96\,\text{oz} \div 12 = 8\,\text{oz}\). 3. Method 2: Find the total number of containers: \(3 \times 12 = 36\). Then divide: \(288\,\text{oz} \div 36 = 8\,\text{oz}\).

Each container holds \(8\,\text{oz}\) of strawberries.

- Evaluate both quotients in each comparison. - Use related multiplication equations. - Compare the results.

1. a) \(150 \div 30 = 5\) and \(200 \div 40 = 5\), so the quotients are equal. 2. b) \(180 \div 20 = 9\) and \(240 \div 30 = 8\), so \(9 > 8\). 3. c) \(400 \div 80 = 5\) and \(450 \div 50 = 9\), so \(5 < 9\). 4. d) \(90 \div 10 = 9\) and \(100 \div 20 = 5\), so \(9 > 5\).

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

- Rewrite each division as a related multiplication fact. - Determine how many times the divisor fits into the dividend. - Decompose a dividend into tens and ones when that makes the division easier. - Compare all five quotients.

1. \(48 \div 12 = 4\), because \(12 \times 4 = 48\). 2. \(60 \div 15 = 4\), because \(15 \times 4 = 60\). 3. \(72 \div 18 = 4\), because \(18 \times 4 = 72\). 4. \(80 \div 16 = 5\), because \(16 \times 5 = 80\). 5. \(96 \div 24 = 4\), because \(24 \times 4 = 96\). 6. Only d) has quotient \(5\); all the others have quotient \(4\).

d) \(80 \div 16\) is the outlier. The quotients are \(4\), \(4\), \(4\), \(5\), and \(4\).

- 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) \(4\)

- 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), the final quotient digit is determined by the displayed partial dividend \(93\). - In b), compare the middle partial dividend with the divisor before deciding whether a quotient digit can be omitted.

1. In a), after \(78-75=3\), bring down \(4\) to get \(34\). Subtract \(25\), leaving \(9\), and bring down \(3\) to get \(93\). Since \(25\times3=75\), the missing quotient digit is \(3\) and the final subtrahend is \(75\). The result is \(313\text{ R }18\). 2. In b), \(150-148=2\). Bringing down the next \(0\) gives \(20\), which is smaller than \(37\), so the tens quotient digit is \(0\) and the subtrahend is \(0\). Bringing down the last \(0\) gives \(200\), and \(37\times5=185\), 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: \(3\); missing subtrahend: \(75\); \(7843\div25=313\text{ R }18\), because \(25\times3=75\) is the greatest multiple of \(25\) not exceeding the displayed \(93\). b) Missing middle quotient digit: \(0\); \(15{,}000\div37=405\text{ R }15\). The \(0\) must be written because the displayed partial dividend \(20<37\), preserving the tens place. 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 that partial dividend. - The chosen multiplier is both the missing quotient digit and the factor that determines the missing subtrahend.

1. The first quotient digit is \(7\) because \(24\times7=168\), leaving \(20\); bringing down \(1\) gives the displayed partial dividend \(201\). 2. The greatest multiple of \(24\) not exceeding \(201\) is \(24\times8=192\). Therefore the missing middle quotient digit is \(8\) and the missing subtrahend is \(192\). 3. Subtracting leaves \(9\); bring down \(6\) to get \(96\). Since \(96=24\times4\), the quotient is \(784\).

The missing quotient digit is \(8\) and the missing subtrahend is \(192\), because \(24\times8=192\le201\) while \(24\times9=216>201\). Subtracting leaves \(9\), so bringing down \(6\) gives \(96\). The quotient is \(784\).

- Find how many full pans can be filled. - Decide whether a nonzero remainder requires another pan. - The question asks for enough pans for every muffin.

1. Divide: \(400\div18=22\) remainder \(4\). 2. Twenty-two full pans hold \(22\times18=396\) muffins, leaving \(4\) muffins. 3. Those remaining muffins require one more pan, so \(22+1=23\) pans are needed.

\(23\) pans

- 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 \(\$130\).

- Simplify the division by removing the same factor of \(100\) from both numbers. - Multiply the quotient by the original divisor to check the result.

1. Divide both the dividend and divisor by \(100\): \(522{,}000 \div 1800 = 5220 \div 18\). 2. Divide: \(5220 \div 18 = 290\). 3. Check: \(290 \times 1800 = 522{,}000\).

\(290\)

- 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.

- Rewrite each division as an unknown-factor multiplication equation. - Use repeated addition or test small factors. - Use the final digit to rule out factors that cannot work. - Check by multiplying the quotient and divisor.

1. a) \(15 \times 5 = 75\), so the quotient is \(5\). 2. b) \(12 \times 8 = 96\), so the quotient is \(8\). 3. c) \(13 \times 5 = 65\), so the quotient is \(5\). 4. d) \(14 \times 6 = 84\), so the quotient is \(6\). 5. e) \(18 \times 5 = 90\), so the quotient is \(5\).

a) \(5\) b) \(8\) c) \(5\) d) \(6\) e) \(5\)

- Calculate all six quotients and organize them by dividend. - Compare the divisors \(10\), \(20\), and \(40\). - Compare each quotient with the next one. - Think about sharing the same amount among twice as many groups.

1. For \(240\): \(240 \div 10 = 24\), \(240 \div 20 = 12\), and \(240 \div 40 = 6\). 2. For \(480\): \(480 \div 10 = 48\), \(480 \div 20 = 24\), and \(480 \div 40 = 12\). 3. The divisors double from \(10\) to \(20\) to \(40\), while each quotient is halved.

a) For \(240\): \(24\), \(12\), \(6\). For \(480\): \(48\), \(24\), \(12\). b) When the divisor doubles, the quotient is halved.

- In part a, divide the total cost equally among all participants. - In part b, first find the total amount paid by the children. - How much of the trip cost is still unpaid? - Divide that remaining amount among the adults.

1. For part a, divide the total cost by the number of people: \(\$3000\div100=\$30\). 2. For part b, the children pay \(20\times\$10=\$200\) altogether. 3. The adults must cover \(\$3000-\$200=\$2800\). 4. Each adult pays \(\$2800\div80=\$35\).

a) Each person would pay \(\$30\). b) Each adult must pay \(\$35\).

- 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. - Convert the full amount of yarn to centimeters. - Find the number of bracelets, then use that number to find each color's total.

1. Find the yarn used for one bracelet: \(20\,\text{cm} + 30\,\text{cm} = 50\,\text{cm}\). 2. Convert the total: \(10\,\text{m} = 1000\,\text{cm}\). 3. Find the number of bracelets: \(1000\,\text{cm} \div 50\,\text{cm} = 20\). 4. Find the green yarn: \(20 \times 20\,\text{cm} = 400\,\text{cm} = 4\,\text{m}\). 5. Find the yellow yarn: \(20 \times 30\,\text{cm} = 600\,\text{cm} = 6\,\text{m}\).

The class uses \(4\,\text{m}\) of green yarn and \(6\,\text{m}\) of yellow yarn.

- First find how many groups of \(80\) are in \(480\). - Then find how many groups of \(40\) are in \(480\). - Compare the two numbers of steps.

1. For part a), \(480 \div 80 = 6\), so \(80\) must be subtracted \(6\) times. 2. For part b), \(480 \div 40 = 12\). 3. Since \(12 = 2 \times 6\), subtracting \(40\) takes exactly twice as many steps. The statement is true.

a) \(6\) times b) The statement is true because \(480 \div 40 = 12\), and \(12\) is twice \(6\).

- 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, check how bringing down the next dividend digit creates the following displayed partial dividend.

1. The first subtraction is \(74-66=8\); bringing down \(2\) gives the displayed partial dividend \(82\). 2. The greatest multiple of \(33\) not exceeding \(82\) is \(66=33\times2\). Therefore the missing quotient digit is \(2\) and the missing subtrahend is \(66\). 3. Subtracting gives \(16\); bringing down the final \(5\) gives \(165\), matching the next displayed partial dividend. 4. The quotient is \(225\).

The missing quotient digit is \(2\) and the missing subtrahend is \(66\), because \(33\times2=66\le82<99\). Then \(82-66=16\), and bringing down \(5\) produces the displayed next partial dividend \(165\). 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 \(0\).

- 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) \(36\,\text{m}\)

- 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.

- Use the displayed middle partial dividend rather than recomputing from the start. - Select the greatest one-digit multiple of \(48\) that fits in that partial dividend. - Check that the remainder, followed by the brought-down final digit, produces the next displayed partial dividend.

1. The first quotient digit is \(3\), since \(48\times3=144\). After subtracting and bringing down \(3\), the displayed partial dividend is \(273\). 2. The greatest multiple of \(48\) not exceeding \(273\) is \(48\times5=240\). The missing quotient digit is \(5\) and the missing subtrahend is \(240\). 3. Subtract: \(273-240=33\). Bring down the final \(6\) to get \(336\), matching the next displayed partial dividend. 4. Since \(336=48\times7\), the quotient is \(357\).

The missing quotient digit is \(5\) and the missing subtrahend is \(240\), because \(48\times5=240\le273<288\). Then \(273-240=33\), and bringing down \(6\) gives the displayed \(336\). The quotient is \(357\).