Free math worksheets

- Think of division as finding a missing factor. - What number multiplied by \(12\) equals \(96\)? - Check your quotient by multiplying it by \(12\).

1. Divide the total number of rolls by the number in each bag: \(96 \div 12\). 2. Use multiplication to find the quotient. Since \(12 \times 8 = 96\), it follows that \(96 \div 12 = 8\).

The baker can fill \(8\) bags.

- Area equals the product of the two side lengths. - One side and the area are known. - Use the inverse relationship between multiplication and division.

1. The known side is \(12\) units and the area is \(96\) square units. 2. The missing side is \(96 \div 12 = 8\) units. 3. Since \(12 \times 8 = 96\), the rectangle shows the factor relationship behind the division.

The missing side is \(8\) units, because the diagram shows the other side as \(12\) and \(96\div12=8\). Equivalently, \(12\times8=96\), so the rectangle models the division.

- Use the stated estimate of \(20\) feet for each car-and-gap unit. - Divide the total traffic-jam length by that amount. - Choose the matching answer.

1. The stated estimating model uses about \(20\) feet of jam length per car-and-gap unit. 2. Divide the total length by that estimate: \(1000 \div 20 = 50\).

b) \(50\)

- 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 \(7\)

- 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) \(>\)

- For part a, what number multiplied by \(15\) equals \(150\)? - For part b, break \(150\) into numbers that are easy to divide by \(25\). - Check each quotient by multiplication.

1. For part a, divide the total number of tulips by the number in each bouquet: \(150 \div 15\). Since \(15 \times 10 = 150\), the quotient is \(10\). 2. For part b, calculate \(150 \div 25\). 3. Break \(150\) into compatible numbers: \(150 = 100 + 50\). 4. Divide each part: \(100 \div 25 = 4\) and \(50 \div 25 = 2\). Add the partial quotients: \(4 + 2 = 6\).

a) The florist can make \(10\) bouquets. b) The florist can make \(6\) bouquets.

- Decide what each divisor represents in the two parts. - For part a, share the students equally among \(10\) groups. - For part b, count how many groups of \(20\) are in \(240\).

1. For part a, divide the number of students by the number of groups: \(240 \div 10 = 24\). 2. For part b, divide the number of students by the size of each group: \(240 \div 20 = 12\).

a) Each group has \(24\) students. b) The school forms \(12\) groups.

- First find the exact number of crates needed. - Divide \(450\) by \(50\). - Compare your quotient with \(10\).

1. Divide the total number of apples by the number in each crate: \(450 \div 50 = 9\). 2. Compare the result with \(10\): \(9 < 10\). 3. The seller is correct because exactly \(9\) crates are needed.

Yes. The seller needs exactly \(9\) crates, and \(9 < 10\).

- 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) \(6\)

- Count by \(25\) until you reach \(175\). - Write a multiplication or division equation for the equal groups. - Check that the number of boxes times \(25\) equals \(175\).

1. Divide the total number of bulbs by the number in each box: \(175 \div 25\). 2. Repeated addition reaches \(175\) in \(7\) equal steps: \(25 + 25 + 25 + 25 + 25 + 25 + 25 = 175\). 3. Therefore, \(175 \div 25 = 7\).

The nursery needs \(7\) boxes.

- Divide the total number of bulbs by \(20\). - Simplify the division by dividing both numbers by \(10\). - Check the quotient by multiplying it by \(20\).

1. Divide the total number of bulbs by the number in each row: \(800 \div 20\). 2. Divide both numbers by \(10\) to write an equivalent division problem: \(80 \div 2 = 40\). 3. Therefore, \(800 \div 20 = 40\).

The gardener needs \(40\) rows.

- Divide the total number of juice boxes by \(50\). - Simplify by dividing both numbers by \(10\). - Check that the number of cartons times \(50\) equals \(600\).

1. Divide the total number of juice boxes by the number in each carton: \(600 \div 50\). 2. Divide both numbers by \(10\): \(60 \div 5 = 12\). 3. Therefore, \(600 \div 50 = 12\).

The school receives \(12\) cartons.

- 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 \(4\).

- Find the area of each rectangle from its side lengths. - Add the two areas and compare the total with \(672\). - Add the widths to find the complete quotient.

1. The first rectangle has area \(24 \times 20 = 480\). 2. The second rectangle has area \(24 \times 8 = 192\). 3. The combined area is \(480 + 192 = 672\). 4. The combined width is \(20 + 8 = 28\), so \(672 \div 24 = 28\).

The first rectangle represents \(24\times20=480\), and the second represents \(24\times8=192\). Since \(480+192=672\), the partial quotients are \(20+8=28\), so \(672\div24=28\).

- Use the two requested multiples of \(18\) as lower and upper benchmarks. - Compare the dividend \(936\) with those products. - Convert that comparison into bounds for the quotient without completing the division.

1. \(18\times50=900\). 2. \(18\times60=1080\). 3. Since \(900<936<1080\), dividing the positive quantities by \(18\) gives \(50<936\div18<60\).

a) \(18\times50=900\) and \(18\times60=1080\). b) \(900<936<1080\), so \(50<936\div18<60\).

- Use the same area equation for each rectangle. - Divide \(240\) by the shown side length in each panel. - Compare the two factor pairs of \(240\).

1. For panel a), the missing side is \(240 \div 12 = 20\) units. 2. For panel b), the missing side is \(240 \div 15 = 16\) units. 3. With the same area, increasing one side requires the other side to decrease so the product remains \(240\).

Panel a): \(20\) units Panel b): \(16\) units The longer shown side pairs with a shorter missing side because both side-length products must equal \(240\).

- 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 \(30\).

- 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) \(>\)

- 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) \(20\)

- 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: \(180\div 6=30\) \(720\div 24=30\) \(60\div 2=30\)

- 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) \(=\)

- First find how many pinecones they collect together in one hour. - Once you know the hourly rate, how can you find the amount for three hours? - How many groups of the one-hour amount make \(200\)?

1. Find their combined hourly rate: \(25 + 15 = 40\) pinecones per hour. 2. Find the number collected in three hours: \(3 \times 40 = 120\). 3. Find the time needed to collect two hundred pinecones: \(200 \div 40 = 5\) hours.

a) They collect \(120\) pinecones in \(3\) hours. b) They must collect together for \(5\) hours.

- First find the total number of adventure books. - Which operation finds a quantity that is a given amount less than another quantity? - Once you know the total, divide it into groups of \(30\).

1. Find the number of adventure books: \(145 - 25 = 120\). 2. Divide by the number of books each shelf holds: \(120 \div 30 = 4\).

The library needs \(4\) shelves for the adventure books.

- First find how many fish the seals receive altogether. - Once you know the total, divide the fish among the buckets. - Think about how multiples of \(20\) can help.

1. Find the number of fish for the seals: \(120 + 40 = 160\). 2. Divide by the number of fish in each bucket: \(160 \div 20 = 8\).

The zoo needs \(8\) buckets for the seals.

- 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) \(6\)

- 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) \(>\)

- Write each two-digit divisor as a product of two one-digit factors. - Divide by one factor and then by the other factor. - Check by multiplying the quotient by the original divisor.

1. Part a: Since \(14 = 7 \times 2\), the missing divisor is \(2\). Then \(5600 \div 7 = 800\) and \(800 \div 2 = 400\). 2. Part b: Since \(15 = 5 \times 3\), the missing divisor is \(3\). Then \(4500 \div 5 = 900\) and \(900 \div 3 = 300\). 3. Part c: Since \(16 = 8 \times 2\), the missing divisor is \(8\). Then \(6400 \div 8 = 800\) and \(800 \div 2 = 400\). 4. Part d: Since \(18 = 9 \times 2\), the missing divisor is \(2\). Then \(9000 \div 9 = 1000\) and \(1000 \div 2 = 500\).

a) Missing divisor: \(2\); quotient: \(400\) b) Missing divisor: \(3\); quotient: \(300\) c) Missing divisor: \(8\); quotient: \(400\) d) Missing divisor: \(2\); quotient: \(500\)

- Write \(12\) as a product of the two divisors used by Lucas and Mia. - Carry out the divisions in the order shown and compare the quotients. - Factor \(14\) into two one-digit factors.

1. Part a: Both students factor \(12\) as \(6 \times 2\). Dividing successively by \(6\) and \(2\), in either order, is equivalent to dividing by \(12\). 2. Part b, Lucas: \(6000 \div 6 = 1000\), and \(1000 \div 2 = 500\). 3. Part b, Mia: \(6000 \div 2 = 3000\), and \(3000 \div 6 = 500\). 4. Part c: Since \(14 = 7 \times 2\), Lucas can calculate \(8400 \div 7 \div 2 = 1200 \div 2 = 600\).

a) Both methods work because \(12 = 6 \times 2\), and the two factors may be used as successive divisors in either order. b) Both methods give \(500\). c) \(8400 \div 7 \div 2 = 1200 \div 2 = 600\)

- 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 \(4\).

- Use the relationship \(\text{dividend} = \text{divisor} \times \text{quotient} + \text{remainder}\). - Choose a divisor first, then work backward to find the dividend. - A remainder must be less than the divisor.

1. For a quotient of \(9\) and remainder \(4\), the dividend must have the form \(9d + 4\), where the divisor \(d\) is greater than \(4\). 2. Choose \(d = 5\): \(9 \times 5 + 4 = 49\), so \(49 \div 5 = 9\text{ R }4\). 3. Choose \(d = 10\): \(9 \times 10 + 4 = 94\), so \(94 \div 10 = 9\text{ R }4\). 4. Choose \(d = 20\): \(9 \times 20 + 4 = 184\), so \(184 \div 20 = 9\text{ R }4\). 5. The divisor must be greater than the remainder, so \(d > 4\).

One possible set is \(49 \div 5 = 9\text{ R }4\), \(94 \div 10 = 9\text{ R }4\), and \(184 \div 20 = 9\text{ R }4\). The divisor must be greater than \(4\).

- 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.

- 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 how many marbles fit in one bag. - How many groups of that size make \(180\) marbles? - Dividing both numbers by \(10\) may make the second division easier.

1. Find the number of marbles in one bag: \(120 \div 6 = 20\). 2. Divide the new total by the number in one bag: \(180 \div 20 = 9\).

The factory needs \(9\) bags.

- Divide \(600\) by each box capacity. - Simplify each division by removing a common factor of \(10\). - Explain how a larger group size affects the number of groups.

1. For part a, divide: \(600 \div 20 = 30\). 2. For part b, divide: \(600 \div 60 = 10\). 3. A box holding \(60\) prizes has a greater capacity than one holding \(20\), so the same total is divided into fewer groups.

a) \(30\) boxes are needed. b) \(10\) boxes are needed. c) Fewer boxes are needed because each larger box holds more prizes.

- Find what remains after removing \(40\) groups of \(24\). - Express that remainder as another multiple of \(24\). - Add the partial quotients to get the full quotient.

1. \(24 \times 40 = 960\), so the remainder after the first partial quotient is \(1248 - 960 = 288\). 2. \(288 = 24 \times 12\), so the missing partial quotient is \(12\). 3. Add the partial quotients: \(40 + 12 = 52\).

a) \(12\) b) \(52\)

- Compare the given area with an easy multiple of \(24\). - If \(24 \times 20\) is already too large, what does that tell you about the missing side? - Search downward for an exact factor of \(432\).

1. Since \(24 \times 20 = 480\), a missing side of \(20\) would already make an area larger than \(432\). 2. Try a smaller factor: \(24 \times 18 = 432\). 3. Therefore, the missing side is \(18\) units, and Mateo's estimate is too large.

The diagram shows a side of \(24\) units. Since \(24\times20=480>432\), the missing side must be less than \(20\). Because \(24\times18=432\), the missing side is \(18\) units.

- Find the weight of the contents by subtracting the empty chest's weight. - How much do all the gold coins weigh? - How much weight remains for the silver coins? - How many silver coins have that total weight?

1. Find the weight of the contents: \(950\,\text{g} - 350\,\text{g} = 600\,\text{g}\). 2. Find the total weight of the gold coins: \(4 \times 60\,\text{g} = 240\,\text{g}\). 3. Find the weight of the silver coins: \(600\,\text{g} - 240\,\text{g} = 360\,\text{g}\). 4. Find the number of silver coins: \(360\,\text{g} \div 40\,\text{g} = 9\). 5. Find the total number of coins: \(4 + 9 = 13\).

The treasure chest contains \(13\) coins altogether.

- Rewrite the division using multiplication and the remainder. - Use the ones digits to determine the missing digit in the divisor. - Check that the remainder is less than the divisor.

1. Use the relationship \(\text{dividend} = \text{divisor} \times \text{quotient} + \text{remainder}\). 2. The product of the divisor and \(23\) must end in \(9\), because adding the remainder \(6\) then gives a number ending in \(5\). 3. Since the ones digit of \(23\) is \(3\), the missing ones digit of the divisor must be \(3\), because \(3 \times 3\) ends in \(9\). 4. Compute \(13 \times 23 = 299\), and add the remainder: \(299 + 6 = 305\). 5. Therefore, the missing digits are \(0\) and \(3\).

\(305 \div 13 = 23\) remainder \(6\)

- Multiply each partial quotient by \(24\). - Check whether each pair of partial products totals \(1008\). - Compare the sums of the partial quotients.

1. Lin's decomposition gives \(24 \times 30 = 720\) and \(24 \times 12 = 288\). Their sum is \(1008\), so \(30 + 12 = 42\) groups. 2. Omar's decomposition gives \(24 \times 40 = 960\) and \(24 \times 2 = 48\). Their sum is also \(1008\), so \(40 + 2 = 42\) groups. 3. Both methods partition the same dividend into multiples of \(24\), so both produce quotient \(42\).

Lin: \(24\times30=720\) and \(24\times12=288\), with \(720+288=1008\); her partial quotients total \(30+12=42\). Omar: \(24\times40=960\) and \(24\times2=48\), with \(960+48=1008\); his partial quotients total \(40+2=42\). Both decompositions therefore partition \(1008\) into multiples of \(24\), so both give quotient \(42\).