Free Math Worksheets

- Write a multiplication equation for \(40\) groups of \(5\). - Use the fact \(4 \times 5 = 20\). - Explain how making one factor ten times as great affects the product.

1. The basic fact is \(4 \times 5 = 20\). 2. Since \(40\) is ten times \(4\), \(40 \times 5\) is ten times \(20\). 3. Therefore, \(40 \times 5 = 200\).

There are \(200\) individual items in \(40\) groups of \(5\).

- “Four times” means multiply by \(4\). - First calculate \(4 \times 2\). - Since \(200\) is \(100\) times \(2\), multiply the basic product by \(100\).

1. Multiply: \(4 \times 200 = 800\).

\(800\)

- How many pieces of candy are in each bag? - There are \(6\) equal groups. - Multiply the number of bags by the number of pieces in each bag.

1. Multiply the number of bags by the number of pieces in each bag: \(6 \times 70\). 2. Use the basic fact \(6 \times 7 = 42\), then multiply by \(10\): \(42 \times 10 = 420\).

The teacher bought \(420\) pieces of candy in all.

- Compare \(8\) with \(80\). - Use the corresponding one-digit multiplication fact first. - Then use the place value of the multiple of \(10\).

1. Since \(80\) is ten times \(8\), \(3 \times 80\) is ten times \(3 \times 8\). Thus \(3 \times 8 = 24\) and \(3 \times 80 = 240\). 2. Use the basic fact \(5 \times 9 = 45\): \(5 \times 90 = 450\). 3. Use the basic fact \(7 \times 4 = 28\): \(7 \times 40 = 280\).

\(3 \times 80\) is ten times \(3 \times 8\): \(240\) compared with \(24\). \(5 \times 90 = 450\) \(7 \times 40 = 280\)

- Find the related basic fact first. - Use place value to compare the one-digit factor with the multiple of \(10\). - Explain why the product is ten times as great.

1. \(4 \times 5 = 20\). Since \(50\) is ten times \(5\), \(4 \times 50 = 200\). 2. \(9 \times 3 = 27\). Since \(30\) is ten times \(3\), \(9 \times 30 = 270\). 3. \(6 \times 7 = 42\). Since \(70\) is ten times \(7\), \(6 \times 70 = 420\).

a) Basic fact: \(20\); product: \(200\) b) Basic fact: \(27\); product: \(270\) c) Basic fact: \(42\); product: \(420\)

- What operation finds the total cost for equal-price notebooks? - Use a related one-digit multiplication fact and place value. - Be sure to give the answer in cents.

1. Multiply the number of notebooks by the cost of each notebook: \(8 \times 90\). 2. Use the basic fact \(8 \times 9 = 72\). 3. Since \(90\) is \(9\) tens, multiply \(72\) by \(10\): \(72 \times 10 = 720\). 4. The total cost is \(720\) cents.

Work: \(8 \times 90 = 720\). The teacher pays \(720\) cents in all.

- Use the related basic multiplication fact without the zero. - How does multiplying one factor by \(10\) change the product?

1. Use \(3 \times 5 = 15\) and multiply the product by \(10\): \(30 \times 5 = 150\). The commutative equation is \(5 \times 30 = 150\). 2. Use \(8 \times 2 = 16\) and multiply the product by \(10\): \(80 \times 2 = 160\). The commutative equation is \(2 \times 80 = 160\). 3. Use \(4 \times 9 = 36\) and multiply the product by \(10\): \(40 \times 9 = 360\). The commutative equation is \(9 \times 40 = 360\).

a) \(30 \times 5 = 150\); \(5 \times 30 = 150\) b) \(80 \times 2 = 160\); \(2 \times 80 = 160\) c) \(40 \times 9 = 360\); \(9 \times 40 = 360\)

- In part a), use the commutative property. - For the other parts, evaluate both products. - Compare the products after calculating.

1. a) The factors are reversed, so the products are equal: \(7 \times 40 = 40 \times 7 = 280\). 2. b) \(60 \times 3 = 180\) and \(3 \times 50 = 150\), so \(180 > 150\). 3. c) \(9 \times 20 = 180\) and \(20 \times 8 = 160\), so \(180 > 160\).

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

- Which factor must move to make a commutative fact? - Evaluate the expression with both known factors. - Use the related basic fact and then account for the factor of \(10\).

1. In part a, switch the factors: \(6 \times 80 = 80 \times 6\). Since \(6 \times 8 = 48\), the product is \(480\). 2. In part b, switch the factors: \(50 \times 7 = 7 \times 50\). Since \(7 \times 5 = 35\), the product is \(350\). 3. In part c, switch the factors: \(4 \times 90 = 90 \times 4\). Since \(4 \times 9 = 36\), the product is \(360\).

a) \(6 \times 80 = 80 \times 6 = 480\) b) \(50 \times 7 = 7 \times 50 = 350\) c) \(4 \times 90 = 90 \times 4 = 360\)

- What changes from the first expression to the other two? - How are \(3\) and \(30\) related? - In each larger expression, which basic factor has been multiplied by \(10\)?

1. The basic fact is \(7 \times 3 = 21\). 2. In part b, one factor is ten times as large, so the product is \(21 \times 10 = 210\). 3. In part c, the other factor is ten times as large, so the product is also \(21 \times 10 = 210\). 4. Both larger expressions come from \(7 \times 3\) with exactly one basic factor multiplied by \(10\), so each product is \(10\) times \(21\).

a) \(21\) b) \(210\) c) \(210\) Parts b and c are equal because each uses \(7 \times 3\) with one factor multiplied by \(10\).

- Relate each product to a basic multiplication fact. - Use place value to account for the multiple of \(10\). - Compare the completed products.

1. a) \(4 \times 80 = 320\) and \(40 \times 8 = 320\), so the products are equal. 2. b) \(6 \times 50 = 300\) and \(5 \times 60 = 300\), so the products are equal. 3. c) \(90 \times 2 = 180\) and \(20 \times 8 = 160\), so \(180 > 160\).

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

- Identify a related basic fact with factors from \(1\) to \(10\). - Use place value to scale that fact by \(10\). - Check each factor by multiplication.

1. For a), \(3 \times 4 = 12\), so \(3 \times 40 = 120\). The missing factor is \(40\). 2. For b), \(6 \times 8 = 48\), so \(60 \times 8 = 480\). The missing factor is \(8\). 3. For c), \(9 \times 4 = 36\), so \(90 \times 4 = 360\). The missing factor is \(90\).

a) \(40\) b) \(8\) c) \(90\)

- Start with the basic fact in part a). - Compare how the first factor and product change in each equation. - Decide when the missing factor must stay the same and when it must become ten times as great.

1. In a), \(36 \div 4 = 9\), so the missing factor is \(9\). 2. In b), both the first factor and the product are ten times those in a), so the missing factor remains \(9\): \(40 \times 9 = 360\). 3. In c), the product is ten times the product in a) while the first factor is unchanged, so the missing factor is ten times as great: \(4 \times 90 = 360\).

a) \(9\) b) \(9\) c) \(90\)

- Use the basic fact \(8 \times 7 = 56\). - Compare which factor or product becomes ten times as great. - Use place value rather than counting written zeros.

1. In a), \(8 \times 7 = 56\), so the missing factor is \(7\). 2. In b), the product is ten times as great while the first factor stays the same, so the missing factor is \(70\): \(8 \times 70 = 560\). 3. In c), both the first factor and the product are ten times those in a), so the missing factor remains \(7\): \(80 \times 7 = 560\). 4. Parts a) and c) have the same missing factor; the missing factor in b) is ten times as great.

a) \(7\) b) \(70\) c) \(7\) The missing factors in a) and c) are equal, and the missing factor in b) is ten times as great.

- Start with the related basic fact. - Compare which factor or product is ten times as great. - Check each answer by multiplication.

1. In c), \(6 \times 8 = 48\), so the missing factor is \(8\). 2. In a), the product is ten times as great while the factor \(6\) stays the same, so the missing factor is \(80\): \(6 \times 80 = 480\). 3. In b), both the known factor and product are ten times those in c), so the missing factor remains \(8\): \(60 \times 8 = 480\).

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

- Increase the first factor by \(1\) in each new expression. - Compare expressions with the same first factor across the two sequences. - Relate \(80\) to \(40\).

1. Sequence A continues with \(6 \times 40 = 240\), \(7 \times 40 = 280\), and \(8 \times 40 = 320\). The products increase by \(40\). 2. Sequence B continues with \(6 \times 80 = 480\), \(7 \times 80 = 560\), and \(8 \times 80 = 640\). The products increase by \(80\). 3. Each product in Sequence B is twice the corresponding product in Sequence A because \(80\) is twice \(40\).

Sequence A: \(6 \times 40 = 240\), \(7 \times 40 = 280\), \(8 \times 40 = 320\) Sequence B: \(6 \times 80 = 480\), \(7 \times 80 = 560\), \(8 \times 80 = 640\) Each product in Sequence B is twice the corresponding product in Sequence A.

- Evaluate each product. - Relate each expression to a basic multiplication fact and use place value. - Match expressions with equal results.

1. Evaluate the products: \(3 \times 80 = 240\), \(5 \times 40 = 200\), \(6 \times 40 = 240\), \(4 \times 50 = 200\), \(2 \times 90 = 180\), and \(3 \times 60 = 180\). 2. Group expressions with equal products.

\(3 \times 80\) and \(6 \times 40\): \(240\) \(5 \times 40\) and \(4 \times 50\): \(200\) \(2 \times 90\) and \(3 \times 60\): \(180\)

- Use division to find an unknown factor. - Relate the equations to basic multiplication facts and use place value. - For part e), find one-digit factors of \(400\).

1. a) \(280 \div 4 = 70\). 2. b) \(480 \div 60 = 8\). 3. c) \(630 \div 9 = 70\). 4. d) \(490 \div 70 = 7\). 5. e) Examples include \(5 \times 80 = 400\) and \(8 \times 50 = 400\).

a) \(70\) b) \(8\) c) \(70\) d) \(7\) e) One possible answer is \(5 \times 80 = 400\) and \(8 \times 50 = 400\).

- Start with the one-digit multiplication fact. - Compare each factor with its related multiple of \(10\). - When one factor is multiplied by \(10\) and the other factor stays the same, the product is also multiplied by \(10\).

1. The products are a) \(24\) and \(240\), b) \(21\) and \(210\), c) \(16\) and \(160\), and d) \(45\) and \(450\). 2. In every pair, one factor is ten times as large while the other factor stays fixed. Therefore the second product is ten times the first product.

a) \(24\) and \(240\) b) \(21\) and \(210\) c) \(16\) and \(160\) d) \(45\) and \(450\) Multiplying one factor by \(10\) makes the product ten times as large.

- Start with the basic multiplication fact. - Use place value to explain how making one factor ten times as great changes the product. - Compare parts b) and c) using the commutative property.

1. The basic fact is \(4 \times 7 = 28\). 2. Since \(70\) is ten times \(7\), \(4 \times 70 = 280\). 3. Since \(40\) is ten times \(4\), \(40 \times 7 = 280\).

a) \(28\) b) \(280\) c) \(280\)

- Write each riddle as a multiplication equation with a missing factor. - Use division to find the missing factor. - Compare how the known factor and product change from one equation to another.

1. In a), \(21 \div 3 = 7\). 2. In b), both \(30\) and \(210\) are ten times the corresponding numbers in a), so the missing factor remains \(7\): \(7 \times 30 = 210\). 3. In c), the product is ten times as great while the known factor remains \(3\), so the missing factor is ten times as great: \(70 \times 3 = 210\).

a) \(7\) b) \(7\) c) \(70\)

- Identify a related basic multiplication fact. - Use place value to scale the basic fact by \(10\). - Check each completed equation by multiplication.

1. In a), \(5 \times 40 = 200\). 2. In b), \(7 \times 5 = 35\), so \(7 \times 50 = 350\). 3. In c), \(8 \times 7 = 56\), so \(8 \times 70 = 560\). 4. In d), \(9 \times 9 = 81\), so \(9 \times 90 = 810\).

a) \(5\) b) \(50\) c) \(8\) d) \(90\)

- First use the related basic multiplication fact without the factor of \(10\). - For a multiple of \(10\), multiply the basic product by \(10\). - Use division to find an unknown factor. - Check whether one result can be related to another by doubling a factor.

1. a) \(40 \times 2 = 80\). 2. b) \(40 \times 5 = 200\). 3. c) \(70 \times 3 = 210\). 4. d) \(160 \div 20 = 8\). 5. e) \(300 \div 6 = 50\).

a) \(80\) b) \(200\) c) \(210\) d) \(8\) e) \(50\)

- Relate each expression to a basic multiplication fact. - Use place value to account for the multiple of \(10\). - Check by repeated addition or a related fact.

1. \(20 \times 7 = 140\). 2. \(40 \times 5 = 200\). 3. \(3 \times 60 = 180\). 4. \(8 \times 20 = 160\). 5. \(50 \times 4 = 200\).

\(140\), \(200\), \(180\), \(160\), \(200\)

- Evaluate both products in each comparison. - Use related multiplication facts and place value. - Compare the results.

1. a) \(40 \times 3 = 120\) and \(20 \times 6 = 120\), so the products are equal. 2. b) \(70 \times 4 = 280\) and \(50 \times 6 = 300\), so \(280 < 300\). 3. c) \(30 \times 9 = 270\) and \(40 \times 7 = 280\), so \(270 < 280\). 4. d) \(60 \times 5 = 300\) and \(3 \times 100 = 300\), so the products are equal.

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

- Find each gardener's total separately. - Use a multiplication fact and place value for each product. - Compare the two products.

1. Find Tim's total: \(6 \times 40 = 240\). 2. Find Mia's total: \(4 \times 60 = 240\). 3. Compare: \(240 = 240\), so they plant the same number.

Tim and Mia each plant \(240\) tulip bulbs, so they plant the same number.

- Evaluate \(30 \times 8\). - Evaluate \(80 \times 3\). - Compare the products. - Notice how the digits \(3\) and \(8\) switch roles in the two expressions.

1. \(30 \times 8 = 240\). 2. \(80 \times 3 = 240\). 3. Since \(240 = 240\), the two quantities are equal.

They are equal. Both products are \(240\).

- Notice the factor of \(10\) in each related pair. - First solve the one-digit multiplication fact. - The second factor in the related expression is ten times as large. - Use place value to scale the product by \(10\).

1. \(4 \times 2 = 8\). Since \(20\) is ten times \(2\), \(4 \times 20 = 80\). 2. \(3 \times 5 = 15\), so \(3 \times 50 = 150\). 3. \(6 \times 3 = 18\), so \(6 \times 30 = 180\). 4. \(8 \times 4 = 32\), so \(8 \times 40 = 320\).

1) \(8\) and \(80\) 2) \(15\) and \(150\) 3) \(18\) and \(180\) 4) \(32\) and \(320\)

- Match the nonzero digits to a basic multiplication fact. - Use division to find an unknown factor. - Account for the factor of \(10\) in the multiple-of-\(10\) number. - Check how the factor of \(10\) changes the product.

1. a) \(7 \times 30 = 210\). 2. b) \(450 \div 5 = 90\). 3. c) \(90 \times 4 = 360\). 4. d) \(120 \div 60 = 2\).

a) \(210\) b) \(90\) c) \(360\) d) \(2\)

- Count how many times the same addend appears. - Write the number of groups as one factor. - Solve the related basic fact, such as \(5 \times 6\), and then multiply that product by \(10\).

1. a) Five groups of \(60\): \(5 \times 60 = 300\). 2. b) Four groups of \(90\): \(4 \times 90 = 360\). 3. c) Seven groups of \(40\): \(7 \times 40 = 280\).

a) \(5 \times 60 = 300\) b) \(4 \times 90 = 360\) c) \(7 \times 40 = 280\)

- Rewrite repeated addition as multiplication. - Compare the number and size of the equal groups. - You may be able to compare without evaluating every full product. - For part c), compare the basic facts \(4 \times 6\) and \(5 \times 5\).

1. a) \(5 \times 70 = 350\), while four groups of \(70\) equal \(280\), so \(350 > 280\). 2. b) Three groups of \(80\) equal \(3 \times 80\), so the expressions are equal. 3. c) \(4 \times 60 = 240\), while five groups of \(50\) equal \(250\), so \(240 < 250\).

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

- Relate each equation to a basic multiplication fact. - Use place value to explain how a factor that is ten times as great changes the product. - In part b), the missing value is a one-digit number.

1. a) Since \(7 \times 7 = 49\), \(70 \times 7 = 490\). The missing value is \(70\). 2. b) Since \(8 \times 3 = 24\), \(80 \times 3 = 240\). The missing value is \(3\). 3. c) Since \(6 \times 5 = 30\), multiplying \(60\) by \(5\) gives \(300\). 4. d) Since \(9 \times 9 = 81\), \(90 \times 9 = 810\). The missing value is \(90\).

a) \(70\) b) \(3\) c) \(300\) d) \(90\)

- Compare \(4\) with \(40\) and \(7\) with \(70\). - Use place value to describe the factor-of-\(10\) change. - Think of another factor pair for \(280\) that includes a multiple of \(10\).

1. \(7 \times 4 = 28\), \(7 \times 40 = 280\), and \(70 \times 4 = 280\). 2. Multiplying either factor by \(10\) makes the product ten times as large. By the commutative property, \(7 \times 40\) and \(70 \times 4\) have the same product. 3. Another example is \(4 \times 70 = 280\).

a) \(28\), \(280\), \(280\) b) The last two products are equal and are ten times \(28\). c) One example is \(4 \times 70 = 280\).

- Notice how the first factor changes. - Compare \(30\) with \(60\). - Find the difference between consecutive products in each sequence.

1. Sequence a) continues with \(5 \times 60 = 300\), \(4 \times 60 = 240\), and \(3 \times 60 = 180\). The products decrease by \(60\). 2. Sequence b) continues with \(5 \times 30 = 150\), \(4 \times 30 = 120\), and \(3 \times 30 = 90\). The products decrease by \(30\). 3. Each product in b) is half the matching product in a) because \(30\) is half of \(60\).

a) \(5 \times 60 = 300\), \(4 \times 60 = 240\), \(3 \times 60 = 180\) b) \(5 \times 30 = 150\), \(4 \times 30 = 120\), \(3 \times 30 = 90\) The products decrease by \(60\) in a) and by \(30\) in b). Each b) product is half the matching a) product.

- Use related multiplication facts with factors from \(1\) to \(9\). - Think of each multiple of ten as a basic fact scaled by \(10\). - Check that each equation has one one-digit factor and one multiple-of-ten factor.

1. For \(150\), examples are \(3 \times 50 = 150\) and \(5 \times 30 = 150\). 2. For \(240\), examples include \(3 \times 80 = 240\), \(4 \times 60 = 240\), \(6 \times 40 = 240\), and \(8 \times 30 = 240\). 3. For \(320\), examples are \(4 \times 80 = 320\) and \(8 \times 40 = 320\). 4. For \(450\), examples are \(5 \times 90 = 450\) and \(9 \times 50 = 450\).

Sample answers: \(150 = 3 \times 50 = 5 \times 30\) \(240 = 4 \times 60 = 8 \times 30\) \(320 = 4 \times 80 = 8 \times 40\) \(450 = 5 \times 90 = 9 \times 50\)

- Evaluate both products in each comparison. - Use related basic multiplication facts. - Keep track of how many tens each expression represents. - Compare the completed products.

1. a) \(60 \times 4 = 240\) and \(30 \times 8 = 240\), so the products are equal. 2. b) \(50 \times 7 = 350\) and \(90 \times 4 = 360\), so \(350 < 360\). 3. c) \(80 \times 5 = 400\) and \(40 \times 9 = 360\), so \(400 > 360\). 4. d) \(20 \times 9 = 180\) and \(60 \times 3 = 180\), so the products are equal.

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

- Evaluate both sides of each comparison. - Use related basic facts and place value. - Estimate before calculating when helpful.

1. a) \(60 \times 3 = 180\), so \(180 < 200\). 2. b) \(40 \times 5 = 200\) and \(2 \times 100 = 200\), so the values are equal. 3. c) \(80 \times 4 = 320\) and \(70 \times 5 = 350\), so \(320 < 350\). 4. d) \(300 \times 3 = 900\) and \(50 \times 2 = 100\), so \(900 > 100\). 5. e) \(90 \times 2 = 180\) and \(60 \times 3 = 180\), so the values are equal.

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

- Find each class's total separately. - Use place value to multiply a one-digit number by a multiple of \(10\). - Compare the two products.

1. Find Class A's total: \(3 \times 240 = 720\). 2. Find Class B's total: \(4 \times 180 = 720\). 3. Since \(720 = 720\), the classes collect the same number.

Both classes collect \(720\) marbles, so they collect the same number.

- First find the total number of seats. - Use place value and a related multiplication fact for the multiple of \(10\). - Compare the number of seats with the number of students.

1. Find the total number of seats: \(7 \times 40 = 280\). 2. Compare: \(280 < 300\), so there are not enough seats. 3. The theater is short by \(300 - 280 = 20\) seats.

No. The theater has \(280\) seats, so it needs \(20\) more seats for all \(300\) students.