Free math worksheets

- 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) \(>\)

- Remove one factor of \(10\) from the tens factor or product to look for a basic fact. - Keep track of which factor is being scaled by \(10\). - Check the completed equation after identifying the related basic fact.

1. For a), \(3\times4=12\), so \(3\times40=120\). The missing factor is \(40\). 2. For b), \(6\times8=48\), so \(60\times8=480\). The missing factor is \(8\). 3. For c), \(9\times4=36\), so \(90\times4=360\). The missing factor is \(90\).

a) Basic fact \(3\times4=12\); missing factor \(40\) b) Basic fact \(6\times8=48\); missing factor \(8\) c) Basic fact \(9\times4=36\); missing factor \(90\)

- Start with the first equation and use a related basic multiplication fact to identify its missing factor. - Compare how the product changes from part a) to part b) while the first factor stays the same. - In part c), compare both the first factor and the product with part a).

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.

- 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 matching 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 matching 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 \(3 \times 60\): \(180\)

- 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) \(50\)

- 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) \(=\)

- 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. \(50 \times 4 = 200\).

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

- 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.

- Describe \(40\) as a number of tens. - Use the related one-digit multiplication fact to count the total tens. - Convert the total number of tens to a whole number.

1. Each group has \(4\) tens. 2. Five groups of \(4\) tens make \(20\) tens. 3. Twenty tens equal \(200\), so \(5\times40=200\).

\(5\times4\) tens \(=20\) tens \(=200\). There are \(200\) items altogether.

- 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 \(240\).

- 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) \(2\)

- Use related one-digit multiplication facts. - Keep track of the tens in each multiple-of-\(10\) factor. - Compare the products only after you know their values.

1. a) \(60\times3=180\), so \(180<200\). 2. b) \(40\times5=200\) and \(4\times50=200\), so the products are equal. 3. c) \(80\times4=320\) and \(70\times5=350\), so \(320<350\). 4. d) \(90\times3=270\) and \(40\times6=240\), so \(270>240\). 5. e) \(90\times2=180\) and \(60\times3=180\), so the products are equal.

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

- Read the value of one row from its place-value columns. - Count how many equal rows are shown. - Describe the total number of tens before writing the whole number.

1. Each row contains \(2\) tens, which is \(20\). 2. There are \(4\) equal rows, so the chart represents \(4\times20\). 3. Four groups of \(2\) tens make \(8\) tens, and \(8\) tens equal \(80\).

The chart represents \(4\times20=80\). It shows \(4\) groups of \(2\) tens, which make \(8\) tens, or \(80\).

- 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. - Before solving, decide whether the empty box is a factor or the product. - For a missing factor, use place value to decide whether the result should be one digit or a multiple of \(10\).

1. a) Since \(7 \times 7 = 49\), \(70 \times 7 = 490\). The missing value is \(70\), a multiple of \(10\). 2. b) Since \(8 \times 3 = 24\), \(80 \times 3 = 240\). The missing value is \(3\), a one-digit number. 3. c) Since \(6 \times 5 = 30\), \(60 \times 5 = 300\). The missing value is \(300\), the product. 4. d) Since \(9 \times 9 = 81\), \(90 \times 9 = 810\). The missing value is \(90\), a multiple of \(10\).

a) \(70\); a multiple of \(10\) b) \(3\); a one-digit number c) \(300\); the product d) \(90\); a multiple of \(10\)

- Solve the basic fact in part a) first. - Compare both the known factor and the product in each later equation with part a). - Decide whether the missing factor must stay the same or scale by \(10\).

1. In a), \(4\times9=36\), 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 stays \(9\): \(40\times9=360\). 3. In c), the first factor stays \(4\) while the product becomes ten times as large, so the missing factor must become ten times as large: \(4\times90=360\).

a) \(9\), because \(4\times9=36\). b) \(9\); both the first factor and the product are \(10\) times those in a), so the missing factor stays the same. c) \(90\); the first factor stays the same while the product is \(10\) times as large, so the missing factor is \(10\) times as large.

- Describe \(80\) as a number of tens. - Use the basic fact to count how many tens there are altogether. - Convert that number of tens back to a whole number.

1. \(80\) is \(8\) tens. 2. Six groups of \(8\) tens make \(48\) tens. 3. \(48\) tens equal \(480\), so \(6\times80=480\). 4. The factor of \(10\) comes from the value of the tens, not from a rule about writing a zero.

\(6\times80=480\) because \(80\) is \(8\) tens, so \(6\) groups of \(8\) tens are \(48\) tens, or \(480\).

- 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 the related basic multiplication fact to find each unknown factor. - For part e), treat a factor pair as the two factor values, regardless of order. - Find a second pair with different factor values rather than reversing the first equation.

1. a) \(280\div4=70\). 2. b) \(480\div60=8\). 3. c) \(630\div9=70\). 4. d) \(490\div70=7\). 5. e) Two different factor pairs are \(5\) with \(80\), and \(8\) with \(50\): \(5\times80=400\) and \(8\times50=400\).

a) \(70\) b) \(8\) c) \(70\) d) \(7\) e) One possible answer is \(5\times80=400\) and \(8\times50=400\).

- Count the rows in the chart and read the value represented by one row. - Describe the total as a number of tens before converting it to a whole number. - Compare what multiplication does with what simply writing two digits next to each other does.

1. The chart has \(3\) rows. Each row has \(4\) tens and \(0\) ones, so each row represents \(40\). The multiplication expression is \(3\times40\). 2. Three groups of \(4\) tens make \(12\) tens, and \(12\) tens equal \(120\). 3. Kai's method joins written digits instead of combining equal groups by place value. Therefore, \(3\times40=120\), not \(34\).

a) \(3\times40\) b) \(3\times4\) tens \(=12\) tens \(=120\) c) Kai joined digits instead of multiplying equal groups; the correct product is \(120\).

- Count the rows in each panel to find the number of equal groups. - Read the number of tens in one row to find the size of each group. - Compare the total number of tens represented by the two panels.

1. Panel a) has \(4\) rows, and each row has \(7\) tens, so it represents \(4\times70\). 2. Panel b) has \(7\) rows, and each row has \(4\) tens, so it represents \(7\times40\). 3. Four groups of \(7\) tens make \(28\) tens, and seven groups of \(4\) tens also make \(28\) tens. 4. Therefore, both products are \(280\).

1) Panel a): \(4\times70\); panel b): \(7\times40\) 2) Both products are \(280\). 3) Both charts represent \(28\) tens, so the products are equal.

- Compare each factor with the basic fact. - Ask whether exactly one factor became ten times as large. - A product changes by the same factor when one factor changes and the other stays fixed.

1. a) is true because changing \(6\) to \(60\) makes that factor ten times as large, so the product becomes \(420\). 2. b) is true for the same reason when \(7\) changes to \(70\). 3. c) is false because \(7\times6=42\), not \(420\). Changing either \(7\) to \(70\) or \(6\) to \(60\) makes the product \(420\).

a) True b) True c) False; for example, \(70\times6=420\) or \(7\times60=420\).

- Think of each multiple of \(10\) as a basic factor scaled by \(10\). - After finding one factor pair, look for different factor values for the second pair. - Reversing an equation does not create a new factor pair.

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

Sample answers: \(150=3\times50=5\times30\) \(240=4\times60=8\times30\) \(320=4\times80=8\times40\) \(450=5\times90=9\times50\)

- Replace the multiple of \(10\) with a number of tens. - Compare the number of tens with \(20\) tens before writing the standard numeral. - Include the related one-digit multiplication fact in each response.

1. \(4\times2=8\), so \(4\times20\) is \(8\) tens. That is less than \(20\) tens, and the exact product is \(80\). 2. \(5\times4=20\), so \(5\times40\) is \(20\) tens. It equals \(20\) tens, and the exact product is \(200\). 3. \(6\times3=18\), so \(6\times30\) is \(18\) tens. That is less than \(20\) tens, and the exact product is \(180\). 4. \(8\times4=32\), so \(8\times40\) is \(32\) tens. That is greater than \(20\) tens, and the exact product is \(320\).

\(4\times20\): \(4\times2=8\); 8 tens is less than 20 tens; exact product \(80\). \(5\times40\): \(5\times4=20\); 20 tens equals 20 tens; exact product \(200\). \(6\times30\): \(6\times3=18\); 18 tens is less than 20 tens; exact product \(180\). \(8\times40\): \(8\times4=32\); 32 tens is greater than 20 tens; exact product \(320\).

- 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.

- Compare the value of \(4\) with the value of \(40\). - Keep one factor fixed when describing how the product changes. - For part c), use basic facts for \(24\) and then scale one factor by \(10\). - Make sure your two equations do not use the same two factors in reverse order.

1. \(7\times4=28\) and \(7\times40=280\). 2. The factor \(40\) is \(10\) times \(4\), so with the other factor unchanged, the product \(280\) is \(10\) times \(28\). 3. Two valid equations for \(240\) with different factor pairs are \(3\times80=240\) and \(4\times60=240\). Other valid choices include \(6\times40\) and \(8\times30\).

a) \(28\) and \(280\) b) \(40\) is \(10\) times \(4\), so the product is also \(10\) times as great: \(280=10\times28\). c) One valid pair of answers is \(3\times80=240\) and \(4\times60=240\); the factor pairs are different.

- Describe \(360\) as a number of tens. - Look for basic multiplication facts whose product is the number of tens you need. - Both numbers in the basic fact must be one-digit numbers. - Use the factor-pair list to decide when your answer is complete.

1. Think of \(360\) as \(36\) tens. The one-digit factor and the number of tens must therefore form a basic multiplication fact with product \(36\). 2. The factor pairs of \(36\) in which both factors are from \(1\) through \(9\) are \(4\times9\), \(6\times6\), and \(9\times4\). 3. Turn the second factor in each basic fact into that many tens: \(4\times90=360\), \(6\times60=360\), and \(9\times40=360\). 4. There are no other one-digit factor pairs of \(36\), so there are no other equations that meet both rules.

\(4\times90=360\) \(6\times60=360\) \(9\times40=360\) This list is complete because \(4\times9\), \(6\times6\), and \(9\times4\) are all the factor pairs of \(36\) with both factors from \(1\) through \(9\); converting the second factor to tens gives every allowed equation for \(360\).

- Start near the lower boundary rather than testing factors randomly. - Notice how much the product changes when the one-digit factor increases by \(1\). - Once you cross the upper boundary, think about what that tells you about all larger factors.

1. \(4\times70=280\), which is too small. 2. The next three factors work: \(5\times70=350\), \(6\times70=420\), and \(7\times70=490\). 3. \(8\times70=560\), which is too large. 4. Products increase by \(70\) each time the one-digit factor increases by \(1\). Therefore, factors below \(5\) give products below the interval, and factors above \(7\) give products above it.

The factors are \(5\), \(6\), and \(7\): \(5\times70=350\) \(6\times70=420\) \(7\times70=490\) No other one-digit factor works because \(4\times70=280\) is already below \(300\), \(8\times70=560\) is already above \(500\), and the products change by \(70\) as the factor changes by \(1\).