Practice · Estimate a product to bound an unknown · Sensim Math

Variant 1 easy

Find every one-digit number that can go in the $\square$ .

$\square \times 37 > 10 \times 30$

Your answer

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 37 is greater than 10 times 30.

Givens

The inequality is (square) x 37 > 10 x 30.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (10 x 30 = 300). Then estimate: each step up in the square adds about 37. Test the boundary digit to find where the product first passes 300, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 10 by 30 to get the number the left side must beat.

10 \times 30 = 300

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 37 passes 300

#6 Guess and Check 3.OA.C.7

Since 37 is close to 40, (square) x 37 is roughly (square) times 40. Test near the boundary: 8 x 37 = 296 (not over 300) and 9 x 37 = 333 (over 300).

8 \times 37 = 296 < 300,\quad 9 \times 37 = 333 > 300

Estimating 37 as about 40 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 9 up, the product only grows, so 9 all satisfy the inequality. Digits 8 and below give 296 or less, which is not over 300.

9 \times 37 = 333

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 9

4 · Reviewdoes it hold up?

The boundary 8 x 37 = 296 is just under 300 and 9 x 37 = 333 is just over, so 9 being the smallest that works makes sense; all one-digit values are 9.

Another way: Divide instead: 300 / 37 is between 8 and 9, so the square must be at least 9. The one-digit numbers 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 10 x 30 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 37 against 300.

💡Takeaway. Treat 37 as almost 40 and you can guess the boundary -- Grade 3 estimating points right to 9!

Variant 2 easy

Find every one-digit number that can go in the $\square$ .

$\square \times 43 > 30 \times 10$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 43 is greater than 30 times 10.

Givens

The inequality is (square) x 43 > 30 x 10.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (30 x 10 = 300). Then estimate: each step up in the square adds about 43. Test the boundary digit to find where the product first passes 300, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 30 by 10 to get the number the left side must beat.

30 \times 10 = 300

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 43 passes 300

#6 Guess and Check 3.OA.C.7

Since 43 is close to 40, (square) x 43 is roughly (square) times 40. Test near the boundary: 6 x 43 = 258 (not over 300) and 7 x 43 = 301 (over 300).

6 \times 43 = 258 < 300,\quad 7 \times 43 = 301 > 300

Estimating 43 as about 40 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 7 up, the product only grows, so 7, 8, and 9 all satisfy the inequality. Digits 6 and below give 258 or less, which is not over 300.

7 \times 43 = 301,\ 8 \times 43 = 344,\ 9 \times 43 = 387

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 6 x 43 = 258 is just under 300 and 7 x 43 = 301 is just over, so 7 being the smallest that works makes sense; all one-digit values are 7, 8, and 9.

Another way: Divide instead: 300 / 43 is between 6 and 7, so the square must be at least 7. The one-digit numbers 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 30 x 10 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 43 against 300.

💡Takeaway. Treat 43 as almost 40 and you can guess the boundary -- Grade 3 estimating points right to 7, 8, and 9!

Variant 3 easy

Find every one-digit number that can go in the $\square$ .

$\square \times 48 > 10 \times 20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 48 is greater than 10 times 20.

Givens

The inequality is (square) x 48 > 10 x 20.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (10 x 20 = 200). Then estimate: each step up in the square adds about 48. Test the boundary digit to find where the product first passes 200, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 10 by 20 to get the number the left side must beat.

10 \times 20 = 200

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 48 passes 200

#6 Guess and Check 3.OA.C.7

Since 48 is close to 50, (square) x 48 is roughly (square) times 50. Test near the boundary: 4 x 48 = 192 (not over 200) and 5 x 48 = 240 (over 200).

4 \times 48 = 192 < 200,\quad 5 \times 48 = 240 > 200

Estimating 48 as about 50 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 5 up, the product only grows, so 5, 6, 7, 8, and 9 all satisfy the inequality. Digits 4 and below give 192 or less, which is not over 200.

5 \times 48 = 240,\ 6 \times 48 = 288,\ 7 \times 48 = 336,\ 8 \times 48 = 384,\ 9 \times 48 = 432

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 5, 6, 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 4 x 48 = 192 is just under 200 and 5 x 48 = 240 is just over, so 5 being the smallest that works makes sense; all one-digit values are 5, 6, 7, 8, and 9.

Another way: Divide instead: 200 / 48 is between 4 and 5, so the square must be at least 5. The one-digit numbers 5, 6, 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 10 x 20 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 48 against 200.

💡Takeaway. Treat 48 as almost 50 and you can guess the boundary -- Grade 3 estimating points right to 5, 6, 7, 8, and 9!

Variant 4 medium

Find every one-digit number that can go in the $\square$ .

$\square \times 62 > 30 \times 10$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 62 is greater than 30 times 10.

Givens

The inequality is (square) x 62 > 30 x 10.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (30 x 10 = 300). Then estimate: each step up in the square adds about 62. Test the boundary digit to find where the product first passes 300, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 30 by 10 to get the number the left side must beat.

30 \times 10 = 300

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 62 passes 300

#6 Guess and Check 3.OA.C.7

Since 62 is close to 60, (square) x 62 is roughly (square) times 60. Test near the boundary: 4 x 62 = 248 (not over 300) and 5 x 62 = 310 (over 300).

4 \times 62 = 248 < 300,\quad 5 \times 62 = 310 > 300

Estimating 62 as about 60 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 5 up, the product only grows, so 5, 6, 7, 8, and 9 all satisfy the inequality. Digits 4 and below give 248 or less, which is not over 300.

5 \times 62 = 310,\ 6 \times 62 = 372,\ 7 \times 62 = 434,\ 8 \times 62 = 496,\ 9 \times 62 = 558

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 5, 6, 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 4 x 62 = 248 is just under 300 and 5 x 62 = 310 is just over, so 5 being the smallest that works makes sense; all one-digit values are 5, 6, 7, 8, and 9.

Another way: Divide instead: 300 / 62 is between 4 and 5, so the square must be at least 5. The one-digit numbers 5, 6, 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 30 x 10 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 62 against 300.

💡Takeaway. Treat 62 as almost 60 and you can guess the boundary -- Grade 3 estimating points right to 5, 6, 7, 8, and 9!

Variant 5 medium

Find every one-digit number that can go in the $\square$ .

$\square \times 66 > 20 \times 20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 66 is greater than 20 times 20.

Givens

The inequality is (square) x 66 > 20 x 20.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (20 x 20 = 400). Then estimate: each step up in the square adds about 66. Test the boundary digit to find where the product first passes 400, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 20 by 20 to get the number the left side must beat.

20 \times 20 = 400

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 66 passes 400

#6 Guess and Check 3.OA.C.7

Since 66 is close to 70, (square) x 66 is roughly (square) times 70. Test near the boundary: 6 x 66 = 396 (not over 400) and 7 x 66 = 462 (over 400).

6 \times 66 = 396 < 400,\quad 7 \times 66 = 462 > 400

Estimating 66 as about 70 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 7 up, the product only grows, so 7, 8, and 9 all satisfy the inequality. Digits 6 and below give 396 or less, which is not over 400.

7 \times 66 = 462,\ 8 \times 66 = 528,\ 9 \times 66 = 594

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 6 x 66 = 396 is just under 400 and 7 x 66 = 462 is just over, so 7 being the smallest that works makes sense; all one-digit values are 7, 8, and 9.

Another way: Divide instead: 400 / 66 is between 6 and 7, so the square must be at least 7. The one-digit numbers 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 20 x 20 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 66 against 400.

💡Takeaway. Treat 66 as almost 70 and you can guess the boundary -- Grade 3 estimating points right to 7, 8, and 9!

Variant 6 medium

Find every one-digit number that can go in the $\square$ .

$\square \times 78 > 20 \times 20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 78 is greater than 20 times 20.

Givens

The inequality is (square) x 78 > 20 x 20.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (20 x 20 = 400). Then estimate: each step up in the square adds about 78. Test the boundary digit to find where the product first passes 400, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 20 by 20 to get the number the left side must beat.

20 \times 20 = 400

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 78 passes 400

#6 Guess and Check 3.OA.C.7

Since 78 is close to 80, (square) x 78 is roughly (square) times 80. Test near the boundary: 5 x 78 = 390 (not over 400) and 6 x 78 = 468 (over 400).

5 \times 78 = 390 < 400,\quad 6 \times 78 = 468 > 400

Estimating 78 as about 80 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 6 up, the product only grows, so 6, 7, 8, and 9 all satisfy the inequality. Digits 5 and below give 390 or less, which is not over 400.

6 \times 78 = 468,\ 7 \times 78 = 546,\ 8 \times 78 = 624,\ 9 \times 78 = 702

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 6, 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 5 x 78 = 390 is just under 400 and 6 x 78 = 468 is just over, so 6 being the smallest that works makes sense; all one-digit values are 6, 7, 8, and 9.

Another way: Divide instead: 400 / 78 is between 5 and 6, so the square must be at least 6. The one-digit numbers 6, 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 20 x 20 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 78 against 400.

💡Takeaway. Treat 78 as almost 80 and you can guess the boundary -- Grade 3 estimating points right to 6, 7, 8, and 9!

Variant 7 medium

Find every one-digit number that can go in the $\square$ .

$\square \times 85 > 10 \times 50$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 85 is greater than 10 times 50.

Givens

The inequality is (square) x 85 > 10 x 50.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (10 x 50 = 500). Then estimate: each step up in the square adds about 85. Test the boundary digit to find where the product first passes 500, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 10 by 50 to get the number the left side must beat.

10 \times 50 = 500

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 85 passes 500

#6 Guess and Check 3.OA.C.7

Since 85 is close to 80, (square) x 85 is roughly (square) times 80. Test near the boundary: 5 x 85 = 425 (not over 500) and 6 x 85 = 510 (over 500).

5 \times 85 = 425 < 500,\quad 6 \times 85 = 510 > 500

Estimating 85 as about 80 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 6 up, the product only grows, so 6, 7, 8, and 9 all satisfy the inequality. Digits 5 and below give 425 or less, which is not over 500.

6 \times 85 = 510,\ 7 \times 85 = 595,\ 8 \times 85 = 680,\ 9 \times 85 = 765

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 6, 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 5 x 85 = 425 is just under 500 and 6 x 85 = 510 is just over, so 6 being the smallest that works makes sense; all one-digit values are 6, 7, 8, and 9.

Another way: Divide instead: 500 / 85 is between 5 and 6, so the square must be at least 6. The one-digit numbers 6, 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 10 x 50 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 85 against 500.

💡Takeaway. Treat 85 as almost 80 and you can guess the boundary -- Grade 3 estimating points right to 6, 7, 8, and 9!

Variant 8 medium

Find every one-digit number that can go in the $\square$ .

$\square \times 88 > 30 \times 20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 88 is greater than 30 times 20.

Givens

The inequality is (square) x 88 > 30 x 20.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (30 x 20 = 600). Then estimate: each step up in the square adds about 88. Test the boundary digit to find where the product first passes 600, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 30 by 20 to get the number the left side must beat.

30 \times 20 = 600

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 88 passes 600

#6 Guess and Check 3.OA.C.7

Since 88 is close to 90, (square) x 88 is roughly (square) times 90. Test near the boundary: 6 x 88 = 528 (not over 600) and 7 x 88 = 616 (over 600).

6 \times 88 = 528 < 600,\quad 7 \times 88 = 616 > 600

Estimating 88 as about 90 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 7 up, the product only grows, so 7, 8, and 9 all satisfy the inequality. Digits 6 and below give 528 or less, which is not over 600.

7 \times 88 = 616,\ 8 \times 88 = 704,\ 9 \times 88 = 792

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 6 x 88 = 528 is just under 600 and 7 x 88 = 616 is just over, so 7 being the smallest that works makes sense; all one-digit values are 7, 8, and 9.

Another way: Divide instead: 600 / 88 is between 6 and 7, so the square must be at least 7. The one-digit numbers 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 30 x 20 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 88 against 600.

💡Takeaway. Treat 88 as almost 90 and you can guess the boundary -- Grade 3 estimating points right to 7, 8, and 9!

Variant 9 hard

Find every one-digit number that can go in the $\square$ .

$\square \times 91 > 40 \times 10$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 91 is greater than 40 times 10.

Givens

The inequality is (square) x 91 > 40 x 10.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (40 x 10 = 400). Then estimate: each step up in the square adds about 91. Test the boundary digit to find where the product first passes 400, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 40 by 10 to get the number the left side must beat.

40 \times 10 = 400

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 91 passes 400

#6 Guess and Check 3.OA.C.7

Since 91 is close to 90, (square) x 91 is roughly (square) times 90. Test near the boundary: 4 x 91 = 364 (not over 400) and 5 x 91 = 455 (over 400).

4 \times 91 = 364 < 400,\quad 5 \times 91 = 455 > 400

Estimating 91 as about 90 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 5 up, the product only grows, so 5, 6, 7, 8, and 9 all satisfy the inequality. Digits 4 and below give 364 or less, which is not over 400.

5 \times 91 = 455,\ 6 \times 91 = 546,\ 7 \times 91 = 637,\ 8 \times 91 = 728,\ 9 \times 91 = 819

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 5, 6, 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 4 x 91 = 364 is just under 400 and 5 x 91 = 455 is just over, so 5 being the smallest that works makes sense; all one-digit values are 5, 6, 7, 8, and 9.

Another way: Divide instead: 400 / 91 is between 4 and 5, so the square must be at least 5. The one-digit numbers 5, 6, 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 40 x 10 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 91 against 400.

💡Takeaway. Treat 91 as almost 90 and you can guess the boundary -- Grade 3 estimating points right to 5, 6, 7, 8, and 9!

Variant 10 hard

Find every one-digit number that can go in the $\square$ .

$\square \times 95 > 20 \times 30$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 95 is greater than 20 times 30.

Givens

The inequality is (square) x 95 > 20 x 30.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (20 x 30 = 600). Then estimate: each step up in the square adds about 95. Test the boundary digit to find where the product first passes 600, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 20 by 30 to get the number the left side must beat.

20 \times 30 = 600

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 95 passes 600

#6 Guess and Check 3.OA.C.7

Since 95 is close to 100, (square) x 95 is roughly (square) times 100. Test near the boundary: 6 x 95 = 570 (not over 600) and 7 x 95 = 665 (over 600).

6 \times 95 = 570 < 600,\quad 7 \times 95 = 665 > 600

Estimating 95 as about 100 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 7 up, the product only grows, so 7, 8, and 9 all satisfy the inequality. Digits 6 and below give 570 or less, which is not over 600.

7 \times 95 = 665,\ 8 \times 95 = 760,\ 9 \times 95 = 855

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 6 x 95 = 570 is just under 600 and 7 x 95 = 665 is just over, so 7 being the smallest that works makes sense; all one-digit values are 7, 8, and 9.

Another way: Divide instead: 600 / 95 is between 6 and 7, so the square must be at least 7. The one-digit numbers 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 20 x 30 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 95 against 600.

💡Takeaway. Treat 95 as almost 100 and you can guess the boundary -- Grade 3 estimating points right to 7, 8, and 9!

Variant 11 hard

Find every one-digit number that can go in the $\square$ .

$\square \times 95 > 20 \times 20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Find every one-digit number that can replace the square so that (square) times 95 is greater than 20 times 20.

Givens

The inequality is (square) x 95 > 20 x 20.
The square must be a one-digit number.

Unknowns

All one-digit numbers that make the inequality true.

Constraints

The square is a single digit (0 through 9).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #8 Analyze the Units

First compute the fixed right side (20 x 20 = 400). Then estimate: each step up in the square adds about 95. Test the boundary digit to find where the product first passes 400, then list every digit at or above it.

3 · Execute3 carry out the plan

1Compute the right side

#8 Analyze the Units 3.NBT.A.3

Multiply 20 by 20 to get the number the left side must beat.

20 \times 20 = 400

Multiplying two multiples of 10 is a small product with zeros tacked on.

2Estimate where (square) x 95 passes 400

#6 Guess and Check 3.OA.C.7

Since 95 is close to 100, (square) x 95 is roughly (square) times 100. Test near the boundary: 4 x 95 = 380 (not over 400) and 5 x 95 = 475 (over 400).

4 \times 95 = 380 < 400,\quad 5 \times 95 = 475 > 400

Estimating 95 as about 100 quickly shows which digit is the turning point.

3List all digits that work

#6 Guess and Check 3.OA.C.7

From 5 up, the product only grows, so 5, 6, 7, 8, and 9 all satisfy the inequality. Digits 4 and below give 380 or less, which is not over 400.

5 \times 95 = 475,\ 6 \times 95 = 570,\ 7 \times 95 = 665,\ 8 \times 95 = 760,\ 9 \times 95 = 855

Once a digit works, every larger digit works too, since the product keeps increasing.

Answer: 5, 6, 7, 8, and 9

4 · Reviewdoes it hold up?

The boundary 4 x 95 = 380 is just under 400 and 5 x 95 = 475 is just over, so 5 being the smallest that works makes sense; all one-digit values are 5, 6, 7, 8, and 9.

Another way: Divide instead: 400 / 95 is between 4 and 5, so the square must be at least 5. The one-digit numbers 5, 6, 7, 8, and 9 follow directly.

Standardsmin grade 3

3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 — Computing 20 x 20 and estimating with multiples of 10.
3.OA.C.7 Fluently multiply and divide within 100 — Testing one-digit multiples of 95 against 400.

💡Takeaway. Treat 95 as almost 100 and you can guess the boundary -- Grade 3 estimating points right to 5, 6, 7, 8, and 9!

Estimate a product to bound an unknown · 11 practice problems

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