Practice · Bound an inequality at equality · Sensim Math

Variant 1 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$3\times9<4\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 4 times the $\square$ is larger than 3 times 9.

Givens

The inequality is 3 x 9 < 4 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 4 x $\square$ greater than 27.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 4 x $\square$ first passes 27; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 3 times 9 so the inequality becomes a comparison with a single number.

3 \times 9 = 27

3 times 9 is a basic multiplication fact, so the left side is just the number 27.

2Find where 4 times the $\square$ passes 27

#6 Guess and Check 3.OA.A.4

Test multiples of 4 in order: 4x6 = 24 is not greater than 27, but 4x7 = 28 is greater than 27. So the

\square

must be 7 or larger.

4 \times 6 = 24,\ 4 \times 7 = 28

Checking the 4-times facts shows exactly when the right side becomes bigger than 27.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 7 and above (and at most 9) makes the right side larger than 27, so collect them.

\square \in \{7, 8, 9\}

Since bigger

\square

values only make 4 x

\square

larger, all of 7, 8, 9 keep the inequality true.

Answer: 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 4x7 = 28 > 27, true; and the value just below, 4x6 = 24, is not greater than 27, so 7 is correctly the cutoff and 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 27 divided by 4 is between 6 and 7, so the $\square$ must be more than 6, giving 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 4 x $\square$ greater than 27.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 3 x 9 and the 4-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 2 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$6\times9<8\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 8 times the $\square$ is larger than 6 times 9.

Givens

The inequality is 6 x 9 < 8 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 8 x $\square$ greater than 54.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 8 x $\square$ first passes 54; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 6 times 9 so the inequality becomes a comparison with a single number.

6 \times 9 = 54

6 times 9 is a basic multiplication fact, so the left side is just the number 54.

2Find where 8 times the $\square$ passes 54

#6 Guess and Check 3.OA.A.4

Test multiples of 8 in order: 8x6 = 48 is not greater than 54, but 8x7 = 56 is greater than 54. So the

\square

must be 7 or larger.

8 \times 6 = 48,\ 8 \times 7 = 56

Checking the 8-times facts shows exactly when the right side becomes bigger than 54.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 7 and above (and at most 9) makes the right side larger than 54, so collect them.

\square \in \{7, 8, 9\}

Since bigger

\square

values only make 8 x

\square

larger, all of 7, 8, 9 keep the inequality true.

Answer: 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 8x7 = 56 > 54, true; and the value just below, 8x6 = 48, is not greater than 54, so 7 is correctly the cutoff and 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 54 divided by 8 is between 6 and 7, so the $\square$ must be more than 6, giving 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 8 x $\square$ greater than 54.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 6 x 9 and the 8-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 3 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$7\times7<9\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 9 times the $\square$ is larger than 7 times 7.

Givens

The inequality is 7 x 7 < 9 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 9 x $\square$ greater than 49.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 9 x $\square$ first passes 49; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 7 times 7 so the inequality becomes a comparison with a single number.

7 \times 7 = 49

7 times 7 is a basic multiplication fact, so the left side is just the number 49.

2Find where 9 times the $\square$ passes 49

#6 Guess and Check 3.OA.A.4

Test multiples of 9 in order: 9x5 = 45 is not greater than 49, but 9x6 = 54 is greater than 49. So the

\square

must be 6 or larger.

9 \times 5 = 45,\ 9 \times 6 = 54

Checking the 9-times facts shows exactly when the right side becomes bigger than 49.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 6 and above (and at most 9) makes the right side larger than 49, so collect them.

\square \in \{6, 7, 8, 9\}

Since bigger

\square

values only make 9 x

\square

larger, all of 6, 7, 8, 9 keep the inequality true.

Answer: 6, 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 9x6 = 54 > 49, true; and the value just below, 9x5 = 45, is not greater than 49, so 6 is correctly the cutoff and 6, 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 49 divided by 9 is between 5 and 6, so the $\square$ must be more than 5, giving 6, 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 9 x $\square$ greater than 49.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 7 x 7 and the 9-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 4 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$5\times8<9\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 9 times the $\square$ is larger than 5 times 8.

Givens

The inequality is 5 x 8 < 9 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 9 x $\square$ greater than 40.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 9 x $\square$ first passes 40; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 5 times 8 so the inequality becomes a comparison with a single number.

5 \times 8 = 40

5 times 8 is a basic multiplication fact, so the left side is just the number 40.

2Find where 9 times the $\square$ passes 40

#6 Guess and Check 3.OA.A.4

Test multiples of 9 in order: 9x4 = 36 is not greater than 40, but 9x5 = 45 is greater than 40. So the

\square

must be 5 or larger.

9 \times 4 = 36,\ 9 \times 5 = 45

Checking the 9-times facts shows exactly when the right side becomes bigger than 40.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 5 and above (and at most 9) makes the right side larger than 40, so collect them.

\square \in \{5, 6, 7, 8, 9\}

Since bigger

\square

values only make 9 x

\square

larger, all of 5, 6, 7, 8, 9 keep the inequality true.

Answer: 5, 6, 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 9x5 = 45 > 40, true; and the value just below, 9x4 = 36, is not greater than 40, so 5 is correctly the cutoff and 5, 6, 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 40 divided by 9 is between 4 and 5, so the $\square$ must be more than 4, giving 5, 6, 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 9 x $\square$ greater than 40.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 5 x 8 and the 9-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 5 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$4\times9<5\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 5 times the $\square$ is larger than 4 times 9.

Givens

The inequality is 4 x 9 < 5 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 5 x $\square$ greater than 36.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 5 x $\square$ first passes 36; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 4 times 9 so the inequality becomes a comparison with a single number.

4 \times 9 = 36

4 times 9 is a basic multiplication fact, so the left side is just the number 36.

2Find where 5 times the $\square$ passes 36

#6 Guess and Check 3.OA.A.4

Test multiples of 5 in order: 5x7 = 35 is not greater than 36, but 5x8 = 40 is greater than 36. So the

\square

must be 8 or larger.

5 \times 7 = 35,\ 5 \times 8 = 40

Checking the 5-times facts shows exactly when the right side becomes bigger than 36.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 8 and above (and at most 9) makes the right side larger than 36, so collect them.

\square \in \{8, 9\}

Since bigger

\square

values only make 5 x

\square

larger, all of 8, 9 keep the inequality true.

Answer: 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 5x8 = 40 > 36, true; and the value just below, 5x7 = 35, is not greater than 36, so 8 is correctly the cutoff and 8, 9 all work.

Another way: You could find where the two sides are equal: 36 divided by 5 is between 7 and 8, so the $\square$ must be more than 7, giving 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 5 x $\square$ greater than 36.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 9 and the 5-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 6 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$5\times9<7\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 7 times the $\square$ is larger than 5 times 9.

Givens

The inequality is 5 x 9 < 7 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 7 x $\square$ greater than 45.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 7 x $\square$ first passes 45; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 5 times 9 so the inequality becomes a comparison with a single number.

5 \times 9 = 45

5 times 9 is a basic multiplication fact, so the left side is just the number 45.

2Find where 7 times the $\square$ passes 45

#6 Guess and Check 3.OA.A.4

Test multiples of 7 in order: 7x6 = 42 is not greater than 45, but 7x7 = 49 is greater than 45. So the

\square

must be 7 or larger.

7 \times 6 = 42,\ 7 \times 7 = 49

Checking the 7-times facts shows exactly when the right side becomes bigger than 45.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 7 and above (and at most 9) makes the right side larger than 45, so collect them.

\square \in \{7, 8, 9\}

Since bigger

\square

values only make 7 x

\square

larger, all of 7, 8, 9 keep the inequality true.

Answer: 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 7x7 = 49 > 45, true; and the value just below, 7x6 = 42, is not greater than 45, so 7 is correctly the cutoff and 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 45 divided by 7 is between 6 and 7, so the $\square$ must be more than 6, giving 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 7 x $\square$ greater than 45.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 5 x 9 and the 7-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 7 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$4\times7<8\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 8 times the $\square$ is larger than 4 times 7.

Givens

The inequality is 4 x 7 < 8 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 8 x $\square$ greater than 28.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 8 x $\square$ first passes 28; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 4 times 7 so the inequality becomes a comparison with a single number.

4 \times 7 = 28

4 times 7 is a basic multiplication fact, so the left side is just the number 28.

2Find where 8 times the $\square$ passes 28

#6 Guess and Check 3.OA.A.4

Test multiples of 8 in order: 8x3 = 24 is not greater than 28, but 8x4 = 32 is greater than 28. So the

\square

must be 4 or larger.

8 \times 3 = 24,\ 8 \times 4 = 32

Checking the 8-times facts shows exactly when the right side becomes bigger than 28.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 4 and above (and at most 9) makes the right side larger than 28, so collect them.

\square \in \{4, 5, 6, 7, 8, 9\}

Since bigger

\square

values only make 8 x

\square

larger, all of 4, 5, 6, 7, 8, 9 keep the inequality true.

Answer: 4, 5, 6, 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 8x4 = 32 > 28, true; and the value just below, 8x3 = 24, is not greater than 28, so 4 is correctly the cutoff and 4, 5, 6, 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 28 divided by 8 is between 3 and 4, so the $\square$ must be more than 3, giving 4, 5, 6, 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 8 x $\square$ greater than 28.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 7 and the 8-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 8 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$3\times8<6\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 6 times the $\square$ is larger than 3 times 8.

Givens

The inequality is 3 x 8 < 6 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 6 x $\square$ greater than 24.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 6 x $\square$ first passes 24; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 3 times 8 so the inequality becomes a comparison with a single number.

3 \times 8 = 24

3 times 8 is a basic multiplication fact, so the left side is just the number 24.

2Find where 6 times the $\square$ passes 24

#6 Guess and Check 3.OA.A.4

Test multiples of 6 in order: 6x4 = 24 is not greater than 24, but 6x5 = 30 is greater than 24. So the

\square

must be 5 or larger.

6 \times 4 = 24,\ 6 \times 5 = 30

Checking the 6-times facts shows exactly when the right side becomes bigger than 24.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 5 and above (and at most 9) makes the right side larger than 24, so collect them.

\square \in \{5, 6, 7, 8, 9\}

Since bigger

\square

values only make 6 x

\square

larger, all of 5, 6, 7, 8, 9 keep the inequality true.

Answer: 5, 6, 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 6x5 = 30 > 24, true; and the value just below, 6x4 = 24, is not greater than 24, so 5 is correctly the cutoff and 5, 6, 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 24 divided by 6 is between 4 and 5, so the $\square$ must be more than 4, giving 5, 6, 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 6 x $\square$ greater than 24.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 3 x 8 and the 6-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 9 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$7\times8<9\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 9 times the $\square$ is larger than 7 times 8.

Givens

The inequality is 7 x 8 < 9 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 9 x $\square$ greater than 56.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 9 x $\square$ first passes 56; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 7 times 8 so the inequality becomes a comparison with a single number.

7 \times 8 = 56

7 times 8 is a basic multiplication fact, so the left side is just the number 56.

2Find where 9 times the $\square$ passes 56

#6 Guess and Check 3.OA.A.4

Test multiples of 9 in order: 9x6 = 54 is not greater than 56, but 9x7 = 63 is greater than 56. So the

\square

must be 7 or larger.

9 \times 6 = 54,\ 9 \times 7 = 63

Checking the 9-times facts shows exactly when the right side becomes bigger than 56.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 7 and above (and at most 9) makes the right side larger than 56, so collect them.

\square \in \{7, 8, 9\}

Since bigger

\square

values only make 9 x

\square

larger, all of 7, 8, 9 keep the inequality true.

Answer: 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 9x7 = 63 > 56, true; and the value just below, 9x6 = 54, is not greater than 56, so 7 is correctly the cutoff and 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 56 divided by 9 is between 6 and 7, so the $\square$ must be more than 6, giving 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 9 x $\square$ greater than 56.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 7 x 8 and the 9-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 10 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$6\times6<8\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 8 times the $\square$ is larger than 6 times 6.

Givens

The inequality is 6 x 6 < 8 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 8 x $\square$ greater than 36.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 8 x $\square$ first passes 36; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 6 times 6 so the inequality becomes a comparison with a single number.

6 \times 6 = 36

6 times 6 is a basic multiplication fact, so the left side is just the number 36.

2Find where 8 times the $\square$ passes 36

#6 Guess and Check 3.OA.A.4

Test multiples of 8 in order: 8x4 = 32 is not greater than 36, but 8x5 = 40 is greater than 36. So the

\square

must be 5 or larger.

8 \times 4 = 32,\ 8 \times 5 = 40

Checking the 8-times facts shows exactly when the right side becomes bigger than 36.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 5 and above (and at most 9) makes the right side larger than 36, so collect them.

\square \in \{5, 6, 7, 8, 9\}

Since bigger

\square

values only make 8 x

\square

larger, all of 5, 6, 7, 8, 9 keep the inequality true.

Answer: 5, 6, 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 8x5 = 40 > 36, true; and the value just below, 8x4 = 32, is not greater than 36, so 5 is correctly the cutoff and 5, 6, 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 36 divided by 8 is between 4 and 5, so the $\square$ must be more than 4, giving 5, 6, 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 8 x $\square$ greater than 36.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 6 x 6 and the 8-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 11 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$8\times4<7\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 7 times the $\square$ is larger than 8 times 4.

Givens

The inequality is 8 x 4 < 7 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 7 x $\square$ greater than 32.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 7 x $\square$ first passes 32; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 8 times 4 so the inequality becomes a comparison with a single number.

8 \times 4 = 32

8 times 4 is a basic multiplication fact, so the left side is just the number 32.

2Find where 7 times the $\square$ passes 32

#6 Guess and Check 3.OA.A.4

Test multiples of 7 in order: 7x4 = 28 is not greater than 32, but 7x5 = 35 is greater than 32. So the

\square

must be 5 or larger.

7 \times 4 = 28,\ 7 \times 5 = 35

Checking the 7-times facts shows exactly when the right side becomes bigger than 32.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 5 and above (and at most 9) makes the right side larger than 32, so collect them.

\square \in \{5, 6, 7, 8, 9\}

Since bigger

\square

values only make 7 x

\square

larger, all of 5, 6, 7, 8, 9 keep the inequality true.

Answer: 5, 6, 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 7x5 = 35 > 32, true; and the value just below, 7x4 = 28, is not greater than 32, so 5 is correctly the cutoff and 5, 6, 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 32 divided by 7 is between 4 and 5, so the $\square$ must be more than 4, giving 5, 6, 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 7 x $\square$ greater than 32.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 8 x 4 and the 7-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Variant 12 medium

Find every number from $1$ to $9$ that can go in the $\square$ to make the statement true.

$6\times8<9\times\square$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We need every digit from 1 to 9 that can fill the $\square$ so that 9 times the $\square$ is larger than 6 times 8.

Givens

The inequality is 6 x 8 < 9 x $\square$ .
The $\square$ must be a whole number from 1 to 9.

Unknowns

All values from 1 to 9 that make 9 x $\square$ greater than 48.

Constraints

The $\square$ is a single whole number between 1 and 9.
The relationship must be strictly greater than (not equal).

2 · Planchoose the strategy

#6 Guess and Check · also uses: #2 Make a Systematic List

First compute the fixed left side, then test the $\square$ values in order from small to large to find where 9 x $\square$ first passes 48; listing the candidates keeps every value from 1 to 9 accounted for.

3 · Execute3 carry out the plan

1Compute the left side

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

Work out 6 times 8 so the inequality becomes a comparison with a single number.

6 \times 8 = 48

6 times 8 is a basic multiplication fact, so the left side is just the number 48.

2Find where 9 times the $\square$ passes 48

#6 Guess and Check 3.OA.A.4

Test multiples of 9 in order: 9x5 = 45 is not greater than 48, but 9x6 = 54 is greater than 48. So the

\square

must be 6 or larger.

9 \times 5 = 45,\ 9 \times 6 = 54

Checking the 9-times facts shows exactly when the right side becomes bigger than 48.

3List all $\square$ values that work

#2 Make a Systematic List 3.OA.A.4

Every

\square

value 6 and above (and at most 9) makes the right side larger than 48, so collect them.

\square \in \{6, 7, 8, 9\}

Since bigger

\square

values only make 9 x

\square

larger, all of 6, 7, 8, 9 keep the inequality true.

Answer: 6, 7, 8, 9

4 · Reviewdoes it hold up?

Checking the smallest answer: 9x6 = 54 > 48, true; and the value just below, 9x5 = 45, is not greater than 48, so 6 is correctly the cutoff and 6, 7, 8, 9 all work.

Another way: You could find where the two sides are equal: 48 divided by 9 is between 5 and 6, so the $\square$ must be more than 5, giving 6, 7, 8, 9.

Standardsmin grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding which $\square$ values keep 9 x $\square$ greater than 48.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 6 x 8 and the 9-times facts used in the comparison.

💡Takeaway. Turn one side into a number, then test the times-table facts -- this is just Grade 3 multiplication you already know!

Bound an inequality at equality · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?