Practice · Compare decimals from the highest place down · Sensim Math

Variant 1 easy

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$4.6 < 4.\boxed{A}$

$4.6 > \boxed{B}.6$

Here A is the tenths digit of the decimal $4.\square$ , and B is the ones digit of the decimal $\square.6$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 4.6 < 4.A and blank B in 4.6 > B.6 so both inequalities hold. A is the tenths digit of 4.A; B is the ones digit of B.6. I want the smallest A that works and the largest B that works.

Givens

4.6 < 4.A, where A is a tenths digit (a single digit 1-9).
4.6 > B.6, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 4), so the tenths digit decides; for B the tenths places are equal (both 6), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 4.6 < 4.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 4, so the ones place is tied. To decide, compare the tenths: we need A > 6. Among digits 1-9 the values greater than 6 are 7, 8, 9.

4.6 < 4.A \Rightarrow A > 6

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 7, 8, 9. The smallest of these is 7.

A_{\min} = 7

Just above 6 is 7, so 7 is the smallest tenths digit that still beats 4.6.

3Solve 4.6 > B.6 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 6, so the tenths place is tied. To decide, compare the ones: we need 4 > B, that is B < 4. Among digits 1-9 the values less than 4 are 1, 2, 3.

4.6 > B.6 \Rightarrow B < 4

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1, 2, 3. The largest of these is 3.

B_{\max} = 3

Just below 4 is 3, so 3 is the largest ones digit that keeps B.6 under 4.6.

Answer: A = 7, B = 3

4 · Reviewdoes it hold up?

Check A = 7: 4.6 < 4.7 is true. Check B = 3: 4.6 > 3.6 is true. Both inequalities hold, and trying A = 6 (4.6 < 4.6 false) or B = 4 (4.6 > 4.6 false) shows 7 and 3 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 7 and the largest passing B is 3.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 4.6 with 4.A and 4.6 with B.6 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Variant 2 hard

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$3.8 < 3.\boxed{A}$

$3.7 > \boxed{B}.7$

Here A is the tenths digit of the decimal $3.\square$ , and B is the ones digit of the decimal $\square.7$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 3.8 < 3.A and blank B in 3.7 > B.7 so both inequalities hold. A is the tenths digit of 3.A; B is the ones digit of B.7. I want the smallest A that works and the largest B that works.

Givens

3.8 < 3.A, where A is a tenths digit (a single digit 1-9).
3.7 > B.7, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 3), so the tenths digit decides; for B the tenths places are equal (both 7), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 3.8 < 3.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 3, so the ones place is tied. To decide, compare the tenths: we need A > 8. Among digits 1-9 the values greater than 8 are 9.

3.8 < 3.A \Rightarrow A > 8

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 9. The smallest of these is 9.

A_{\min} = 9

Just above 8 is 9, so 9 is the smallest tenths digit that still beats 3.8.

3Solve 3.7 > B.7 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 7, so the tenths place is tied. To decide, compare the ones: we need 3 > B, that is B < 3. Among digits 1-9 the values less than 3 are 1, 2.

3.7 > B.7 \Rightarrow B < 3

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1, 2. The largest of these is 2.

B_{\max} = 2

Just below 3 is 2, so 2 is the largest ones digit that keeps B.7 under 3.7.

Answer: A = 9, B = 2

4 · Reviewdoes it hold up?

Check A = 9: 3.8 < 3.9 is true. Check B = 2: 3.7 > 2.7 is true. Both inequalities hold, and trying A = 8 (3.8 < 3.8 false) or B = 3 (3.7 > 3.7 false) shows 9 and 2 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 9 and the largest passing B is 2.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 3.8 with 3.A and 3.7 with B.7 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Variant 3 medium

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$2.4 < 2.\boxed{A}$

$2.9 > \boxed{B}.9$

Here A is the tenths digit of the decimal $2.\square$ , and B is the ones digit of the decimal $\square.9$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 2.4 < 2.A and blank B in 2.9 > B.9 so both inequalities hold. A is the tenths digit of 2.A; B is the ones digit of B.9. I want the smallest A that works and the largest B that works.

Givens

2.4 < 2.A, where A is a tenths digit (a single digit 1-9).
2.9 > B.9, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 2), so the tenths digit decides; for B the tenths places are equal (both 9), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 2.4 < 2.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 2, so the ones place is tied. To decide, compare the tenths: we need A > 4. Among digits 1-9 the values greater than 4 are 5, 6, 7, 8, 9.

2.4 < 2.A \Rightarrow A > 4

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 5, 6, 7, 8, 9. The smallest of these is 5.

A_{\min} = 5

Just above 4 is 5, so 5 is the smallest tenths digit that still beats 2.4.

3Solve 2.9 > B.9 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 9, so the tenths place is tied. To decide, compare the ones: we need 2 > B, that is B < 2. Among digits 1-9 the values less than 2 are 1.

2.9 > B.9 \Rightarrow B < 2

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1. The largest of these is 1.

B_{\max} = 1

Just below 2 is 1, so 1 is the largest ones digit that keeps B.9 under 2.9.

Answer: A = 5, B = 1

4 · Reviewdoes it hold up?

Check A = 5: 2.4 < 2.5 is true. Check B = 1: 2.9 > 1.9 is true. Both inequalities hold, and trying A = 4 (2.4 < 2.4 false) or B = 2 (2.9 > 2.9 false) shows 5 and 1 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 5 and the largest passing B is 1.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 2.4 with 2.A and 2.9 with B.9 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Variant 4 medium

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$8.7 < 8.\boxed{A}$

$8.3 > \boxed{B}.3$

Here A is the tenths digit of the decimal $8.\square$ , and B is the ones digit of the decimal $\square.3$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 8.7 < 8.A and blank B in 8.3 > B.3 so both inequalities hold. A is the tenths digit of 8.A; B is the ones digit of B.3. I want the smallest A that works and the largest B that works.

Givens

8.7 < 8.A, where A is a tenths digit (a single digit 1-9).
8.3 > B.3, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 8), so the tenths digit decides; for B the tenths places are equal (both 3), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 8.7 < 8.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 8, so the ones place is tied. To decide, compare the tenths: we need A > 7. Among digits 1-9 the values greater than 7 are 8, 9.

8.7 < 8.A \Rightarrow A > 7

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 8, 9. The smallest of these is 8.

A_{\min} = 8

Just above 7 is 8, so 8 is the smallest tenths digit that still beats 8.7.

3Solve 8.3 > B.3 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 3, so the tenths place is tied. To decide, compare the ones: we need 8 > B, that is B < 8. Among digits 1-9 the values less than 8 are 1, 2, 3, 4, 5, 6, 7.

8.3 > B.3 \Rightarrow B < 8

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1, 2, 3, 4, 5, 6, 7. The largest of these is 7.

B_{\max} = 7

Just below 8 is 7, so 7 is the largest ones digit that keeps B.3 under 8.3.

Answer: A = 8, B = 7

4 · Reviewdoes it hold up?

Check A = 8: 8.7 < 8.8 is true. Check B = 7: 8.3 > 7.3 is true. Both inequalities hold, and trying A = 7 (8.7 < 8.7 false) or B = 8 (8.3 > 8.3 false) shows 8 and 7 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 8 and the largest passing B is 7.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 8.7 with 8.A and 8.3 with B.3 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Variant 5 medium

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$7.8 < 7.\boxed{A}$

$7.1 > \boxed{B}.1$

Here A is the tenths digit of the decimal $7.\square$ , and B is the ones digit of the decimal $\square.1$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 7.8 < 7.A and blank B in 7.1 > B.1 so both inequalities hold. A is the tenths digit of 7.A; B is the ones digit of B.1. I want the smallest A that works and the largest B that works.

Givens

7.8 < 7.A, where A is a tenths digit (a single digit 1-9).
7.1 > B.1, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 7), so the tenths digit decides; for B the tenths places are equal (both 1), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 7.8 < 7.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 7, so the ones place is tied. To decide, compare the tenths: we need A > 8. Among digits 1-9 the values greater than 8 are 9.

7.8 < 7.A \Rightarrow A > 8

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 9. The smallest of these is 9.

A_{\min} = 9

Just above 8 is 9, so 9 is the smallest tenths digit that still beats 7.8.

3Solve 7.1 > B.1 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 1, so the tenths place is tied. To decide, compare the ones: we need 7 > B, that is B < 7. Among digits 1-9 the values less than 7 are 1, 2, 3, 4, 5, 6.

7.1 > B.1 \Rightarrow B < 7

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1, 2, 3, 4, 5, 6. The largest of these is 6.

B_{\max} = 6

Just below 7 is 6, so 6 is the largest ones digit that keeps B.1 under 7.1.

Answer: A = 9, B = 6

4 · Reviewdoes it hold up?

Check A = 9: 7.8 < 7.9 is true. Check B = 6: 7.1 > 6.1 is true. Both inequalities hold, and trying A = 8 (7.8 < 7.8 false) or B = 7 (7.1 > 7.1 false) shows 9 and 6 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 9 and the largest passing B is 6.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 7.8 with 7.A and 7.1 with B.1 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Variant 6 medium

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$6.1 < 6.\boxed{A}$

$6.5 > \boxed{B}.5$

Here A is the tenths digit of the decimal $6.\square$ , and B is the ones digit of the decimal $\square.5$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 6.1 < 6.A and blank B in 6.5 > B.5 so both inequalities hold. A is the tenths digit of 6.A; B is the ones digit of B.5. I want the smallest A that works and the largest B that works.

Givens

6.1 < 6.A, where A is a tenths digit (a single digit 1-9).
6.5 > B.5, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 6), so the tenths digit decides; for B the tenths places are equal (both 5), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 6.1 < 6.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 6, so the ones place is tied. To decide, compare the tenths: we need A > 1. Among digits 1-9 the values greater than 1 are 2, 3, 4, 5, 6, 7, 8, 9.

6.1 < 6.A \Rightarrow A > 1

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 2, 3, 4, 5, 6, 7, 8, 9. The smallest of these is 2.

A_{\min} = 2

Just above 1 is 2, so 2 is the smallest tenths digit that still beats 6.1.

3Solve 6.5 > B.5 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 5, so the tenths place is tied. To decide, compare the ones: we need 6 > B, that is B < 6. Among digits 1-9 the values less than 6 are 1, 2, 3, 4, 5.

6.5 > B.5 \Rightarrow B < 6

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1, 2, 3, 4, 5. The largest of these is 5.

B_{\max} = 5

Just below 6 is 5, so 5 is the largest ones digit that keeps B.5 under 6.5.

Answer: A = 2, B = 5

4 · Reviewdoes it hold up?

Check A = 2: 6.1 < 6.2 is true. Check B = 5: 6.5 > 5.5 is true. Both inequalities hold, and trying A = 1 (6.1 < 6.1 false) or B = 6 (6.5 > 6.5 false) shows 2 and 5 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 2 and the largest passing B is 5.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 6.1 with 6.A and 6.5 with B.5 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Variant 7 easy

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$5.2 < 5.\boxed{A}$

$5.8 > \boxed{B}.8$

Here A is the tenths digit of the decimal $5.\square$ , and B is the ones digit of the decimal $\square.8$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 5.2 < 5.A and blank B in 5.8 > B.8 so both inequalities hold. A is the tenths digit of 5.A; B is the ones digit of B.8. I want the smallest A that works and the largest B that works.

Givens

5.2 < 5.A, where A is a tenths digit (a single digit 1-9).
5.8 > B.8, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 5), so the tenths digit decides; for B the tenths places are equal (both 8), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 5.2 < 5.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 5, so the ones place is tied. To decide, compare the tenths: we need A > 2. Among digits 1-9 the values greater than 2 are 3, 4, 5, 6, 7, 8, 9.

5.2 < 5.A \Rightarrow A > 2

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 3, 4, 5, 6, 7, 8, 9. The smallest of these is 3.

A_{\min} = 3

Just above 2 is 3, so 3 is the smallest tenths digit that still beats 5.2.

3Solve 5.8 > B.8 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 8, so the tenths place is tied. To decide, compare the ones: we need 5 > B, that is B < 5. Among digits 1-9 the values less than 5 are 1, 2, 3, 4.

5.8 > B.8 \Rightarrow B < 5

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1, 2, 3, 4. The largest of these is 4.

B_{\max} = 4

Just below 5 is 4, so 4 is the largest ones digit that keeps B.8 under 5.8.

Answer: A = 3, B = 4

4 · Reviewdoes it hold up?

Check A = 3: 5.2 < 5.3 is true. Check B = 4: 5.8 > 4.8 is true. Both inequalities hold, and trying A = 2 (5.2 < 5.2 false) or B = 5 (5.8 > 5.8 false) shows 3 and 4 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 3 and the largest passing B is 4.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 5.2 with 5.A and 5.8 with B.8 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Variant 8 easy

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$3.5 < 3.\boxed{A}$

$3.2 > \boxed{B}.2$

Here A is the tenths digit of the decimal $3.\square$ , and B is the ones digit of the decimal $\square.2$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 3.5 < 3.A and blank B in 3.2 > B.2 so both inequalities hold. A is the tenths digit of 3.A; B is the ones digit of B.2. I want the smallest A that works and the largest B that works.

Givens

3.5 < 3.A, where A is a tenths digit (a single digit 1-9).
3.2 > B.2, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 3), so the tenths digit decides; for B the tenths places are equal (both 2), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 3.5 < 3.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 3, so the ones place is tied. To decide, compare the tenths: we need A > 5. Among digits 1-9 the values greater than 5 are 6, 7, 8, 9.

3.5 < 3.A \Rightarrow A > 5

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 6, 7, 8, 9. The smallest of these is 6.

A_{\min} = 6

Just above 5 is 6, so 6 is the smallest tenths digit that still beats 3.5.

3Solve 3.2 > B.2 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 2, so the tenths place is tied. To decide, compare the ones: we need 3 > B, that is B < 3. Among digits 1-9 the values less than 3 are 1, 2.

3.2 > B.2 \Rightarrow B < 3

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1, 2. The largest of these is 2.

B_{\max} = 2

Just below 3 is 2, so 2 is the largest ones digit that keeps B.2 under 3.2.

Answer: A = 6, B = 2

4 · Reviewdoes it hold up?

Check A = 6: 3.5 < 3.6 is true. Check B = 2: 3.2 > 2.2 is true. Both inequalities hold, and trying A = 5 (3.5 < 3.5 false) or B = 3 (3.2 > 3.2 false) shows 6 and 2 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 6 and the largest passing B is 2.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 3.5 with 3.A and 3.2 with B.2 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Variant 9 hard

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$5.5 < 5.\boxed{A}$

$5.6 > \boxed{B}.6$

Here A is the tenths digit of the decimal $5.\square$ , and B is the ones digit of the decimal $\square.6$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 5.5 < 5.A and blank B in 5.6 > B.6 so both inequalities hold. A is the tenths digit of 5.A; B is the ones digit of B.6. I want the smallest A that works and the largest B that works.

Givens

5.5 < 5.A, where A is a tenths digit (a single digit 1-9).
5.6 > B.6, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 5), so the tenths digit decides; for B the tenths places are equal (both 6), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 5.5 < 5.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 5, so the ones place is tied. To decide, compare the tenths: we need A > 5. Among digits 1-9 the values greater than 5 are 6, 7, 8, 9.

5.5 < 5.A \Rightarrow A > 5

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 6, 7, 8, 9. The smallest of these is 6.

A_{\min} = 6

Just above 5 is 6, so 6 is the smallest tenths digit that still beats 5.5.

3Solve 5.6 > B.6 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 6, so the tenths place is tied. To decide, compare the ones: we need 5 > B, that is B < 5. Among digits 1-9 the values less than 5 are 1, 2, 3, 4.

5.6 > B.6 \Rightarrow B < 5

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1, 2, 3, 4. The largest of these is 4.

B_{\max} = 4

Just below 5 is 4, so 4 is the largest ones digit that keeps B.6 under 5.6.

Answer: A = 6, B = 4

4 · Reviewdoes it hold up?

Check A = 6: 5.5 < 5.6 is true. Check B = 4: 5.6 > 4.6 is true. Both inequalities hold, and trying A = 5 (5.5 < 5.5 false) or B = 5 (5.6 > 5.6 false) shows 6 and 4 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 6 and the largest passing B is 4.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 5.5 with 5.A and 5.6 with B.6 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Variant 10 hard

Using the digits 1 through 9, fill in the blanks A and B so that both inequalities below are true. Find the smallest possible value of A and the largest possible value of B.

$9.3 < 9.\boxed{A}$

$9.4 > \boxed{B}.4$

Here A is the tenths digit of the decimal $9.\square$ , and B is the ones digit of the decimal $\square.4$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Using single digits 1 through 9, I must fill blank A in 9.3 < 9.A and blank B in 9.4 > B.4 so both inequalities hold. A is the tenths digit of 9.A; B is the ones digit of B.4. I want the smallest A that works and the largest B that works.

Givens

9.3 < 9.A, where A is a tenths digit (a single digit 1-9).
9.4 > B.4, where B is a ones digit (a single digit 1-9).

Unknowns

The smallest possible value of A.
The largest possible value of B.

Constraints

A and B are each one of the digits 1, 2, ..., 9.
Compare decimals by their highest place first.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #5 Look for a Pattern

Each blank has only the 9 digits to test, so I can reason place-by-place. For A the ones places are equal (both 9), so the tenths digit decides; for B the tenths places are equal (both 4), so the ones digit decides. Testing the boundary digit confirms the smallest A and largest B.

3 · Execute4 carry out the plan

1Solve 9.3 < 9.A by the tenths place

#6 Guess and Check 4.NF.C.7

Both numbers have ones digit 9, so the ones place is tied. To decide, compare the tenths: we need A > 3. Among digits 1-9 the values greater than 3 are 4, 5, 6, 7, 8, 9.

9.3 < 9.A \Rightarrow A > 3

When the whole-number parts match, the bigger decimal is the one with the bigger tenths digit.

2Pick the smallest A

#5 Look for a Pattern 4.NF.C.7

The valid digits for A are 4, 5, 6, 7, 8, 9. The smallest of these is 4.

A_{\min} = 4

Just above 3 is 4, so 4 is the smallest tenths digit that still beats 9.3.

3Solve 9.4 > B.4 by the ones place

#6 Guess and Check 4.NF.C.7

Both numbers have the same tenths digit 4, so the tenths place is tied. To decide, compare the ones: we need 9 > B, that is B < 9. Among digits 1-9 the values less than 9 are 1, 2, 3, 4, 5, 6, 7, 8.

9.4 > B.4 \Rightarrow B < 9

With equal tenths, the smaller number is the one with the smaller ones digit.

4Pick the largest B

#5 Look for a Pattern 4.NF.C.7

The valid digits for B are 1, 2, 3, 4, 5, 6, 7, 8. The largest of these is 8.

B_{\max} = 8

Just below 9 is 8, so 8 is the largest ones digit that keeps B.4 under 9.4.

Answer: A = 4, B = 8

4 · Reviewdoes it hold up?

Check A = 4: 9.3 < 9.4 is true. Check B = 8: 9.4 > 8.4 is true. Both inequalities hold, and trying A = 3 (9.3 < 9.3 false) or B = 9 (9.4 > 9.4 false) shows 4 and 8 are the boundary digits.

Another way: Make a systematic list (tool 2): for A list the outcomes of each tenths digit and for B list each ones digit, marking which satisfy the inequality; the smallest passing A is 4 and the largest passing B is 8.

Standardsmin grade 4

4.NF.C.7 Compare two decimals to hundredths by reasoning about their size — Comparing 9.3 with 9.A and 9.4 with B.4 place by place to bound A and B.

💡Takeaway. Compare decimals from the highest place down: same ones? look at tenths; same tenths? look at ones - pure Grade 4 place value!

Compare decimals from the highest place down · 10 practice problems

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