← Match denominators or numerators to compare · Compare Fractions and Decimals by Structure

Match denominators or numerators to compare · 12 practice problems

4.NF.A.13.NF.A.34.NF.A.2

0 / 12 solved

Show all answers (parent / teacher)

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{1}{6} < \dfrac{3}{\square} < \dfrac{3}{5}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 3 and a hidden denominator must sit strictly between 1 over 6 and 3 over 5. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{3}{\square}$ .
It must be greater than $\frac{1}{6}$ .
It must be less than $\frac{3}{5}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 3; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 3

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 3. The value does not change -- only the way it is written.

\frac{1}{6} = \frac{3}{18},\quad \frac{3}{5} = \frac{3}{5}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

5 < \square < 18

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below 18 is 17.

\frac{1}{6} < \frac{3}{17} < \frac{3}{5}

The answer is 17.

Answer: 17

4 · Reviewdoes it hold up?

Test the next one up: with 18 in the box the middle fraction would be at or below $\frac{1}{6}$ , so 17 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 3 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 3.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 2 easy

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{1}{3} < \dfrac{6}{\square} < \dfrac{3}{4}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 6 and a hidden denominator must sit strictly between 1 over 3 and 3 over 4. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{6}{\square}$ .
It must be greater than $\frac{1}{3}$ .
It must be less than $\frac{3}{4}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 6; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 6

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 6. The value does not change -- only the way it is written.

\frac{1}{3} = \frac{6}{18},\quad \frac{3}{4} = \frac{6}{8}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

8 < \square < 18

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below 18 is 17.

\frac{1}{3} < \frac{6}{17} < \frac{3}{4}

The answer is 17.

Answer: 17

4 · Reviewdoes it hold up?

Test the next one up: with 18 in the box the middle fraction would be at or below $\frac{1}{3}$ , so 17 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 6 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 6.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 3 easy

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{1}{4} < \dfrac{5}{\square} < \dfrac{5}{6}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 5 and a hidden denominator must sit strictly between 1 over 4 and 5 over 6. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{5}{\square}$ .
It must be greater than $\frac{1}{4}$ .
It must be less than $\frac{5}{6}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 5; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 5

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 5. The value does not change -- only the way it is written.

\frac{1}{4} = \frac{5}{20},\quad \frac{5}{6} = \frac{5}{6}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

6 < \square < 20

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below 20 is 19.

\frac{1}{4} < \frac{5}{19} < \frac{5}{6}

The answer is 19.

Answer: 19

4 · Reviewdoes it hold up?

Test the next one up: with 20 in the box the middle fraction would be at or below $\frac{1}{4}$ , so 19 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 5 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 5.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 4 easy

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{2}{7} < \dfrac{8}{\square} < \dfrac{4}{5}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 8 and a hidden denominator must sit strictly between 2 over 7 and 4 over 5. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{8}{\square}$ .
It must be greater than $\frac{2}{7}$ .
It must be less than $\frac{4}{5}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 8; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 8

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 8. The value does not change -- only the way it is written.

\frac{2}{7} = \frac{8}{28},\quad \frac{4}{5} = \frac{8}{10}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

10 < \square < 28

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below 28 is 27.

\frac{2}{7} < \frac{8}{27} < \frac{4}{5}

The answer is 27.

Answer: 27

4 · Reviewdoes it hold up?

Test the next one up: with 28 in the box the middle fraction would be at or below $\frac{2}{7}$ , so 27 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 8 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 8.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 5 medium

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{2}{5} < \dfrac{4}{\square} < \dfrac{8}{9}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 4 and a hidden denominator must sit strictly between 2 over 5 and 8 over 9. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{4}{\square}$ .
It must be greater than $\frac{2}{5}$ .
It must be less than $\frac{8}{9}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 4; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 4

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 4. The value does not change -- only the way it is written.

\frac{2}{5} = \frac{4}{10},\quad \frac{8}{9} = \frac{4}{\frac{9}{2}}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

\frac{9}{2} < \square < 10

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below 10 is 9.

\frac{2}{5} < \frac{4}{9} < \frac{8}{9}

The answer is 9.

Answer: 9

4 · Reviewdoes it hold up?

Test the next one up: with 10 in the box the middle fraction would be at or below $\frac{2}{5}$ , so 9 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 4 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 4.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 6 medium

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{2}{9} < \dfrac{7}{\square} < \dfrac{7}{8}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 7 and a hidden denominator must sit strictly between 2 over 9 and 7 over 8. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{7}{\square}$ .
It must be greater than $\frac{2}{9}$ .
It must be less than $\frac{7}{8}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 7; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 7

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 7. The value does not change -- only the way it is written.

\frac{2}{9} = \frac{7}{\frac{63}{2}},\quad \frac{7}{8} = \frac{7}{8}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

8 < \square < \frac{63}{2}

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below

\frac{63}{2}

is 31.

\frac{2}{9} < \frac{7}{31} < \frac{7}{8}

The answer is 31.

Answer: 31

4 · Reviewdoes it hold up?

Test the next one up: with 32 in the box the middle fraction would be at or below $\frac{2}{9}$ , so 31 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 7 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 7.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 7 medium

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{1}{2} < \dfrac{9}{\square} < \dfrac{9}{10}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 9 and a hidden denominator must sit strictly between 1 over 2 and 9 over 10. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{9}{\square}$ .
It must be greater than $\frac{1}{2}$ .
It must be less than $\frac{9}{10}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 9; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 9

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 9. The value does not change -- only the way it is written.

\frac{1}{2} = \frac{9}{18},\quad \frac{9}{10} = \frac{9}{10}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

10 < \square < 18

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below 18 is 17.

\frac{1}{2} < \frac{9}{17} < \frac{9}{10}

The answer is 17.

Answer: 17

4 · Reviewdoes it hold up?

Test the next one up: with 18 in the box the middle fraction would be at or below $\frac{1}{2}$ , so 17 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 9 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 9.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 8 medium

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{3}{8} < \dfrac{10}{\square} < \dfrac{5}{6}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 10 and a hidden denominator must sit strictly between 3 over 8 and 5 over 6. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{10}{\square}$ .
It must be greater than $\frac{3}{8}$ .
It must be less than $\frac{5}{6}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 10; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 10

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 10. The value does not change -- only the way it is written.

\frac{3}{8} = \frac{10}{\frac{80}{3}},\quad \frac{5}{6} = \frac{10}{12}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

12 < \square < \frac{80}{3}

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below

\frac{80}{3}

is 26.

\frac{3}{8} < \frac{10}{26} < \frac{5}{6}

The answer is 26.

Answer: 26

4 · Reviewdoes it hold up?

Test the next one up: with 27 in the box the middle fraction would be at or below $\frac{3}{8}$ , so 26 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 10 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 10.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 9 hard

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{3}{10} < \dfrac{6}{\square} < \dfrac{2}{3}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 6 and a hidden denominator must sit strictly between 3 over 10 and 2 over 3. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{6}{\square}$ .
It must be greater than $\frac{3}{10}$ .
It must be less than $\frac{2}{3}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 6

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 6. The value does not change -- only the way it is written.

\frac{3}{10} = \frac{6}{20},\quad \frac{2}{3} = \frac{6}{9}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

9 < \square < 20

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below 20 is 19.

\frac{3}{10} < \frac{6}{19} < \frac{2}{3}

The answer is 19.

Answer: 19

4 · Reviewdoes it hold up?

Test the next one up: with 20 in the box the middle fraction would be at or below $\frac{3}{10}$ , so 19 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 6 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 6.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 10 hard

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{2}{11} < \dfrac{5}{\square} < \dfrac{5}{7}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 5 and a hidden denominator must sit strictly between 2 over 11 and 5 over 7. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{5}{\square}$ .
It must be greater than $\frac{2}{11}$ .
It must be less than $\frac{5}{7}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 5

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 5. The value does not change -- only the way it is written.

\frac{2}{11} = \frac{5}{\frac{55}{2}},\quad \frac{5}{7} = \frac{5}{7}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

7 < \square < \frac{55}{2}

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below

\frac{55}{2}

is 27.

\frac{2}{11} < \frac{5}{27} < \frac{5}{7}

The answer is 27.

Answer: 27

4 · Reviewdoes it hold up?

Test the next one up: with 28 in the box the middle fraction would be at or below $\frac{2}{11}$ , so 27 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 5 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 5.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 11 hard

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{4}{9} < \dfrac{12}{\square} < \dfrac{6}{7}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 12 and a hidden denominator must sit strictly between 4 over 9 and 6 over 7. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{12}{\square}$ .
It must be greater than $\frac{4}{9}$ .
It must be less than $\frac{6}{7}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 12; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 12

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 12. The value does not change -- only the way it is written.

\frac{4}{9} = \frac{12}{27},\quad \frac{6}{7} = \frac{12}{14}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

14 < \square < 27

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below 27 is 26.

\frac{4}{9} < \frac{12}{26} < \frac{6}{7}

The answer is 26.

Answer: 26

4 · Reviewdoes it hold up?

Test the next one up: with 27 in the box the middle fraction would be at or below $\frac{4}{9}$ , so 26 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 12 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 12.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Variant 12 hard

Find the largest natural number that can go in the box so that the inequality holds.

$\dfrac{5}{12} < \dfrac{14}{\square} < \dfrac{7}{9}$

Your answer

Show solution

1 · Understandwhat's really being asked

A fraction with numerator 14 and a hidden denominator must sit strictly between 5 over 12 and 7 over 9. We want the largest whole number the denominator can be.

Givens

The middle fraction is $\frac{14}{\square}$ .
It must be greater than $\frac{5}{12}$ .
It must be less than $\frac{7}{9}$ .

Unknowns

The largest natural number the box can hold.

Constraints

Both inequalities are strict, so neither endpoint may be reached.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check

Three fractions with different numerators are hard to compare at once. Give all three the numerator 14; then only the denominators differ, and the comparison becomes a size question about whole numbers.

3 · Execute3 carry out the plan

1Give the outer fractions the numerator 14

#9 Solve an Easier Related Problem 4.NF.A.1

Scale each outer fraction so its top is 14. The value does not change -- only the way it is written.

\frac{5}{12} = \frac{14}{\frac{168}{5}},\quad \frac{7}{9} = \frac{14}{18}

All three fractions now count the same-sized numerator.

2Compare fractions that share a numerator

#9 Solve an Easier Related Problem 3.NF.A.3

With equal tops, the fraction with the bigger bottom is the smaller one -- the whole is cut into more pieces. So the inequality flips into a range for the box.

18 < \square < \frac{168}{5}

The box is squeezed between two numbers.

3Pick and check the largest natural number

#6 Guess and Check 4.NF.A.2

The largest whole number strictly below

\frac{168}{5}

is 33.

\frac{5}{12} < \frac{14}{33} < \frac{7}{9}

The answer is 33.

Answer: 33

4 · Reviewdoes it hold up?

Test the next one up: with 34 in the box the middle fraction would be at or below $\frac{5}{12}$ , so 33 really is the largest that fits.

Another way: Common denominators work too, but they make three big numbers to juggle; matching the numerator to 14 keeps the comparison to one column.

Standardsmin grade 4

4.NF.A.1 Explain why a fraction is equivalent to another fraction — Rewriting both outer fractions with numerator 14.
3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Comparing same-numerator fractions by their denominators.
4.NF.A.2 Compare two fractions with different numerators and different denominators — Confirming the chosen denominator satisfies both inequalities.

💡Takeaway. Matching numerators is as good as matching denominators -- and then the bigger bottom is the smaller fraction.

Where next?

← Back to the problem: Match denominators or numerators to compare