← 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

Generated variants — 12

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

Variant 1 easy answer: 17

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

16<3□<35\dfrac{1}{6} < \dfrac{3}{\square} < \dfrac{3}{5}

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 3□\frac{3}{\square}.
  • It must be greater than 16\frac{1}{6}.
  • It must be less than 35\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.
16=318,35=35\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<□<185 < \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.
16<317<35\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 16\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 answer: 17

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

13<6□<34\dfrac{1}{3} < \dfrac{6}{\square} < \dfrac{3}{4}

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 6□\frac{6}{\square}.
  • It must be greater than 13\frac{1}{3}.
  • It must be less than 34\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.
13=618,34=68\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<□<188 < \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.
13<617<34\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 13\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 answer: 19

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

14<5□<56\dfrac{1}{4} < \dfrac{5}{\square} < \dfrac{5}{6}

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 5□\frac{5}{\square}.
  • It must be greater than 14\frac{1}{4}.
  • It must be less than 56\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.
14=520,56=56\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<□<206 < \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.
14<519<56\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 14\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 answer: 27

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

27<8□<45\dfrac{2}{7} < \dfrac{8}{\square} < \dfrac{4}{5}

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 8□\frac{8}{\square}.
  • It must be greater than 27\frac{2}{7}.
  • It must be less than 45\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.
27=828,45=810\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<□<2810 < \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.
27<827<45\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 27\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 answer: 9

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

25<4□<89\dfrac{2}{5} < \dfrac{4}{\square} < \dfrac{8}{9}

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 4□\frac{4}{\square}.
  • It must be greater than 25\frac{2}{5}.
  • It must be less than 89\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.
25=410,89=492\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.
92<□<10\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.
25<49<89\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 25\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 answer: 31

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

29<7□<78\dfrac{2}{9} < \dfrac{7}{\square} < \dfrac{7}{8}

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 7□\frac{7}{\square}.
  • It must be greater than 29\frac{2}{9}.
  • It must be less than 78\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.
29=7632,78=78\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<□<6328 < \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 632\frac{63}{2} is 31.
29<731<78\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 29\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 answer: 17

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

12<9□<910\dfrac{1}{2} < \dfrac{9}{\square} < \dfrac{9}{10}

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 9□\frac{9}{\square}.
  • It must be greater than 12\frac{1}{2}.
  • It must be less than 910\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.
12=918,910=910\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<□<1810 < \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.
12<917<910\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 12\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 answer: 26

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

38<10□<56\dfrac{3}{8} < \dfrac{10}{\square} < \dfrac{5}{6}

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 10□\frac{10}{\square}.
  • It must be greater than 38\frac{3}{8}.
  • It must be less than 56\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.
38=10803,56=1012\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<□<80312 < \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 803\frac{80}{3} is 26.
38<1026<56\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 38\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 answer: 19

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

310<6□<23\dfrac{3}{10} < \dfrac{6}{\square} < \dfrac{2}{3}

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 6□\frac{6}{\square}.
  • It must be greater than 310\frac{3}{10}.
  • It must be less than 23\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

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.
310=620,23=69\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<□<209 < \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.
310<619<23\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 310\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 answer: 27

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

211<5□<57\dfrac{2}{11} < \dfrac{5}{\square} < \dfrac{5}{7}

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 5□\frac{5}{\square}.
  • It must be greater than 211\frac{2}{11}.
  • It must be less than 57\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

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.
211=5552,57=57\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<□<5527 < \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 552\frac{55}{2} is 27.
211<527<57\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 211\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 answer: 26

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

49<12□<67\dfrac{4}{9} < \dfrac{12}{\square} < \dfrac{6}{7}

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 12□\frac{12}{\square}.
  • It must be greater than 49\frac{4}{9}.
  • It must be less than 67\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.
49=1227,67=1214\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<□<2714 < \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.
49<1226<67\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 49\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 answer: 33

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

512<14□<79\dfrac{5}{12} < \dfrac{14}{\square} < \dfrac{7}{9}

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 14□\frac{14}{\square}.
  • It must be greater than 512\frac{5}{12}.
  • It must be less than 79\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.
512=141685,79=1418\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<□<168518 < \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 1685\frac{168}{5} is 33.
512<1433<79\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 512\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.