← Smaller denominator means larger unit fraction · Compare Fractions and Decimals by Structure

Smaller denominator means larger unit fraction · 12 practice problems

5.NF.B.44.NF.A.23.NF.A.3

Generated variants — 12

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

Variant 1 easy answer: 5

Find the greatest natural number that can go in ■\blacksquare.

14×19<1■×16\frac{1}{4} \times \frac{1}{9} < \frac{1}{\blacksquare} \times \frac{1}{6}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 4 times one over 9; on the right, one over the hidden number times one over 6. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 14×19\frac{1}{4} \times \frac{1}{9}.
  • The right side is 1■×16\frac{1}{\blacksquare} \times \frac{1}{6}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
14×19=136\frac{1}{4} \times \frac{1}{9} = \frac{1}{36}
The left side is one 36th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×16=1■×6\frac{1}{\blacksquare} \times \frac{1}{6} = \frac{1}{\blacksquare \times 6}
The right side is one over 6 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×6<36\blacksquare \times 6 < 36
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 6 has to stay under 36, and 5 x 6 = 30 does while 6 x 6 = 36 does not.
■=5\blacksquare = 5
The answer is 5.
Answer: 5
4 · Reviewdoes it hold up?

Put it back: 136\frac{1}{36} against 130\frac{1}{30} -- and 30 is less than 36, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 36 divided by 6 is 6, and the greatest whole number below that is 5.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 2 easy answer: 7

Find the greatest natural number that can go in ■\blacksquare.

16×111<1■×19\frac{1}{6} \times \frac{1}{11} < \frac{1}{\blacksquare} \times \frac{1}{9}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 6 times one over 11; on the right, one over the hidden number times one over 9. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 16×111\frac{1}{6} \times \frac{1}{11}.
  • The right side is 1■×19\frac{1}{\blacksquare} \times \frac{1}{9}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
16×111=166\frac{1}{6} \times \frac{1}{11} = \frac{1}{66}
The left side is one 66th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×19=1■×9\frac{1}{\blacksquare} \times \frac{1}{9} = \frac{1}{\blacksquare \times 9}
The right side is one over 9 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×9<66\blacksquare \times 9 < 66
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 9 has to stay under 66, and 7 x 9 = 63 does while 8 x 9 = 72 does not.
■=7\blacksquare = 7
The answer is 7.
Answer: 7
4 · Reviewdoes it hold up?

Put it back: 166\frac{1}{66} against 163\frac{1}{63} -- and 63 is less than 66, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 66 divided by 9 is 7.33333, and the greatest whole number below that is 7.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 3 easy answer: 5

Find the greatest natural number that can go in ■\blacksquare.

17×110<1■×112\frac{1}{7} \times \frac{1}{10} < \frac{1}{\blacksquare} \times \frac{1}{12}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 7 times one over 10; on the right, one over the hidden number times one over 12. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 17×110\frac{1}{7} \times \frac{1}{10}.
  • The right side is 1■×112\frac{1}{\blacksquare} \times \frac{1}{12}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
17×110=170\frac{1}{7} \times \frac{1}{10} = \frac{1}{70}
The left side is one 70th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×112=1■×12\frac{1}{\blacksquare} \times \frac{1}{12} = \frac{1}{\blacksquare \times 12}
The right side is one over 12 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×12<70\blacksquare \times 12 < 70
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 12 has to stay under 70, and 5 x 12 = 60 does while 6 x 12 = 72 does not.
■=5\blacksquare = 5
The answer is 5.
Answer: 5
4 · Reviewdoes it hold up?

Put it back: 170\frac{1}{70} against 160\frac{1}{60} -- and 60 is less than 70, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 70 divided by 12 is 5.83333, and the greatest whole number below that is 5.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 4 easy answer: 7

Find the greatest natural number that can go in ■\blacksquare.

15×112<1■×18\frac{1}{5} \times \frac{1}{12} < \frac{1}{\blacksquare} \times \frac{1}{8}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 5 times one over 12; on the right, one over the hidden number times one over 8. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 15×112\frac{1}{5} \times \frac{1}{12}.
  • The right side is 1■×18\frac{1}{\blacksquare} \times \frac{1}{8}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
15×112=160\frac{1}{5} \times \frac{1}{12} = \frac{1}{60}
The left side is one 60th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×18=1■×8\frac{1}{\blacksquare} \times \frac{1}{8} = \frac{1}{\blacksquare \times 8}
The right side is one over 8 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×8<60\blacksquare \times 8 < 60
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 8 has to stay under 60, and 7 x 8 = 56 does while 8 x 8 = 64 does not.
■=7\blacksquare = 7
The answer is 7.
Answer: 7
4 · Reviewdoes it hold up?

Put it back: 160\frac{1}{60} against 156\frac{1}{56} -- and 56 is less than 60, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 60 divided by 8 is 7.5, and the greatest whole number below that is 7.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 5 medium answer: 6

Find the greatest natural number that can go in ■\blacksquare.

17×112<1■×113\frac{1}{7} \times \frac{1}{12} < \frac{1}{\blacksquare} \times \frac{1}{13}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 7 times one over 12; on the right, one over the hidden number times one over 13. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 17×112\frac{1}{7} \times \frac{1}{12}.
  • The right side is 1■×113\frac{1}{\blacksquare} \times \frac{1}{13}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
17×112=184\frac{1}{7} \times \frac{1}{12} = \frac{1}{84}
The left side is one 84th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×113=1■×13\frac{1}{\blacksquare} \times \frac{1}{13} = \frac{1}{\blacksquare \times 13}
The right side is one over 13 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×13<84\blacksquare \times 13 < 84
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 13 has to stay under 84, and 6 x 13 = 78 does while 7 x 13 = 91 does not.
■=6\blacksquare = 6
The answer is 6.
Answer: 6
4 · Reviewdoes it hold up?

Put it back: 184\frac{1}{84} against 178\frac{1}{78} -- and 78 is less than 84, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 84 divided by 13 is 6.46154, and the greatest whole number below that is 6.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 6 medium answer: 8

Find the greatest natural number that can go in ■\blacksquare.

13×114<1■×15\frac{1}{3} \times \frac{1}{14} < \frac{1}{\blacksquare} \times \frac{1}{5}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 3 times one over 14; on the right, one over the hidden number times one over 5. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 13×114\frac{1}{3} \times \frac{1}{14}.
  • The right side is 1■×15\frac{1}{\blacksquare} \times \frac{1}{5}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
13×114=142\frac{1}{3} \times \frac{1}{14} = \frac{1}{42}
The left side is one 42th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×15=1■×5\frac{1}{\blacksquare} \times \frac{1}{5} = \frac{1}{\blacksquare \times 5}
The right side is one over 5 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×5<42\blacksquare \times 5 < 42
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 5 has to stay under 42, and 8 x 5 = 40 does while 9 x 5 = 45 does not.
■=8\blacksquare = 8
The answer is 8.
Answer: 8
4 · Reviewdoes it hold up?

Put it back: 142\frac{1}{42} against 140\frac{1}{40} -- and 40 is less than 42, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 42 divided by 5 is 8.4, and the greatest whole number below that is 8.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 7 medium answer: 5

Find the greatest natural number that can go in ■\blacksquare.

19×110<1■×115\frac{1}{9} \times \frac{1}{10} < \frac{1}{\blacksquare} \times \frac{1}{15}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 9 times one over 10; on the right, one over the hidden number times one over 15. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 19×110\frac{1}{9} \times \frac{1}{10}.
  • The right side is 1■×115\frac{1}{\blacksquare} \times \frac{1}{15}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
19×110=190\frac{1}{9} \times \frac{1}{10} = \frac{1}{90}
The left side is one 90th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×115=1■×15\frac{1}{\blacksquare} \times \frac{1}{15} = \frac{1}{\blacksquare \times 15}
The right side is one over 15 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×15<90\blacksquare \times 15 < 90
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 15 has to stay under 90, and 5 x 15 = 75 does while 6 x 15 = 90 does not.
■=5\blacksquare = 5
The answer is 5.
Answer: 5
4 · Reviewdoes it hold up?

Put it back: 190\frac{1}{90} against 175\frac{1}{75} -- and 75 is less than 90, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 90 divided by 15 is 6, and the greatest whole number below that is 5.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 8 medium answer: 12

Find the greatest natural number that can go in ■\blacksquare.

15×118<1■×17\frac{1}{5} \times \frac{1}{18} < \frac{1}{\blacksquare} \times \frac{1}{7}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 5 times one over 18; on the right, one over the hidden number times one over 7. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 15×118\frac{1}{5} \times \frac{1}{18}.
  • The right side is 1■×17\frac{1}{\blacksquare} \times \frac{1}{7}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
15×118=190\frac{1}{5} \times \frac{1}{18} = \frac{1}{90}
The left side is one 90th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×17=1■×7\frac{1}{\blacksquare} \times \frac{1}{7} = \frac{1}{\blacksquare \times 7}
The right side is one over 7 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×7<90\blacksquare \times 7 < 90
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 7 has to stay under 90, and 12 x 7 = 84 does while 13 x 7 = 91 does not.
■=12\blacksquare = 12
The answer is 12.
Answer: 12
4 · Reviewdoes it hold up?

Put it back: 190\frac{1}{90} against 184\frac{1}{84} -- and 84 is less than 90, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 90 divided by 7 is 12.8571, and the greatest whole number below that is 12.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 9 hard answer: 5

Find the greatest natural number that can go in ■\blacksquare.

18×115<1■×120\frac{1}{8} \times \frac{1}{15} < \frac{1}{\blacksquare} \times \frac{1}{20}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 8 times one over 15; on the right, one over the hidden number times one over 20. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 18×115\frac{1}{8} \times \frac{1}{15}.
  • The right side is 1■×120\frac{1}{\blacksquare} \times \frac{1}{20}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
18×115=1120\frac{1}{8} \times \frac{1}{15} = \frac{1}{120}
The left side is one 120th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×120=1■×20\frac{1}{\blacksquare} \times \frac{1}{20} = \frac{1}{\blacksquare \times 20}
The right side is one over 20 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×20<120\blacksquare \times 20 < 120
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 20 has to stay under 120, and 5 x 20 = 100 does while 6 x 20 = 120 does not.
■=5\blacksquare = 5
The answer is 5.
Answer: 5
4 · Reviewdoes it hold up?

Put it back: 1120\frac{1}{120} against 1100\frac{1}{100} -- and 100 is less than 120, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 120 divided by 20 is 6, and the greatest whole number below that is 5.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 10 hard answer: 7

Find the greatest natural number that can go in ■\blacksquare.

14×121<1■×111\frac{1}{4} \times \frac{1}{21} < \frac{1}{\blacksquare} \times \frac{1}{11}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 4 times one over 21; on the right, one over the hidden number times one over 11. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 14×121\frac{1}{4} \times \frac{1}{21}.
  • The right side is 1■×111\frac{1}{\blacksquare} \times \frac{1}{11}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
14×121=184\frac{1}{4} \times \frac{1}{21} = \frac{1}{84}
The left side is one 84th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×111=1■×11\frac{1}{\blacksquare} \times \frac{1}{11} = \frac{1}{\blacksquare \times 11}
The right side is one over 11 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×11<84\blacksquare \times 11 < 84
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 11 has to stay under 84, and 7 x 11 = 77 does while 8 x 11 = 88 does not.
■=7\blacksquare = 7
The answer is 7.
Answer: 7
4 · Reviewdoes it hold up?

Put it back: 184\frac{1}{84} against 177\frac{1}{77} -- and 77 is less than 84, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 84 divided by 11 is 7.63636, and the greatest whole number below that is 7.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 11 hard answer: 5

Find the greatest natural number that can go in ■\blacksquare.

110×113<1■×124\frac{1}{10} \times \frac{1}{13} < \frac{1}{\blacksquare} \times \frac{1}{24}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 10 times one over 13; on the right, one over the hidden number times one over 24. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 110×113\frac{1}{10} \times \frac{1}{13}.
  • The right side is 1■×124\frac{1}{\blacksquare} \times \frac{1}{24}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
110×113=1130\frac{1}{10} \times \frac{1}{13} = \frac{1}{130}
The left side is one 130th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×124=1■×24\frac{1}{\blacksquare} \times \frac{1}{24} = \frac{1}{\blacksquare \times 24}
The right side is one over 24 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×24<130\blacksquare \times 24 < 130
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 24 has to stay under 130, and 5 x 24 = 120 does while 6 x 24 = 144 does not.
■=5\blacksquare = 5
The answer is 5.
Answer: 5
4 · Reviewdoes it hold up?

Put it back: 1130\frac{1}{130} against 1120\frac{1}{120} -- and 120 is less than 130, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 130 divided by 24 is 5.41667, and the greatest whole number below that is 5.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.
Variant 12 hard answer: 9

Find the greatest natural number that can go in ■\blacksquare.

16×125<1■×116\frac{1}{6} \times \frac{1}{25} < \frac{1}{\blacksquare} \times \frac{1}{16}

Show solution
1 · Understandwhat's really being asked

Two products of unit fractions are compared. On the left, one over 6 times one over 25; on the right, one over the hidden number times one over 16. The left must be the smaller. We want the biggest whole number that keeps it so.

Givens
  • The left side is 16×125\frac{1}{6} \times \frac{1}{25}.
  • The right side is 1■×116\frac{1}{\blacksquare} \times \frac{1}{16}.
  • The left side must be strictly less than the right.
Unknowns
  • The greatest natural number the box can hold.
Constraints
  • The box is a natural number, so at least 1.
2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #6 Guess and Check#5 Look for a Pattern

Multiply each side out first. Both become single unit fractions, and unit fractions compare by their denominators alone -- the larger denominator being the smaller fraction.

3 · Execute4 carry out the plan

1Multiply the left side

#9 Solve an Easier Related Problem 5.NF.B.4
Multiplying unit fractions multiplies the denominators.
16×125=1150\frac{1}{6} \times \frac{1}{25} = \frac{1}{150}
The left side is one 150th.

2Multiply the right side

#9 Solve an Easier Related Problem 5.NF.B.4
The same on the right, leaving the box inside the denominator.
1■×116=1■×16\frac{1}{\blacksquare} \times \frac{1}{16} = \frac{1}{\blacksquare \times 16}
The right side is one over 16 times the box.

3Compare two unit fractions

#9 Solve an Easier Related Problem 4.NF.A.2
With 1 on top of both, the smaller fraction is the one with the bigger bottom. The left is smaller, so its denominator must be the bigger one.
■×16<150\blacksquare \times 16 < 150
The comparison flipped into one about whole numbers.

4Find the biggest whole number

#6 Guess and Check 3.NF.A.3
The box times 16 has to stay under 150, and 9 x 16 = 144 does while 10 x 16 = 160 does not.
■=9\blacksquare = 9
The answer is 9.
Answer: 9
4 · Reviewdoes it hold up?

Put it back: 1150\frac{1}{150} against 1144\frac{1}{144} -- and 144 is less than 150, so the right side is indeed the larger fraction.

Another way: Dividing gives the bound straight away: 150 divided by 16 is 9.375, and the greatest whole number below that is 9.

Standardsmin grade 5
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each pair of unit fractions into one.
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators — Turning the fraction comparison into one about denominators.
  • 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Checking the chosen number and the next one up.
💡Takeaway. Unit fractions compare backwards: the bigger the bottom, the smaller the piece.