← Fraction of a number: divide then multiply · Multiplicative Comparison and Unit Rate

Fraction of a number: divide then multiply · 12 practice problems

3.NF.A.13.OA.A.2

Generated variants — 12

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

Variant 1 easy answer: Cole (the piece is 10 m, the longest)

A ribbon $24\ \text{m}$ long is cut by Ava, Ben, and Cole, each taking the length they need to wrap a gift. Ava cut $\frac{1}{4}$ of the whole ribbon, Ben cut $\frac{1}{3}$ of the whole ribbon, and Cole cut $\frac{5}{12}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 24 m ribbon is shared. Ava takes 1/4 of the whole, Ben takes 1/3 of the whole, Cole takes 5/12 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 24 m.
Ava cut 1/4 of 24 m.
Ben cut 1/3 of 24 m.
Cole cut 5/12 of 24 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 24 m.
A fraction of 24 m is found by dividing 24 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 24 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Ava's piece: 1/4 of 24

#7 Identify Subproblems 3.NF.A.1

Divide 24 into 4 equal parts, then take 1 part.

24 \div 4 = 6,\quad 6 \times 1 = 6

A fraction of a length means split into equal parts and take that many.

2Ben's piece: 1/3 of 24

#7 Identify Subproblems 3.NF.A.1

Divide 24 into 3 equal parts, then take 1 part.

24 \div 3 = 8,\quad 8 \times 1 = 8

Same split-and-take rule with denominator 3 and numerator 1.

3Cole's piece: 5/12 of 24

#7 Identify Subproblems 3.NF.A.1

Divide 24 into 12 equal parts, then take 5 parts.

24 \div 12 = 2,\quad 2 \times 5 = 10

Same split-and-take rule with denominator 12 and numerator 5.

4Compare the three lengths

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

List them: Ava 6 m, Ben 8 m, Cole 10 m. The largest is 10 m, so Cole cut the longest piece.

6 < 8 < 10 \Rightarrow \text{Cole}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Cole (the piece is 10 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 6 + 8 + 10 = 24 m, which is less than the 24 m ribbon, so the shares are sensible (some ribbon is left over). Cole's 10 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 24: Ava 6/24, Ben 8/24, Cole 10/24; the largest fraction of the same whole belongs to Cole.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 24 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 24 m, take your parts, then see whose pile is biggest!

Variant 2 easy answer: Noah (the piece is 9 m, the longest)

A ribbon $30\ \text{m}$ long is cut by Liam, Mia, and Noah, each taking the length they need to wrap a gift. Liam cut $\frac{1}{6}$ of the whole ribbon, Mia cut $\frac{4}{15}$ of the whole ribbon, and Noah cut $\frac{3}{10}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 30 m ribbon is shared. Liam takes 1/6 of the whole, Mia takes 4/15 of the whole, Noah takes 3/10 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 30 m.
Liam cut 1/6 of 30 m.
Mia cut 4/15 of 30 m.
Noah cut 3/10 of 30 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 30 m.
A fraction of 30 m is found by dividing 30 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 30 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Liam's piece: 1/6 of 30

#7 Identify Subproblems 3.NF.A.1

Divide 30 into 6 equal parts, then take 1 part.

30 \div 6 = 5,\quad 5 \times 1 = 5

A fraction of a length means split into equal parts and take that many.

2Mia's piece: 4/15 of 30

#7 Identify Subproblems 3.NF.A.1

Divide 30 into 15 equal parts, then take 4 parts.

30 \div 15 = 2,\quad 2 \times 4 = 8

Same split-and-take rule with denominator 15 and numerator 4.

3Noah's piece: 3/10 of 30

#7 Identify Subproblems 3.NF.A.1

Divide 30 into 10 equal parts, then take 3 parts.

30 \div 10 = 3,\quad 3 \times 3 = 9

Same split-and-take rule with denominator 10 and numerator 3.

4Compare the three lengths

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

List them: Liam 5 m, Mia 8 m, Noah 9 m. The largest is 9 m, so Noah cut the longest piece.

5 < 8 < 9 \Rightarrow \text{Noah}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Noah (the piece is 9 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 5 + 8 + 9 = 22 m, which is less than the 30 m ribbon, so the shares are sensible (some ribbon is left over). Noah's 9 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 30: Liam 5/30, Mia 8/30, Noah 9/30; the largest fraction of the same whole belongs to Noah.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 30 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 30 m, take your parts, then see whose pile is biggest!

Variant 3 easy answer: Gabe (the piece is 15 m, the longest)

A ribbon $36\ \text{m}$ long is cut by Emma, Finn, and Gabe, each taking the length they need to wrap a gift. Emma cut $\frac{1}{6}$ of the whole ribbon, Finn cut $\frac{2}{9}$ of the whole ribbon, and Gabe cut $\frac{5}{12}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 36 m ribbon is shared. Emma takes 1/6 of the whole, Finn takes 2/9 of the whole, Gabe takes 5/12 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 36 m.
Emma cut 1/6 of 36 m.
Finn cut 2/9 of 36 m.
Gabe cut 5/12 of 36 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 36 m.
A fraction of 36 m is found by dividing 36 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 36 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Emma's piece: 1/6 of 36

#7 Identify Subproblems 3.NF.A.1

Divide 36 into 6 equal parts, then take 1 part.

36 \div 6 = 6,\quad 6 \times 1 = 6

A fraction of a length means split into equal parts and take that many.

2Finn's piece: 2/9 of 36

#7 Identify Subproblems 3.NF.A.1

Divide 36 into 9 equal parts, then take 2 parts.

36 \div 9 = 4,\quad 4 \times 2 = 8

Same split-and-take rule with denominator 9 and numerator 2.

3Gabe's piece: 5/12 of 36

#7 Identify Subproblems 3.NF.A.1

Divide 36 into 12 equal parts, then take 5 parts.

36 \div 12 = 3,\quad 3 \times 5 = 15

Same split-and-take rule with denominator 12 and numerator 5.

4Compare the three lengths

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

List them: Emma 6 m, Finn 8 m, Gabe 15 m. The largest is 15 m, so Gabe cut the longest piece.

6 < 8 < 15 \Rightarrow \text{Gabe}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Gabe (the piece is 15 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 6 + 8 + 15 = 29 m, which is less than the 36 m ribbon, so the shares are sensible (some ribbon is left over). Gabe's 15 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 36: Emma 6/36, Finn 8/36, Gabe 15/36; the largest fraction of the same whole belongs to Gabe.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 36 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 36 m, take your parts, then see whose pile is biggest!

Variant 4 easy answer: Jude (the piece is 16 m, the longest)

A ribbon $40\ \text{m}$ long is cut by Hana, Ivy, and Jude, each taking the length they need to wrap a gift. Hana cut $\frac{1}{8}$ of the whole ribbon, Ivy cut $\frac{3}{10}$ of the whole ribbon, and Jude cut $\frac{2}{5}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 40 m ribbon is shared. Hana takes 1/8 of the whole, Ivy takes 3/10 of the whole, Jude takes 2/5 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 40 m.
Hana cut 1/8 of 40 m.
Ivy cut 3/10 of 40 m.
Jude cut 2/5 of 40 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 40 m.
A fraction of 40 m is found by dividing 40 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 40 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Hana's piece: 1/8 of 40

#7 Identify Subproblems 3.NF.A.1

Divide 40 into 8 equal parts, then take 1 part.

40 \div 8 = 5,\quad 5 \times 1 = 5

A fraction of a length means split into equal parts and take that many.

2Ivy's piece: 3/10 of 40

#7 Identify Subproblems 3.NF.A.1

Divide 40 into 10 equal parts, then take 3 parts.

40 \div 10 = 4,\quad 4 \times 3 = 12

Same split-and-take rule with denominator 10 and numerator 3.

3Jude's piece: 2/5 of 40

#7 Identify Subproblems 3.NF.A.1

Divide 40 into 5 equal parts, then take 2 parts.

40 \div 5 = 8,\quad 8 \times 2 = 16

Same split-and-take rule with denominator 5 and numerator 2.

4Compare the three lengths

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

List them: Hana 5 m, Ivy 12 m, Jude 16 m. The largest is 16 m, so Jude cut the longest piece.

5 < 12 < 16 \Rightarrow \text{Jude}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Jude (the piece is 16 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 5 + 12 + 16 = 33 m, which is less than the 40 m ribbon, so the shares are sensible (some ribbon is left over). Jude's 16 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 40: Hana 5/40, Ivy 12/40, Jude 16/40; the largest fraction of the same whole belongs to Jude.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 40 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 40 m, take your parts, then see whose pile is biggest!

Variant 5 medium answer: Hugo (the piece is 9 m, the longest)

A ribbon $42\ \text{m}$ long is cut by Fiza, Gus, and Hugo, each taking the length they need to wrap a gift. Fiza cut $\frac{1}{7}$ of the whole ribbon, Gus cut $\frac{1}{6}$ of the whole ribbon, and Hugo cut $\frac{3}{14}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 42 m ribbon is shared. Fiza takes 1/7 of the whole, Gus takes 1/6 of the whole, Hugo takes 3/14 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 42 m.
Fiza cut 1/7 of 42 m.
Gus cut 1/6 of 42 m.
Hugo cut 3/14 of 42 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 42 m.
A fraction of 42 m is found by dividing 42 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 42 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Fiza's piece: 1/7 of 42

#7 Identify Subproblems 3.NF.A.1

Divide 42 into 7 equal parts, then take 1 part.

42 \div 7 = 6,\quad 6 \times 1 = 6

A fraction of a length means split into equal parts and take that many.

2Gus's piece: 1/6 of 42

#7 Identify Subproblems 3.NF.A.1

Divide 42 into 6 equal parts, then take 1 part.

42 \div 6 = 7,\quad 7 \times 1 = 7

Same split-and-take rule with denominator 6 and numerator 1.

3Hugo's piece: 3/14 of 42

#7 Identify Subproblems 3.NF.A.1

Divide 42 into 14 equal parts, then take 3 parts.

42 \div 14 = 3,\quad 3 \times 3 = 9

Same split-and-take rule with denominator 14 and numerator 3.

4Compare the three lengths

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

List them: Fiza 6 m, Gus 7 m, Hugo 9 m. The largest is 9 m, so Hugo cut the longest piece.

6 < 7 < 9 \Rightarrow \text{Hugo}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Hugo (the piece is 9 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 6 + 7 + 9 = 22 m, which is less than the 42 m ribbon, so the shares are sensible (some ribbon is left over). Hugo's 9 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 42: Fiza 6/42, Gus 7/42, Hugo 9/42; the largest fraction of the same whole belongs to Hugo.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 42 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 42 m, take your parts, then see whose pile is biggest!

Variant 6 medium answer: Rosa (the piece is 18 m, the longest)

A ribbon $45\ \text{m}$ long is cut by Quinn, Rosa, and Sam, each taking the length they need to wrap a gift. Quinn cut $\frac{1}{9}$ of the whole ribbon, Rosa cut $\frac{2}{5}$ of the whole ribbon, and Sam cut $\frac{1}{3}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 45 m ribbon is shared. Quinn takes 1/9 of the whole, Rosa takes 2/5 of the whole, Sam takes 1/3 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 45 m.
Quinn cut 1/9 of 45 m.
Rosa cut 2/5 of 45 m.
Sam cut 1/3 of 45 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 45 m.
A fraction of 45 m is found by dividing 45 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 45 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Quinn's piece: 1/9 of 45

#7 Identify Subproblems 3.NF.A.1

Divide 45 into 9 equal parts, then take 1 part.

45 \div 9 = 5,\quad 5 \times 1 = 5

A fraction of a length means split into equal parts and take that many.

2Rosa's piece: 2/5 of 45

#7 Identify Subproblems 3.NF.A.1

Divide 45 into 5 equal parts, then take 2 parts.

45 \div 5 = 9,\quad 9 \times 2 = 18

Same split-and-take rule with denominator 5 and numerator 2.

3Sam's piece: 1/3 of 45

#7 Identify Subproblems 3.NF.A.1

Divide 45 into 3 equal parts, then take 1 part.

45 \div 3 = 15,\quad 15 \times 1 = 15

Same split-and-take rule with denominator 3 and numerator 1.

4Compare the three lengths

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

List them: Quinn 5 m, Rosa 18 m, Sam 15 m. The largest is 18 m, so Rosa cut the longest piece.

5 < 15 < 18 \Rightarrow \text{Rosa}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Rosa (the piece is 18 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 5 + 18 + 15 = 38 m, which is less than the 45 m ribbon, so the shares are sensible (some ribbon is left over). Rosa's 18 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 45: Quinn 5/45, Rosa 18/45, Sam 15/45; the largest fraction of the same whole belongs to Rosa.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 45 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 45 m, take your parts, then see whose pile is biggest!

Variant 7 medium answer: Pia (the piece is 20 m, the longest)

A ribbon $48\ \text{m}$ long is cut by Nina, Omar, and Pia, each taking the length they need to wrap a gift. Nina cut $\frac{1}{6}$ of the whole ribbon, Omar cut $\frac{1}{4}$ of the whole ribbon, and Pia cut $\frac{5}{12}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 48 m ribbon is shared. Nina takes 1/6 of the whole, Omar takes 1/4 of the whole, Pia takes 5/12 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 48 m.
Nina cut 1/6 of 48 m.
Omar cut 1/4 of 48 m.
Pia cut 5/12 of 48 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 48 m.
A fraction of 48 m is found by dividing 48 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 48 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Nina's piece: 1/6 of 48

#7 Identify Subproblems 3.NF.A.1

Divide 48 into 6 equal parts, then take 1 part.

48 \div 6 = 8,\quad 8 \times 1 = 8

A fraction of a length means split into equal parts and take that many.

2Omar's piece: 1/4 of 48

#7 Identify Subproblems 3.NF.A.1

Divide 48 into 4 equal parts, then take 1 part.

48 \div 4 = 12,\quad 12 \times 1 = 12

Same split-and-take rule with denominator 4 and numerator 1.

3Pia's piece: 5/12 of 48

#7 Identify Subproblems 3.NF.A.1

Divide 48 into 12 equal parts, then take 5 parts.

48 \div 12 = 4,\quad 4 \times 5 = 20

Same split-and-take rule with denominator 12 and numerator 5.

4Compare the three lengths

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

List them: Nina 8 m, Omar 12 m, Pia 20 m. The largest is 20 m, so Pia cut the longest piece.

8 < 12 < 20 \Rightarrow \text{Pia}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Pia (the piece is 20 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 8 + 12 + 20 = 40 m, which is less than the 48 m ribbon, so the shares are sensible (some ribbon is left over). Pia's 20 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 48: Nina 8/48, Omar 12/48, Pia 20/48; the largest fraction of the same whole belongs to Pia.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 48 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 48 m, take your parts, then see whose pile is biggest!

Variant 8 medium answer: Vera (the piece is 15 m, the longest)

A ribbon $50\ \text{m}$ long is cut by Tara, Uri, and Vera, each taking the length they need to wrap a gift. Tara cut $\frac{1}{10}$ of the whole ribbon, Uri cut $\frac{1}{5}$ of the whole ribbon, and Vera cut $\frac{3}{10}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 50 m ribbon is shared. Tara takes 1/10 of the whole, Uri takes 1/5 of the whole, Vera takes 3/10 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 50 m.
Tara cut 1/10 of 50 m.
Uri cut 1/5 of 50 m.
Vera cut 3/10 of 50 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 50 m.
A fraction of 50 m is found by dividing 50 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 50 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Tara's piece: 1/10 of 50

#7 Identify Subproblems 3.NF.A.1

Divide 50 into 10 equal parts, then take 1 part.

50 \div 10 = 5,\quad 5 \times 1 = 5

A fraction of a length means split into equal parts and take that many.

2Uri's piece: 1/5 of 50

#7 Identify Subproblems 3.NF.A.1

Divide 50 into 5 equal parts, then take 1 part.

50 \div 5 = 10,\quad 10 \times 1 = 10

Same split-and-take rule with denominator 5 and numerator 1.

3Vera's piece: 3/10 of 50

#7 Identify Subproblems 3.NF.A.1

Divide 50 into 10 equal parts, then take 3 parts.

50 \div 10 = 5,\quad 5 \times 3 = 15

Same split-and-take rule with denominator 10 and numerator 3.

4Compare the three lengths

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

List them: Tara 5 m, Uri 10 m, Vera 15 m. The largest is 15 m, so Vera cut the longest piece.

5 < 10 < 15 \Rightarrow \text{Vera}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Vera (the piece is 15 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 5 + 10 + 15 = 30 m, which is less than the 50 m ribbon, so the shares are sensible (some ribbon is left over). Vera's 15 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 50: Tara 5/50, Uri 10/50, Vera 15/50; the largest fraction of the same whole belongs to Vera.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 50 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 50 m, take your parts, then see whose pile is biggest!

Variant 9 hard answer: Esme (the piece is 18 m, the longest)

A ribbon $54\ \text{m}$ long is cut by Cleo, Dara, and Esme, each taking the length they need to wrap a gift. Cleo cut $\frac{1}{9}$ of the whole ribbon, Dara cut $\frac{2}{9}$ of the whole ribbon, and Esme cut $\frac{1}{3}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 54 m ribbon is shared. Cleo takes 1/9 of the whole, Dara takes 2/9 of the whole, Esme takes 1/3 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 54 m.
Cleo cut 1/9 of 54 m.
Dara cut 2/9 of 54 m.
Esme cut 1/3 of 54 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 54 m.
A fraction of 54 m is found by dividing 54 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 54 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Cleo's piece: 1/9 of 54

#7 Identify Subproblems 3.NF.A.1

Divide 54 into 9 equal parts, then take 1 part.

54 \div 9 = 6,\quad 6 \times 1 = 6

A fraction of a length means split into equal parts and take that many.

2Dara's piece: 2/9 of 54

#7 Identify Subproblems 3.NF.A.1

Divide 54 into 9 equal parts, then take 2 parts.

54 \div 9 = 6,\quad 6 \times 2 = 12

Same split-and-take rule with denominator 9 and numerator 2.

3Esme's piece: 1/3 of 54

#7 Identify Subproblems 3.NF.A.1

Divide 54 into 3 equal parts, then take 1 part.

54 \div 3 = 18,\quad 18 \times 1 = 18

Same split-and-take rule with denominator 3 and numerator 1.

4Compare the three lengths

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

List them: Cleo 6 m, Dara 12 m, Esme 18 m. The largest is 18 m, so Esme cut the longest piece.

6 < 12 < 18 \Rightarrow \text{Esme}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Esme (the piece is 18 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 6 + 12 + 18 = 36 m, which is less than the 54 m ribbon, so the shares are sensible (some ribbon is left over). Esme's 18 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 54: Cleo 6/54, Dara 12/54, Esme 18/54; the largest fraction of the same whole belongs to Esme.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 54 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 54 m, take your parts, then see whose pile is biggest!

Variant 10 hard answer: Maya (the piece is 18 m, the longest)

A ribbon $60\ \text{m}$ long is cut by Kate, Leo, and Maya, each taking the length they need to wrap a gift. Kate cut $\frac{1}{5}$ of the whole ribbon, Leo cut $\frac{1}{4}$ of the whole ribbon, and Maya cut $\frac{3}{10}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 60 m ribbon is shared. Kate takes 1/5 of the whole, Leo takes 1/4 of the whole, Maya takes 3/10 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 60 m.
Kate cut 1/5 of 60 m.
Leo cut 1/4 of 60 m.
Maya cut 3/10 of 60 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 60 m.
A fraction of 60 m is found by dividing 60 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 60 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Kate's piece: 1/5 of 60

#7 Identify Subproblems 3.NF.A.1

Divide 60 into 5 equal parts, then take 1 part.

60 \div 5 = 12,\quad 12 \times 1 = 12

A fraction of a length means split into equal parts and take that many.

2Leo's piece: 1/4 of 60

#7 Identify Subproblems 3.NF.A.1

Divide 60 into 4 equal parts, then take 1 part.

60 \div 4 = 15,\quad 15 \times 1 = 15

Same split-and-take rule with denominator 4 and numerator 1.

3Maya's piece: 3/10 of 60

#7 Identify Subproblems 3.NF.A.1

Divide 60 into 10 equal parts, then take 3 parts.

60 \div 10 = 6,\quad 6 \times 3 = 18

Same split-and-take rule with denominator 10 and numerator 3.

4Compare the three lengths

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

List them: Kate 12 m, Leo 15 m, Maya 18 m. The largest is 18 m, so Maya cut the longest piece.

12 < 15 < 18 \Rightarrow \text{Maya}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Maya (the piece is 18 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 12 + 15 + 18 = 45 m, which is less than the 60 m ribbon, so the shares are sensible (some ribbon is left over). Maya's 18 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 60: Kate 12/60, Leo 15/60, Maya 18/60; the largest fraction of the same whole belongs to Maya.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 60 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 60 m, take your parts, then see whose pile is biggest!

Variant 11 hard answer: Yuki (the piece is 16 m, the longest)

A ribbon $72\ \text{m}$ long is cut by Wade, Xena, and Yuki, each taking the length they need to wrap a gift. Wade cut $\frac{1}{8}$ of the whole ribbon, Xena cut $\frac{1}{6}$ of the whole ribbon, and Yuki cut $\frac{2}{9}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 72 m ribbon is shared. Wade takes 1/8 of the whole, Xena takes 1/6 of the whole, Yuki takes 2/9 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 72 m.
Wade cut 1/8 of 72 m.
Xena cut 1/6 of 72 m.
Yuki cut 2/9 of 72 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 72 m.
A fraction of 72 m is found by dividing 72 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 72 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Wade's piece: 1/8 of 72

#7 Identify Subproblems 3.NF.A.1

Divide 72 into 8 equal parts, then take 1 part.

72 \div 8 = 9,\quad 9 \times 1 = 9

A fraction of a length means split into equal parts and take that many.

2Xena's piece: 1/6 of 72

#7 Identify Subproblems 3.NF.A.1

Divide 72 into 6 equal parts, then take 1 part.

72 \div 6 = 12,\quad 12 \times 1 = 12

Same split-and-take rule with denominator 6 and numerator 1.

3Yuki's piece: 2/9 of 72

#7 Identify Subproblems 3.NF.A.1

Divide 72 into 9 equal parts, then take 2 parts.

72 \div 9 = 8,\quad 8 \times 2 = 16

Same split-and-take rule with denominator 9 and numerator 2.

4Compare the three lengths

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

List them: Wade 9 m, Xena 12 m, Yuki 16 m. The largest is 16 m, so Yuki cut the longest piece.

9 < 12 < 16 \Rightarrow \text{Yuki}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Yuki (the piece is 16 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 9 + 12 + 16 = 37 m, which is less than the 72 m ribbon, so the shares are sensible (some ribbon is left over). Yuki's 16 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 72: Wade 9/72, Xena 12/72, Yuki 16/72; the largest fraction of the same whole belongs to Yuki.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 72 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 72 m, take your parts, then see whose pile is biggest!

Variant 12 hard answer: Bram (the piece is 18 m, the longest)

A ribbon $90\ \text{m}$ long is cut by Zane, Anya, and Bram, each taking the length they need to wrap a gift. Zane cut $\frac{1}{9}$ of the whole ribbon, Anya cut $\frac{2}{15}$ of the whole ribbon, and Bram cut $\frac{1}{5}$ of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

1 · Understandwhat's really being asked

A 90 m ribbon is shared. Zane takes 1/9 of the whole, Anya takes 2/15 of the whole, Bram takes 1/5 of the whole. I need to find each person's length and decide who cut the longest piece.

Givens

The whole ribbon is 90 m.
Zane cut 1/9 of 90 m.
Anya cut 2/15 of 90 m.
Bram cut 1/5 of 90 m.

Unknowns

Who cut the longest piece (and the lengths to compare).

Constraints

Each fraction is of the same whole, 90 m.
A fraction of 90 m is found by dividing 90 into the denominator's equal parts, then taking the numerator's worth.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #2 Make a Systematic List

Each person's length is one fraction-of-a-number subproblem (divide 90 by the denominator, multiply by the numerator). Listing the three results side by side makes the comparison straightforward.

3 · Execute4 carry out the plan

1Zane's piece: 1/9 of 90

#7 Identify Subproblems 3.NF.A.1

Divide 90 into 9 equal parts, then take 1 part.

90 \div 9 = 10,\quad 10 \times 1 = 10

A fraction of a length means split into equal parts and take that many.

2Anya's piece: 2/15 of 90

#7 Identify Subproblems 3.NF.A.1

Divide 90 into 15 equal parts, then take 2 parts.

90 \div 15 = 6,\quad 6 \times 2 = 12

Same split-and-take rule with denominator 15 and numerator 2.

3Bram's piece: 1/5 of 90

#7 Identify Subproblems 3.NF.A.1

Divide 90 into 5 equal parts, then take 1 part.

90 \div 5 = 18,\quad 18 \times 1 = 18

Same split-and-take rule with denominator 5 and numerator 1.

4Compare the three lengths

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

List them: Zane 10 m, Anya 12 m, Bram 18 m. The largest is 18 m, so Bram cut the longest piece.

10 < 12 < 18 \Rightarrow \text{Bram}

Once all three are plain meter amounts, comparing is just ordering whole numbers.

Answer: Bram (the piece is 18 m, the longest)

4 · Reviewdoes it hold up?

The three pieces total 10 + 12 + 18 = 40 m, which is less than the 90 m ribbon, so the shares are sensible (some ribbon is left over). Bram's 18 m is clearly the largest.

Another way: Compare the fractions directly (tool 15) by a common denominator of 90: Zane 10/90, Anya 12/90, Bram 18/90; the largest fraction of the same whole belongs to Bram.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Computing each person's length as a fraction of the 90 m whole.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing by each denominator and comparing the resulting lengths.

💡Takeaway. This only needs Grade 3 fraction sense: split the 90 m, take your parts, then see whose pile is biggest!