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← 3-2 · 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 answer: Yuki (the piece is 16 m, the longest)

A ribbon 72 m72\ \text{m} long is cut by Wade, Xena, and Yuki, each taking the length they need to wrap a gift. Wade cut 18\frac{1}{8} of the whole ribbon, Xena cut 16\frac{1}{6} of the whole ribbon, and Yuki cut 29\frac{2}{9} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 72 into 8 equal parts, then take 1 part.
72÷8=9,9×1=972 \div 8 = 9,\quad 9 \times 1 = 9
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 72 into 6 equal parts, then take 1 part.
72÷6=12,12×1=1272 \div 6 = 12,\quad 12 \times 1 = 12
Same split-and-take rule with denominator 6 and numerator 1.
#7 Identify Subproblems 3.NF.A.1
Divide 72 into 9 equal parts, then take 2 parts.
72÷9=8,8×2=1672 \div 9 = 8,\quad 8 \times 2 = 16
Same split-and-take rule with denominator 9 and numerator 2.
#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<16Yuki9 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 72 m, take your parts, then see whose pile is biggest!
Variant 2 answer: Vera (the piece is 15 m, the longest)

A ribbon 50 m50\ \text{m} long is cut by Tara, Uri, and Vera, each taking the length they need to wrap a gift. Tara cut 110\frac{1}{10} of the whole ribbon, Uri cut 15\frac{1}{5} of the whole ribbon, and Vera cut 310\frac{3}{10} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 50 into 10 equal parts, then take 1 part.
50÷10=5,5×1=550 \div 10 = 5,\quad 5 \times 1 = 5
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 50 into 5 equal parts, then take 1 part.
50÷5=10,10×1=1050 \div 5 = 10,\quad 10 \times 1 = 10
Same split-and-take rule with denominator 5 and numerator 1.
#7 Identify Subproblems 3.NF.A.1
Divide 50 into 10 equal parts, then take 3 parts.
50÷10=5,5×3=1550 \div 10 = 5,\quad 5 \times 3 = 15
Same split-and-take rule with denominator 10 and numerator 3.
#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<15Vera5 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 50 m, take your parts, then see whose pile is biggest!
Variant 3 answer: Maya (the piece is 18 m, the longest)

A ribbon 60 m60\ \text{m} long is cut by Kate, Leo, and Maya, each taking the length they need to wrap a gift. Kate cut 15\frac{1}{5} of the whole ribbon, Leo cut 14\frac{1}{4} of the whole ribbon, and Maya cut 310\frac{3}{10} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 60 into 5 equal parts, then take 1 part.
60÷5=12,12×1=1260 \div 5 = 12,\quad 12 \times 1 = 12
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 60 into 4 equal parts, then take 1 part.
60÷4=15,15×1=1560 \div 4 = 15,\quad 15 \times 1 = 15
Same split-and-take rule with denominator 4 and numerator 1.
#7 Identify Subproblems 3.NF.A.1
Divide 60 into 10 equal parts, then take 3 parts.
60÷10=6,6×3=1860 \div 10 = 6,\quad 6 \times 3 = 18
Same split-and-take rule with denominator 10 and numerator 3.
#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<18Maya12 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 60 m, take your parts, then see whose pile is biggest!
Variant 4 answer: Hugo (the piece is 9 m, the longest)

A ribbon 42 m42\ \text{m} long is cut by Fiza, Gus, and Hugo, each taking the length they need to wrap a gift. Fiza cut 17\frac{1}{7} of the whole ribbon, Gus cut 16\frac{1}{6} of the whole ribbon, and Hugo cut 314\frac{3}{14} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 42 into 7 equal parts, then take 1 part.
42÷7=6,6×1=642 \div 7 = 6,\quad 6 \times 1 = 6
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 42 into 6 equal parts, then take 1 part.
42÷6=7,7×1=742 \div 6 = 7,\quad 7 \times 1 = 7
Same split-and-take rule with denominator 6 and numerator 1.
#7 Identify Subproblems 3.NF.A.1
Divide 42 into 14 equal parts, then take 3 parts.
42÷14=3,3×3=942 \div 14 = 3,\quad 3 \times 3 = 9
Same split-and-take rule with denominator 14 and numerator 3.
#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<9Hugo6 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 42 m, take your parts, then see whose pile is biggest!
Variant 5 answer: Jude (the piece is 16 m, the longest)

A ribbon 40 m40\ \text{m} long is cut by Hana, Ivy, and Jude, each taking the length they need to wrap a gift. Hana cut 18\frac{1}{8} of the whole ribbon, Ivy cut 310\frac{3}{10} of the whole ribbon, and Jude cut 25\frac{2}{5} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 40 into 8 equal parts, then take 1 part.
40÷8=5,5×1=540 \div 8 = 5,\quad 5 \times 1 = 5
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 40 into 10 equal parts, then take 3 parts.
40÷10=4,4×3=1240 \div 10 = 4,\quad 4 \times 3 = 12
Same split-and-take rule with denominator 10 and numerator 3.
#7 Identify Subproblems 3.NF.A.1
Divide 40 into 5 equal parts, then take 2 parts.
40÷5=8,8×2=1640 \div 5 = 8,\quad 8 \times 2 = 16
Same split-and-take rule with denominator 5 and numerator 2.
#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<16Jude5 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 40 m, take your parts, then see whose pile is biggest!
Variant 6 answer: Gabe (the piece is 15 m, the longest)

A ribbon 36 m36\ \text{m} long is cut by Emma, Finn, and Gabe, each taking the length they need to wrap a gift. Emma cut 16\frac{1}{6} of the whole ribbon, Finn cut 29\frac{2}{9} of the whole ribbon, and Gabe cut 512\frac{5}{12} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 36 into 6 equal parts, then take 1 part.
36÷6=6,6×1=636 \div 6 = 6,\quad 6 \times 1 = 6
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 36 into 9 equal parts, then take 2 parts.
36÷9=4,4×2=836 \div 9 = 4,\quad 4 \times 2 = 8
Same split-and-take rule with denominator 9 and numerator 2.
#7 Identify Subproblems 3.NF.A.1
Divide 36 into 12 equal parts, then take 5 parts.
36÷12=3,3×5=1536 \div 12 = 3,\quad 3 \times 5 = 15
Same split-and-take rule with denominator 12 and numerator 5.
#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<15Gabe6 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 36 m, take your parts, then see whose pile is biggest!
Variant 7 answer: Cole (the piece is 10 m, the longest)

A ribbon 24 m24\ \text{m} long is cut by Ava, Ben, and Cole, each taking the length they need to wrap a gift. Ava cut 14\frac{1}{4} of the whole ribbon, Ben cut 13\frac{1}{3} of the whole ribbon, and Cole cut 512\frac{5}{12} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 24 into 4 equal parts, then take 1 part.
24÷4=6,6×1=624 \div 4 = 6,\quad 6 \times 1 = 6
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 24 into 3 equal parts, then take 1 part.
24÷3=8,8×1=824 \div 3 = 8,\quad 8 \times 1 = 8
Same split-and-take rule with denominator 3 and numerator 1.
#7 Identify Subproblems 3.NF.A.1
Divide 24 into 12 equal parts, then take 5 parts.
24÷12=2,2×5=1024 \div 12 = 2,\quad 2 \times 5 = 10
Same split-and-take rule with denominator 12 and numerator 5.
#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<10Cole6 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 24 m, take your parts, then see whose pile is biggest!
Variant 8 answer: Pia (the piece is 20 m, the longest)

A ribbon 48 m48\ \text{m} long is cut by Nina, Omar, and Pia, each taking the length they need to wrap a gift. Nina cut 16\frac{1}{6} of the whole ribbon, Omar cut 14\frac{1}{4} of the whole ribbon, and Pia cut 512\frac{5}{12} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 48 into 6 equal parts, then take 1 part.
48÷6=8,8×1=848 \div 6 = 8,\quad 8 \times 1 = 8
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 48 into 4 equal parts, then take 1 part.
48÷4=12,12×1=1248 \div 4 = 12,\quad 12 \times 1 = 12
Same split-and-take rule with denominator 4 and numerator 1.
#7 Identify Subproblems 3.NF.A.1
Divide 48 into 12 equal parts, then take 5 parts.
48÷12=4,4×5=2048 \div 12 = 4,\quad 4 \times 5 = 20
Same split-and-take rule with denominator 12 and numerator 5.
#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<20Pia8 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 48 m, take your parts, then see whose pile is biggest!
Variant 9 answer: Bram (the piece is 18 m, the longest)

A ribbon 90 m90\ \text{m} long is cut by Zane, Anya, and Bram, each taking the length they need to wrap a gift. Zane cut 19\frac{1}{9} of the whole ribbon, Anya cut 215\frac{2}{15} of the whole ribbon, and Bram cut 15\frac{1}{5} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 90 into 9 equal parts, then take 1 part.
90÷9=10,10×1=1090 \div 9 = 10,\quad 10 \times 1 = 10
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 90 into 15 equal parts, then take 2 parts.
90÷15=6,6×2=1290 \div 15 = 6,\quad 6 \times 2 = 12
Same split-and-take rule with denominator 15 and numerator 2.
#7 Identify Subproblems 3.NF.A.1
Divide 90 into 5 equal parts, then take 1 part.
90÷5=18,18×1=1890 \div 5 = 18,\quad 18 \times 1 = 18
Same split-and-take rule with denominator 5 and numerator 1.
#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<18Bram10 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 90 m, take your parts, then see whose pile is biggest!
Variant 10 answer: Esme (the piece is 18 m, the longest)

A ribbon 54 m54\ \text{m} long is cut by Cleo, Dara, and Esme, each taking the length they need to wrap a gift. Cleo cut 19\frac{1}{9} of the whole ribbon, Dara cut 29\frac{2}{9} of the whole ribbon, and Esme cut 13\frac{1}{3} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 54 into 9 equal parts, then take 1 part.
54÷9=6,6×1=654 \div 9 = 6,\quad 6 \times 1 = 6
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 54 into 9 equal parts, then take 2 parts.
54÷9=6,6×2=1254 \div 9 = 6,\quad 6 \times 2 = 12
Same split-and-take rule with denominator 9 and numerator 2.
#7 Identify Subproblems 3.NF.A.1
Divide 54 into 3 equal parts, then take 1 part.
54÷3=18,18×1=1854 \div 3 = 18,\quad 18 \times 1 = 18
Same split-and-take rule with denominator 3 and numerator 1.
#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<18Esme6 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 54 m, take your parts, then see whose pile is biggest!
Variant 11 answer: Rosa (the piece is 18 m, the longest)

A ribbon 45 m45\ \text{m} long is cut by Quinn, Rosa, and Sam, each taking the length they need to wrap a gift. Quinn cut 19\frac{1}{9} of the whole ribbon, Rosa cut 25\frac{2}{5} of the whole ribbon, and Sam cut 13\frac{1}{3} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 45 into 9 equal parts, then take 1 part.
45÷9=5,5×1=545 \div 9 = 5,\quad 5 \times 1 = 5
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 45 into 5 equal parts, then take 2 parts.
45÷5=9,9×2=1845 \div 5 = 9,\quad 9 \times 2 = 18
Same split-and-take rule with denominator 5 and numerator 2.
#7 Identify Subproblems 3.NF.A.1
Divide 45 into 3 equal parts, then take 1 part.
45÷3=15,15×1=1545 \div 3 = 15,\quad 15 \times 1 = 15
Same split-and-take rule with denominator 3 and numerator 1.
#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<18Rosa5 < 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)

Review

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.

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.

Standards · min 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.
💡 This only needs Grade 3 fraction sense: split the 45 m, take your parts, then see whose pile is biggest!
Variant 12 answer: Noah (the piece is 9 m, the longest)

A ribbon 30 m30\ \text{m} long is cut by Liam, Mia, and Noah, each taking the length they need to wrap a gift. Liam cut 16\frac{1}{6} of the whole ribbon, Mia cut 415\frac{4}{15} of the whole ribbon, and Noah cut 310\frac{3}{10} of the whole ribbon. Who cut the longest piece of ribbon?

Show solution

Understand

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.

Plan

#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.

Execute

#7 Identify Subproblems 3.NF.A.1
Divide 30 into 6 equal parts, then take 1 part.
30÷6=5,5×1=530 \div 6 = 5,\quad 5 \times 1 = 5
A fraction of a length means split into equal parts and take that many.
#7 Identify Subproblems 3.NF.A.1
Divide 30 into 15 equal parts, then take 4 parts.
30÷15=2,2×4=830 \div 15 = 2,\quad 2 \times 4 = 8
Same split-and-take rule with denominator 15 and numerator 4.
#7 Identify Subproblems 3.NF.A.1
Divide 30 into 10 equal parts, then take 3 parts.
30÷10=3,3×3=930 \div 10 = 3,\quad 3 \times 3 = 9
Same split-and-take rule with denominator 10 and numerator 3.
#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<9Noah5 < 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)

Review

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.

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.

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