Practice · Whole as 1: find remaining fraction · Sensim Math

Variant 1 easy

Out of $48$ apples, $\dfrac{1}{6}$ are placed in box A, and $\dfrac{3}{5}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

48 apples are sorted. Box A takes 1 of every 6. Box B then takes 3 of every 5 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 48 apples to begin with.
Box A takes $\frac{1}{6}$ of them.
Box B takes $\frac{3}{5}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 48.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{1}{6}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 48 into 6 equal parts and give 1 of them to box A.

48 \div 6 \times 1 = 8

Box A holds 8 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{5}{6}

of the whole.

48 - 8 = 40

40 apples are still loose.

3Box B: take $\frac{3}{5}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 40 are split into 5 parts and 3 of them go to box B. Note this is 5ths of 40, not of 48.

40 \div 5 \times 3 = 24

Box B holds 24 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 40 that were loose.

40 - 24 = 16

16 apples are left.

Answer: 16 apples

4 · Reviewdoes it hold up?

Everything adds back: 8 + 24 + 16 = 48.

Another way: Treat the whole as 1: $\frac{5}{6}$ survives box A and $\frac{2}{5}$ of that survives box B, so $\frac{5}{6}$ x $\frac{2}{5}$ = $\frac{1}{3}$ of 48 is 16 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 2 easy

Out of $54$ apples, $\dfrac{2}{9}$ are placed in box A, and $\dfrac{1}{2}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

54 apples are sorted. Box A takes 2 of every 9. Box B then takes 1 of every 2 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 54 apples to begin with.
Box A takes $\frac{2}{9}$ of them.
Box B takes $\frac{1}{2}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 54.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{2}{9}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 54 into 9 equal parts and give 2 of them to box A.

54 \div 9 \times 2 = 12

Box A holds 12 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{7}{9}

of the whole.

54 - 12 = 42

42 apples are still loose.

3Box B: take $\frac{1}{2}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 42 are split into 2 parts and 1 of them go to box B. Note this is 2ths of 42, not of 54.

42 \div 2 \times 1 = 21

Box B holds 21 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 42 that were loose.

42 - 21 = 21

21 apples are left.

Answer: 21 apples

4 · Reviewdoes it hold up?

Everything adds back: 12 + 21 + 21 = 54.

Another way: Treat the whole as 1: $\frac{7}{9}$ survives box A and $\frac{1}{2}$ of that survives box B, so $\frac{7}{9}$ x $\frac{1}{2}$ = $\frac{7}{18}$ of 54 is 21 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 3 easy

Out of $60$ apples, $\dfrac{1}{3}$ are placed in box A, and $\dfrac{3}{4}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

60 apples are sorted. Box A takes 1 of every 3. Box B then takes 3 of every 4 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 60 apples to begin with.
Box A takes $\frac{1}{3}$ of them.
Box B takes $\frac{3}{4}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 60.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{1}{3}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 60 into 3 equal parts and give 1 of them to box A.

60 \div 3 \times 1 = 20

Box A holds 20 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{2}{3}

of the whole.

60 - 20 = 40

40 apples are still loose.

3Box B: take $\frac{3}{4}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 40 are split into 4 parts and 3 of them go to box B. Note this is 4ths of 40, not of 60.

40 \div 4 \times 3 = 30

Box B holds 30 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 40 that were loose.

40 - 30 = 10

10 apples are left.

Answer: 10 apples

4 · Reviewdoes it hold up?

Everything adds back: 20 + 30 + 10 = 60.

Another way: Treat the whole as 1: $\frac{2}{3}$ survives box A and $\frac{1}{4}$ of that survives box B, so $\frac{2}{3}$ x $\frac{1}{4}$ = $\frac{1}{6}$ of 60 is 10 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 4 easy

Out of $66$ apples, $\dfrac{1}{11}$ are placed in box A, and $\dfrac{2}{5}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

66 apples are sorted. Box A takes 1 of every 11. Box B then takes 2 of every 5 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 66 apples to begin with.
Box A takes $\frac{1}{11}$ of them.
Box B takes $\frac{2}{5}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 66.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{1}{11}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 66 into 11 equal parts and give 1 of them to box A.

66 \div 11 \times 1 = 6

Box A holds 6 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{10}{11}

of the whole.

66 - 6 = 60

60 apples are still loose.

3Box B: take $\frac{2}{5}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 60 are split into 5 parts and 2 of them go to box B. Note this is 5ths of 60, not of 66.

60 \div 5 \times 2 = 24

Box B holds 24 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 60 that were loose.

60 - 24 = 36

36 apples are left.

Answer: 36 apples

4 · Reviewdoes it hold up?

Everything adds back: 6 + 24 + 36 = 66.

Another way: Treat the whole as 1: $\frac{10}{11}$ survives box A and $\frac{3}{5}$ of that survives box B, so $\frac{10}{11}$ x $\frac{3}{5}$ = $\frac{6}{11}$ of 66 is 36 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 5 medium

Out of $72$ apples, $\dfrac{1}{4}$ are placed in box A, and $\dfrac{1}{3}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

72 apples are sorted. Box A takes 1 of every 4. Box B then takes 1 of every 3 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 72 apples to begin with.
Box A takes $\frac{1}{4}$ of them.
Box B takes $\frac{1}{3}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 72.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{1}{4}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 72 into 4 equal parts and give 1 of them to box A.

72 \div 4 \times 1 = 18

Box A holds 18 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{3}{4}

of the whole.

72 - 18 = 54

54 apples are still loose.

3Box B: take $\frac{1}{3}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 54 are split into 3 parts and 1 of them go to box B. Note this is 3ths of 54, not of 72.

54 \div 3 \times 1 = 18

Box B holds 18 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 54 that were loose.

54 - 18 = 36

36 apples are left.

Answer: 36 apples

4 · Reviewdoes it hold up?

Everything adds back: 18 + 18 + 36 = 72.

Another way: Treat the whole as 1: $\frac{3}{4}$ survives box A and $\frac{2}{3}$ of that survives box B, so $\frac{3}{4}$ x $\frac{2}{3}$ = $\frac{1}{2}$ of 72 is 36 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 6 medium

Out of $75$ apples, $\dfrac{2}{15}$ are placed in box A, and $\dfrac{3}{13}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

75 apples are sorted. Box A takes 2 of every 15. Box B then takes 3 of every 13 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 75 apples to begin with.
Box A takes $\frac{2}{15}$ of them.
Box B takes $\frac{3}{13}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 75.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{2}{15}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 75 into 15 equal parts and give 2 of them to box A.

75 \div 15 \times 2 = 10

Box A holds 10 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{13}{15}

of the whole.

75 - 10 = 65

65 apples are still loose.

3Box B: take $\frac{3}{13}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 65 are split into 13 parts and 3 of them go to box B. Note this is 13ths of 65, not of 75.

65 \div 13 \times 3 = 15

Box B holds 15 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 65 that were loose.

65 - 15 = 50

50 apples are left.

Answer: 50 apples

4 · Reviewdoes it hold up?

Everything adds back: 10 + 15 + 50 = 75.

Another way: Treat the whole as 1: $\frac{13}{15}$ survives box A and $\frac{10}{13}$ of that survives box B, so $\frac{13}{15}$ x $\frac{10}{13}$ = $\frac{2}{3}$ of 75 is 50 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 7 medium

Out of $80$ apples, $\dfrac{1}{4}$ are placed in box A, and $\dfrac{2}{3}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

80 apples are sorted. Box A takes 1 of every 4. Box B then takes 2 of every 3 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 80 apples to begin with.
Box A takes $\frac{1}{4}$ of them.
Box B takes $\frac{2}{3}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 80.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{1}{4}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 80 into 4 equal parts and give 1 of them to box A.

80 \div 4 \times 1 = 20

Box A holds 20 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{3}{4}

of the whole.

80 - 20 = 60

60 apples are still loose.

3Box B: take $\frac{2}{3}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 60 are split into 3 parts and 2 of them go to box B. Note this is 3ths of 60, not of 80.

60 \div 3 \times 2 = 40

Box B holds 40 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 60 that were loose.

60 - 40 = 20

20 apples are left.

Answer: 20 apples

4 · Reviewdoes it hold up?

Everything adds back: 20 + 40 + 20 = 80.

Another way: Treat the whole as 1: $\frac{3}{4}$ survives box A and $\frac{1}{3}$ of that survives box B, so $\frac{3}{4}$ x $\frac{1}{3}$ = $\frac{1}{4}$ of 80 is 20 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 8 medium

Out of $84$ apples, $\dfrac{1}{7}$ are placed in box A, and $\dfrac{5}{6}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

84 apples are sorted. Box A takes 1 of every 7. Box B then takes 5 of every 6 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 84 apples to begin with.
Box A takes $\frac{1}{7}$ of them.
Box B takes $\frac{5}{6}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 84.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{1}{7}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 84 into 7 equal parts and give 1 of them to box A.

84 \div 7 \times 1 = 12

Box A holds 12 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{6}{7}

of the whole.

84 - 12 = 72

72 apples are still loose.

3Box B: take $\frac{5}{6}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 72 are split into 6 parts and 5 of them go to box B. Note this is 6ths of 72, not of 84.

72 \div 6 \times 5 = 60

Box B holds 60 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 72 that were loose.

72 - 60 = 12

12 apples are left.

Answer: 12 apples

4 · Reviewdoes it hold up?

Everything adds back: 12 + 60 + 12 = 84.

Another way: Treat the whole as 1: $\frac{6}{7}$ survives box A and $\frac{1}{6}$ of that survives box B, so $\frac{6}{7}$ x $\frac{1}{6}$ = $\frac{1}{7}$ of 84 is 12 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 9 hard

Out of $90$ apples, $\dfrac{2}{5}$ are placed in box A, and $\dfrac{2}{3}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

90 apples are sorted. Box A takes 2 of every 5. Box B then takes 2 of every 3 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 90 apples to begin with.
Box A takes $\frac{2}{5}$ of them.
Box B takes $\frac{2}{3}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 90.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{2}{5}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 90 into 5 equal parts and give 2 of them to box A.

90 \div 5 \times 2 = 36

Box A holds 36 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{3}{5}

of the whole.

90 - 36 = 54

54 apples are still loose.

3Box B: take $\frac{2}{3}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 54 are split into 3 parts and 2 of them go to box B. Note this is 3ths of 54, not of 90.

54 \div 3 \times 2 = 36

Box B holds 36 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 54 that were loose.

54 - 36 = 18

18 apples are left.

Answer: 18 apples

4 · Reviewdoes it hold up?

Everything adds back: 36 + 36 + 18 = 90.

Another way: Treat the whole as 1: $\frac{3}{5}$ survives box A and $\frac{1}{3}$ of that survives box B, so $\frac{3}{5}$ x $\frac{1}{3}$ = $\frac{1}{5}$ of 90 is 18 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 10 hard

Out of $100$ apples, $\dfrac{3}{10}$ are placed in box A, and $\dfrac{4}{7}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

100 apples are sorted. Box A takes 3 of every 10. Box B then takes 4 of every 7 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 100 apples to begin with.
Box A takes $\frac{3}{10}$ of them.
Box B takes $\frac{4}{7}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 100.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{3}{10}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 100 into 10 equal parts and give 3 of them to box A.

100 \div 10 \times 3 = 30

Box A holds 30 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{7}{10}

of the whole.

100 - 30 = 70

70 apples are still loose.

3Box B: take $\frac{4}{7}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 70 are split into 7 parts and 4 of them go to box B. Note this is 7ths of 70, not of 100.

70 \div 7 \times 4 = 40

Box B holds 40 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 70 that were loose.

70 - 40 = 30

30 apples are left.

Answer: 30 apples

4 · Reviewdoes it hold up?

Everything adds back: 30 + 40 + 30 = 100.

Another way: Treat the whole as 1: $\frac{7}{10}$ survives box A and $\frac{3}{7}$ of that survives box B, so $\frac{7}{10}$ x $\frac{3}{7}$ = $\frac{3}{10}$ of 100 is 30 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 11 hard

Out of $120$ apples, $\dfrac{1}{5}$ are placed in box A, and $\dfrac{3}{8}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

120 apples are sorted. Box A takes 1 of every 5. Box B then takes 3 of every 8 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 120 apples to begin with.
Box A takes $\frac{1}{5}$ of them.
Box B takes $\frac{3}{8}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 120.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{1}{5}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 120 into 5 equal parts and give 1 of them to box A.

120 \div 5 \times 1 = 24

Box A holds 24 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{4}{5}

of the whole.

120 - 24 = 96

96 apples are still loose.

3Box B: take $\frac{3}{8}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 96 are split into 8 parts and 3 of them go to box B. Note this is 8ths of 96, not of 120.

96 \div 8 \times 3 = 36

Box B holds 36 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 96 that were loose.

96 - 36 = 60

60 apples are left.

Answer: 60 apples

4 · Reviewdoes it hold up?

Everything adds back: 24 + 36 + 60 = 120.

Another way: Treat the whole as 1: $\frac{4}{5}$ survives box A and $\frac{5}{8}$ of that survives box B, so $\frac{4}{5}$ x $\frac{5}{8}$ = $\frac{1}{2}$ of 120 is 60 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Variant 12 hard

Out of $144$ apples, $\dfrac{5}{12}$ are placed in box A, and $\dfrac{3}{7}$ of the rest are placed in box B.

How many apples are left after filling both boxes?

Your answer

Show solution

1 · Understandwhat's really being asked

144 apples are sorted. Box A takes 5 of every 12. Box B then takes 3 of every 7 of whatever is still loose. We need how many are still loose at the end.

Givens

There are 144 apples to begin with.
Box A takes $\frac{5}{12}$ of them.
Box B takes $\frac{3}{7}$ of the rest.

Unknowns

How many apples are left.

Constraints

Box B's fraction is of the leftover, not of the original 144.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Draw one bar for the whole batch. Shade box A's share, then shade box B's share of only the unshaded part. What stays blank is the answer, and the picture makes clear which whole each fraction refers to.

3 · Execute4 carry out the plan

1Box A: take $\frac{5}{12}$ of the whole bar

#1 Draw a Diagram 5.NF.B.6

Split the 144 into 12 equal parts and give 5 of them to box A.

144 \div 12 \times 5 = 60

Box A holds 60 apples.

2Find the rest (the remaining bar)

#7 Identify Subproblems 5.NF.B.6

Taking box A's share away leaves the part box B will draw from -- that is

\frac{7}{12}

of the whole.

144 - 60 = 84

84 apples are still loose.

3Box B: take $\frac{3}{7}$ of the rest

#1 Draw a Diagram 5.NF.B.4

The 84 are split into 7 parts and 3 of them go to box B. Note this is 7ths of 84, not of 144.

84 \div 7 \times 3 = 36

Box B holds 36 apples.

4What is left after both boxes

#7 Identify Subproblems 5.NF.B.6

Take box B's share off the 84 that were loose.

84 - 36 = 48

48 apples are left.

Answer: 48 apples

4 · Reviewdoes it hold up?

Everything adds back: 60 + 36 + 48 = 144.

Another way: Treat the whole as 1: $\frac{7}{12}$ survives box A and $\frac{4}{7}$ of that survives box B, so $\frac{7}{12}$ x $\frac{4}{7}$ = $\frac{1}{3}$ of 144 is 48 -- one multiplication instead of four steps.

Standardsmin grade 5

5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Taking each box's share and subtracting it from what was there.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of the leftover rather than of the original.

💡Takeaway. Ask what each fraction is a fraction OF. Box B's share comes from the leftover, so it is smaller than it looks.

Whole as 1: find remaining fraction · 12 practice problems

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