Practice · Bouncing ball: repeated fraction-ratio heights · Sensim Math

Variant 1 easy

A ball bounces back up to $\dfrac{2}{3}$ of the height from which it falls. The ball is dropped straight down from a height of $12\text{ ft}$ above the floor. Find the height the ball reaches on its second bounce.

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1 · Understandwhat's really being asked

A ball dropped from 12 ft rebounds to $\dfrac{2}{3}$ of whatever height it fell from. We need how high it goes after 2 bounces.

Givens

The drop height is 12 ft.
Each bounce reaches $\dfrac{2}{3}$ of the height it fell from.

Unknowns

The height of the second bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 2 times.

3 · Execute3 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 12 ft, so it comes back up to

\dfrac{2}{3}

of 12 ft. Solving this one easier bounce shows what every later bounce does.

12 \times \dfrac{2}{3} = 8

The first bounce reaches 8 ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{2}{3}

. Because

\dfrac{2}{3}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{2}{3}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{2}{3}

again.

8 \times \dfrac{2}{3} = \dfrac{16}{3}

The second bounce reaches

5\frac{1}{3}

ft.

Answer:

\frac{16}{3}

ft (=

5\frac{1}{3}

ft ≈ 5.33 ft)

4 · Reviewdoes it hold up?

Multiplying 2 times by $\dfrac{2}{3}$ at once gives the same thing: 12 x (2/3)^2 = $\frac{16}{3}$ ft, about 5.33 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (8 ft) and larger than zero -- 5.33 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 2 easy

A ball bounces back up to $\dfrac{2}{3}$ of the height from which it falls. The ball is dropped straight down from a height of $25\text{ ft}$ above the floor. Find the height the ball reaches on its second bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 25 ft rebounds to $\dfrac{2}{3}$ of whatever height it fell from. We need how high it goes after 2 bounces.

Givens

The drop height is 25 ft.
Each bounce reaches $\dfrac{2}{3}$ of the height it fell from.

Unknowns

The height of the second bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 2 times.

3 · Execute3 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 25 ft, so it comes back up to

\dfrac{2}{3}

of 25 ft. Solving this one easier bounce shows what every later bounce does.

25 \times \dfrac{2}{3} = \dfrac{50}{3}

The first bounce reaches

16\frac{2}{3}

ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{2}{3}

. Because

\dfrac{2}{3}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{2}{3}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{2}{3}

again.

\dfrac{50}{3} \times \dfrac{2}{3} = \dfrac{100}{9}

The second bounce reaches

11\frac{1}{9}

ft.

Answer:

\frac{100}{9}

ft (=

11\frac{1}{9}

ft ≈ 11.11 ft)

4 · Reviewdoes it hold up?

Multiplying 2 times by $\dfrac{2}{3}$ at once gives the same thing: 25 x (2/3)^2 = $\frac{100}{9}$ ft, about 11.11 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce ( $16\frac{2}{3}$ ft) and larger than zero -- 11.11 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 3 easy

A ball bounces back up to $\dfrac{2}{3}$ of the height from which it falls. The ball is dropped straight down from a height of $30\text{ ft}$ above the floor. Find the height the ball reaches on its second bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 30 ft rebounds to $\dfrac{2}{3}$ of whatever height it fell from. We need how high it goes after 2 bounces.

Givens

The drop height is 30 ft.
Each bounce reaches $\dfrac{2}{3}$ of the height it fell from.

Unknowns

The height of the second bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 2 times.

3 · Execute3 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 30 ft, so it comes back up to

\dfrac{2}{3}

of 30 ft. Solving this one easier bounce shows what every later bounce does.

30 \times \dfrac{2}{3} = 20

The first bounce reaches 20 ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{2}{3}

. Because

\dfrac{2}{3}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{2}{3}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{2}{3}

again.

20 \times \dfrac{2}{3} = \dfrac{40}{3}

The second bounce reaches

13\frac{1}{3}

ft.

Answer:

\frac{40}{3}

ft (=

13\frac{1}{3}

ft ≈ 13.33 ft)

4 · Reviewdoes it hold up?

Multiplying 2 times by $\dfrac{2}{3}$ at once gives the same thing: 30 x (2/3)^2 = $\frac{40}{3}$ ft, about 13.33 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (20 ft) and larger than zero -- 13.33 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 4 easy

A ball bounces back up to $\dfrac{5}{6}$ of the height from which it falls. The ball is dropped straight down from a height of $33\text{ ft}$ above the floor. Find the height the ball reaches on its second bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 33 ft rebounds to $\dfrac{5}{6}$ of whatever height it fell from. We need how high it goes after 2 bounces.

Givens

The drop height is 33 ft.
Each bounce reaches $\dfrac{5}{6}$ of the height it fell from.

Unknowns

The height of the second bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 2 times.

3 · Execute3 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 33 ft, so it comes back up to

\dfrac{5}{6}

of 33 ft. Solving this one easier bounce shows what every later bounce does.

33 \times \dfrac{5}{6} = \dfrac{55}{2}

The first bounce reaches

27\frac{1}{2}

ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{5}{6}

. Because

\dfrac{5}{6}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{5}{6}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{5}{6}

again.

\dfrac{55}{2} \times \dfrac{5}{6} = \dfrac{275}{12}

The second bounce reaches

22\frac{11}{12}

ft.

Answer:

\frac{275}{12}

ft (=

22\frac{11}{12}

ft ≈ 22.92 ft)

4 · Reviewdoes it hold up?

Multiplying 2 times by $\dfrac{5}{6}$ at once gives the same thing: 33 x (5/6)^2 = $\frac{275}{12}$ ft, about 22.92 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce ( $27\frac{1}{2}$ ft) and larger than zero -- 22.92 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 5 medium

A ball bounces back up to $\dfrac{2}{3}$ of the height from which it falls. The ball is dropped straight down from a height of $35\text{ ft}$ above the floor. Find the height the ball reaches on its second bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 35 ft rebounds to $\dfrac{2}{3}$ of whatever height it fell from. We need how high it goes after 2 bounces.

Givens

The drop height is 35 ft.
Each bounce reaches $\dfrac{2}{3}$ of the height it fell from.

Unknowns

The height of the second bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 2 times.

3 · Execute3 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 35 ft, so it comes back up to

\dfrac{2}{3}

of 35 ft. Solving this one easier bounce shows what every later bounce does.

35 \times \dfrac{2}{3} = \dfrac{70}{3}

The first bounce reaches

23\frac{1}{3}

ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{2}{3}

. Because

\dfrac{2}{3}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{2}{3}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{2}{3}

again.

\dfrac{70}{3} \times \dfrac{2}{3} = \dfrac{140}{9}

The second bounce reaches

15\frac{5}{9}

ft.

Answer:

\frac{140}{9}

ft (=

15\frac{5}{9}

ft ≈ 15.56 ft)

4 · Reviewdoes it hold up?

Multiplying 2 times by $\dfrac{2}{3}$ at once gives the same thing: 35 x (2/3)^2 = $\frac{140}{9}$ ft, about 15.56 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce ( $23\frac{1}{3}$ ft) and larger than zero -- 15.56 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 6 medium

A ball bounces back up to $\dfrac{2}{3}$ of the height from which it falls. The ball is dropped straight down from a height of $18\text{ ft}$ above the floor. Find the height the ball reaches on its third bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 18 ft rebounds to $\dfrac{2}{3}$ of whatever height it fell from. We need how high it goes after 3 bounces.

Givens

The drop height is 18 ft.
Each bounce reaches $\dfrac{2}{3}$ of the height it fell from.

Unknowns

The height of the third bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 3 times.

3 · Execute4 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 18 ft, so it comes back up to

\dfrac{2}{3}

of 18 ft. Solving this one easier bounce shows what every later bounce does.

18 \times \dfrac{2}{3} = 12

The first bounce reaches 12 ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{2}{3}

. Because

\dfrac{2}{3}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{2}{3}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{2}{3}

again.

12 \times \dfrac{2}{3} = 8

The second bounce reaches 8 ft.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4

Take the second bounce height and multiply by

\dfrac{2}{3}

again.

8 \times \dfrac{2}{3} = \dfrac{16}{3}

The third bounce reaches

5\frac{1}{3}

ft.

Answer:

\frac{16}{3}

ft (=

5\frac{1}{3}

ft ≈ 5.33 ft)

4 · Reviewdoes it hold up?

Multiplying 3 times by $\dfrac{2}{3}$ at once gives the same thing: 18 x (2/3)^3 = $\frac{16}{3}$ ft, about 5.33 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (12 ft) and larger than zero -- 5.33 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 7 medium

A ball bounces back up to $\dfrac{2}{3}$ of the height from which it falls. The ball is dropped straight down from a height of $21\text{ ft}$ above the floor. Find the height the ball reaches on its third bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 21 ft rebounds to $\dfrac{2}{3}$ of whatever height it fell from. We need how high it goes after 3 bounces.

Givens

The drop height is 21 ft.
Each bounce reaches $\dfrac{2}{3}$ of the height it fell from.

Unknowns

The height of the third bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 3 times.

3 · Execute4 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 21 ft, so it comes back up to

\dfrac{2}{3}

of 21 ft. Solving this one easier bounce shows what every later bounce does.

21 \times \dfrac{2}{3} = 14

The first bounce reaches 14 ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{2}{3}

. Because

\dfrac{2}{3}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{2}{3}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{2}{3}

again.

14 \times \dfrac{2}{3} = \dfrac{28}{3}

The second bounce reaches

9\frac{1}{3}

ft.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4

Take the second bounce height and multiply by

\dfrac{2}{3}

again.

\dfrac{28}{3} \times \dfrac{2}{3} = \dfrac{56}{9}

The third bounce reaches

6\frac{2}{9}

ft.

Answer:

\frac{56}{9}

ft (=

6\frac{2}{9}

ft ≈ 6.22 ft)

4 · Reviewdoes it hold up?

Multiplying 3 times by $\dfrac{2}{3}$ at once gives the same thing: 21 x (2/3)^3 = $\frac{56}{9}$ ft, about 6.22 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (14 ft) and larger than zero -- 6.22 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 8 medium

A ball bounces back up to $\dfrac{5}{6}$ of the height from which it falls. The ball is dropped straight down from a height of $24\text{ ft}$ above the floor. Find the height the ball reaches on its third bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 24 ft rebounds to $\dfrac{5}{6}$ of whatever height it fell from. We need how high it goes after 3 bounces.

Givens

The drop height is 24 ft.
Each bounce reaches $\dfrac{5}{6}$ of the height it fell from.

Unknowns

The height of the third bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 3 times.

3 · Execute4 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 24 ft, so it comes back up to

\dfrac{5}{6}

of 24 ft. Solving this one easier bounce shows what every later bounce does.

24 \times \dfrac{5}{6} = 20

The first bounce reaches 20 ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{5}{6}

. Because

\dfrac{5}{6}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{5}{6}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{5}{6}

again.

20 \times \dfrac{5}{6} = \dfrac{50}{3}

The second bounce reaches

16\frac{2}{3}

ft.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4

Take the second bounce height and multiply by

\dfrac{5}{6}

again.

\dfrac{50}{3} \times \dfrac{5}{6} = \dfrac{125}{9}

The third bounce reaches

13\frac{8}{9}

ft.

Answer:

\frac{125}{9}

ft (=

13\frac{8}{9}

ft ≈ 13.89 ft)

4 · Reviewdoes it hold up?

Multiplying 3 times by $\dfrac{5}{6}$ at once gives the same thing: 24 x (5/6)^3 = $\frac{125}{9}$ ft, about 13.89 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (20 ft) and larger than zero -- 13.89 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 9 hard

A ball bounces back up to $\dfrac{5}{6}$ of the height from which it falls. The ball is dropped straight down from a height of $40\text{ ft}$ above the floor. Find the height the ball reaches on its third bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 40 ft rebounds to $\dfrac{5}{6}$ of whatever height it fell from. We need how high it goes after 3 bounces.

Givens

The drop height is 40 ft.
Each bounce reaches $\dfrac{5}{6}$ of the height it fell from.

Unknowns

The height of the third bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 3 times.

3 · Execute4 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 40 ft, so it comes back up to

\dfrac{5}{6}

of 40 ft. Solving this one easier bounce shows what every later bounce does.

40 \times \dfrac{5}{6} = \dfrac{100}{3}

The first bounce reaches

33\frac{1}{3}

ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{5}{6}

. Because

\dfrac{5}{6}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{5}{6}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{5}{6}

again.

\dfrac{100}{3} \times \dfrac{5}{6} = \dfrac{250}{9}

The second bounce reaches

27\frac{7}{9}

ft.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4

Take the second bounce height and multiply by

\dfrac{5}{6}

again.

\dfrac{250}{9} \times \dfrac{5}{6} = \dfrac{625}{27}

The third bounce reaches

23\frac{4}{27}

ft.

Answer:

\frac{625}{27}

ft (=

23\frac{4}{27}

ft ≈ 23.15 ft)

4 · Reviewdoes it hold up?

Multiplying 3 times by $\dfrac{5}{6}$ at once gives the same thing: 40 x (5/6)^3 = $\frac{625}{27}$ ft, about 23.15 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce ( $33\frac{1}{3}$ ft) and larger than zero -- 23.15 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 10 hard

A ball bounces back up to $\dfrac{2}{3}$ of the height from which it falls. The ball is dropped straight down from a height of $45\text{ ft}$ above the floor. Find the height the ball reaches on its third bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 45 ft rebounds to $\dfrac{2}{3}$ of whatever height it fell from. We need how high it goes after 3 bounces.

Givens

The drop height is 45 ft.
Each bounce reaches $\dfrac{2}{3}$ of the height it fell from.

Unknowns

The height of the third bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 3 times.

3 · Execute4 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 45 ft, so it comes back up to

\dfrac{2}{3}

of 45 ft. Solving this one easier bounce shows what every later bounce does.

45 \times \dfrac{2}{3} = 30

The first bounce reaches 30 ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{2}{3}

. Because

\dfrac{2}{3}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{2}{3}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{2}{3}

again.

30 \times \dfrac{2}{3} = 20

The second bounce reaches 20 ft.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4

Take the second bounce height and multiply by

\dfrac{2}{3}

again.

20 \times \dfrac{2}{3} = \dfrac{40}{3}

The third bounce reaches

13\frac{1}{3}

ft.

Answer:

\frac{40}{3}

ft (=

13\frac{1}{3}

ft ≈ 13.33 ft)

4 · Reviewdoes it hold up?

Multiplying 3 times by $\dfrac{2}{3}$ at once gives the same thing: 45 x (2/3)^3 = $\frac{40}{3}$ ft, about 13.33 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (30 ft) and larger than zero -- 13.33 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 11 hard

A ball bounces back up to $\dfrac{5}{6}$ of the height from which it falls. The ball is dropped straight down from a height of $48\text{ ft}$ above the floor. Find the height the ball reaches on its third bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 48 ft rebounds to $\dfrac{5}{6}$ of whatever height it fell from. We need how high it goes after 3 bounces.

Givens

The drop height is 48 ft.
Each bounce reaches $\dfrac{5}{6}$ of the height it fell from.

Unknowns

The height of the third bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 3 times.

3 · Execute4 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 48 ft, so it comes back up to

\dfrac{5}{6}

of 48 ft. Solving this one easier bounce shows what every later bounce does.

48 \times \dfrac{5}{6} = 40

The first bounce reaches 40 ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{5}{6}

. Because

\dfrac{5}{6}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{5}{6}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{5}{6}

again.

40 \times \dfrac{5}{6} = \dfrac{100}{3}

The second bounce reaches

33\frac{1}{3}

ft.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4

Take the second bounce height and multiply by

\dfrac{5}{6}

again.

\dfrac{100}{3} \times \dfrac{5}{6} = \dfrac{250}{9}

The third bounce reaches

27\frac{7}{9}

ft.

Answer:

\frac{250}{9}

ft (=

27\frac{7}{9}

ft ≈ 27.78 ft)

4 · Reviewdoes it hold up?

Multiplying 3 times by $\dfrac{5}{6}$ at once gives the same thing: 48 x (5/6)^3 = $\frac{250}{9}$ ft, about 27.78 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (40 ft) and larger than zero -- 27.78 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Variant 12 hard

A ball bounces back up to $\dfrac{5}{6}$ of the height from which it falls. The ball is dropped straight down from a height of $60\text{ ft}$ above the floor. Find the height the ball reaches on its fourth bounce.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A ball dropped from 60 ft rebounds to $\dfrac{5}{6}$ of whatever height it fell from. We need how high it goes after 4 bounces.

Givens

The drop height is 60 ft.
Each bounce reaches $\dfrac{5}{6}$ of the height it fell from.

Unknowns

The height of the fourth bounce.

Constraints

The same fraction applies to every bounce, not just the first.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem#8 Analyze the Units

Do one bounce first, notice that the next bounce does exactly the same thing to a smaller number, and then repeat it 4 times.

3 · Execute5 carry out the plan

1Work out the first bounce on its own

#9 Solve an Easier Related Problem 5.NF.B.6

The ball falls 60 ft, so it comes back up to

\dfrac{5}{6}

of 60 ft. Solving this one easier bounce shows what every later bounce does.

60 \times \dfrac{5}{6} = 50

The first bounce reaches 50 ft.

2See the repeating rule

#5 Look for a Pattern 5.NF.B.5

Every bounce starts from the height of the one before it, so every bounce multiplies by the same

\dfrac{5}{6}

. Because

\dfrac{5}{6}

is less than 1, each bounce is shorter than the last.

\text{next height} = \text{this height} \times \dfrac{5}{6}

The same multiplier, over and over.

3Apply the rule for the second bounce

#5 Look for a Pattern 5.NF.B.4

Take the first bounce height and multiply by

\dfrac{5}{6}

again.

50 \times \dfrac{5}{6} = \dfrac{125}{3}

The second bounce reaches

41\frac{2}{3}

ft.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4

Take the second bounce height and multiply by

\dfrac{5}{6}

again.

\dfrac{125}{3} \times \dfrac{5}{6} = \dfrac{625}{18}

The third bounce reaches

34\frac{13}{18}

ft.

5Apply the rule for the fourth bounce

#5 Look for a Pattern 5.NF.B.4

Take the third bounce height and multiply by

\dfrac{5}{6}

again.

\dfrac{625}{18} \times \dfrac{5}{6} = \dfrac{3125}{108}

The fourth bounce reaches

28\frac{101}{108}

ft.

Answer:

\frac{3125}{108}

ft (=

28\frac{101}{108}

ft ≈ 28.94 ft)

4 · Reviewdoes it hold up?

Multiplying 4 times by $\dfrac{5}{6}$ at once gives the same thing: 60 x (5/6)^4 = $\frac{3125}{108}$ ft, about 28.94 ft.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (50 ft) and larger than zero -- 28.94 ft sits there.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying each bounce height by the rebound fraction.
5.NF.B.5 Interpret multiplication as scaling or resizing — Reading a fraction less than 1 as something that shrinks a height.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Setting the first bounce up from the real-world drop height.

💡Takeaway. When the same rule happens again and again, do it once carefully and then repeat it -- each bounce just multiplies by the same fraction.

Bouncing ball: repeated fraction-ratio heights · 12 practice problems

Generated variants — 12

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