← Bouncing ball: repeated fraction-ratio heights · Part-Whole Fraction Reasoning

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

5.NF.B.45.NF.B.55.NF.B.6

Generated variants — 12

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

Variant 1 easy answer: 163\frac{16}{3} m (= 5135\frac{1}{3} m ≈ 5.33 m)

A ball bounces back up to 23\dfrac{2}{3} of the height from which it falls. The ball is dropped straight down from a height of 12 m12\text{ m} above the floor. Find the height the ball reaches on its second bounce.

12 m 1st 2nd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 12 m rebounds to 23\dfrac{2}{3} of whatever height it fell from. We need how high it goes after 2 bounces.

Givens
  • The drop height is 12 m.
  • Each bounce reaches 23\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 m, so it comes back up to 23\dfrac{2}{3} of 12 m. Solving this one easier bounce shows what every later bounce does.
12×23=812 \times \dfrac{2}{3} = 8
The first bounce reaches 8 m.

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 23\dfrac{2}{3}. Because 23\dfrac{2}{3} is less than 1, each bounce is shorter than the last.
next height=this height×23\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 23\dfrac{2}{3} again.
8×23=1638 \times \dfrac{2}{3} = \dfrac{16}{3}
The second bounce reaches 5 1/3 m.
Answer: 163\frac{16}{3} m (= 5135\frac{1}{3} m ≈ 5.33 m)
4 · Reviewdoes it hold up?

Multiplying 2 times by 23\dfrac{2}{3} at once gives the same thing: 12 x (2/3)^2 = 16/3 m, about 5.33 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (8 m) and larger than zero -- 5.33 m 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 answer: 1009\frac{100}{9} m (= 111911\frac{1}{9} m ≈ 11.11 m)

A ball bounces back up to 23\dfrac{2}{3} of the height from which it falls. The ball is dropped straight down from a height of 25 m25\text{ m} above the floor. Find the height the ball reaches on its second bounce.

25 m 1st 2nd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 25 m rebounds to 23\dfrac{2}{3} of whatever height it fell from. We need how high it goes after 2 bounces.

Givens
  • The drop height is 25 m.
  • Each bounce reaches 23\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 m, so it comes back up to 23\dfrac{2}{3} of 25 m. Solving this one easier bounce shows what every later bounce does.
25×23=50325 \times \dfrac{2}{3} = \dfrac{50}{3}
The first bounce reaches 16 2/3 m.

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 23\dfrac{2}{3}. Because 23\dfrac{2}{3} is less than 1, each bounce is shorter than the last.
next height=this height×23\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 23\dfrac{2}{3} again.
503×23=1009\dfrac{50}{3} \times \dfrac{2}{3} = \dfrac{100}{9}
The second bounce reaches 11 1/9 m.
Answer: 1009\frac{100}{9} m (= 111911\frac{1}{9} m ≈ 11.11 m)
4 · Reviewdoes it hold up?

Multiplying 2 times by 23\dfrac{2}{3} at once gives the same thing: 25 x (2/3)^2 = 100/9 m, about 11.11 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (16 2/3 m) and larger than zero -- 11.11 m 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 answer: 403\frac{40}{3} m (= 131313\frac{1}{3} m ≈ 13.33 m)

A ball bounces back up to 23\dfrac{2}{3} of the height from which it falls. The ball is dropped straight down from a height of 30 m30\text{ m} above the floor. Find the height the ball reaches on its second bounce.

30 m 1st 2nd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 30 m rebounds to 23\dfrac{2}{3} of whatever height it fell from. We need how high it goes after 2 bounces.

Givens
  • The drop height is 30 m.
  • Each bounce reaches 23\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 m, so it comes back up to 23\dfrac{2}{3} of 30 m. Solving this one easier bounce shows what every later bounce does.
30×23=2030 \times \dfrac{2}{3} = 20
The first bounce reaches 20 m.

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 23\dfrac{2}{3}. Because 23\dfrac{2}{3} is less than 1, each bounce is shorter than the last.
next height=this height×23\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 23\dfrac{2}{3} again.
20×23=40320 \times \dfrac{2}{3} = \dfrac{40}{3}
The second bounce reaches 13 1/3 m.
Answer: 403\frac{40}{3} m (= 131313\frac{1}{3} m ≈ 13.33 m)
4 · Reviewdoes it hold up?

Multiplying 2 times by 23\dfrac{2}{3} at once gives the same thing: 30 x (2/3)^2 = 40/3 m, about 13.33 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (20 m) and larger than zero -- 13.33 m 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 answer: 27512\frac{275}{12} m (= 22111222\frac{11}{12} m ≈ 22.92 m)

A ball bounces back up to 56\dfrac{5}{6} of the height from which it falls. The ball is dropped straight down from a height of 33 m33\text{ m} above the floor. Find the height the ball reaches on its second bounce.

33 m 1st 2nd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 33 m rebounds to 56\dfrac{5}{6} of whatever height it fell from. We need how high it goes after 2 bounces.

Givens
  • The drop height is 33 m.
  • Each bounce reaches 56\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 m, so it comes back up to 56\dfrac{5}{6} of 33 m. Solving this one easier bounce shows what every later bounce does.
33×56=55233 \times \dfrac{5}{6} = \dfrac{55}{2}
The first bounce reaches 27 1/2 m.

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 56\dfrac{5}{6}. Because 56\dfrac{5}{6} is less than 1, each bounce is shorter than the last.
next height=this height×56\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 56\dfrac{5}{6} again.
552×56=27512\dfrac{55}{2} \times \dfrac{5}{6} = \dfrac{275}{12}
The second bounce reaches 22 11/12 m.
Answer: 27512\frac{275}{12} m (= 22111222\frac{11}{12} m ≈ 22.92 m)
4 · Reviewdoes it hold up?

Multiplying 2 times by 56\dfrac{5}{6} at once gives the same thing: 33 x (5/6)^2 = 275/12 m, about 22.92 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (27 1/2 m) and larger than zero -- 22.92 m 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 answer: 1409\frac{140}{9} m (= 155915\frac{5}{9} m ≈ 15.56 m)

A ball bounces back up to 23\dfrac{2}{3} of the height from which it falls. The ball is dropped straight down from a height of 35 m35\text{ m} above the floor. Find the height the ball reaches on its second bounce.

35 m 1st 2nd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 35 m rebounds to 23\dfrac{2}{3} of whatever height it fell from. We need how high it goes after 2 bounces.

Givens
  • The drop height is 35 m.
  • Each bounce reaches 23\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 m, so it comes back up to 23\dfrac{2}{3} of 35 m. Solving this one easier bounce shows what every later bounce does.
35×23=70335 \times \dfrac{2}{3} = \dfrac{70}{3}
The first bounce reaches 23 1/3 m.

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 23\dfrac{2}{3}. Because 23\dfrac{2}{3} is less than 1, each bounce is shorter than the last.
next height=this height×23\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 23\dfrac{2}{3} again.
703×23=1409\dfrac{70}{3} \times \dfrac{2}{3} = \dfrac{140}{9}
The second bounce reaches 15 5/9 m.
Answer: 1409\frac{140}{9} m (= 155915\frac{5}{9} m ≈ 15.56 m)
4 · Reviewdoes it hold up?

Multiplying 2 times by 23\dfrac{2}{3} at once gives the same thing: 35 x (2/3)^2 = 140/9 m, about 15.56 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (23 1/3 m) and larger than zero -- 15.56 m 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 answer: 163\frac{16}{3} m (= 5135\frac{1}{3} m ≈ 5.33 m)

A ball bounces back up to 23\dfrac{2}{3} of the height from which it falls. The ball is dropped straight down from a height of 18 m18\text{ m} above the floor. Find the height the ball reaches on its third bounce.

18 m 1st 2nd 3rd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 18 m rebounds to 23\dfrac{2}{3} of whatever height it fell from. We need how high it goes after 3 bounces.

Givens
  • The drop height is 18 m.
  • Each bounce reaches 23\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 m, so it comes back up to 23\dfrac{2}{3} of 18 m. Solving this one easier bounce shows what every later bounce does.
18×23=1218 \times \dfrac{2}{3} = 12
The first bounce reaches 12 m.

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 23\dfrac{2}{3}. Because 23\dfrac{2}{3} is less than 1, each bounce is shorter than the last.
next height=this height×23\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 23\dfrac{2}{3} again.
12×23=812 \times \dfrac{2}{3} = 8
The second bounce reaches 8 m.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4
Take the second bounce height and multiply by 23\dfrac{2}{3} again.
8×23=1638 \times \dfrac{2}{3} = \dfrac{16}{3}
The third bounce reaches 5 1/3 m.
Answer: 163\frac{16}{3} m (= 5135\frac{1}{3} m ≈ 5.33 m)
4 · Reviewdoes it hold up?

Multiplying 3 times by 23\dfrac{2}{3} at once gives the same thing: 18 x (2/3)^3 = 16/3 m, about 5.33 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (12 m) and larger than zero -- 5.33 m 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 answer: 569\frac{56}{9} m (= 6296\frac{2}{9} m ≈ 6.22 m)

A ball bounces back up to 23\dfrac{2}{3} of the height from which it falls. The ball is dropped straight down from a height of 21 m21\text{ m} above the floor. Find the height the ball reaches on its third bounce.

21 m 1st 2nd 3rd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 21 m rebounds to 23\dfrac{2}{3} of whatever height it fell from. We need how high it goes after 3 bounces.

Givens
  • The drop height is 21 m.
  • Each bounce reaches 23\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 m, so it comes back up to 23\dfrac{2}{3} of 21 m. Solving this one easier bounce shows what every later bounce does.
21×23=1421 \times \dfrac{2}{3} = 14
The first bounce reaches 14 m.

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 23\dfrac{2}{3}. Because 23\dfrac{2}{3} is less than 1, each bounce is shorter than the last.
next height=this height×23\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 23\dfrac{2}{3} again.
14×23=28314 \times \dfrac{2}{3} = \dfrac{28}{3}
The second bounce reaches 9 1/3 m.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4
Take the second bounce height and multiply by 23\dfrac{2}{3} again.
283×23=569\dfrac{28}{3} \times \dfrac{2}{3} = \dfrac{56}{9}
The third bounce reaches 6 2/9 m.
Answer: 569\frac{56}{9} m (= 6296\frac{2}{9} m ≈ 6.22 m)
4 · Reviewdoes it hold up?

Multiplying 3 times by 23\dfrac{2}{3} at once gives the same thing: 21 x (2/3)^3 = 56/9 m, about 6.22 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (14 m) and larger than zero -- 6.22 m 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 answer: 1259\frac{125}{9} m (= 138913\frac{8}{9} m ≈ 13.89 m)

A ball bounces back up to 56\dfrac{5}{6} of the height from which it falls. The ball is dropped straight down from a height of 24 m24\text{ m} above the floor. Find the height the ball reaches on its third bounce.

24 m 1st 2nd 3rd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 24 m rebounds to 56\dfrac{5}{6} of whatever height it fell from. We need how high it goes after 3 bounces.

Givens
  • The drop height is 24 m.
  • Each bounce reaches 56\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 m, so it comes back up to 56\dfrac{5}{6} of 24 m. Solving this one easier bounce shows what every later bounce does.
24×56=2024 \times \dfrac{5}{6} = 20
The first bounce reaches 20 m.

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 56\dfrac{5}{6}. Because 56\dfrac{5}{6} is less than 1, each bounce is shorter than the last.
next height=this height×56\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 56\dfrac{5}{6} again.
20×56=50320 \times \dfrac{5}{6} = \dfrac{50}{3}
The second bounce reaches 16 2/3 m.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4
Take the second bounce height and multiply by 56\dfrac{5}{6} again.
503×56=1259\dfrac{50}{3} \times \dfrac{5}{6} = \dfrac{125}{9}
The third bounce reaches 13 8/9 m.
Answer: 1259\frac{125}{9} m (= 138913\frac{8}{9} m ≈ 13.89 m)
4 · Reviewdoes it hold up?

Multiplying 3 times by 56\dfrac{5}{6} at once gives the same thing: 24 x (5/6)^3 = 125/9 m, about 13.89 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (20 m) and larger than zero -- 13.89 m 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 answer: 62527\frac{625}{27} m (= 2342723\frac{4}{27} m ≈ 23.15 m)

A ball bounces back up to 56\dfrac{5}{6} of the height from which it falls. The ball is dropped straight down from a height of 40 m40\text{ m} above the floor. Find the height the ball reaches on its third bounce.

40 m 1st 2nd 3rd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 40 m rebounds to 56\dfrac{5}{6} of whatever height it fell from. We need how high it goes after 3 bounces.

Givens
  • The drop height is 40 m.
  • Each bounce reaches 56\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 m, so it comes back up to 56\dfrac{5}{6} of 40 m. Solving this one easier bounce shows what every later bounce does.
40×56=100340 \times \dfrac{5}{6} = \dfrac{100}{3}
The first bounce reaches 33 1/3 m.

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 56\dfrac{5}{6}. Because 56\dfrac{5}{6} is less than 1, each bounce is shorter than the last.
next height=this height×56\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 56\dfrac{5}{6} again.
1003×56=2509\dfrac{100}{3} \times \dfrac{5}{6} = \dfrac{250}{9}
The second bounce reaches 27 7/9 m.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4
Take the second bounce height and multiply by 56\dfrac{5}{6} again.
2509×56=62527\dfrac{250}{9} \times \dfrac{5}{6} = \dfrac{625}{27}
The third bounce reaches 23 4/27 m.
Answer: 62527\frac{625}{27} m (= 2342723\frac{4}{27} m ≈ 23.15 m)
4 · Reviewdoes it hold up?

Multiplying 3 times by 56\dfrac{5}{6} at once gives the same thing: 40 x (5/6)^3 = 625/27 m, about 23.15 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (33 1/3 m) and larger than zero -- 23.15 m 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 answer: 403\frac{40}{3} m (= 131313\frac{1}{3} m ≈ 13.33 m)

A ball bounces back up to 23\dfrac{2}{3} of the height from which it falls. The ball is dropped straight down from a height of 45 m45\text{ m} above the floor. Find the height the ball reaches on its third bounce.

45 m 1st 2nd 3rd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 45 m rebounds to 23\dfrac{2}{3} of whatever height it fell from. We need how high it goes after 3 bounces.

Givens
  • The drop height is 45 m.
  • Each bounce reaches 23\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 m, so it comes back up to 23\dfrac{2}{3} of 45 m. Solving this one easier bounce shows what every later bounce does.
45×23=3045 \times \dfrac{2}{3} = 30
The first bounce reaches 30 m.

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 23\dfrac{2}{3}. Because 23\dfrac{2}{3} is less than 1, each bounce is shorter than the last.
next height=this height×23\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 23\dfrac{2}{3} again.
30×23=2030 \times \dfrac{2}{3} = 20
The second bounce reaches 20 m.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4
Take the second bounce height and multiply by 23\dfrac{2}{3} again.
20×23=40320 \times \dfrac{2}{3} = \dfrac{40}{3}
The third bounce reaches 13 1/3 m.
Answer: 403\frac{40}{3} m (= 131313\frac{1}{3} m ≈ 13.33 m)
4 · Reviewdoes it hold up?

Multiplying 3 times by 23\dfrac{2}{3} at once gives the same thing: 45 x (2/3)^3 = 40/3 m, about 13.33 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (30 m) and larger than zero -- 13.33 m 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 answer: 2509\frac{250}{9} m (= 277927\frac{7}{9} m ≈ 27.78 m)

A ball bounces back up to 56\dfrac{5}{6} of the height from which it falls. The ball is dropped straight down from a height of 48 m48\text{ m} above the floor. Find the height the ball reaches on its third bounce.

48 m 1st 2nd 3rd floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 48 m rebounds to 56\dfrac{5}{6} of whatever height it fell from. We need how high it goes after 3 bounces.

Givens
  • The drop height is 48 m.
  • Each bounce reaches 56\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 m, so it comes back up to 56\dfrac{5}{6} of 48 m. Solving this one easier bounce shows what every later bounce does.
48×56=4048 \times \dfrac{5}{6} = 40
The first bounce reaches 40 m.

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 56\dfrac{5}{6}. Because 56\dfrac{5}{6} is less than 1, each bounce is shorter than the last.
next height=this height×56\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 56\dfrac{5}{6} again.
40×56=100340 \times \dfrac{5}{6} = \dfrac{100}{3}
The second bounce reaches 33 1/3 m.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4
Take the second bounce height and multiply by 56\dfrac{5}{6} again.
1003×56=2509\dfrac{100}{3} \times \dfrac{5}{6} = \dfrac{250}{9}
The third bounce reaches 27 7/9 m.
Answer: 2509\frac{250}{9} m (= 277927\frac{7}{9} m ≈ 27.78 m)
4 · Reviewdoes it hold up?

Multiplying 3 times by 56\dfrac{5}{6} at once gives the same thing: 48 x (5/6)^3 = 250/9 m, about 27.78 m.

Another way: Because every bounce shrinks the height, the answer has to be smaller than the first bounce (40 m) and larger than zero -- 27.78 m 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 answer: 3125108\frac{3125}{108} m (= 2810110828\frac{101}{108} m ≈ 28.94 m)

A ball bounces back up to 56\dfrac{5}{6} of the height from which it falls. The ball is dropped straight down from a height of 60 m60\text{ m} above the floor. Find the height the ball reaches on its fourth bounce.

60 m 1st 2nd 3rd 4th floor
Show solution
1 · Understandwhat's really being asked

A ball dropped from 60 m rebounds to 56\dfrac{5}{6} of whatever height it fell from. We need how high it goes after 4 bounces.

Givens
  • The drop height is 60 m.
  • Each bounce reaches 56\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 m, so it comes back up to 56\dfrac{5}{6} of 60 m. Solving this one easier bounce shows what every later bounce does.
60×56=5060 \times \dfrac{5}{6} = 50
The first bounce reaches 50 m.

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 56\dfrac{5}{6}. Because 56\dfrac{5}{6} is less than 1, each bounce is shorter than the last.
next height=this height×56\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 56\dfrac{5}{6} again.
50×56=125350 \times \dfrac{5}{6} = \dfrac{125}{3}
The second bounce reaches 41 2/3 m.

4Apply the rule for the third bounce

#5 Look for a Pattern 5.NF.B.4
Take the second bounce height and multiply by 56\dfrac{5}{6} again.
1253×56=62518\dfrac{125}{3} \times \dfrac{5}{6} = \dfrac{625}{18}
The third bounce reaches 34 13/18 m.

5Apply the rule for the fourth bounce

#5 Look for a Pattern 5.NF.B.4
Take the third bounce height and multiply by 56\dfrac{5}{6} again.
62518×56=3125108\dfrac{625}{18} \times \dfrac{5}{6} = \dfrac{3125}{108}
The fourth bounce reaches 28 101/108 m.
Answer: 3125108\frac{3125}{108} m (= 2810110828\frac{101}{108} m ≈ 28.94 m)
4 · Reviewdoes it hold up?

Multiplying 4 times by 56\dfrac{5}{6} at once gives the same thing: 60 x (5/6)^4 = 3125/108 m, about 28.94 m.

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