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A ball is dropped from a height of 20 feet above the ground, and after

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twobagels
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A ball is dropped from a height of 20 feet above the ground, and after [#permalink] New post   Updated on: 07 May 2021, 10:14
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A ball is dropped from a height of 20 feet above the ground, and after each bounce it rebounds to one fourth of its previous height. What is the total distance, in feet travelled by the ball?

A. 26.67
B. 29.33
C. 33.33
D. 36.67
E. 37.33
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C

Originally posted by twobagels on 17 Apr 2021, 16:40.
Last edited by twobagels on 07 May 2021, 10:14, edited 1 time in total.
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Re: A ball is dropped from a height of 20 feet above the ground, and after [#permalink] New post   17 Apr 2021, 20:49
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twobagels wrote:
A ball is dropped from a height of 20 feet above the ground, and after each bounce it rebounds to one fourth of its previous height. What is the total distance, in feet travelled by the ball?

A. 26.66
B. 29.33
C. 33.33
D. 36.67
E. 37.33

1st drop = 20
2nd rebounce and drop = \(\frac{20}{4}*2\) = 10
3rd rebounce and drop = \(\frac{5}{4}*2\) = 2.5
4th rebounce and drop = \(\frac{1.25}{4}*2\) = 0.6225 and so on...

Total = 20+10+2.5+0.6225+...= 33.1225+...
nearest is C

IMO C
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Re: A ball is dropped from a height of 20 feet above the ground, and after [#permalink] New post   19 Apr 2021, 01:38
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Given: A ball is dropped from a height of 20 feet above the ground, after each bounce it rebounds to one fourth of its previous height.
The total distance travelled by the ball =?

We know that, Sum of an infinite Geometric Progression (G.P.) = a/1-r

Distance covered while coming down = \(20 + \frac{20}{4} + \frac{20}{4^2} + .. = \frac{20}{1-1/4} = \frac{20}{3/4} = \frac{80}{3}\) feet [infinite G.P. with a = 20 and r = \(\frac{1}{4}\)]

Distance covered while going up = \(5 + \frac{5}{4} + \frac{5}{4^2} + .. = \frac{5}{1-1/4} = \frac{5}{3/4} = \frac{20}{3 }\) feet [infinite G.P. with a = 5 and r = \(\frac{1}{4}\)]

Total distance covered = \(\frac{80}{3}+ \frac{20}{3} = \frac{100}{3} \approx 33.33\) feet.

So, correct answer is option C.
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Re: A ball is dropped from a height of 20 feet above the ground, and after [#permalink] New post   20 Apr 2021, 02:07
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\(20 + \frac{1}{4} * 20 + \frac{1}{4} * 20 + (\frac{1}{4})^2 * 20 + ....\)

\(20 + 5 + 5 + \frac{5}{4} + \frac{5}{4} + \frac{5}{16} + \frac{5}{16}+ .....\)

\(20 + 10 + \frac{10}{4} + \frac{10}{16} + .....\)

\(20 + [10 + \frac{10}{4} + \frac{10}{16} + ......\)This forms G.P. with a common ratio of \(\frac{1}{4} = 0.25\) and first term as 10]

Sum of G.P. (infinity) = \(\frac{10 }{ 1 - 0.25} = \frac{40 }{ 3}\)

\(20 + \frac{40}{3} = \frac{100}{3} = 33.33\)

Answer C
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Re: A ball is dropped from a height of 20 feet above the ground, and after [#permalink] New post   12 Jul 2021, 23:50
20+1/4∗20+1/4∗20+(1/4)2∗20+....20+1/4∗20+1/4∗20+(1/4)2∗20+....

20+10+10/4+10/16+.....20+10+10/4+10/16+.....

20+[10+10/4+10/16+......20+[10+10/4+10/16+......This forms G.P. with a common ratio of 1/4=0.25
as the common ratio and first term as 10]

a/1-r (sum of infinite series ) = 10/1−0.25=403101−0.25=40/3

20+40/3=100/3=33.33
Hence IMO C
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Re: A ball is dropped from a height of 20 feet above the ground, and after [#permalink]
12 Jul 2021, 23:50
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