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After being dropped a certain ball always bounces back to 2/5 of the

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Bunuel
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After being dropped a certain ball always bounces back to 2/5 of the [#permalink] New post   22 May 2016, 13:54
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After being dropped a certain ball always bounces back to 2/5 of the height of its previous bounce. After the first bounce it reaches a height of 125 inches. How high (in inches) will it reach after its fourth bounce?

A. 20
B. 15
C. 8
D. 5
E. 3.2
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adidhar
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Re: After being dropped a certain ball always bounces back to 2/5 of the [#permalink] New post   18 May 2017, 14:29
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1st bounce = 125
2nd bounce = 2/5*125 = 50
3rd Bounce = 2/5*50 = 20
4th bounce = 2/5*20 = 8

Ans = c
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After being dropped a certain ball always bounces back to 2/5 of the [#permalink] New post   22 May 2016, 14:43
Uhm in this case the calculations are a bit tricky,
the initial height was \(125/0,4\) -> \(125*\frac{10}{4}\)->\(\frac{1250}{4}\) -> \(312,5\)
at the fourth drop the height will have been reduced by \(0,4*0,4*0,4*0,4= 0,2^8 = 0,0256 ≈ 2,5\)%
\(3,125*\frac{10}{4}\) -> let's round it up -> \(\frac{32}{4} -> 8\)

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Re: After being dropped a certain ball always bounces back to 2/5 of the [#permalink] New post   04 Jun 2021, 00:35
Each bounce will impact its first bouncing height by 2/5,

so, if we want to find it's height after nth bounce -> X*(2/5)^(n-1)

here, x = 125, and n = 4

Substitute to get 8

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After being dropped a certain ball always bounces back to 2/5 of the [#permalink] New post   04 Jun 2021, 01:14
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Solution:

Given that after each bounce it reaches 2/5 of the previous height,

=>After the second bounce it will reach 2/5 * 125.

=>After the third bounce it will reach 2/5 * 2/5 * 125.

=>After the fourth bounce it will reach 2/5 * 2/5 * 2/5 *125

= 8/125 * 125

=8 (option c)

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Re: After being dropped a certain ball always bounces back to 2/5 of the [#permalink] New post   01 Nov 2023, 13:39
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Re: After being dropped a certain ball always bounces back to 2/5 of the [#permalink]
01 Nov 2023, 13:39
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