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A ball is dropped from 248 feet above the ground
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ashikaverma13
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A ball is dropped from 248 feet above the ground [#permalink] New post  Updated on: 19 Apr 2017, 04:55
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55% (hard)

Question Stats:

62% (01:47) correct 38% (01:22) wrong based on 73 sessions

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A ball is dropped from 248 feet above the ground and after the second bounce it rises to the height of 62 feet. If the height to which the ball rises after each bounce is always the same fraction of the height reached on its previous bounce, what is this fraction?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

I have an explanation but it's not quite clear. kindly explain your solution in detail. Thanks!


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Originally posted by ashikaverma13 on 17 Apr 2017, 11:34.
Last edited by ashikaverma13 on 19 Apr 2017, 04:55, edited 1 time in total.
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quantumliner
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Re: A ball is dropped from 248 feet above the ground [#permalink] New post  17 Apr 2017, 12:01
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Let us say

The Height from which the ball was first dropped as H0 = 248 feet
The Height reached by the ball after second bounce is H1
The Height reached by the ball after second bounce is H2 = 62 feet

Given , Height the ball reaches after bounce is a fraction of the height the ball reached in the previous bounce

The Height reached by the ball after second bounce = H2 = X of H1
==> 62 = X * H1 ...................Equation 1

The Height reached by the ball after first bounce = H1 = X of H0

==> H1 = X * 248 ...................Equation 2

Substituting value of H1 from Equation 2 to Equation 1

62 = X * H1 = X * X * 248

X^2 = 62/248 = 31/124 = 1/4

X = 1/2

Answer is A. 1/2


This can be solved by Geometric Progression formula as well

Term n = Term 1 * R ^ (n -1)

Here

R is nothing but the fraction, the question refers to

Term 1 is the Height from which the ball is dropped from = 248 feet
Term 2 is the Height reached by the ball after first bounce
.
.
.
Term n can be Term 3, which is the height the ball reached after the second bounce = 62 feet

Putting the above in equation

62 = 248 * x ^ (3-2)

62 = 248 * X ^ 2

62/248 = X ^ 2

X ^ 2 = 1/4

X = 1/2
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attari92
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Re: A ball is dropped from 248 feet above the ground [#permalink] New post  17 Apr 2017, 12:11
distances travel by the ball on bounces will make a geometric progression.
consider only the bounces as it's what req.
assume a=248 ;ar^2 =62 and that implies r=1/2-> req. fraction
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ashikaverma13
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Joined: 19 Aug 2016
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Re: A ball is dropped from 248 feet above the ground [#permalink] New post  17 Apr 2017, 22:54
quantumliner wrote:
Let us say

The Height from which the ball was first dropped as H0 = 248 feet
The Height reached by the ball after second bounce is H1
The Height reached by the ball after second bounce is H2 = 62 feet

Given , Height the ball reaches after bounce is a fraction of the height the ball reached in the previous bounce

The Height reached by the ball after second bounce = H2 = X of H1
==> 62 = X * H1 ...................Equation 1

The Height reached by the ball after first bounce = H1 = X of H0

==> H1 = X * 248 ...................Equation 2

Substituting value of H1 from Equation 2 to Equation 1

62 = X * H1 = X * X * 248

X^2 = 62/248 = 31/124 = 1/4

X = 1/2

Answer is A. 1/2


This can be solved by Geometric Progression formula as well

Term n = Term 1 * R ^ (n -1)

Here

R is nothing but the fraction, the question refers to

Term 1 is the Height from which the ball is dropped from = 248 feet
Term 2 is the Height reached by the ball after first bounce
.
.
.
Term n can be Term 3, which is the height the ball reached after the second bounce = 62 feet

Putting the above in equation

62 = 248 * x ^ (3-2)

62 = 248 * X ^ 2

62/248 = X ^ 2

X ^ 2 = 1/4

X = 1/2


Hey, thanks! I understood it now.

your clear explanation makes it so much simpler. much appreciated.
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Bunuel
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Posts: 75019
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Re: A ball is dropped from 248 feet above the ground [#permalink] New post  18 Apr 2017, 01:50
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ashikaverma13 wrote:
A ball is dropped from 248 feet above the ground and after the second bounce it rises to the height of 62 feet. If the height to which the ball rises after each bounce is always the same fraction of the height reached on its previous bounce, what is this fraction?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

I have an explanation but it's not quite clear. kindly explain your solution in detail. Thanks!


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ashikaverma13
Current Student
Joined: 19 Aug 2016
Posts: 140
Location: India
Schools: ISB '20 (WD), Schulich Sept 18 Intake (A), Rotman '20 (S)
GMAT 1: 640 Q47 V31
GPA: 3.82
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Re: A ball is dropped from 248 feet above the ground [#permalink] New post  19 Apr 2017, 04:32
Bunuel wrote:

Similar questions to practice:
https://gmatclub.com/forum/a-certain-ba ... 09106.html
https://gmatclub.com/forum/an-object-th ... 13837.html
https://gmatclub.com/forum/a-ball-is-dr ... 38317.html
https://gmatclub.com/forum/after-being- ... 18935.html
https://gmatclub.com/forum/height-to-wh ... 20917.html
https://gmatclub.com/forum/a-ball-drops ... 34726.html
https://gmatclub.com/forum/one-ball-wil ... 20703.html


Bunuel. Helpful and amazing as always! thanks for these!
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cjk52793
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Joined: 16 Jan 2020
Posts: 5
Re: A ball is dropped from 248 feet above the ground [#permalink] New post  26 Jan 2020, 03:25
An easy way I solved this: I noticed that 248 is 4x62. I rewrote it to the heights of 8 and 2 with 2 bounces. It was clear at this point that each bounce was reduced by 1/2.

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Gknight5603
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Joined: 25 Oct 2019
Posts: 128
Re: A ball is dropped from 248 feet above the ground [#permalink] New post  26 Jan 2020, 03:44
the question is telling that the height achieved by the ball after each bounce is in geometric progression.
248 X 62......
the option A satisfies this progression....
248/2= 124
124/2= 62
so 1/2 is the common ratio of the geometric progression.

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Re: A ball is dropped from 248 feet above the ground [#permalink]
26 Jan 2020, 03:44
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