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ashikaverma13
GMAT 1: 640 Q47 V31
Posts: 123
Updated on: 19 Apr 2017, 05:55
Originally posted by ashikaverma13 on 17 Apr 2017, 12:34.
Last edited by ashikaverma13 on 19 Apr 2017, 05:55, edited 1 time in total.
Last edited by ashikaverma13 on 19 Apr 2017, 05:55, edited 1 time in total.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
60% (01:44) correct
40%
(01:35)
wrong
based on 208
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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!
Hit 'kudos' if you find the question helpful.
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!
Hit 'kudos' if you find the question helpful.
17 Apr 2017, 13:01
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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
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
17 Apr 2017, 13:11
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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
consider only the bounces as it's what req.
assume a=248 ;ar^2 =62 and that implies r=1/2-> req. fraction
