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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
51% (01:52) correct
49%
(01:51)
wrong
based on 175
sessions
History
Date
Time
Result
Not Attempted Yet
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
A. 26.67
B. 29.33
C. 33.33
D. 36.67
E. 37.33
ramlala
Joined: 22 Aug 2020
Last visit: 13 Dec 2022
Posts: 468
Given Kudos: 30
Location: India
Concentration: International Business, Finance
Schools: IIMA PGPX'23 INSEAD Jan'23 intake Sloan (MIT) ISB HEC Said
GPA: 4
WE:Project Management (Energy)
17 Apr 2021, 20:49
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twobagels
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
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.
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.
