An engineer designed a ball so that when it was dropped, it : Problem Solving (PS)

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An engineer designed a ball so that when it was dropped, it

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Senior Manager

Joined: 10 Jul 2013

Posts: 314

An engineer designed a ball so that when it was dropped, it [#permalink]

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21 Aug 2013, 17:06

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Difficulty:

55% (hard)

Question Stats:

66% (02:10) correct

34% (02:24) wrong

based on 144 sessions

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An engineer designed a ball so that when it was dropped, it rose with each bounce exactly one-half as high as it had fallen. The engineer dropped the ball from a 16-meter platform and caught it after it had traveled 46.5 meters. How many times did the ball bounce?
A. 5
B. 6
C. 7
D. 8
E. 9

Spoiler: OA

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Asif vai.....

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Posts: 50712

Re: An engineer designed a ball so that when it was dropped, it [#permalink]

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22 Aug 2013, 03:43

Asifpirlo wrote:

Down = 16;
Up = 8 --> sum 24;
Down = 8 --> sum 32;
Up = 4 --> sum 36;
Down = 4 --> sum 40;
Up = 2 --> sum=42;
Down = 2 --> sum=44;
Up = 1 --> sum=45;
Down = 1 --> sum=46;
Up = 0.5 --> sum=46.5.

5 times down.

Answer: A.
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Re: An engineer designed a ball so that when it was dropped, it [#permalink]

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21 Aug 2013, 19:01

Asifpirlo wrote:

Ans: 5

Division of total diatance travelled will be 16+16+8+4+2+0.5

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Re: An engineer designed a ball so that when it was dropped, it [#permalink]

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21 Aug 2013, 19:20

Asifpirlo wrote:

Easy enough to visualize...16 ft travelled on the drop, then the ball rises 8 ft (half) and then falls the same distance, then rises 4 ft and falls, and so on. You know that it must travel 30.5 meters starting with the first bounce-- 16 + 8 + 4 + 2 + 0.5 for 5 bounces.

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Re: An engineer designed a ball so that when it was dropped, it [#permalink]

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12 Sep 2018, 05:58

Asifpirlo wrote:

1st Drop = 16m
1st Rise = 8m
2nd Drop = 8m
2nd Rise = 4m
3rd Drop = 4m
3rd Rise = 2m
4th Drop = 2m
4th Rise = 1m
5th Drop = 1m
5th Rise = 0.5m

Add all these - 16 + 8 + 8 + 4 + 4 + 2 + 2 + 1 + 1 + 0.50 = 46.50, Answer must be (A) 5
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