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Oscar is running in a straight line away from Nancy at the rate of 20

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Oscar is running in a straight line away from Nancy at the rate of 20  [#permalink]

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New post 06 May 2017, 12:42
1
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A
B
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D
E

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Question Stats:

97% (01:03) correct 3% (01:58) wrong based on 48 sessions

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Oscar is running in a straight line away from Nancy at the rate of 20 feet per second. Nancy is chasing Oscar at the rate of 25 feet per second. If Oscar has a 100-foot head start, how long, in seconds, will it take Nancy to catch Oscar?

A. 4
B. 5
C. 10
D. 20
E. 100

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Re: Oscar is running in a straight line away from Nancy at the rate of 20  [#permalink]

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New post 06 May 2017, 12:54
Formula used
Time = Distance to catch/speed

Since the relative speed is 5ft/s (as they are travelling in the same direction), the time taken for Nancy to catch Oscar is
100ft/(5 ft/s) or 20 seconds(Option D)
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Oscar is running in a straight line away from Nancy at the rate of 20  [#permalink]

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New post 06 May 2017, 20:23
Here the rate at which Nancy runs is 5 more than the rate of Oscar, so to cover the earlier distance of 100 feet, it takes 100/5 = 20 seconds
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Re: Oscar is running in a straight line away from Nancy at the rate of 20  [#permalink]

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New post 07 May 2017, 03:25
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relative distance/ relative speed= meeting time
100/(25-20)=20 sec
answer: D
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Re: Oscar is running in a straight line away from Nancy at the rate of 20  [#permalink]

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New post 07 May 2017, 07:09
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Time taken by Nancy to catch Oscar= \(\frac{Distance}{Relative Speed}\)

Distance = \(\frac{100}{(25-20)}\)= \(\frac{100}{5}\)=20

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Re: Oscar is running in a straight line away from Nancy at the rate of 20  [#permalink]

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New post 03 Jul 2017, 11:38
Is relative distance the only option to solve such questions?
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Re: Oscar is running in a straight line away from Nancy at the rate of 20   [#permalink] 03 Jul 2017, 11:38
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Oscar is running in a straight line away from Nancy at the rate of 20

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