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# Tough and tricky 5: Race

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Math Expert
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Tough and tricky 5: Race [#permalink]

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11 Oct 2009, 18:14
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Question Stats:

45% (00:00) correct 55% (06:18) wrong based on 23 sessions

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A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
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Re: Tough and tricky 5: Race [#permalink]

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12 Oct 2009, 03:55
solve these two simultaneous equations

432/b - 480/a = 6 - I
480/a - 336/b = 2 - II

I - II gives you
so we get 96/b = 8
therefore b = 12

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Re: Tough and tricky 5: Race [#permalink]

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12 Oct 2009, 07:37
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Re: Tough and tricky 5: Race [#permalink]

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12 Oct 2009, 09:26
Expert's post
Set of Tough & tricky questions is now combined in one thread: tough-tricky-set-of-problms-85211.html

You can continue discussions and see the solutions there.
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Re: Tough and tricky 5: Race [#permalink]

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12 Oct 2009, 09:35
1/10th of a minute = 6 seconds
Similarly 1/30th of a minute is 2 seconds
Also in first race A reaches before B whereas in second race B reaches before. Hence the order of equations
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Re: Tough and tricky 5: Race [#permalink]

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09 May 2011, 03:39
432/Sb -480/Sa = 6

480/Sa - 336/Sb = 2

gives Sb = 12
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Re: Tough and tricky 5: Race   [#permalink] 09 May 2011, 03:39
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# Tough and tricky 5: Race

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