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# A and B ran, at their respective constant rates, a race of 480 m. In

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A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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11 Oct 2009, 18:14
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Question Stats:

59% (03:33) correct 41% (03:44) wrong based on 62 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-and-b-ran-at-their-respective-constant-rates-a-race-of-480-m-in-106921.html
[Reveal] Spoiler: OA

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Re: A and B ran, at their respective constant rates, a race of 480 m. In [#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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Kudos [?]: 71 [0], given: 17

Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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12 Oct 2009, 07:37
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Kudos [?]: 85537 [0], given: 10237

Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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12 Oct 2009, 09:26
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: A and B ran, at their respective constant rates, a race of 480 m. In [#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: A and B ran, at their respective constant rates, a race of 480 m. In [#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: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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24 Jul 2016, 06:07
Hello from the GMAT Club BumpBot!

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Re: A and B ran, at their respective constant rates, a race of 480 m. In [#permalink]

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24 Jul 2016, 08:18
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Bunuel wrote:
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

Let $$x$$ be the speed of B.

Write the equation:

(480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

$$\frac{480-48}{x}-6=\frac{480-144}{x}+2$$ --> $$x=12$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-and-b-ran-at-their-respective-constant-rates-a-race-of-480-m-in-106921.html
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Re: A and B ran, at their respective constant rates, a race of 480 m. In   [#permalink] 24 Jul 2016, 08:18
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