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RACE

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RACE [#permalink]  05 Sep 2010, 18:16
Hi all!
Am new to the forum. I'm working through question 6 and the answer explanation has me lost. I'm having trouble putting the numbers together in a formula that makes sense and am not following the variables posted in the answer. Any help you can offer would be greatly appreciated.
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Re: RACE [#permalink]  05 Sep 2010, 18:33
cheetarah1980 wrote:
Hi all!
Am new to the forum. I'm working through question 6 and the answer explanation has me lost. I'm having trouble putting the numbers together in a formula that makes sense and am not following the variables posted in the answer. Any help you can offer would be greatly appreciated.

Hi, and welcome to Gmat Club.

Are you talking about the RACE question, if yes then below is solution to it:

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.

Hope it's clear.
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Re: RACE [#permalink]  15 Oct 2010, 10:01
"in both heats A runs with constant rate, so the time for first and second heats are the same"
I did'nt understand this part
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Re: RACE [#permalink]  15 Oct 2010, 10:07
JoyLibs wrote:
"in both heats A runs with constant rate, so the time for first and second heats are the same"
I did'nt understand this part

In both races A runs the whole distance of 480m with the same constant speed, thus covers both in the same time.
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Re: RACE [#permalink]  15 Oct 2010, 10:56
Hi cheetarah1980

To solve this problem , you need to understand the folllowing

1. In both cases , A runs 480 m and B runs < than 480 m
2. In first race , A takes less time than B and wins
3. In second race , B takes less time than A and wins

Let a and b be the constant speed of A and B respectively

by condition ,

432/b - 480/a = 6 ---(1) ----> As 480-48 = 432 which B covers
480/a - 336/b = 2 --(2) -----> As 480 - 144 = 336 which B covers

Adding (1) and (2) we get 432/b - 336/b = 8 which gives b = 12 m/s

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Re: RACE [#permalink]  17 Oct 2010, 13:28
a possibly quicker (and easier to my simple mind) method;

conditions: both run at the same rate in both heats!

First heat, B has a 48m head start; second heat, B has 144m headstart; a net difference of 96m.

now 96m difference creates a difference of 8 seconds in total in relation to speed of A (480m and speed of A is really there to create reference point, we don't really care about his exact speed).

or simply, the time taken for B is 8 seconds less (6+2 seconds), for a net distance of 96m. or in another word, a length of 96m reduced time taken by B by 8 seconds.

which means, speed of B = 96/8 = 12m/s.
Re: RACE   [#permalink] 17 Oct 2010, 13:28
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