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A marathon-Hard Q.

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A marathon-Hard Q.

by Ozlemg » Sun Aug 21, 2011 5:13 am
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

OA:A
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by pgkkumar » Sun Aug 21, 2011 5:30 am
Let the speed of A be "y" m/s and the speed of B be "x" m/s. We can write equations based on the results of the two heats as:

(480-48)/x - 480/y = 60/10 - - - - (1)

480/y - (480-144)/x = 60/30 - - - - (2)

Add (1) & (2). You get,

(480-48-480+144)/x = 8 or 96/x = 8 i.e., x is 12 m/s
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by dendude » Sun Aug 21, 2011 5:48 am
Here's how I solved it,
Distance = 480m

1st heat - A runs 480m whereas B runs only 432m because of the 48m headstart,
SpeedA = 480/ta1; SpeedB = 432/tb1
Here ta1 = tb1 - 6 (since A beats B by 1/10th of a min = 6secs)

2nd heat - A runs 480m whereas B runs only 336m because of the 144m headstart,
SpeedA = 480/ta2; SpeedB = 432/tb2
Here ta1 = tb2 + 2 (since A loses to B by 1/30th of a min = 2secs)

Both A and B run with the same speed in both races, so their speeds can be equated from heat1 and heat2
So, A's speed in terms of B would be,
480/(432/SpeedB)-6 1st heat
and 480/(336/SpeedB)+2 2nd heat

Equate these two and solve,
8*SpeedB = 96
SpeedB = 12m/sec

Ans A
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by GMATGuruNY » Sun Aug 21, 2011 11:58 am
I posted a solution here:

https://www.beatthegmat.com/difficult-ma ... -t429.html
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by saketk » Mon Aug 22, 2011 4:33 am
Hi I used a bit unconventional method. Although you can use the method suggested by Mitch and other guys, I'll explain you what I did.
Note the numbers here. First one is 48 and the next one is 144. Both these numbers are multiple of 12.
So I plugged 12 In the question and found A's speed from equation 1. Then I plugged these two values in the second equation and hey! Both they satisfied the condition :)
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