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Rate/distance problem

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Rate/distance problem

by ksc1940 » Sun Jul 29, 2012 1:15 pm
This is a rate/distance problem from the gmat math tough problems file. I understand the math behind it, but when i solve for the speed of B, i get 1/12, NOT 12. Trying to figure out what i'm doing wrong here. I copied and pasted the question as well as the solution that they provide.

A and B ran 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

Ans: race 1 :- ta = tb-6 ( because A beats B by 6 sec)
race 2 :- Ta = tb+2 ( because A looses to B by 2 sec)

By the formula D= S * T
we get two equations
480/Sa = 432/Sb -6 ------------1)
480/Sa = 336/Sb +2------------2)
Equating these two equations we get Sb = 12
ta,Sa stand for time taken by A and speed of A resp.
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by eagleeye » Sun Jul 29, 2012 2:13 pm
ksc1940 wrote:This is a rate/distance problem from the gmat math tough problems file. I understand the math behind it, but when i solve for the speed of B, i get 1/12, NOT 12. Trying to figure out what i'm doing wrong here. I copied and pasted the question as well as the solution that they provide.

A and B ran 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

Ans: race 1 :- ta = tb-6 ( because A beats B by 6 sec)
race 2 :- Ta = tb+2 ( because A looses to B by 2 sec)

By the formula D= S * T
we get two equations
480/Sa = 432/Sb -6 ------------1)
480/Sa = 336/Sb +2------------2)
Equating these two equations we get Sb = 12
ta,Sa stand for time taken by A and speed of A resp.
Sb =x
Subtract the two equations
0 = 432/x -6 -(336/x + 2)
0= (432-336)/x -8
96/x = 8
x = 96/8 = 12
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by NicoleWhite » Sun Jul 29, 2012 4:26 pm
Hello eagleeye,

Could you by chance provide a more detailed solution? I am having a hard time following the poster's method. I also do not see where the "head starts" of 48m and 144m are taken into account.

Thanks!
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by eagleeye » Sun Jul 29, 2012 6:12 pm
NicoleWhite wrote:Hello eagleeye,

Could you by chance provide a more detailed solution? I am having a hard time following the poster's method. I also do not see where the "head starts" of 48m and 144m are taken into account.

Thanks!
A and B ran 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?

Let's assume B's speed is x. Now distance traveled by B = 480-48 = 432
Since time taken by A to run both races remains constant, let time taken by A to run 480m = t.
Now in the first case, B takes 1/10th of a minute more = 1/10*60 = 6 seconds more than A to finish the race (which is a distance of 432m for B). Time taken by B = t+6.

In the second case, B finishes the race earlier than A by 1/30th of a minute = 60*1/30 = 2seconds less than A. In this case distance traveled by B = 480-144 = 336. Time taken by B = t-2.

Then we have, in the first case, for B, time = t+6 = distance/speed = 432/x

Similarly, in the second case, for B, time = t - 2 = 336/x

Subtracting the bottom equation from the top one gives us:

(t+6 - (t-2)) = 432/x - 336/x
=> 8 = (432-336)/x = 96/x
=> x = 96/8 = 12m/s.

A is correct. Cheers!
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