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A and B ran, at their respective constant rates, a race of 4

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A and B ran, at their respective constant rates, a race of 4 [#permalink] New post 05 Sep 2010, 17:16
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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


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.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 27 Jun 2013, 00:42, edited 1 time in total.
Edited the question.
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Re: RACE [#permalink] New post 05 Sep 2010, 17:33
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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.

Answer: A.

Hope it's clear.
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Re: RACE [#permalink] New post 15 Oct 2010, 09:01
"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] New post 15 Oct 2010, 09:07
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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] New post 15 Oct 2010, 09:56
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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

Hope this clarifies your doubt
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Re: RACE [#permalink] New post 17 Oct 2010, 12:28
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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.
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Re: A and B ran, at their respective constant rates, a race of 4 [#permalink] New post 08 Jan 2014, 05:08
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Re: RACE [#permalink] New post 17 Jan 2014, 11:30
Bunuel wrote:
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.

Answer: A.

Hope it's clear.


Hi Bunuel,

Can you please explain why have you subtracted 6 and added 2. I understand in the second heat B won. But are we finding the total time on either sides?
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Re: RACE [#permalink] New post 18 Jan 2014, 02:37
Expert's post
Vidhi1 wrote:
Bunuel wrote:
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.

Answer: A.

Hope it's clear.


Hi Bunuel,

Can you please explain why have you subtracted 6 and added 2. I understand in the second heat B won. But are we finding the total time on either sides?


In both heats A runs with constant rate, thus the times for first and second heats for A are the same.

In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute: Time of A = \frac{480-48}{x}-6 ((480-48)/x is the time of B, which is 6 seconds more than time of A, thus we need to subtract 6 from time of B to get time of A).

In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute: Time of A = \frac{480-144}{x}+2 ((480-144)/x is the time of B, which is 2 seconds less than time of A, thus we need to add 2 to time of B to get time of A).

Now, we can equate.

Hope it's clear.
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Re: RACE [#permalink] New post 18 Jan 2014, 23:07
Bunuel wrote:
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.

Answer: A.

Hope it's clear.


I did that and got x=24 :roll: what did i do wrong? On the test I would have looked at the answer choices, looked for a relationship and said...ah 12 is half of 24...there were two races \frac{24}{2}=12 but while I have time, I'd like to learn as many concepts as possible
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Re: RACE [#permalink] New post 19 Jan 2014, 08:20
Expert's post
TroyfontaineMacon wrote:
Bunuel wrote:
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.

Answer: A.

Hope it's clear.


I did that and got x=24 :roll: what did i do wrong? On the test I would have looked at the answer choices, looked for a relationship and said...ah 12 is half of 24...there were two races \frac{24}{2}=12 but while I have time, I'd like to learn as many concepts as possible


The solution of (480-48)/x - 6 = (480-144)/x + 2 is x=12. x is the rate of B. Why did you multiply it by 2?
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Re: A and B ran, at their respective constant rates, a race of 4 [#permalink] New post 19 Jan 2014, 19:49
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 in m/s

Time taken by B to run the distance that is difference between 144m and 48m = (1/30th of minute) + (1/10th of minute)

Or,\frac{(144m-48m)}{x} = 6s + 2s

Or,x = \frac{96m}{8s} = 12 m/s

Answer: (A)
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Re: A and B ran, at their respective constant rates, a race of 4 [#permalink] New post 20 Jan 2014, 10:16
1. heat: B * (t + 6s) = A*t - 48meter; Bt + 6B = At - 48meter
2. heat: B * (t - 2s) = A*t -144meter; Bt - 2B = At - 144meter

Heat 2 - Heat 1: -8B = 98meter; B = \frac{98}{-8} = [fraction]-12[/fraction]

Where do I have the mixup about the positive/negativ failure?
At the Real test I would have just clicked 12 seconds :D
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Re: A and B ran, at their respective constant rates, a race of 4   [#permalink] 20 Jan 2014, 10:16
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