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

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Difficulty:

45% (medium)

Question Stats:

48% (04:45) correct
51% (03:59) wrong based on 39 sessions

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.

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)

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

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.

Re: A and B ran, at their respective constant rates, a race of 4 [#permalink]
08 Jan 2014, 05:08

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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?

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).

(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 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

(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 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?
_________________

Re: A and B ran, at their respective constant rates, a race of 4 [#permalink]
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)