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

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BTGmoderatorLU: Moderator; Posts: 2650; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Thu Dec 14, 2017 12:04 pm

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A and B ran, at either 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

The OA is A.

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. Can any expert help me with this PS question, please? Thanks!

Source: Beat The GMAT — Problem Solving | Thu Dec 14, 2017 12:04 pm

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by GMATGuruNY » Thu Dec 14, 2017 1:02 pm

LUANDATO wrote:A and B ran, at either 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

Head start in the second heat - head start in the first heat = 144-48 = 96 meters.
Thus, B travels an additional 96 meters in the first heat.

In the second heat, B travels for 2 seconds LESS than A (with the result that B WINS by 2 seconds).
In the first heat, B travels for 6 seconds MORE than A (with the result that B LOSES by 6 seconds).
Since 2+6 = 8, B travels for 8 seconds LONGER in the first heat than he does in the second heat.

Since in these 8 seconds B travels an additional 96 meters, his rate = d/t = 96/8 = 12 meters per second.

The correct answer is A.

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Re: A and B ran, at their respective constant rates, a race of..

by Scott@TargetTestPrep » Thu Apr 02, 2020 10:15 am

BTGmoderatorLU wrote: ↑
Thu Dec 14, 2017 12:04 pm
A and B ran, at either 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

The OA is A.

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. Can any expert help me with this PS question, please? Thanks!

We can let a and b be the rates of A and B, in m/s, respectively. Since 1/10 minute = 6 seconds and 1/30 minute = 2 seconds, we can now create the equations:

480/a + 6 = (480 - 48)/b

and

480/a - 2 = (480 - 144)/b

Subtracting the two equations, we have:

8 = (480 - 48)/b - (480 - 144)/b

8 = 432/b - 336/b

8 = 96/b

8b = 96

b = 12

Answer: A

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