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

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

by rakeshd347 » Wed Oct 02, 2013 6:30 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 is A

Are these sort of questions common on gmat. What would you rate the difficulty level of these questions if they are common on GMAT.
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by Brent@GMATPrepNow » Wed Oct 02, 2013 6:47 am
rakeshd347 wrote: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 is A

Are these sort of questions common on gmat. What would you rate the difficulty level of these questions if they are common on GMAT.
This question is pushing the limits of the GMAT (difficulty-wise, and time-wise), but I think it's still within scope. That said, it's definitely 700+ level.


Here's one approach. . .
Let A = A's speed in meters per second
Let B = B's speed in meters per second

A gives B a head start of 48 m and beats him by 1/10th of a minute.
1/10th of a minute = 6 seconds
A's travel time is 6 seconds less than B's travel time
So, (A's travel time) = (B's travel time) - 6

A traveled 480 meters and B traveled 432 meters.
Travel time = distance/speed, so . . .
(480/A) = (432/B) - 6

A gives B a head start of 144 m and is beaten by 1/30th of a minute.
1/30th of a minute = 2 seconds
A's travel time is 2 seconds more than B's travel time
So, (A's travel time) = (B's travel time) + 2

A traveled 480 meters and B traveled 336 meters.
Travel time = distance/speed, so . . .
(480/A) = (336/B) + 2

IMPORTANT: Since both equations are set equal to 480/A, we can set them equal to each other.
(432/B) - 6 = (336/B) + 2
Multiply both sides by B: 432 - 6B = 336 + 2B
Rearrange: 96 = 8B
Solve: B = 12

Answer: A

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by GMATGuruNY » Wed Oct 02, 2013 7:44 am
rakeshd347 wrote: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 is A
This question is harder than what you're likely to see on the GMAT, but it offers a useful take-away:
When faced with a hard PS problem, consider PLUGGING IN THE ANSWERS.
Here, the answers represent B's rate in meters per second.

In the first heat, B gets a head start of 48 meters.
Thus, the distance traveled by B in the first heat = 480-48 = 432 meters.
In the second heat, B gets a head start of 144 meters.
Thus, the distance traveled by B in the second heat = 480-144 = 336 meters.

In each heat, the difference between B's number of seconds and A's number of seconds is an INTEGER VALUE:
1/10 of a minute = 6 seconds in the first heat.
1/30 of a minute = 2 seconds in the second heat.
Thus, the correct rate for B must be a factor of both of B's distances (432 and 336).
Only A and C satisfy this constraint.

Answer choice C: B's rate = 16 meters per second
First heat:
Time for B to travel 432 meters = d/r = 432/16 = 27 seconds.
Since A wins by 6 seconds, A's time to travel the entire 480 meters = 27-6 = 21 seconds.
Since A's rate (21 seconds) is not a factor of A's distance (480 meters), eliminate C.

The correct answer is A.

Answer choice A: B's rate = 12 meters per second.
First heat:
Time for B to travel 432 meters = d/r = 432/12 = 36 seconds.
Since A wins by 6 seconds, A's time to travel the entire 480 meters = 36-6 = 30 seconds.
Thus, A's rate = d/t = 480/30 = 16 meters per second.

Second heat:
Time for B to travel 336 meters = d/r = 336/12 = 28 seconds.
Time for A to travel the entire 480 meters = d/r = 480/16 = 30 seconds.
Time difference = 30-28 = 2 seconds.
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by GMATGuruNY » Wed Oct 02, 2013 8:14 am
rakeshd347 wrote: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 is A

Are these sort of questions common on gmat. What would you rate the difficulty level of these questions if they are common on GMAT.
Alternate approach:

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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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