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A and B ran, at their respective constant rates, a race [#permalink]
26 Dec 2010, 07:47

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Question Stats:

65% (03:49) correct
35% (02:43) wrong based on 152 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

Re: Speed & Time Problems [#permalink]
26 Dec 2010, 08:08

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surendar26 wrote:

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 1/10 th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30 th of a minute. What is B’s speed in m/s?

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)

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?

what does giving head start mean in this question???? _________________

First of all, a "head start" is a term used frequently in American pop culture. If I have a "head start" in a race, that means that, for whatever reason, I have been given permission to walk beyond the starting line and start out already at a certain distance into the race. Suppose the race is from the 0 meters mark to the 100 meters mark. The standard participants will start at 0 meters and end at 100 meters. If I am given a "head start", I am allowed to start, say, at the 20 meter mark, and during the race, I have to run only from 20 meters to 100 meters. In other words, it's an advantage given to me, usually because I am perceived as being less able to compete well on my own. It's similar to the idea of a "handicap" in a sport like golf -- you can read more about that here: http://en.wikipedia.org/wiki/Handicapping

So, in the question you describe: In the first heat, A runs the full 480 meter, and B (with a head start of 48 m) runs a total distance of 480 - 48 = 432 meters. In that heat, A beat B by 1/10 of a minute, i.e. 6 seconds. It took B six seconds longer to finish.

In the second heat, A runs the full 480 m, and B (now with a head start of 144 m) runs a total distance of 480 - 144 = 336 meters. In that heat, B beat A by 1/30 of a minute, i.e. 2 seconds. It took B 2 seconds fewer to finish.

D = RT, so T = D/R

We will let t be the time it takes A to run the 480. Let vA be A's speed, and vB be B's speed. Then, we have

(1) t = 480/vA

(2) t + 6 = 432/vB

(3) t - 2 = 336/vB

Subtract equation (3) from equation (2), and we are left with: 8 = 96/vB --> 8vB = 96 ---> vB = 12 m/s

Does that make sense? Please let me know if you have any questions.

Mike _________________

Mike McGarry Magoosh Test Prep

Last edited by mikemcgarry on 11 Jan 2012, 08:44, edited 2 times in total.

We will let t be the time it takes A to run the 480. Let vA be A's speed, and vB be B's speed. Then, we have

(1) t = 480/vA

(2) t + 6 = 432/vB

(3) t + 2 = 336/vB

Mike, I think the 3rd equation should be t-2 =336/vB, because it is stated in the question that A got beaten by B. So according to that the answer should come out to be 12m/s.

Re: A and B ran, at their respective constant rates, a race [#permalink]
24 Feb 2012, 09:09

surendar26 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

Hi!

B ran 96m less in second heat (144-48), which allowed him to “gain back” 8 seconds (from 6 loss to 2 seconds win). So, 96/8 = 12m/s.

Re: A and B ran, at their respective constant rates, a race [#permalink]
19 Apr 2013, 02:39

Expert's post

captainhunchy 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

I initially solved the equation like this and got the wrong answer, but don't know why??

Time of B (heat 1) = (480+48)/A + 6 ; where A = rate of A Time of B (heat 2) = (480+144)/A - 2

Since B's time is constant in both heats I will set them equal to each other.

(480+48)/A + 6 = (480+144)/A -2

528/A + 6 = 624/A -2 A = 12m/s

I know this is wrong because, this means B's rate can't be 12ms... But I don't get why the equation is wrong.

In the first heat B covers 480-48=432 meters and in the second heat B covers 480-144=336 meters, thus the times of B in two heats cannot be the same. In both heats A runs 480 meters, so the times of A in two heats are the same.

Re: A and B ran, at their respective constant rates, a race [#permalink]
19 Apr 2013, 10:47

Experts a question for you.. I am aiming for 650+ score and quant is an area i can improve on.. What should be an ideal time for solving a question like this. I am at currently around 2m 45 secs. _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

Re: A and B ran, at their respective constant rates, a race [#permalink]
25 Jun 2014, 10:04

How can you add or subtract time to a rate. Based on the explanation you are equating the two rates of B but you are also subtracting 6 seconds from the rate and adding 2 seconds to the rate. I am looking at this through the D = RT formula and don't know how you can do R = (D/T) - 6.

Appreciate the help.

Bunuel wrote:

surendar26 wrote:

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 1/10 th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30 th of a minute. What is B’s speed in m/s?

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)

Re: A and B ran, at their respective constant rates, a race [#permalink]
25 Jun 2014, 14:15

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In first case, B was given 48m headstart and he by 6 seconds. (1/10th of a minute) In the second case, B was given head start of 144m and he wins by 2 sec (1/30th of a minute)

Hence we know this time difference of 8 seconds (from loosing by 6 sec to winning by 2 seconds), is due to the fact that B travelled 144-48 = 96m more in first case. Now B travelled this 96m in 8 sec and hence, speed of B = 96/8 = 12m/s

Am I correct by approaching the problem this way or is there any thing I missed ?

Re: A and B ran, at their respective constant rates, a race [#permalink]
25 Jun 2014, 15:33

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Time taken by A in both the cases are the same. In the first case it is 6 seconds less than that of B and in the second case it is 2 seconds more than that of B.

A and B ran, at their respective constant rates, a race of 480 m [#permalink]
09 May 2015, 22:36

Expert's post

Hi All,

This is an old series of posts (most of them are over 10 years old), but this question can be solved in a couple of different ways. Since I don't want to do lots of formulaic math if I can avoid it (since it takes so long), I'm going to use the built-in patterns to save some time.

While the prompt doesn't state it, we're meant to assume that the two runners run at constant speeds. We're given some comparative data to work with:

1) Each FULL race is 480m 2) When runnner A gives runner B a 48m head start, runner A WINS by 1/10th of a minute (meaning 6 seconds). 3) When runnner A gives runner B a 144m head start, runner A LOSES by 1/30th of a minute (meaning 2 seconds).

We're asked for runner B's speed in meters/second.

We can use the DIFFERENCES in distance and time to figure out speed.

Since the difference in distances is 144-48 = 96 meters and the difference in times is (6 second WIN) - (2 second LOSS) = 8 seconds, we can figure out B's rate....it's 96/8 = 12 m/sec.

If you're skeptical of this conclusion, then you can use it to verify the speed of Runner A....

In the 1st race... Running 12m/sec, runner B would run 432m in.... D = (R)(T) 432 = (12)(T) 432/12 = T 36 seconds = T

Since runner A WINS by 6 seconds, runner A needs 30 seconds to complete 480m D = (R)(T) 480 = (R)(30) 480/30 = R 16 meters/sec = R

In the 2nd race....

Running 12m/sec, runner B would run 336m in.... D = (R)(T) 336 = (12)(T) 336/12 = T 28 seconds = T

Since runner A runs at a constant rate, we know that it takes runner A 30 seconds to run a 480m race. Runner A LOSES by 2 seconds, which "fits" this information (runner B ran 336m in 28 seconds while runner A ran 480m in 30 seconds.....the difference is a 2 second LOSS).

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...