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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 [#permalink] New post 26 Dec 2010, 07:47
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jan 2012, 01:29, edited 1 time in total.
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Re: Speed & Time Problems [#permalink] New post 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)

\frac{480-48}{x}-6=\frac{480-144}{x}+2 --> x=12.

Answer: A.

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time problem [#permalink] New post 10 Jan 2012, 05:40
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????
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Re: time problem [#permalink] New post 10 Jan 2012, 17:40
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Hi, there. I'm happy to help with this. :)

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 :)
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Last edited by mikemcgarry on 11 Jan 2012, 08:44, edited 2 times in total.
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Re: time problem [#permalink] New post 10 Jan 2012, 19:41
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mikemcgarry wrote:
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.
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Re: time problem [#permalink] New post 11 Jan 2012, 08:43
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subhajeet

Good catch! I went back and corrected the error. Thanks for catching that.

Mike :-)
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Re: time problem [#permalink] New post 11 Jan 2012, 20:49
mikemcgarry wrote:
subhajeet

Good catch! I went back and corrected the error. Thanks for catching that.

Mike :-)


Hey Mike... Thanks :-D nyways ur doing a great job here helping us to achieve our dream
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Re: A and B ran, at their respective constant rates, a race [#permalink] New post 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.
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Re: A and B ran, at their respective constant rates, a race [#permalink] New post 24 Feb 2012, 09:54
1. Distance of B = 432
if A takes t mins , B takes = (t + 1/10)

2. Distance of B = 336
if A takes t mins , B takes = (t - 1/30)

Therefore to run (432 - 336) = 96 m , B took time (1/30 + 1/10)

i.e 96 = (1/30 + 1/10) * Speed of B

This is speed of B in m/min . Dividing by 60 , speed of B in m/sec = 12 .

Hence : A
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Re: A and B ran, at their respective constant rates, a race [#permalink] New post 18 Apr 2013, 20:41
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.
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Re: A and B ran, at their respective constant rates, a race [#permalink] New post 19 Apr 2013, 02:39
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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.

Check here: speed-time-problems-106921.html#p841786

Hope it helps.
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Re: A and B ran, at their respective constant rates, a race [#permalink] New post 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.
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Re: A and B ran, at their respective constant rates, a race [#permalink] New post 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)

\frac{480-48}{x}-6=\frac{480-144}{x}+2 --> x=12.

Answer: A.

P.S. Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

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

(432 / s2) - 6 = (336/s2) +2
s2=12
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Re: A and B ran, at their respective constant rates, a race   [#permalink] 25 Jun 2014, 15:33
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