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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]

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26 Dec 2010, 08:47
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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, 02:29, edited 1 time in total.
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Re: Speed & Time Problems [#permalink]

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26 Dec 2010, 09:08
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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)

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

So please provide answer choices for PS questions.
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10 Jan 2012, 06:40
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]

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10 Jan 2012, 18:40
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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, 09:44, edited 2 times in total.
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Re: time problem [#permalink]

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10 Jan 2012, 20:41
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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]

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11 Jan 2012, 09:43
Expert's post
subhajeet

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

Mike
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Re: time problem [#permalink]

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11 Jan 2012, 21:49
mikemcgarry wrote:
subhajeet

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

Mike

Hey Mike... Thanks 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]

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24 Feb 2012, 10:09
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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]

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24 Feb 2012, 10:54
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]

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18 Apr 2013, 21:41
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]

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19 Apr 2013, 03:39
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.

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]

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19 Apr 2013, 11:47
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]

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25 Jun 2014, 11:04
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)

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

So please provide answer choices for PS questions.
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Re: A and B ran, at their respective constant rates, a race [#permalink]

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25 Jun 2014, 15:15
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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]

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25 Jun 2014, 16:33
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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.

(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]

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29 Aug 2014, 23:14
Is this a 600-700 level question or 700+ ?
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A and B ran, at their respective constant rates, a race of 480 m [#permalink]

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09 May 2015, 23:36
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).

[Reveal] Spoiler:
A

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

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15 Oct 2015, 21:17
Time= Distance/Speed
First heat,
Ta= 480/Sa
Tb=432/Sb

480/Sa = 432/Sb - 6 ---1

Second heat,

480/Sa = 336/Sb + 2 ---2

Equating 1 and 2 , we get
Sb=12
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A and B ran, at their respective constant rates, a race [#permalink]

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18 Oct 2015, 22:28
Another way:

when B had 48m handicap - 6 sec beaten by A

when had 144m--- 2 sec beats A

means that 144-48=96m is run in 8 sec by B. So, his speed is 96/8=12m/s

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

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27 Jan 2016, 08:03
Dear Rich

Can you please explain how did you derive the difference in distances & time?

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.
Re: A and B ran, at their respective constant rates, a race   [#permalink] 27 Jan 2016, 08:03

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