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# A and B ran, at their respective constant rates, a race of 480 m. In

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Intern
Joined: 30 Sep 2010
Posts: 19
A and B ran, at their respective constant rates, a race of 480 m. In  [#permalink]

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Updated on: 24 Jul 2016, 08:17
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Question Stats:

59% (03:08) correct 41% (03:10) wrong based on 422 sessions

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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

Originally posted by surendar26 on 26 Dec 2010, 08:47.
Last edited by Bunuel on 24 Jul 2016, 08:17, edited 2 times in total.
Edited the question
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Re: A and B ran, at their respective constant rates, a race of 480 m. In  [#permalink]

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Updated on: 11 Jan 2012, 09:44
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5
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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Originally posted by mikemcgarry on 10 Jan 2012, 18:40.
Last edited by mikemcgarry on 11 Jan 2012, 09:44, edited 2 times in total.
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Re: A and B ran, at their respective constant rates, a race of 480 m. In  [#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.
##### General Discussion
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A and B ran, at their respective constant rates, a race of 480 m. In  [#permalink]

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26 Dec 2010, 09:08
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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

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)

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

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Re: A and B ran, at their respective constant rates, a race of 480 m. In  [#permalink]

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24 Feb 2012, 10:54
1
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 of 480 m. In  [#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 of 480 m. In  [#permalink]

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19 Apr 2013, 03:39
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 of 480 m. In  [#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)

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

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Re: A and B ran, at their respective constant rates, a race of 480 m. In  [#permalink]

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25 Jun 2014, 15:15
1
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 of 480 m. In  [#permalink]

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25 Jun 2014, 16:33
3
2
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 of 480 m. In  [#permalink]

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09 May 2015, 23:36
1
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).

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Re: A and B ran, at their respective constant rates, a race of 480 m. In  [#permalink]

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15 Oct 2015, 21:17
2
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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Re: A and B ran, at their respective constant rates, a race of 480 m. In  [#permalink]

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27 Jan 2016, 16:37
let t=A's time in seconds for each heat, assuming consistent speed of A
B's speed in heat 1=432/t+6 m/s
B's speed in heat 2=336/(t-2} m/s
assuming consistent speed of B, 432/(t+6)=336/(t-2)
t=30 seconds
B's speed=432/36=336/28=12 m/s
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Re: A and B ran, at their respective constant rates, a race of 480 m. In  [#permalink]

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08 Jul 2017, 09:29
3
case 1:, B has been given 48m headstart; A won by 6 seconds.
case 2, B is given headstart of 144m; B won by 2 sec

So, B went on from losing to winning & we know this happened for time difference of 8 seconds (from loosing by 6 sec to winning by 2 seconds), This happened because B traveled 144-48 = 96m more in first case.
Now B traveled 96m in 8 sec and hence, speed of B = 96/8 = 12m/s

Ans is A
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Re: A and B ran, at their respective constant rates, a race of 480 m. In  [#permalink]

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20 Jan 2018, 05:32
A goes 480m in t seconds. In t seconds B goes 432-6x where x is his rate. In the next race b goes 336 in t-2 seconds. So in two seconds b goes (432-6x)-336=96-6x meters. He would also go 2x meters every two seconds. 96-6x=2x so x =12m/s

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Re: A and B ran, at their respective constant rates, a race of 480 m. In &nbs [#permalink] 20 Jan 2018, 05:32
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