In a km race, A gives B a start of 20 seconds and beats him : PS Archive
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# In a km race, A gives B a start of 20 seconds and beats him

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VP
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In a km race, A gives B a start of 20 seconds and beats him [#permalink]

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20 May 2005, 06:06
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In a km race, A gives B a start of 20 seconds and beats him by 40m. However, when he gives B a start of 25 seconds they finish in a dead heat. What is Aâ€™s speed in m/sec?

(1) 12.5 m/sec
(2) 20 m/sec
(3) 8 m/sec
(4) 10 m/sec
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20 May 2005, 12:43
christoph wrote:
In a km race, A gives B a start of 20 seconds and beats him by 40m. However, when he gives B a start of 25 seconds they finish in a dead heat. What is Aâ€™s speed in m/sec?

(1) 12.5 m/sec
(2) 20 m/sec
(3) 8 m/sec
(4) 10 m/sec

Poorly worded.
Are you sure A beats be by 40 minutes? A lot of confusion here with minutes/seconds.
Where is it stated that speed of A and B did not change?
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20 May 2005, 13:12
1000/A + 20 = 960/B

1000/A + 25 = 1000/B

Subtracting, 5 = 40/B => B = 8m/s. => A = 10m/s.
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Director
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20 May 2005, 14:36
Tyr wrote:
christoph wrote:
In a km race, A gives B a start of 20 seconds and beats him by 40m. However, when he gives B a start of 25 seconds they finish in a dead heat. What is Aâ€™s speed in m/sec?

(1) 12.5 m/sec
(2) 20 m/sec
(3) 8 m/sec
(4) 10 m/sec

Poorly worded.
Are you sure A beats be by 40 minutes? A lot of confusion here with minutes/seconds.
Where is it stated that speed of A and B did not change?

The problem is correct . 40m is not 40 minutes. It is 40 meters

Here

A gives B a start of 20 seconds and beats him by 40m means B's speed is 40/20= 2m/s (it means when A travelled 1000mts B travelled only 960mts)

B's speed ---> 2m/s

Let speed of A be s1 and B be s2

From this "A gives B a start of 20 seconds and beats him by 40m "

we have s1(t - 20) = s2*t + 40(Since B is slow and still needs to run 40m)
--------(1)

s1*t = s2(t+25) (Since B gets a start of 25 sec) --(2)

Solve (1) and (2) and substitute s2 =2 and you get s1=8

A's speed = 8m/s
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21 May 2005, 03:55
i agree with Kapslock. from 1 and 2 statements B needed 5 extra sec to run another 40 m. thus, his speed is 8m/sec. a's speed was 10 m/s
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21 May 2005, 07:12
july05 wrote:
i agree with Kapslock. from 1 and 2 statements B needed 5 extra sec to run another 40 m. thus, his speed is 8m/sec. a's speed was 10 m/s

I too arrived at 10 but in a convoluted way. your shortcut is great ( arriving at B's speed without equations)!

HTMG
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22 May 2005, 05:59
Guys can you let me know where I went wrong in my solution?
Thanks
SVP
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22 May 2005, 19:17
gmat2me2 wrote:

A gives B a start of 20 seconds and beats him by 40m means B's speed is 40/20= 2m/s (it means when A travelled 1000mts B travelled only 960mts)

B's speed ---> 2m/s

It is right to say that if A travelled 1000m then B travelled only 960m. But it is incorrect to say that B needed 20 second to run 40m. In fact, when A give B 20s B missed 40m, but when A gives B 25s B they finished together. This means B really needed 25-20=5s for that 40m. Therefore you get B's speed is 8m/s.

You can then solve for A's speed for 10m/s.
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22 May 2005, 20:23
HongHu wrote:
gmat2me2 wrote:

A gives B a start of 20 seconds and beats him by 40m means B's speed is 40/20= 2m/s (it means when A travelled 1000mts B travelled only 960mts)

B's speed ---> 2m/s

It is right to say that if A travelled 1000m then B travelled only 960m. But it is incorrect to say that B needed 20 second to run 40m. In fact, when A give B 20s B missed 40m, but when A gives B 25s B they finished together. This means B really needed 25-20=5s for that 40m. Therefore you get B's speed is 8m/s.

You can then solve for A's speed for 10m/s.

Hmmmm........I know whats the gap there....Thanks anyways HongHu.....
Re: PS - Race   [#permalink] 22 May 2005, 20:23
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