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P can give Q a start of 20 seconds in a kilometer race

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P can give Q a start of 20 seconds in a kilometer race [#permalink] New post 25 Aug 2005, 15:36
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P can give Q a start of 20 seconds in a kilometer race. P can give R a start of 200 meters in the same kilometer race. And Q can give R a start of 20 seconds in the same kilometer race. How long does P take to run the kilometer?


(1) 160 seconds (2) 140 seconds
(3) 200 seconds (4) 240 seconds
(5) 260 seconds
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 [#permalink] New post 26 Aug 2005, 06:42
1)

R starts first, 20 seconds later Q start, 20 seconds later P starts.
By now, R has run 200 meter at the speed of 40/200 = 1/5 second/meter.
To finish the rest of the race (800 meters), R needs:
800/5 = 160 seconds

Since all P,Q,R arrives at the same time, it takes P 160 seconds to finish
1000 meters.
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 [#permalink] New post 26 Aug 2005, 13:53
qpoo wrote:
1)

R starts first, 20 seconds later Q start, 20 seconds later P starts.
By now, R has run 200 meter at the speed of 40/200 = 1/5 second/meter.
To finish the rest of the race (800 meters), R needs:
800/5 = 160 seconds

Since all P,Q,R arrives at the same time, it takes P 160 seconds to finish
1000 meters.


thanks for your explanation qpoo plz may i ask you two questions
where the 4O comes from and how do u see that they arrived at the same time

regards

mandy
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 [#permalink] New post 26 Aug 2005, 15:07
It took me atleast 4 mins to solve it.....

R needs 200 mts advantage over P or 40 sec

so 40 sec for 200mts for R, so 200sec for R to finish and 200-40 = 160 seconds for P...

There are 3 conditions in the problem if the third condition was given second this problem would be a whole lot simpler....
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 [#permalink] New post 26 Aug 2005, 15:28
mandy wrote:
qpoo wrote:
1)

R starts first, 20 seconds later Q start, 20 seconds later P starts.
By now, R has run 200 meter at the speed of 40/200 = 1/5 second/meter.
To finish the rest of the race (800 meters), R needs:
800/5 = 160 seconds

Since all P,Q,R arrives at the same time, it takes P 160 seconds to finish
1000 meters.


thanks for your explanation qpoo plz may i ask you two questions
where the 4O comes from and how do u see that they arrived at the same time

regards

mandy


the stem says: P can give Q a start of 20 seconds in a kilometer race ... Q can give R a start of 20 seconds, thus 20+20 = 40

"P can give Q a start of 20 seconds in a kilometer" means Q starts first, 20 seconds later P starts and they finish at the same time, right ?
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 [#permalink] New post 26 Aug 2005, 15:53
P can give Q a start of 20 seconds in a kilometer race.
P can give R a start of 200 meters in the same kilometer race.
And Q can give R a start of 20 seconds in the same kilometer race.

At the start

1) P <-20s-> Q<-20s-> R Time domain

So even before P starts to run R has run for 40 seconds

2) P <-200m-> R Distance

So, R runs 200m in 40 sec.

By this it must be clear where the 40 sec came from.


Next,

If a problem states that A can give B a start of x sec. It means that A and B will reach the destination at the same time, when B runs for x sec before A starting.

Speed of A in this case should be greater than speed of B.

Same is the case with the distance.

How long does P take to run the kilometer?
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 [#permalink] New post 29 Aug 2005, 13:59
lets say P Q and R are their velocities and t is the time taken by P


1) Pt = Q(t+20) -- p gives q a start of 20 sec
2) 1000/P = 800/R ----P gives R astart of 200 m
3) Q(t+20) = R(t+40) --- q gives R a start of 20 sec

from 1 and 3

Pt = R(t + 40)
P/R = ( t+40) / t

from 2
P/R = 1000/ 800 = 5/4\


hence

(t+40) /t = 5/4

hence t = 160 time taken by P
  [#permalink] 29 Aug 2005, 13:59
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