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# How much time did it take a certain car to travel 400

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Manager
Joined: 13 Apr 2006
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How much time did it take a certain car to travel 400 [#permalink]  19 Jun 2006, 09:07
How much time did it take a certain car to travel 400 kilometers?
1. Car traveled the first 200 kilometers in 2.5 hours
2. If car's average speed was 20 kilometers greater, it would have taken 1 hour less.

Please explain your answer. I got it right, but what is the best equation to use here?
Senior Manager
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B it is

1) obviously insufficient
2) denote s for speed, so the equation : 400/s -1 = 400/(s+20) one variable so we can find the speed , whatever it is. sufficient
Manager
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can we really equate s-1 (denoting time) to s+20 (denoting speed)? I'm missing something...Can you explain?

THANKS!
VP
Joined: 15 Jun 2006
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B is my answer as well
1) is not enough for us since we do not know the speed with which the car drove the second part of the route (we are not said that the speed was constant, although if it would have been said then it would be sufficient as well)
2) is enough:
we can construct an equation (v is fpr speed)
400/v=400/(v+20)+1
400/v-400/(v+20)=1
(400(v+20)-400v)/v(v+20)=1
(400v+8000-400v)/(v^2+20v)=1
8000/(v^2+20v)=1
v^2+20v=8000
v^2+20v-8000=0
a quadratic equation. if we solve it we will get that the speed is v=80
VP
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kuristar wrote:
can we really equate s-1 (denoting time) to s+20 (denoting speed)? I'm missing something...Can you explain?

THANKS!

not (s-1) for time, but (400/s)-1 for time
likewise 400/(s+20) for time
VP
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Stmt1 is not sufficient to compute the total time, as remaining 200 kms can be covered as slowly as possible.

Stmt2 is not sufficient.
Let s be average speed and t be total time taken.
sxt = 400 ...(a)

Given
(s+20)(t-1) = 400 ...(b)

Combining
st -s +20t -20 = 400
400 -s+20t -20 = 400
or s-20t = -20

now s= 400/t
400-20t^2= -20t
20 -t^2 = -t
or t^2+t-20=0

(t+5)(t-4) =0 => t=4

Therefore B.
Senior Manager
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Statmet II is sufficient.. (B)
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