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Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours

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Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours [#permalink] New post 14 Jun 2005, 05:26
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A
B
C
D
E

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Question Stats:

90% (01:21) correct 10% (01:25) wrong based on 8 sessions
Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours and Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X, how many hours did it take Car Y to travel the route?

(A) 2/3
(B) 1
(C) 4/3
(D) 8/5
(E) 3

OPEN DISCUSSION OF THIS QUESTION IS HERE: car-x-and-car-y-traveled-the-same-80-mile-route-if-car-x-to-168130.html
[Reveal] Spoiler: OA
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Re: Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours [#permalink] New post 14 Jun 2005, 05:37
mandy wrote:
. :) Hello
i found B but it seem an ets trap

Car X and car Y traveled the same 80-mile route. If car X took 2 hours and car Y traveled at an average speed that was 50 percent faster than the averages speed of car X, how many hours did it take car Y to travel the route ?
(A) 2/3
(B) 1
(C) 1+ 1/3
(D) 1+3/5
(E) 3


Car x : 80 / 2 = 40 m/hr
Car y = 3/2 x 40 = 60 mil/hr

80 / 60 = 4/3, answer is C
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Re: Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours [#permalink] New post 14 Jun 2005, 05:48
X travelled at an avarage speed of 40 miles per hour

Y travelled 50 % faster i.e 50 % of 40 = 20
so avarage speed of car y = 40 + 20 = 60

time taken by Y = 80/60 = 4/3 i.e 1+1/3

so C is the answer
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Re: Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours [#permalink] New post 14 Jun 2005, 11:39
mandy wrote:
. :) Hello
i found B but it seem an ets trap

Car X and car Y traveled the same 80-mile route. If car X took 2 hours and car Y traveled at an average speed that was 50 percent faster than the averages speed of car X, how many hours did it take car Y to travel the route ?
(A) 2/3
(B) 1
(C) 1+ 1/3
(D) 1+3/5
(E) 3


The distance travelled by both the cars is the same hence.

Let the speed of X be s m/h
Then s*2 = total distance travelled = (1.5s)(time travelled by Y)
so Y = (s*2)/(1.5).

HMTG.
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Re: Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours [#permalink] New post 14 Jun 2005, 14:51
thanks
so the trap was to assume that if y is going 50% faster he will need 50% less time right ?
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Re: Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours [#permalink] New post 22 Oct 2014, 11:27
mandy wrote:
thanks
so the trap was to assume that if y is going 50% faster he will need 50% less time right ?


He is going 50% faster means his speed has increase by 50%. His new speed will be 150%. Had his speed increased by 100%, the time taken would have halved and his new speed be 200% , i.e double of initial speed.
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Re: Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours [#permalink] New post 22 Oct 2014, 11:53
Expert's post
mandy wrote:
Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours and Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X, how many hours did it take Car Y to travel the route?

(A) 2/3
(B) 1
(C) 4/3
(D) 8/5
(E) 3


The speed of car X is (distance)/(time) = 80/2 = 40 miles per hour.

The speed of car Y = 3/2*40 = 60 miles per hour --> (time) = (distance)/(speed) = 80/60 = 4/3 hours.

Answer: C.

Or: to cover the same distance at 3/2 as fast rate 2/3 as much time is needed --> (time)*2/3 = 2*2/3 = 4/3 hours.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: car-x-and-car-y-traveled-the-same-80-mile-route-if-car-x-to-168130.html
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Re: Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours   [#permalink] 22 Oct 2014, 11:53
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Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours

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