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Senior Manager
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CARS [#permalink] New post 14 Jun 2005, 05:26
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 0 sessions
. :) 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
Manager
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Re: CARS [#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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 [#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
Senior Manager
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Re: CARS [#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.
Senior Manager
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 [#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 ?
  [#permalink] 14 Jun 2005, 14:51
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