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Car X and Car Y traveled the same 80-mile route. If Car X to

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Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post Updated on: 01 Mar 2014, 06:50
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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

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Originally posted by seabhi on 01 Mar 2014, 05:02.
Last edited by Bunuel on 01 Mar 2014, 06:50, edited 1 time in total.
Edited the question.
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Re: Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 01 Mar 2014, 06:58
seabhi 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.
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Re: Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 06 Feb 2017, 10:15
As Y traveled 50 percent faster than X
So, if X needs 1.5 hour Y needs 1 hour
if X needs 2 hour Y needs 2/1.5 hour = 4/3 hour

Answer : C
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Re: Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 05 Jun 2017, 07:05
the distance for both the parties is constant so as we can see that speed of y is multiplied by 3/2 then time will also multiplied by reciprocal of 3/2 which will be 2*2/3= 4/3 time taken by y

hence C
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Re: Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 05 Jun 2017, 08:41
seabhi wrote:
Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours and Car Y travelled 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


For Car X -

\(Speed = 40 \ miles/hr\)
\(Time = 2 \ Hours\)

For Car Y -

\(Speed = 60 \ miles/hr\)
\(Time = \frac{80}{60}\)

So, the time required is \(\frac{4}{3}\) Hours, answer will be (C) \(\frac{4}{3}\)
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Re: Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 06 Jun 2017, 16:47
seabhi 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


We are given that Car X traveled 80 miles in 2 hours. Thus, the rate of car X was 80/2 = 40 mph.

We are also given that Car Y traveled 50% faster than Car X. Thus, Car Y traveled at a rate of 1.5 x 40 = 60 mph.

So, it took Car Y 80/60 = 8/6 = 4/3 hours to travel the route.

Answer: C
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Re: Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 28 Jan 2018, 08:48
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seabhi 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


There's a nice rule we can use here.

To set up the rule, recognize that if Y travels 2 times as fast as X, then Y's travel time will be 1/2 of X's.
Similarly, if Y travels 3 times as fast as X, then Y's travel time will be 1/3 of X's.
Or if Y travels 1/4 as fast as X, then Y's travel time will be 4 times X's travel time.

In general, if Y travels a/b times as fast as X, then Y's travel time will be b/a of X's travel time.

So, if Y's speed is 50% more than X's speed, we can say that Y travels 1.5 times as fast as X.
In other words, if Y travels 3/2 times as fast as X, which means Y's travel time will be 2/3 that of X's travel time.

Since X's travel time is 2 hours, Y's travel time will be (2/3)(2) = 4/3 = 1 1/3

Answer: C

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Re: Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 21 Feb 2018, 08:49
Bunuel wrote:
seabhi 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.



Does 3/2 mean 50% more ? :? ---> when you multiply 3/2*40
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Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 21 Feb 2018, 08:59
GMATPrepNow wrote:
seabhi 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


There's a nice rule we can use here.

To set up the rule, recognize that if Y travels 2 times as fast as X, then Y's travel time will be 1/2 of X's.
Similarly, if Y travels 3 times as fast as X, then Y's travel time will be 1/3 of X's.
Or if Y travels 1/4 as fast as X, then Y's travel time will be 4 times X's travel time.

In general, if Y travels a/b times as fast as X, then Y's travel time will be b/a of X's travel time.

So, if Y's speed is 50% more than X's speed, we can say that Y travels 1.5 times as fast as X.
In other words, if Y travels 3/2 times as fast as X, which means Y's travel time will be 2/3 that of X's travel time.

Since X's travel time is 2 hours, Y's travel time will be (2/3)(2) = 4/3 = 1 1/3

Answer: C

Cheers,
Brent



hi Brent yes you :) GMATPrepNow

you say " if Y travels 2 times as fast as X, then Y's travel time will be 1/2 of X's. "

how can that be possible :? if Y travels twice as fast as X that means for example if Y speed is 30 km per hour and X speed is 15 km per hour. no ?

but you say i need to multiply \(\frac{1}{2}\)* 15 = 7.5 :? then i get Y =7.5 :?

pls explain :)
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Re: Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 21 Feb 2018, 09:20
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Top Contributor
dave13 wrote:
GMATPrepNow wrote:
seabhi 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


There's a nice rule we can use here.

To set up the rule, recognize that if Y travels 2 times as fast as X, then Y's travel time will be 1/2 of X's.
Similarly, if Y travels 3 times as fast as X, then Y's travel time will be 1/3 of X's.
Or if Y travels 1/4 as fast as X, then Y's travel time will be 4 times X's travel time.

In general, if Y travels a/b times as fast as X, then Y's travel time will be b/a of X's travel time.

So, if Y's speed is 50% more than X's speed, we can say that Y travels 1.5 times as fast as X.
In other words, if Y travels 3/2 times as fast as X, which means Y's travel time will be 2/3 that of X's travel time.

Since X's travel time is 2 hours, Y's travel time will be (2/3)(2) = 4/3 = 1 1/3

Answer: C

Cheers,
Brent



hi Brent yes you :) GMATPrepNow

you say " if Y travels 2 times as fast as X, then Y's travel time will be 1/2 of X's. "

how can that be possible :? if Y travels twice as fast as X that means for example if Y speed is 30 km per hour and X speed is 15 km per hour. no ?

but you say i need to multiply \(\frac{1}{2}\)* 15 = 7.5 :? then i get Y =7.5 :?

pls explain :)


I think you are mixing up SPEED and TIME

TRUE: If Y travels 2 times as fast as X, then Y's travel time will be 1/2 of X's.

So, if we're talking SPEED, then Y's speed = 30 km per hour and X's speed = 15 km per hour meets the given condition

Now let's compare travel TIMES.
Let's say both cars are traveling 30 km
Then Y's travel TIME = 1 hour and X's travel TIME = 2 hours

Does that help?

Cheers,
Brent
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Car X and Car Y traveled the same 80-mile route. If Car X to  [#permalink]

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New post 21 Feb 2018, 11:13
dave13 wrote:
Bunuel wrote:
seabhi 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.

Does 3/2 mean 50% more ? :? ---> when you multiply 3/2*40

Hi dave13 . Yes, 3/2 means 50 percent more.
Why? 3/2 = 1.50, and 1.50 (or 1.5) is the multiplier for "50 percent faster."

Often fractions are easier than decimals in "percent increase/decrease" problems. Just convert the decimal multiplier to a fraction.

Y's rate is a percent increase of X's rate. Here, 50 percent faster =
Original speed + 50% of original speed
--Original speed (X's speed) = 40
--50 percent of 40 = 20
(Original + 50% of original) = (40+20) = 60 = Y
OR

\(Y=1.5X\)
\(1.5=1\frac{5}{10}=1\frac{1}{2}=\frac{3}{2}\)
\(Y = \frac{3}{2}X\)


Here are some common fractions used in percent increase/ decrease problems: 5/4, 6/5, 2/5, 1/4, 1/2, 3/2. Some are increases, some are decreases. Decreases are harder. mikemcgarry explains the issue you asked about here

Hope that helps. :-)
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Car X and Car Y traveled the same 80-mile route. If Car X to   [#permalink] 21 Feb 2018, 11:13
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