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While on a straight road, Car X and Car Y are traveling at

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While on a straight road, Car X and Car Y are traveling at [#permalink] New post 17 Mar 2012, 17:44
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75% (01:54) correct 25% (01:13) wrong based on 385 sessions
While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y ?

(1) Car X is traveling at 50 miles per hour and Car Y is traveling at 40 miles per hour.
(2) Three minutes ago Car X was 1/2 mile ahead of Car Y.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Dec 2012, 09:38, edited 1 time in total.
Renamed the topic and edited the question.
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Re: GMAT Paper Tests 1 [#permalink] New post 17 Mar 2012, 20:22
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A follows B, Time to catchup = (Distance between A and B)/(SA - SB)

A)
SA - SB is the relative speed = 50-40=10 m/hr
Distance between A and B = 2-1 = 1.

Therefore, Time to catchup = 1/10 hours = 0.1*60 minutes = 6 mintues

Sufficient.

B)
3 minutes for 1/2 mile
Therefore 1 mile will be 6 minutes

Sufficient

Thus D.
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Re: While on a straight road, Car X and Car Y are traveling at [#permalink] New post 01 Aug 2013, 12:00
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While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y ?

Car X is traveling at a greater speed/rate than Y.

(1) Car X is traveling at 50 miles per hour and Car Y is traveling at 40 miles per hour.

Car X is traveling at 5/6 miles/minute.
Car Y is traveling at 4/6 miles/minute.

Every minute, Car X moves away from car Y at a rate of (5/6) - (4/6) = 1/6 miles. Therefore, car X will have moved an additional mile away from car Y in six minutes.
SUFFICIENT

(2) Three minutes ago Car X was 1/2 mile ahead of Car Y.
In three minutes, car X managed to move 1/2 mile further away from Y. Because both cars were moving at constant rates, car x would have moved away from Y at a rate of 1/6 miles per minute, or 3/6 = 1/2 miles in three minutes. We know how long it takes x to move 1/2 mile from Y and because the speed of Y is constant, the rate at which x moves away from y will be constant as well.
SUFFICIENT

(D)
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Re: While on a straight road, car X and car Y are traveling at [#permalink] New post 18 Mar 2012, 06:21
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While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y ?

(1) Car X is traveling at 50 miles per hour and Car Y is traveling at 40 miles per hour --> since we have the rates of both cars and the distance between them we can calculate any other question regarding them. Sufficient.

(2) Three minutes ago Car X was 1/2 mile ahead of Car Y --> car X gains 1/2 mile in every 3 minutes (since now it's 1 mile ahead), hence it'll gain additional 1 mile in next 6 minutes. Sufficient.

Answer: D.
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Re: While on a straight road, car X and car Y are traveling at [#permalink] New post 11 Nov 2012, 00:54
Bro Bunuel,

How to solve this to get the time in secs if we consider this as 2 independent PS questions?

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Re: While on a straight road, car X and car Y are traveling at [#permalink] New post 11 Nov 2012, 01:06
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ok.. i just figured out..

1)

relative speed: 10 miles/ hour = 1/6 miles / min
distance to be covered: 1 mile.

time taken : 1/(1/6) = 6 mins..

2)
find the rate :
distance : 1/2 mile
time taken: 3 mins

rate: 1/6 miles / min

now,
distance=1 mile
rate : 1/6 miles/ min

so time: 6 mins

:lol:

hope it helps! ( Bunuel's statement :) )
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Re: While on a straight road, Car X and Car Y are traveling at [#permalink] New post 22 Oct 2013, 10:43
WholeLottaLove wrote:
While on a straight road, Car X and Car Y are traveling at different constant rates. If Car X is now 1 mile ahead of Car Y, how many minutes from now will Car X be 2 miles ahead of Car Y ?

Car X is traveling at a greater speed/rate than Y.

(1) Car X is traveling at 50 miles per hour and Car Y is traveling at 40 miles per hour.

Car X is traveling at 5/6 miles/minute.
Car Y is traveling at 4/6 miles/minute.

Every minute, Car X moves away from car Y at a rate of (5/6) - (4/6) = 1/6 miles. Therefore, car X will have moved an additional mile away from car Y in six minutes.
SUFFICIENT

(2) Three minutes ago Car X was 1/2 mile ahead of Car Y.
In three minutes, car X managed to move 1/2 mile further away from Y. Because both cars were moving at constant rates, car x would have moved away from Y at a rate of 1/6 miles per minute, or 3/6 = 1/2 miles in three minutes. We know how long it takes x to move 1/2 mile from Y and because the speed of Y is constant, the rate at which x moves away from y will be constant as well.
SUFFICIENT

(D)


Correction in the explanation.
Every minute, Car X moves away from car Y at a rate of (5/6) - (4/6) = 1/3 miles. Therefore, car X will have to travel for 6 mins to move 2 miles farther away from car Y..
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Re: While on a straight road, Car X and Car Y are traveling at [#permalink] New post 18 May 2014, 09:56
Statement 1: Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour.
Notice that we could easily duplicate this scenario in real life.
Start with Car X 1 mile ahead of car Y (given info)
Have Car X drive at 50 mph and car Y at 40mph.
Use a stopwatch to time how long it takes for Car X to be 2 miles ahead of Y.
As you can see, we have enough information to answer the target question
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: 3 minutes ago car X was 1/2 mile ahead of car Y.
If car X is presently 1 mile ahead, we can see that car X gains 1/2 mile every 3 minutes.
At that rate, car X will gain another 1 mile in 6 minutes.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Hence D
Re: While on a straight road, Car X and Car Y are traveling at   [#permalink] 18 May 2014, 09:56
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