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# X and Y start traveling in the same direction, from a particular point

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Math Expert
Joined: 02 Sep 2009
Posts: 54462
X and Y start traveling in the same direction, from a particular point  [#permalink]

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06 Jul 2017, 02:34
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Question Stats:

47% (02:00) correct 53% (01:58) wrong based on 97 sessions

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X and Y start traveling in the same direction, from a particular point, at 7:00 am and 7:45 am, respectively. At what time do they meet?

(1) Speed of Y is double that of X.
(2) After 5 hours, Y is 10 miles ahead of X.

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X and Y start traveling in the same direction, from a particular point  [#permalink]

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Updated on: 06 Jul 2017, 06:53
1
Bunuel wrote:
X and Y start traveling in the same direction, from a particular point, at 7:00 am and 7:45 am, respectively. At what time do they meet?

(1) Speed of Y is double that of X.
(2) After 5 hours, Y is 10 miles ahead of X.

$$Speed = \frac{Distance}{Time}$$

Question: Time of meeting = ?

To answer the question we need the speeds of X and Y alongwith the distance that they are travelling together

Statement 1:Y = 2X
For meeting the distances travelled by them must be same
let X travels for T hours to meet Y
then Y travels (T-0.75) hours to meet X [45 mins = 3/4 Hours = 0.75 Hours)
i.e. Distances travelled to meet = X*T = Y*(T-0.75)
i.e. X*T = 2X*(T-0.75)
i.e. T = 2T-1.5
i.e. T = 1.5 hours after X starts @7:00 AM
i.e. Time = 8:30 AM
SUFFICIENT

Statement 2: After 5 hours, Y is 10 miles ahead of X.
i.e. at relative speed the total distance change in 5 hours between X and Y = (Distance travelled by X in 45 mins+10)
but neither do we know the speed of x and Y nor do we know any relationship between their speeds hence
NOT SUFFICIENT

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Originally posted by GMATinsight on 06 Jul 2017, 03:22.
Last edited by GMATinsight on 06 Jul 2017, 06:53, edited 1 time in total.
Math Expert
Joined: 02 Aug 2009
Posts: 7578
Re: X and Y start traveling in the same direction, from a particular point  [#permalink]

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06 Jul 2017, 05:54
Bunuel wrote:
X and Y start traveling in the same direction, from a particular point, at 7:00 am and 7:45 am, respectively. At what time do they meet?

(1) Speed of Y is double that of X.
(2) After 5 hours, Y is 10 miles ahead of X.

Hi,
A very good Q by Bunuel...

I would find it closer to 700 level as it has that perfect C trap
We know x starts 45 minutes before y..

Let's see the statements
1) Speed of Y is double that of X..
Easy to think this as insufficient..
BUT...
If the speed is double,y will cover the distance in half the time..
X is 45 minutes ahead. How much time would Y take to cover these 45 minutes.

In next 45 minutes, Y will cover a distance that X will cover in 45+45 or 90 minutes.
So both are traveling for 45 minutes but Y is covering up the 45 minutes of early start of X

They meet at 7:45+45= 8:30
Sufficient
2)After 5 hours, Y is 10 km ahead of X..
We require relative speed or distance or anyone's speed.
Insufficient

A
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Posts: 7578
Re: X and Y start traveling in the same direction, from a particular point  [#permalink]

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06 Jul 2017, 05:56
GMATinsight wrote:
Bunuel wrote:
X and Y start traveling in the same direction, from a particular point, at 7:00 am and 7:45 am, respectively. At what time do they meet?

(1) Speed of Y is double that of X.
(2) After 5 hours, Y is 10 miles ahead of X.

$$Speed = \frac{Distance}{Time}$$

Question: Time of meeting = ?

To answer the question we need the speeds of X and Y alongwith the distance that they are travelling together

Statement 1:Y = 2X
We know the ratio of speeds but we don't know the speeds and distance hence
NOT SUFFICIENT

Statement 2: After 5 hours, Y is 10 miles ahead of X.
i.e. at relative speed the total distance change in 5 hours between X and Y = (Distance travelled by X in 45 mins+10)
but neither do we know the speed of x and Y nor do we know any relationship between their speeds hence
NOT SUFFICIENT

Combining the two statements

Now we know that the speed of Y is double the speed of X and
i.e. Relative speed = Y-X = 2X-X = X

(the distance travelled by Y in 5 hours) = (Distance of X in 5 hour and 45 mins) + 10
i.e. 5*Y = 5.75*X + 10
i.e. 5*(2X) = 5.75*X + 10
which gives us the value of X adn Y both i.e. their speeds

Now, Time of meeting = 5 hours - (time to travel 10 miles at their relative speeds)

since we know their speeds so we can canculate the time they will take to travel 10 miles at their relative speeds hence

SUFFICIENT

Hi,

I believe ANSWER should be A
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Re: X and Y start traveling in the same direction, from a particular point  [#permalink]

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06 Jul 2017, 11:05
lets luk at this way..

stat 1 : we know that time to cover up = distance / relative speed...(1)
given relation between speed of Y= 2X
relative speed = 2X - X = X...

distance between x and y in 45 mins will be the distance travelled by X in 45 mins, since x has a headstart of 45 mins..
distance = speed * time
= X * 3/4

substituting in (1)... time = 3/4 hr = 45 mins..
hence 7:45 + 0:45 = 8:30
suff

stat2 : clearly not suff

ans A
Re: X and Y start traveling in the same direction, from a particular point   [#permalink] 06 Jul 2017, 11:05
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