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A car starts from X and moves towards Y. At the same time

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A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 29 Aug 2013, 04:15
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A car starts from X and moves towards Y. At the same time another car starts from Y and moves towards X. Both travel at a constant speed and meet after 4 hours. During that time, the faster car traveled 2/3 of the distance XY. How long would it take the slower car to travel the distance XY?

A 6 hours
B. 8 hours
C. 10 hours
D. 12 hours
E. 15 hours

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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 07 Oct 2013, 02:32
Ans is D
This is how
Say Total Dist. XY = 1
Let constant rate of Car X & Car Y be "x" & "y" respectively
Car X distance traveled = 2/3
Time = 4 hrs
Rate of Car X=2/3 div by 4
i.e x = 1/6

When traveling towards each other
Speed = Dist./Time
(x+y) = 1/4
substituting fr "x" in above eqn. we get
y = 1/12
Therefore Time taken to travel Distance XY = 1 div. 1/12
Ans D) 12
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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 07 Oct 2013, 04:56
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My view on this Question:

If one car travelled 2/3, then the other car must have travelled only 1/3, as they are meeting after 4 hours in a certain point. So:

(1/3)XY= 4*y --> It took the car 4 hours to travel 1/3 of the distance at a constant speed y.

So if we solve this last equation:

XY= 3*4*y= 12*y --> It will take this car 12 hours in total to reach its final destination.

Answer: D
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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 09 Oct 2013, 01:33
SravnaTestPrep wrote:
A car starts from X and moves towards Y. At the same time another car starts from Y and moves towards X. Both travel at a constant speed and meet after 4 hours. During that time, the faster car traveled 2/3 of the distance XY. How long would it take the slower car to travel the distance XY?

A 6 hours
B. 8 hours
C. 10 hours
D. 12 hours
E. 15 hours




D --> Take a smart number (as X travel's 2/3 & Y travel's 1/3 of the distance, i took D = 60 since ,we also have to divide 2/3 & 1/3 by 4 i.e time taken by both car)
D = 60 , therefore , X travels 60*2/3 = 40. Now, since t = 4 , therefore X's speed = 40/4 = 10.
Similarly , Y travels 60*1/3 = 20. Since t=4 , Y's Speed = 20/4 = 5
Therefore total time taken by Y is t = D/S = 60/5 = 12hrs
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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 09 Oct 2013, 08:51
SravnaTestPrep wrote:
A car starts from X and moves towards Y. At the same time another car starts from Y and moves towards X. Both travel at a constant speed and meet after 4 hours. During that time, the faster car traveled 2/3 of the distance XY. How long would it take the slower car to travel the distance XY?

A 6 hours
B. 8 hours
C. 10 hours
D. 12 hours
E. 15 hours


Quick and easy:
It is clear that the slower car is the one starting from Y (car B).
If car B spends 4h to complete d/3 of the traject, then it will spend 8h to complete 2d/3 and 12h to make the whole traject (3d/3).

Answer is D, 12h.
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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 12 Oct 2013, 02:57
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SravnaTestPrep wrote:
A car starts from X and moves towards Y. At the same time another car starts from Y and moves towards X. Both travel at a constant speed and meet after 4 hours. During that time, the faster car traveled 2/3 of the distance XY. How long would it take the slower car to travel the distance XY?

A 6 hours
B. 8 hours
C. 10 hours
D. 12 hours
E. 15 hours


Take XY = 60 km
Car A travelled distance D1 = 2/3*60 = 40km
Car B travelled distance D2 = 1/3*60 = 20km

Car B is slower, travelling 20 km in 4 hours. Car B speed = 20/4 = 5 km/h

Time for Car B to travel whole distance XY = 60 km:
D = S*T
60 = 5*T
T = 60/5 = 12 hours - Answer.

Hope it helps. :)
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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 12 Oct 2013, 04:37
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The faster car travels \(\frac{2}{3}\) of this distance in 4 hours.

The slower car hence travels \(\frac{1}{3}\) of this distance in 4 hours.

So, to travel the total distance , the slower car would take 4*3 = 12 hours

Answer is D
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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 15 Oct 2013, 01:43
Distance is in the ratio of 2:1 so speed will also be 2:1. Time will be in the ratio of 1:2 because Speed is inversely proportional to time keeping distant constant.
The slowest car's time is \(\frac{1}{3}\)
so \(\frac{1}{3}*x=4\)
\(x=12\)
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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 09 Jan 2014, 12:56
SravnaTestPrep wrote:
A car starts from X and moves towards Y. At the same time another car starts from Y and moves towards X. Both travel at a constant speed and meet after 4 hours. During that time, the faster car traveled 2/3 of the distance XY. How long would it take the slower car to travel the distance XY?

A 6 hours
B. 8 hours
C. 10 hours
D. 12 hours
E. 15 hours


4 (x+y) = d

4x = 2/3d

4y = 1/3d

Therefore, y = d/12

D 12 hours
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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 16 Feb 2014, 08:34
Lets solve it "plug value" ,method as it about fraction....
Say total distance 3 miles so faster car traveled 2 mile and slower one is 1 mile..

So the slower one traveled 1 mile in 4 hour(because they met in 4 hours) that makes the rate 1/4...
Now lets assume T hour required to cover the distance so 1/4*T=3 So T=12 hours(d) ..ans...

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Re: A car starts from X and moves towards Y. At the same time  [#permalink]

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New post 07 Sep 2018, 18:01
If we let the distance = 12

Then the faster car have travelled 8 miles \(= \frac{2}{3}* 12 = 8\) miles. As a result the slower car have travelled 4 miles.

Since they meet in 4 hours, then Speed of the slow car \(= \frac{4}{4} = 1\).

On the basis of our assumption that Distance = 12 then Time would be \(= \frac{12}{1} = 12\) hours
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Re: A car starts from X and moves towards Y. At the same time &nbs [#permalink] 07 Sep 2018, 18:01
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