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A car starts from X and moves towards Y. At the same time 29 Aug 2013, 04:15
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D
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25% (medium)

Question Stats:

77% (02:01) correct 23% (02:16) wrong based on 402 sessions

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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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D
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A car starts from X and moves towards Y. At the same time 07 Oct 2013, 02:32
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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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A car starts from X and moves towards Y. At the same time 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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A car starts from X and moves towards Y. At the same time 09 Oct 2013, 01:33
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SravnaTestPrep
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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A car starts from X and moves towards Y. At the same time 09 Oct 2013, 08:51
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SravnaTestPrep
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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A car starts from X and moves towards Y. At the same time 12 Oct 2013, 02:57
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SravnaTestPrep
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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A car starts from X and moves towards Y. At the same time 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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A car starts from X and moves towards Y. At the same time 09 Jan 2014, 12:56
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SravnaTestPrep
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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A car starts from X and moves towards Y. At the same time 16 Feb 2014, 08:34
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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...

Rgds
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A car starts from X and moves towards Y. At the same time 07 Sep 2018, 18:01
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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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A car starts from X and moves towards Y. At the same time 17 Oct 2019, 20:38
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SravnaTestPrep
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

We can let the total distance = 240 miles. So the faster car traveled 2/3 x 240 = 160 miles, and thus the slower car traveled 240 - 160 = 80 miles. Since the two cars meet after 4 hours, it must be the case that the slower car traveled 80 miles in 4 hours, which means the speed of the slower car is 80/4 = 20 miles per hour. Therefore, it would take the slower car 240/20 = 12 hours to travel the entire distance XY.

Answer: D
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A car starts from X and moves towards Y. At the same time 14 Jul 2020, 21:41
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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

Given: 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.
Asked: How long would it take the slower car to travel the distance XY?

Faster car travels distance 2/3 * XY in 4 hours
Slower car travels distance 1/3 * XY in 4 hours

Slower car will travel distance XY in 4*3 = 12 hours

iMO D
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A car starts from X and moves towards Y. At the same time 09 Feb 2025, 22:41
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let,xy = 12( LCM of 3 &4)
A goes = 12*3/4 =8 & B = 4
B's speed = 4/4 = 1
B's time = 12/1
=12
Answer : D
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A car starts from X and moves towards Y. At the same time 09 Feb 2025, 22:59
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If xy is cosidersd as 3 units (for calc sake). Hence slower travelled only 1 unit in 4 hrs hence would take 12hrs to travel 3 units.
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