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Work Distance Problem

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KunalPratap
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Work Distance Problem [#permalink] New post   12 Aug 2011, 04:26
It Takes high speed train "x" hours to travel "z" miles distance from A to B. It takes Regular speed train "y" hours to do same distance. If High speed train leaves from A and Regualr speed train leaves from B at same time . how many more miles High speed train would have already travelled than the regualr train when they pass each other??
1) z( y-x)/ x+y
2) z (x-y)/ x+y
3) z (x+y)/ y-x
4) xy (x+y)/ y-x
5) xy (x+y)/ x-y



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Re: Work Distance Problem [#permalink] New post   12 Aug 2011, 12:04
KunalPratap wrote:
It Takes high speed train "x" hours to travel "z" miles distance from A to B. It takes Regular speed train "y" hours to do same distance. If High speed train leaves from A and Regualr speed train leaves from B at same time . how many more miles High speed train would have already travelled than the regualr train when they pass each other??
1) z( y-x)/ x+y
2) z (x-y)/ x+y
3) z (x+y)/ y-x
4) xy (x+y)/ y-x
5) xy (x+y)/ x-y




Now first thing first, Because y>x (regular traing will take more time than high-speed train to cover the distance), so any answer choice contains x-y should be rejected. That leaves us with A, C, D.

Now you can do the calculation, that will require you
- to find the average speed of both trains
- to reach to a common point, both will take same time and we need to find out the difference of the distances of that common point from Terminals A and B

on the calculation I find A as the right answer
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TomB
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Re: Work Distance Problem [#permalink] New post   12 Aug 2011, 18:38
can anybody solve the problem by using values for X ,Y , Z. This will help a lot
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Re: Work Distance Problem [#permalink] New post   15 Aug 2011, 13:31
TomB wrote:
can anybody solve the problem by using values for X ,Y , Z. This will help a lot


We know the speed of each train is Z/X for the high speed, and Z/Y for the regular (total distance traveled/how much time it used).
Assume that it took the amount of time "T" for them to pass each other.
Thus distance of high speed train traveled = Z/X*T, and distance of regular speed train traveled = Z/Y*T
Z/X*T+Z/Y*T=Z or (1/X+1/Y)*T*Z=Z
therefore T=XY/(X+Y)

Now we can solve Z/X*T-Z/Y*T, which gives us the answer A.
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Re: Work Distance Problem [#permalink] New post   15 Aug 2011, 21:24
Expert Reply
KunalPratap wrote:
It Takes high speed train "x" hours to travel "z" miles distance from A to B. It takes Regular speed train "y" hours to do same distance. If High speed train leaves from A and Regualr speed train leaves from B at same time . how many more miles High speed train would have already travelled than the regualr train when they pass each other??
1) z( y-x)/ x+y
2) z (x-y)/ x+y
3) z (x+y)/ y-x
4) xy (x+y)/ y-x
5) xy (x+y)/ x-y


Using ratios can be very helpful too.
To cover the same distance, time taken by high speed:time taken by regular speed = x:y
Therefore, speed of high speed : speed of regular = y:x

Of the total distance, z, distance covered by high speed:distance covered by regular = y:x (in same time)
Distance covered by high speed = z*y/(y+x)
Distance covered by regular = z*x/(y+x)
Difference = z*(y-x)/(y+x)

OR

Put in some values. Say, distance = 10 km
x = 2 hrs
y = 5 hrs
Speed of high speed:speed of regular = 5:2
Of the 10 km, the high speed will cover 10*(5/7)
Regular will cover 10*(2/7)
Difference = 10*3/7

Put in the values of z, x and y in the options to get the answer.
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Re: Work Distance Problem [#permalink] New post   16 Aug 2011, 03:33
Thanks! This helps.



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
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Re: Work Distance Problem [#permalink]
16 Aug 2011, 03:33
Moderator:
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Senior Moderator - Masters Forum
3137 posts

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