Problem Solving

Two trains continuously travel between Washington DC and Baltimore, which is 500 miles away. They start simultaneously, train A at Washington and train B at Baltimore, and run at 30 and 90 mph respectively. The station turnaround times are negligible. What is the distance between the point where the trains meet for the first time and the point where they meet for the third time?


This is a similar question posted at https://www.beatthegmat.com/trains-t110287.html#469513
but with a different flavour. The first one was relatively easy ? but how do you solve this one ?.

Note : I donot know the answer. The only way i would do this is to manually test the distances and speeds for each section. And that is horrible. Can anyone tell me the way, or if there is no other way and such a problem will not come on gmat.

Since they start together and A's speed is (1/3)rd of B's speed, A always covers (1/3)rd of the distance that B covers.

They first meet at 125 miles from Washington (because 125 + 375 = 500 miles and 125 is (1/3)rd of 375. You may calculate this if you want). Let this point be P.
Washington.....(125).....P............(375)............Baltimore

Let the second place where they meet be x miles from P towards Baltimore.
Train A will go towards Baltimore, while Train B will first go to Washington and then return back to meet Train B.
Then,
A's distance = (1/3)*(B's distance)
x = 1/3 * (125 + 125 + x)
3x = x + 250
x = 125
Let this point be Q. (Notice this is midway Washington and Baltimore)
Washington........(250)........Q.........(250).........Baltimore

Let the third point where they meet be y miles from Q towards Baltimore.
Train A is still moving towards Baltimore. Train B will go to Baltimore and meet Train A while heading towards Washington.
A's distance = (1/3)*(B's distance)
y = 1/3 * (250 + 250 - y)
3y = 500 - y
y = 125
Let this point be R
Washington...........(375)...........R.......(125).......Baltimore

Overall
Washington...(125)...P...(125)...Q...(125)...R...(125)...Baltimore

Distance between first and third point = [spoiler]125 + 125 = 250 miles[/spoiler] (Answer)

Hi aneesh,
Thanks for the logic of finding a relation between A and B's speed. That was a very unique approach. It helped me make quick calculation for 4th and 5th meeting point.

It turns out they meet again at P for the fourth time and then at Q for the 5th time.

Washington...(125)...P...(125)...Q...(125)...R...(125)...Baltimore

Does that mean they will keep meeting at 125 intervals continiously ? If that is true we can even find the difference between their first and 10th meeting point ? :O

The answer for this one is 250 miles, and an easy way to solve this is as follows-

Since the trains move in opposite directions their relative speed is = 30+90 = 120

Lets take Washington as the reference point

When they meet first time, total distance traveled = 500 mi
Distance traveled by train A = 500*30/120 = 125
Therefore,
W-125-P-325-B

When they meet third time, total distance traveled = 5*500 mi
Distance traveled by train A = 5*500*30/120 = 625
Therefore,
W-325-Q-125-B

Thus, distance between P and Q = 250 mi

Hope it helps!!

shantanu86 wrote: When they meet third time, total distance traveled = 5*500 mi

How did you get the no 5*500 mi for when they meet the third time.