Two trains, X and Y, started simultaneously from opposite ends of a 10

relative speed of both
100/5+100/3 ; 800/15
so time taken ; 100/800 * 15 ; 15/8 hrs
so for train A distance ; 20*15/8 ; 75/2 ; 37.5 miles
IMO A

Time taken to meet = Total distance travelled/Relative speed

Speed of train X = 100/5 = 20 miles/hr
Speed of train Y = 100/3 = 33.3 miles/hr
Relative Speed = 20 + 33= 53 miles/hr
Distance between them = 100 miles
Time taken to meet = 100/(53) hr = 1.88 HR

bACK TO TARGET QUESTION'
How many miles had Train X traveled when it met Train Y?

dISTANCE TRAVELLED BY TRAIN X= 20 TIME =1.88
ANS= 20*1.88 =37.7 ANS A

The concept used in these questions is Relative Speed.

If two people walk in opposite directions (either towards each other or away from each other), their speed relative to each other is the sum of their speeds. e.g. If you are walking away from me at a speed of 2 miles/hr and I am walking away from you at a speed of 1 mile/hr, together we are creating a distance of 3 miles in 1 hr between us so our relative speed is 2 + 1 = 3 miles/hr
On the other hand, when two people walk in the same direction, their relative speed is the difference between their speeds.
e.g. if you are walking away from me at 1 mile/hr and I am walking towards you at 2 miles/hr, my speed relative to you is 2-1 = 1 mile/hr.

I used estimation to narrow down the answer.
Speed of X = 100/5 = 20
Speed of Y= 100/3 = 33.33
Relative Speed = 53.33

Relative time taken = 100/53.33 or we can estimate 100/50= 2 hours (approx.) (Actually, it will be less than 2 hours)

In 2 hours X will cover = 20*2=40 miles(approx). Since the time it will take will be slightly less than 2, the distance covered will be slightly less than 40 miles which ends up pointing towards one answer only - A.

both train X and train Y leave at the same instant towards each other with a 100 mile Gap Distance between them.

When X and Y meet, the TIME each traveled will be the SAME.

when the Time Traveled is Constant: Speed is Directly Proportional to Distance Traveled

this means that the RATIO of the Speeds traveled at will be Equivalent to the RATIO of the Distance each travels when they meet each other.


Speed X : Speed Y = (100/5) : (100/3)

..................................... *15............*15

Speed X : Speed Y = 300 : 500 = 3x : 5x

Distance X : Distance Y = 3x : 5x

when X and Y Meet: X will have covered (3x) / (8x) = (3/8) of the entire Gap Distance Between them

(3/8) * 100 miles = 37.5 miles

answer (A)

Given
• Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and travelled toward each other on parallel tracks
• Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours
• Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours.
To find
• The number of miles Train X had travelled when it met Train Y

Approach and Working out
Suppose the trains meet after T hours. So, both trains will have travelled for T hours till they meet.
• Train X:

o Speed = 100/5 = 20 mph.
o Time = T
o So, distance travelled by train X in T hours = 20T miles

• Train Y:

o Speed = (100/3) mph.
o Time = T
o So, distance travelled by train Y in T hours = 100T/3 miles.

• Now, distance travelled by train X + distance travelled by train Y = initial distance between them = 100 miles.

o So, 20T + (100T/3) = 100
o T = 15/8 hours

• Finally, distance travelled by Train X in 15/8 hours = 20 * (15/8) = 37.5 miles

Correct Answer: Option A

Video solution from Quant Reasoning starts at 0:26
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Hi Brent BrentGMATPrepNow, if using the shrinking concept, once we found out the rate and add the two rates together (20 + 33 = 55 mph). So time = 100/55 = 2 something. So the answer is not right. Not sure what did I miss here? Thanks Brent

First off, 100/55 is less than 2, which means it can't equal 2.something
In other words 100/55 = 1.something

So the distance train X travels = (speed)(time) = (20 mph)(1.something) = some number less than 40
Answer: A

Thanks BrentGMATPrepNow for the clarificatin. So the concept is right but my silly calculation here then . Noted.

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