A train X starts from Meerut at 4 p.m. and reaches Ghaziabad at 5 p.m.while another train Y starts from Ghaziabad at 4 p.m. and reaches Meerut at 5:30 p.m. The two trains will cross each other at:

A. 4:36 p.m.
B. 4:42 p.m.
C. 4:48 p.m.
D. 4:50 p.m.
E. 4:52 p.m.

Sol:

Let the distance between two points be "D miles"

Speed of X = D/1= D miles/h
Speed of Y = D/1.5= 2D/3 miles/h

Suppose X meets Y after completing "x miles" traveling for \(t_x\) hours and y traveled \(t_y\) hours

\(t_x=t_y\)

\(time=\frac{Distance}{Speed}\)

\(\frac{x}{D}=\frac{3(D-x)}{2D}\)

\(2x=3(D-x)\)

\(5x=3D\)

\(x=\frac{3}{5}D\)

Now,
x travels D miles in 1 hour
To travel (3/5)D miles, it would need 3/5 hours = 36minutes.

Thus; the trains meet at 4:36PM.
_________________

Going to explain how I calculated it.

The first train travels the entire distance in 1 hour. The other train travels only 2/3 of the distance in one hour. When the trains meet they will together have covered the entire distance, from each others ends. The speeds of the trains are 1 times the distance per hours and 2/3 of the distance per hour. After traveling at their respective speeds for time X both trains will together have covered the entire distance.

Therefore:

1X + (2/3)X = 1.

(5/3)X = 1

X = 1/(5/3)

X = 3/5

Both trains will meet after 3/5 hours, as they will have covered the entire distance together at this point in time. 1/5 of an hour is 12 minutes. 12 x 3 is 36. The trains will meet at 4:36.
_________________