Question

A train X departs from station A at 11.00 a. m. for station B, which is 180 km away. Another train Y departs from station B at 11 : 00 a. m. for station A. Train X travels at an average speed of 70 km/hr and not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 km/hr, but has to stop for 15 minutes at station C, which is 60 kms away from station B enroute to station A. Ignoring the lengths of the trains, what is the distance, to the nearest km, from station A to the point where the train crosses each other ? (2001)

• A. 112
• B. 118
• C. 120
• D. None of these

Answer 1

The correct option is A 112

Approach 1:
Option (a) Time taken by B to cover 60 km $=\frac{60}{50}=\frac{6}{5}$ hours.
Time taken by Y at station C $=\frac{15}{60}=\frac{1}{4}$ hours.
Now distance travelled bt train X in $(\frac{6}{5}+\frac{1}{4})=\frac{29}{20}$ hours $=70\times \frac{29}{20}=101.5$ km.
Distance between X and Y when Y starts from station C =180 -(101.5 +60)=18.5 km.
Relative speed =(70+50)=120 km/hr.
Hence, time taken by them in crossing one another $=\frac{18.5}{120}=0.15$ hours. Now, distance travelled by X in 0.15 hour $=70\times 0.15=10.5$ km.
Therefore, distance of X from station A, when they meet =(101.5+10.5)=112 km.

Approach 2:
By using a table to check the distance travelled by each of X and Y:

$\begin{array}{cccc}Text& Xdistance& Ydistance& Total\\ 12:00& 70& 50& 120\\ 12:12& 14& 10& 144\\ 12:27& 17.5& 0& 161.5\\ Tally& 101.5& & \end{array}$

The last 18.5 kms will have to be divided in the ratio of 70:50. Hence we see that the distance travelled by X is around 112 kms.