A train X departs from station A at 11.00 a. m. for station B, which is 180 km a

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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)

Options:

A . 112

B . 118

C . 120

D . None of these

Answer: Option A
:
A
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{matrix} T e x t & X d i s t a n c e & Y d i s t a n c e & T o t a l 12 : 00 & 70 & 50 & 120 12 : 12 & 14 & 10 & 144 12 : 27 & 17.5 & 0 & 161.5 T a l l y & 101.5 \end{matrix}

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.

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