question_answer

Direction: Study the following information carefully and answer the questions given below:

Train P and Train Q are travelling towards each other from stations X and Y respectively. Train P left station X at 9 : 35 am at speed of 70 km/hr. After half an hour train Q starts at the speed of 80 km/hr from station Y for X. Station X and Y are situated at a distance of m km and both the trains meet each other at 2 : 45 pm the same day.

Calculate the difference between the original time taken, as given in the above information, for train P and Q to meet since the time when P started and the time taken by Train Q to catch Train P if Train P had started in the same direction as that of Q, Train Q having started 1 hr after train P while running in the same direction.

A)  \[\frac{129}{3}hr\]                   

B)  \[\frac{125}{5}hr\]       

C)    \[\frac{226}{3}hr\]       

D)    \[\frac{205}{3}hr\]

E)  None of these

Correct Answer: C

Solution :

In \[\frac{1}{2}hr\]tram P will cover =35 km The time at which both trains meet is 2 : 45 PM. \[\therefore \] Time taken = 2 : 45 PM \[-\] (9 : 35 + 30) AM \[=4:40\,m=4\frac{2}{3}=\frac{14}{3}hours\] Distance between X and Y is m km. Then \[m=35+\frac{14}{3}\times 150=35+700=735\,km\] Total time taken in given condition \[=\left( \frac{1}{2}+\frac{14}{3} \right)hours=\frac{3+28}{6}=\frac{31}{6}hr\] New time taken according to the given condition. \[=\frac{735+1\times 70}{(80-70)}=\frac{735+70}{10}=\frac{805}{10}h\] \[\therefore \] Reqd difference\[=\frac{805}{10}-\frac{31}{6}\] \[=\frac{2415-155}{30}=\frac{2260}{30}=\frac{226}{3}h\]