From a taxi stand, two cabs start at a speed of 74 km/hr at an interval of 28 minutes, both cabs travelling in the same direction. A man coming in the opposite direction towards the taxi stand meets the cabs at an interval of 10 minutes. Find the speed (in km/hr) of the man.​

Correct option is B

Given:

Two cabs from a taxi stand move in the same direction at speed u = 74 km/h.

Time gap between their starts = 28 minutes =$$\dfrac{28}{60}=\dfrac{7}{15}$$​ h.

A man moves towards the stand (opposite to cabs) and meets the two cabs with a time gap of (10) minutes =$$\dfrac{1}{6}$$​ h.

Formula Used:

Distance between the two cabs is constant (same speed): D =u$$\times$$​ start-gap

Time to meet second cab after meeting first =$$\dfrac{\text{initial separation}}{\text{relative speed}}$$​​

Relative speed (opposite directions) = u + v .

Solution:
Let the man’s speed be v km/h .

Distance between cabs:
D = 74$$\times \frac{7}{15}=\frac{518}{15}\ \text{km}.$$​​
When the man meets the first cab, he is D km away from the second cab.
He meets the second cab after $$\frac{1}{6}$$​h, so
​$$\frac{D}{u+v}=\frac{1}{6}\quad\implies\quad u+v=6D$$​​

​$$=6\times\frac{518}{15}=207.2.$$​​
Hence,
v = 207.2 - 74 = 133.2 km/h.