From a taxi stand, two cabs start at a speed of 54 km/hr at an interval of 29 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 12 minutes. Find the speed (in km/hr) of the man.​

Correct option is C

Given:

Speed of each cab $$v_c =$$​ 54 km/h.

Start interval between cabs = 29 min.

The man, moving in the opposite direction, meets the two cabs at an interval of 12 min.

Formula Used:

Separation between cabs (same speed) d = $$v_c \times \text{start interval}$$​​

Relative speed (cab vs man, opposite directions) = $$v_c + v_m.$$​​

Time between successive meetings $$=\dfrac{d}{v_c+v_m}$$​​

Solution:
Convert minutes to hours where needed.
d = $$54 \times \frac{29}{60} = 26.1\ \text{km}$$​​
Given the man meets them 12 min apart = $$\frac{12}{60}=$$​0.2 h,

0.2 $$= \frac{26.1}{54 + v_m}$$​

5$$4 + v_m = \frac{26.1}{0.2} = 130.5$$​​

​$$v_m = 130$$​.5 - 54 = 76.5 km/h
​