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.

Given:

Speed of cabs (V_c) = 74 km/hr

Time interval between cabs starting (T_c) = 28 minutes

Time interval at which the man meets the cabs (T_m) = 10 minutes

Formula Used:

Distance = Speed × Time

Relative speed when moving in opposite directions = Sum of individual speeds

Calculation:

T_c = 28 minutes = \(\frac{28}{60}\) hours = \(\frac{7}{15}\) hours

T_m = 10 minutes = \(\frac{10}{60}\) hours = \(\frac{1}{6}\) hours

The distance between two consecutive cabs is the distance covered by the first cab in the interval before the second cab starts.

Distance between cabs = V_c × T_c

Distance between cabs = \(74 × \frac{28}{60}\) km

Let the speed of the man be V_m km/hr.

Since the man is moving in the opposite direction to the cabs, their relative speed is the sum of their individual speeds (V_c + V_m).

The man meets the two cabs at an interval of T_m (10 minutes). This means the distance between the cabs is covered by their relative speed in 10 minutes.

Distance between cabs = (V_c + V_m) × T_m

Distance between cabs = \((74 + V_m) × \frac{10}{60}\) km

Equate the expressions for the distance between cabs and solve for V_m.

\(74 × \frac{28}{60} = (74 + V_m) × \frac{10}{60}\)

74 × 28 = (74 + V_m) × 10

2072 = 740 + 10V_m

2072 - 740 = 10V_m

1332 = 10V_m

\(V_m = \frac{1332}{10}\)

V_m = 133.2 km/hr

∴ The speed of the man is 133.2 km/hr.