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.

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RRB NTPC Graduate Level CBT-I Official Paper (Held On: 05 Jun, 2025 Shift 2)
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  1. 125.5
  2. 143.1
  3. 133.2
  4. 128.8

Answer (Detailed Solution Below)

Option 3 : 133.2
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Detailed Solution

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Given:

Speed of cabs (Vc) = 74 km/hr

Time interval between cabs starting (Tc) = 28 minutes

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

Formula Used:

Distance = Speed × Time

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

Calculation:

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

Tm = 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 = Vc × Tc

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

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

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

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

Distance between cabs = (Vc + Vm) × Tm

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

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

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

74 × 28 = (74 + Vm) × 10

2072 = 740 + 10Vm

2072 - 740 = 10Vm

1332 = 10Vm

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

Vm = 133.2 km/hr

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

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