Question

A train 110 metres in length passes a man walking at the speed of $6 \mathrm{~km} / \mathrm{hr}$, against it in 6 seconds. The speed of the train in $\mathrm{km}$ per hour is (a) $60 \mathrm{~km} / \mathrm{hr}$ (b) $45 \mathrm{~km} / \mathrm{hr}$ (c) $50 \mathrm{~km} / \mathrm{hr}$ (d) $55 \mathrm{~km} / \mathrm{hr}$

   A train 110 metres in length passes a man walking at the speed of $6 \mathrm{~km} / \mathrm{hr}$, against it in 6 seconds. The speed of the train in $\mathrm{km}$ per hour is
(a) $60 \mathrm{~km} / \mathrm{hr}$
(b) $45 \mathrm{~km} / \mathrm{hr}$
(c) $50 \mathrm{~km} / \mathrm{hr}$
(d) $55 \mathrm{~km} / \mathrm{hr}$

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The Pearson Guide to Objective Arithmetic for Competitive Examinations

The Pearson Guide to Objective Arithmetic for Competitive Examinations

Dinesh Khattar 2nd Edition

Chapter 15, Problem 59

Instant Answer

Step 1

The speed of the man is given as \(6 \mathrm{~km/hr}\). To convert km/hr to m/s, we use the conversion factor \( \frac{5}{18} \). \[ 6 \mathrm{~km/hr} = 6 \times \frac{5}{18} = \frac{30}{18} = \frac{5}{3} \mathrm{~m/s} \approx 1.67 \mathrm{~m/s} \] Show more…

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A train 110 metres in length passes a man walking at the speed of $6 \mathrm{~km} / \mathrm{hr}$, against it in 6 seconds. The speed of the train in $\mathrm{km}$ per hour is (a) $60 \mathrm{~km} / \mathrm{hr}$ (b) $45 \mathrm{~km} / \mathrm{hr}$ (c) $50 \mathrm{~km} / \mathrm{hr}$ (d) $55 \mathrm{~km} / \mathrm{hr}$

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