India’s No.1 Educational Platform For UPSC,PSC And All Competitive Exam

Download Our App

A train passes a 50 meter long platform in 14 seconds and a man standing on platform 10 seconds.The speed of the train is?

124 km/hr
236 km/hr
340 km/hr
445 km/hr

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Distance travelled in 14 sec \\ =  50 + l \\ Distance travelled in 10 sec \\ = l \\ So speed of train \\ = \frac{50}{14 - 10} m/sec \\ = \frac{50}{4} \times \frac{18}{5} km/hr \\ =   45 km/hr \end{aligned}

$\eqalign{ & {\text{Distance travelled in 14 sec}} \cr & {\text{ = 50 + }}l \cr & {\text{Distance travelled in 10 sec}} \cr & {\text{ = }}l \cr & {\text{So speed of train}} \cr & {\text{ = }}\frac{{50}}{{14 - 10}}{\text{m/sec}} \cr & {\text{ = }}\frac{{50}}{4} \times \frac{{18}}{5}{\text{km/hr}} \cr & {\text{ = 45 km/hr}} \cr}$

Workspace Report

Post your Comments

", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Distance travelled in 14 sec \\ =  50 + l \\ Distance travelled in 10 sec \\ = l \\ So speed of train \\ = \frac{50}{14 - 10} m/sec \\ = \frac{50}{4} \times \frac{18}{5} km/hr \\ =   45 km/hr \end{aligned}

$\eqalign{ & {\text{Distance travelled in 14 sec}} \cr & {\text{ = 50 + }}l \cr & {\text{Distance travelled in 10 sec}} \cr & {\text{ = }}l \cr & {\text{So speed of train}} \cr & {\text{ = }}\frac{{50}}{{14 - 10}}{\text{m/sec}} \cr & {\text{ = }}\frac{{50}}{4} \times \frac{{18}}{5}{\text{km/hr}} \cr & {\text{ = 45 km/hr}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

Test

Classes

E-Book