GATE2018 ME-1: GA-8

Question

From the time the front of a train enters a platform, it takes $25$ seconds for the back of the train to leave the platform, while traveling at a constant speed of $54$ km/h. At the same speed, it takes $14$ seconds to pass a man running at $9$ km/h in the same direction as the train. What is the length of the train and that of the platform in meters, respectively?

• A. $210$ and $140$
• B. $162.5$ and $187.5$
• C. $245$ and $130$
• D. $175$ and $200$

Comments

Option elimination method

distance travelled by train in 25 seconds =54*5*25/18m=375m

so length of train+length of platform=375

so option A and option B are eliminated as their sum is not 375

now in 14 seconds distance travelled by train  54*5*14/18m=210m

when the man would be stationary then the length of the train would be 210m but the man is running in the direction of the train so the train travelled more distance to cross a man so its length will be less than 210m  so option C eliminated.

option D is the answer

Answer 1

Speed of train$=54$ km/h
Speed of man $=9$ km/h
Relative speed of train $=54-9=45$ km/h

Time taken by the train to pass the men $=14$ sec

Distance Traveled = Length of train $=$ Relative Speed $\times$ Time
Length of train $=45\times 14 \times \frac{1}{3600} km =175\;m$

Time for traveling the length of the train and length of platform $=25$ sec
Distance traveled $=$ Length of Train $+$ Length of Platform $=$ Speed of train $\times$ Time
$\qquad\qquad=54\times  25 \times \frac{1}{3600} km=375\;m$
Therefore, length of platform  $=375-175=200\;m$

So, (D) is the correct answer.

Answer 2

Train and Running Man

$S_{relative}=\dfrac{D_{train}}{Time}$

$\dfrac{45\times 1000m}{3600sec}=\dfrac{D_{train}}{14sec}$

$D_{train}=175m$

 

Train and Platform

$S=\dfrac{D_{train}+D_{platform}}{Time}$

$\dfrac{54\times 1000m}{3600sec}=\dfrac{175+D_{paltform}}{25sec}$

$D_{paltform}=200m$

 

$Ans:D$