CTET Class I to V 1 Jan 2021 Paper

Question : 73 of 149

Marks: +1, -0

A boy boarded a train on 4th June, 2020 at Nagercoil for Madgaon. The train departed at

22: 30 {\h}

from Nagercoil and reached Madgaon at 4:30

{\h}

on 6th June, 2020. If the distance covered by the train during this time interval is

1145 {\km}

, then average speed of the train was nearly

Solution:

Given, distance between Nagercoil to Madgaon

=1145 {\km}

and time

=1: 30+24: 00+4: 30=30 {\h}

Hence, average speed of the train

={{ \distance }}/{{ \time }}={1145}/{30}

=38.16{\km} \/ {\h} ≈ 38.5 {\km} \/ {\h}

Go to Question:

Prev Question

Next Question