Question

A person is travelling in a train from station A to station B. He covers the first ${\frac{1}{3}}^{rd}$ at ${\frac{3}{4}}^{th}$ usual speed 3rd ${\frac{1}{3}}^{rd}$ at ${\frac{5}{6}}^{th}$ usual speed. When he reaches station C which is at ${\frac{1}{3}}^{rd}$ of distance from A, he is 40 min late than usual time. When he reaches B he will be late by how much time?

• A. 74 min
• B. 80 min
• C. 94 min
• D. 112 min

Answer 1

The correct option is C 94 min

Time taken to reach $C_1$
$\frac{d}{\frac{3}{4}{V}_0}=\frac{4}{3}\frac{d}{V_0}$
As per question,

$\frac{4}{3}\frac{d}{V_0}-\frac{d}{V_0}=40\Rightarrow \frac{d}{30}=40\Rightarrow \frac{d}{V_0}=120$
Time to reach B $=\frac{d}{\frac{3}{4}{V}_0}+\frac{d}{\frac{4}{5}{V}_0}+\frac{d}{\frac{5}{6}{V}_0}=(\frac{4}{3}+\frac{5}{4}+\frac{6}{5})\frac{d}{V_0}=\frac{227}{60}\times 120=454min$
Time to reach station B $=\frac{3d}{V_0}=3\times 120=360m$
(without being late)
$\therefore$ He is late by 454 - 360 = 94~mins