Question

A train, after travelling $70$ km from a station $A$ towards a station $B$, develops a fault in the engine at $C$ and covers the remaining journey to $B$ at $\frac{3}{4}$ of its earlier speed and arrives at $B$ $1$ hour and $20$ minutes late. If the fault had developed $35$ km further on at $D$, it would have arrived $20$ minutes sooner. Find the speed (in km / hr) of the train.

• A. $30$
• B. $45$
• C. $60$
• D. $15$

Answer 1

Answer: (B)

$45$

Let, the train's speed $=$ y
$\therefore$ Total time $=\frac{10+x}{y}$
Case I :
Total time $=\frac{70}{y}+\frac{x}{\frac{3}{4}y}=\frac{70}{y}+\frac{4x}{3y}=\frac{210+4x}{3y}$

$\therefore \frac{210+4x}{3y}=\frac{70+x}{y}+1+\frac{1}{3}$
$\Rightarrow \frac{210+4x}{3y}=\frac{210+3x+3y+y}{3y}$
$\Rightarrow 210+4x=210+3x+4y$
$\Rightarrow x-4y=0$ ......... (i)
Case II :
Total time $=\frac{70+35}{y}+\frac{x-35}{\frac{3}{4}y}$
$=\frac{70+35}{y}+\frac{4x-140}{3y}=\frac{105\times 3+4x-140}{3y}$
$=\frac{315-140+4x}{3y}=\frac{175+4x}{3y}$
$\therefore \frac{175+4x}{3y}=\frac{70+x}{y}+1$
$\Rightarrow \frac{175+4x}{3y}=\frac{70+x+y}{y}$
$\Rightarrow 175+4x=210+3x+3y$
$\Rightarrow 4x-3x=210-175$
$\Rightarrow x-3y=45$ ............. (ii)
We have,
(i) $\Rightarrow x-4y=0$
(i) $\Rightarrow x-3y=45$
$x-4y=0$
$x-3y=0$
____________
$-y=-45$ $\Rightarrow y=45$ km/h
$\therefore$ Speed of train $=$ 45 km/h