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Question

(C) Let the distance between two stations $A$ and $B$ be $x \; km$ and the speed of the train be $y \; km/h$.Since $	ext{Time} = \frac{	ext{Distance

Vedclass · Accepted Answer

$60 \; km, 80 \; km/h$

Vedclass · Answer

(C) Let the distance between two stations $A$ and $B$ be $x \; km$ and the speed of the train be $y \; km/h$.Since $	ext{Time} = \frac{	ext{Distance}}{	ext{Speed}}$,we have:$\frac{x}{y} = \frac{45}{60} = \frac{3}{4} \implies x = \frac{3}{4}y \quad (1)$When the speed is reduced by $5 \; km/h$,the new speed is $(y - 5) \; km/h$ and the time taken is $48$ minutes:$\frac{x}{y - 5} = \frac{48}{60} = \frac{4}{5} \implies x = \frac{4}{5}(y - 5) \quad (2)$Equating $(1)$ and $(2)$:$\frac{3}{4}y = \frac{4}{5}(y - 5)$$15y = 16(y - 5)$$15y = 16y - 80$$y = 80 \; km/h$Substituting $y = 80$ in $(1)$:$x = \frac{3}{4} 	imes 80 = 60 \; km$Thus,the distance is $60 \; km$ and the speed is $80 \; km/h$.

$A$ train covers a distance between two stations $A$ and $B$ in $45$ minutes. If the speed is reduced by $5 \; km/h$,it will cover the same distance in $48$ minutes. What is the distance between the two stations $A$ and $B$ (in $km$)? Also,find the speed of the train.

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