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Question

(A) Let the distance between station $A$ and station $B$ be $x\, km.$
The time taken is $45\, minutes = \frac{45}{60} = \frac{3}{4}\, hours.$
The or

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) Let the distance between station $A$ and station $B$ be $x\, km.$
The time taken is $45\, minutes = \frac{45}{60} = \frac{3}{4}\, hours.$
The original speed of the train is $v = \frac{	ext{Distance}}{	ext{Time}} = \frac{x}{3/4} = \frac{4x}{3}\, km/h.$
According to the problem,if the speed is reduced by $5\, km/h,$ the new speed becomes $(\frac{4x}{3} - 5)\, km/h.$
The new time taken is $48\, minutes = \frac{48}{60} = \frac{4}{5}\, hours.$
Using the formula $	ext{Distance} = 	ext{Speed} 	imes 	ext{Time}$:
$x = (\frac{4x}{3} - 5) 	imes \frac{4}{5}$
Multiplying both sides by $5$:
$5x = (\frac{4x}{3} - 5) 	imes 4$
$5x = \frac{16x}{3} - 20$
Multiplying by $3$ to clear the denominator:
$15x = 16x - 60$
$60 = 16x - 15x$
$x = 60\, km.$
Thus,the distance between the stations is $60\, km.$

$A$ train covers a distance between station $A$ and station $B$ in $45\, minutes.$ If the speed of the train is reduced by $5\, km/h,$ then the same distance is covered in $48\, minutes.$ The distance between stations $A$ and $B$ is (in $km$):

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