Question

After covering a distance of $30km.$ With a uniform speed there is some defect in a train enzine and therefore its speed is reduce to $4/5$ of its original speed consiquetely the train reaches its destination late by $45$ minutes had it happened after covering $18km.$ More the train would have reached $9$ minutes earlier find the speed of the train and distance of journey.

Answer 1

Let,

the speed of the train$=xkm/hr$

the total distance covered by the train $=ykm$

Original time taken to reach the destination =$\frac{y}{x}$ hrs.

Reduced speed of train =$\frac{4x}{5}$ km/hr

According to the first condition, we get

Time taken at original speed + time taken at reduced speed = original time taken $+45$ minutes

$\frac{30}{x}+\frac{y-30}{\frac{4x}{5}}=\frac{y}{x}+\frac{45}{60}$

After solving .we get

$Y-3x$ = $30$ ……..(1)

According to the second condition, we get

$\frac{48}{x}+\frac{y-48}{\frac{4x}{5}}=\frac{y}{x}+36$

After solving $5y-12=240$ ………(2)

Solve equation (1) and (2) ,we get

$x=30$ ,$y$ = $120$

Therefore original speed of train = 30 km/hr and distance of journey = 120 km