Question

After covering a distance of 30 km with a uniform speed there is some defect in a train engine and therefore, its speed is reduced to 4/5 of its original speed. Consequently, the train reaches its destination late by 45 minutes. Had it happened after covering 18 kilometers more,the train would have reached 9 minutes earlier Find the speed of the train and the distance of journey.

• A. Speed = $40$ km/hr, Distance = $160$ km
• B. Speed = $20$ km/hr , Distance= $100$ km
• C. Speed = $30$ km/hr , Distance = $120$ km
• D. Speed = $35$ km/hr, Distance =$140$ km

Answer 1

The correct option is C Speed = $30$ km/hr , Distance = $120$ km

Let the speed of the train be s and total distance travelled be d
Now, $\frac{30}{s}+\frac{d-30}{\frac{4s}{5}}=\frac{d}{s}+\frac{45}{60}$
$\frac{30-d}{s}+\frac{d-30}{\frac{4s}{5}}=\frac{45}{60}$
$\frac{120-4d+5d-150}{4s}=\frac{3}{4}$
$d-30=3s$
$d-3s=30$ (1)
$\frac{48}{s}+\frac{d-48}{\frac{4s}{5}}=\frac{d}{s}+\frac{36}{60}$
$\frac{48-d}{s}+\frac{d-48}{\frac{4s}{5}}=\frac{36}{60}$
$\frac{48\times 4-4d+5d-240}{4s}=\frac{3}{5}$
$5(-48+d)=12s$
$5d-240=12s$ (2)
From (1) and (2)
$5d-240=4d-120$
$d=120$ km
Thus, $3s=120-30=90$
$s=30$ km/hr