Question

A car after travelling a distance of 110 km develops a problem in the engine and proceeds at $\frac{3}{4}th$ of its former speed and arrives at the destination 60 min late. Had the problem developed 30 km further on, the car would have arrived 12 min sooner. Find the original speed of the car.

• A. 45 km/h
• B. 60 km/h
• C. 50 km/h
• D. None of these

Answer 1

The correct option is C 50 km/h

Consider that 30 km which have different speeds in both cases.
Decrease in speed = $\frac{1}{4}\times Originalspeed$.
So, increase in time to cover this distance = $\frac{1}{3}\times Originaltime$
But this is equal to 12 min.
Hence, time taken to cover this 30 km with usual (original) speed = $3\times 12=36min.$
$\therefore$ Original speed = $\frac{30km}{36min}=50kmph$

Alternate Approach:
Let the speed of car be x kmph.
Let the distance between the point of engine failure and destination be y km
If travelled through original speed, time = $\frac{y}{x}hours$
But since speed is 3/4th of original. So, time = $\frac{4y}{3x}hours.$
According to given conditions: $\frac{4y}{3x}-\frac{y}{x}=1hour......\left(i\right)$
Also, $\frac{4(y-30)}{3x}-\frac{(y-30)}{x}=\frac{48}{60}hours......\left(ii\right)$
Solving both the equations, we get x = 50 kmph.