Question

An engineer works at a factory out of town. A car is sent for him from the factory every day and arrives at the railway station at the same time as the train. One day the engineer arrived at the station one hour before his usual time and without waiting for the car, started walking towards factory. On his way he met the car and reached his factory $10$ minutes before the usual time. For how much time (in minutes) did the engineer walk before he met the car? (The car moves with the same speed everyday.)

• A. $105$ min
• B. $55$ min
• C. $75$ min
• D. $45$ min

Answer 1

Answer: (B)

In the figures $S$ is station, $F$ is Factory and $P$ is the place where he meets the car.

Usual day:



Car starts from $F$ at $t=0$, reaches station at $T$ and again reaches at the factory at time $2T$.

This day:



Person reaches ‘$S$’ at $T-60$. Car starts at $t=0$ from $F$. Person walks for time $t$ and reaches point $P$ at time $T-60+t$. At this time car also reaches ${}^{\prime }{P}^{\prime }$ Car comes back at ${}^{\prime }{F}^{\prime }$ at time $(2T-10)$. That means car takes time $T-5$ from $F$ to $P$. That means car reach at ‘$P$’ at time $T-5$.

Now $T-5=T-60+t$

$\Rightarrow t=55$ min.