Question

A teacher works at a college out of town. A car is sent for him from the college every day that arrives at the railway station at the same time as the train. One day the teacher arrived at the station one hour before his usual time and without waiting for the car started to walk towards the college. On his way, he met the car and reached the college 10 minutes before his usual time. The time for which the teacher walked will be
(A) 40 minutes
(B) 45 minutes
(C) 50 minutes
(D) 55 minutes

Answer 1

Hint: We are given in the question that a teacher who reached the railway one hour earlier than his usual routine and so he started walking towards the school and on the way he met the car sent for him to get to school. We are asked to find the time for which the teacher walked. We will assume that the usual time that the teacher reaches at the station is \[t\]. So, the time when he reaches an hour early will be, \[t-60\]. He walks for \[T\] minutes. He meets the car at a time, \[t-60+T\]. The car also reaches from the school to the station at a time \[t\]. So, the total time taken will be \[2t\]. But when the teacher started his journey an hour ago for school, the total time taken by the car to reach school was 10 minutes less, \[2t-10\]. So, the time taken by the car to reach the teacher will be half, that is, \[t-5\]. We will equate the two expressions and we will get the required value, that is, ‘T’.

Complete step by step answer:
According to the given question, we are given the schedule of a teacher from his place to school. One day he reaches an hour early and we are asked to find the time for which the teacher travelled before meeting the car.
Let the usual time at which the teacher arrives at the railway station be \[t\] minutes
One day he reached an hour earlier, so the time will be \[t-60\] mins.
It is given in the question that since he reached early at the station, he did not wait there and started walking towards school and on the way he met the car.
So, let the time for which the teacher walked be \[T\] minutes
Therefore, the time taken by the teacher before he meets the car is \[t-60+T\] minutes.
We know that the car travels from the school to the station and takes time \[t\] minutes and the same amount of time is taken to reach back to school. So, the total time taken to reach the station and back to school is \[t+t=2t\] minutes
On the day that the teacher reached the station early, the total time taken to reach school was 10 minutes early, that is, \[2t-10\] minutes
So, the time taken by the car to meet the teacher who already started walking from the station is, \[\dfrac{2t-10}{2}=t-5\] minutes
So, the time when the teacher and the car will meet is,
\[t-60+T=t-5\]
We will now the above expression and we get,
\[\Rightarrow -60+T=-5\]
So, we get the value of ‘T’, which is the time walked by the teacher before he met the car, as
\[\Rightarrow T=60-5\]
\[\Rightarrow T=55\]minutes

So, the correct answer is “Option D”.

Note: The above assumption of the time taken by the car is made by the fact that the car travels with the same speed always. Also, the labelling of the time taken by the teacher and the car should be carefully written in an expression. And the further calculation to find the time for which the teacher walked before meeting the car should be computed step wise.