Question

a train after travelling 70 km meets with an accident and then proceeds at ¾ of its former speed and arrives at its destination 35 min late. had the accidents occurred 15 km further on it would have reached the destination only 23 min late. the speed of the train is

Answer 1

Given:

• A train after traveling for 70km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late.

To Find:

• Speed of the train?

Solution:

let the two end points between which this train is running be P and Q.

Also suppose that accident occurs at R which is at a distance of & $70km$ from P.

Thus we have

        P---------70 km----------R----------------------------Q

Time taken to travel the distance after the accident  35 minutes more than the usual time.

Means $35 minutes$ more than the usual time for distance RQ.

now, if the accident were to occur at 15 km farther, then train would have reached only 23 minutes late. i.e.

       P--------70 km----------R---------15--------S---------------Q

If we look at both the situation  we will see that (35-23 = 12) minute difference in time is occurring because of this 15 Km  which is traveled by its usual speed (=x) in the later case.

Thus we have

               $\frac{15}{\frac{3}{4} x } - \frac{15}{x} = \frac{12}{60}$

               $\frac{15 \times 4}{3x } - \frac{15}{x} = \frac{12}{60}$

                $\frac{60-45}{3x } = \frac{1}{5}$

                $3x= 75$

               $x = 75km/h$

Thus, the speed of train is $75km/h$

Answer 2

Answer:

25 km

Explanation:

3x =75

x= 75/3

x= 25km/hr