Question

A train travels a distance of 750 km at a constant speed .If the speed increased by 25 km/hr .The time taken for journey reduces by 1 hr .What is the initial speed

Answer 1

Answer:

The initial speed of the train = 125 km per hour

Explanation:

Let S = original speed of the train

T = time taken for the journey

D = total distance = 750 km

Now, S = D/T => D = ST

=> ST = 750 ........Eqn 1

If the speed is increased by 25 km per hour, the time reduces by 1 hour.

=> (S + 25)(T - 1) = 750 ......Eqn 2

(Note that for any pair of "speed" and "time" for the same "distance", we have distance = speed*time)

From Eqn 1 an Eqn 2, we get:

ST = (S+25)(T - 1)

ST = ST - S + 25T - 25

0 = -S + 25T - 25

Now, ST = 750 => T = (750/S)

Substituting for T, we get:

0 = -S + 25(750/S) - 25

0 = -S + 18750/S - 25

18750/S - S - 25 = 0

18750 - S² - 25S = 0

S² + 25S - 18750 = 0

On factorizing:

(S - 125)(S + 150) = 0

=> S = 125 or S = -150

As "S" is speed, negative values are not admissible.

=> S = 125

The original speed of the train = 125 km per hour

Verify:

Original speed = 125, Time = 750/125 = 6

Revised speed = (125+25), Time = 750/150 = 5 = (6 - 1)

When speed increases by 25 km per hour, time reduced by 1 hour (Verified)