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Question

A train take a 2 hours less for a journey of 300 km . If the speed of the is increased by 5 km / hr from its usual speed , find the usual speed of the train.

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Solution

Let x = normal speed


Increased speed =x+5

Distance to travel=300

usual time =distance/usual speed=300/x---(1)

New time =distance/increased speed
=300/x+5 ----(2)

Question says that new time is 2 hour less than usual time.
Ie, new time =old time-2
new time +2=old time---(3)

From equation (1) and (3) ,we will get the formula for new time and old time.
[new time=300/x+5 and old time =300/x]

Substitute these formula in equation (3).

(300/(x+5))+2=300/x

. . . combine left side into a single fraction
(300+2x+2×5)/(x+5) =300/x

(310 + 2x) / (x + 5) = 300 / x

. . . cross multiply

310x + 2x^2 = 300x + 1500

2x^2 + 10x - 1500 = 0

x^2 + 5x - 750 = 0

Solve this quadratic equation.

(x - 25) (x + 30) = 0

x = -30 <== invalid negative time

x = 25 km / h <===== answer

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