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A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.
Let the usual speed of the train be x km/h
Time taken to cover 300 km =
New speed of train = x + 5 km/h
Time taken to cover 90 km =
According to the question
750 = x2 + 5x
x2 + 5x – 750 = 0
Using the splitting middle term – the middle term of the general equation is divided in two such values that:
Product = a.c
For the given equation a = 1 b = 5 c = – 750
= 1. – 750 = – 750
And either of their sum or difference = b
= 5
Thus the two terms are – 25 and 30
Difference = 30 – 25 = 5
Product = 30. – 25 = – 750
x2 + 30x – 25x – 750 = 0
x(x + 30) – 25(x + 30) = 0
(x + 30) (x – 25) = 0
x = – 30 or x = 25 (but x cannot be negative)
x = 25
Hence the usual speed of the train is 25 km/h