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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
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
While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes. To reach the destination, 1500 km away, in time, the pilot increased the speed by 100 km/hour. Find the original speed of the plane. Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?