 ## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 49 of Exercise 10E

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49
##### 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

49

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