## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 48 of Exercise 10E

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48
##### A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Let the original speed of the train be x km/h

Time taken to cover 90 km =

New speed of train = x + 15 km/h

Time taken to cover 90 km =

According to the question

2700 = x2 + 15x

x2 + 15x – 2700 = 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 = 15 c = – 2700

= 1. – 2700 = – 2700

And either of their sum or difference = b

= 15

Thus the two terms are – 45 and 60

Difference = 60 – 45 = 15

Product = 60. – 45 = – 2700

x2 + 60x – 45x – 2700 = 0

x(x + 60) – 45(x + 60) = 0

(x + 60) (x – 45) = 0

x = – 60 or x = 45 (but x cannot be negative)

x = 45

Hence the original speed of the train is 45 km/h

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