 ## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 27 of Exercise 10E

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27
##### The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by . Find the fraction.

Let the denominator of required fraction be x

Numerator of required fraction be = x – 3

Original number = If 1 is added to the denominator, then the new fraction will become According to the given condition,     x2 + x = 15x – 45

x2 – 14x + 45 = 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 = – 14 c = 45

= 1.45 = 45

And either of their sum or difference = b

= – 14

Thus the two terms are – 9 and – 5

Sum = – 9 – 5 = – 14

Product = – 9. – 5 = – 45

x2 – 14x + 45 = 0

x2 – 9x – 5x + 45 = 0

x(x – 9) – 5(x – 9) = 0

(x – 9)(x – 5) = 0

x = 9 or x = 5

Case I: x = 5 Case II: x = 9 (Rejected because this does not satisfy the condition given)

Hence the required fraction is 27

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