This question has been taken from Book:- ML aggarwal, Avichal publication, class10th, quadratic equation in one variable, chapter 5, exercise 5.5
This is an important ques.
In a certain positive fraction, the denominator is greater than the numerator by 3.
If 1 is subtracted from both the numerator and denominator,
the fraction is decreased by 1/14. Find the fraction.
Question no.9 , ML Aggarwal, chapter 5, exercise 5.5, quadratic equation in one variable, ICSE board,
Solution:
Let the numerator be ‘x’
Denominator be ‘x+3’
So the fraction is x/(x+3)
According to the question,
By cross multiplying, we get
(x – 1) (14x + 42) = (x + 2) (13x – 3)
14x2 + 42x – 14x – 42 = 13x2 – 3x + 26x – 6
14x2 + 42x – 14x – 42 – 13x2 + 3x – 26x + 6 = 0
x2 + 5x – 36 = 0
let us factorize,
x2 + 9x – 4x – 36 = 0
x(x + 9) – 4 (x + 9) = 0
(x + 9) (x – 4) = 0
So,
(x + 9) = 0 or (x – 4) = 0
x = -9 or x = 4
So the value of x = 4 [since, -9 is a negative number]
When substitute the value of x = 4 in the fraction x/(x+3), we get
4/(4+3) = 4/7
∴ The required fraction is = 4/7