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 and asked in 2014
A two digit positive number is such that the product of its digits is 6.
If 9 is added to the number, the digits interchange their places.
Find the number.
Question no.12 , ML Aggarwal, chapter 5, exercise 5.5, quadratic equation in one variable, ICSE board,
Solution:
Let us consider 2-digit number be ‘xy’ = 10x + y
Reversed digits = yx = 10y + x
So according to the question,
10x + y + 9 = 10y + x
It is given that,
xy = 6
y = 6/x
so, by substituting the value in above equation, we get
10x + 6/x + 9 = 10(6/x) + x
By taking LCM,
10x2 + 6 + 9x = 60 + x2
10x2 + 6 + 9x – 60 – x2 = 0
9x2 + 9x – 54 = 0
Divide by 9, we get
x2 + x – 6 = 0
let us factorize,
x2 + 3x – 2x – 6 = 0
x(x + 3) – 2 (x + 3) = 0
(x + 3) (x – 2) = 0
So,
(x + 3) = 0 or (x – 2) = 0
x = -3 or x = 2
Value of x = 2 [since, -3 is a negative value]
Now, substitute the value of x in y = 6/x, we get
y = 6/2 = 3
∴ 2-digit number = 10x + y = 10(2) + 3 = 23