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,

10x

^{2}+ 6 + 9x = 60 + x^{2}10x

^{2}+ 6 + 9x – 60 – x^{2}= 09x

^{2}+ 9x – 54 = 0Divide by 9, we get

x

^{2}+ x – 6 = 0let us factorize,

x

^{2}+ 3x – 2x – 6 = 0x(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