Question

A two-digit number is such that the product of its digit is 20. If 9 is added to the number, the digits interchange their places. Find the number.

Answer 1

Hint- Here we will proceed by assuming x and y be the number at unit place and digit place. Then we will use algebraic identity formula i.e. ${\left( {x + y} \right)^2} = {\left( {x - y} \right)^2} + 4xy$ to solve the equation and get the required answer.

Complete step-by-step solution -
Let the $x$ be the number at unit place.
And let $y$ be the number at unit digit place.
So number is- $10x + y $
We are given that the product of digits is 20.
$ \Rightarrow xy = 20..........\left( 1 \right)$
Now, original number + 9 = reverse number. (given)
$\Rightarrow x + 10y + 9 = 10x + y $
$\Rightarrow 9x – 9y = 9 $
$\Rightarrow x – y = 1 $……………. (2)
using algebraic identity formula ${\left( {x + y} \right)^2} = {\left( {x - y} \right)^2} + 4xy$
we get-
${\left( {x + y} \right)^2} = 1 + 4\left( {20} \right)$
$\Rightarrow x + y = \sqrt {81} $
$\Rightarrow x + y = 9 $……………. (3)
Solving equation 2 and equation 3,
$\left( {x - y = 1} \right) - \left( {x + y = 9} \right)$
We get-
$\Rightarrow 2x = 10$
$\Rightarrow x = 5 $
Now putting value of x in equation 2,
We get-
$\Rightarrow 5 – y = 1$
$\Rightarrow y = 4 $
The required number is $ x + 10y $
= 5 + 10(4)
= 45
Therefore, the required number is 45.

Note- While solving this question, we can assume any variables instead of x and y to form equations. We must know all the algebraic identities to solve this question and similar problems as here we used one of them i.e. ${\left( {x + y} \right)^2} = {\left( {x - y} \right)^2} + 4xy$.