Question

The difference between a 2-digit number and the number obtained by interchanging its digits is 63. What is the difference between the digits of the number?

Answer 1

Hint: In this question firstly we will assume a 2-digit number. After that as given in the question we will interchange its digit which will get another 2 digit number, then take the difference of both the numbers and equate it with 63(given in the question) to get the desired result.

Complete step-by-step answer:
Let a 2-digit number be \[10x + y\]
We are given a difference between a 2-digit number and the number obtained by interchanging its digits is 63.
Therefore, the reversed 2-digit number will be \[10y + x\]
\[ \Rightarrow \left( {10x + y} \right) - \left( {10y + x} \right) = 63\]
On solving the above expression further we get,
\[10x + y - 10y - x = 63\]
\[ \Rightarrow 10x - x + y - 10y = 63\]
\[ \Rightarrow 9x - 9y = 63\]
Taking 9 throughout common we get,
\[9\left( {x - y} \right) = 63\]
\[ \Rightarrow \left( {x - y} \right) = \dfrac{{63}}{9}\]
\[ \Rightarrow \left( {x - y} \right) = 7\]
As x and y two different digits of a number \[10x + y\],\[\left( {x - y} \right) = 7\]is the difference between the digits of the number.
Hence, the difference between two digits is 7.


Note: In this type of question we should remember that when we want to interchange the digits of a number the place value of the digits will be changed.
For example: If the 2-digit number is \[10x + y\]. In this number x is placed at ten’s place and y is placed at one’s place. If we interchange the digits of the number \[10x + y\] i.e. \[10y + x\], x will be placed at one’s place and y will be placed at ten’s place.