Question

The sum of the digits of a two-digit number is 12. If  the new number formed by reversing the digits is greater than the original number by 54, Find the original number. Check your solution.

Answer 1

Hint: In order to solve this problem assume the variables for the two digit number. Let's suppose a,b are variables. Then $ab$ is the two digit number with the digits $a$ in tens place and $b$ in units place. The value of $ab$ will be equal to $(a \times 10)+ (b \times 1)$. Eg: $62$, here $a=6$ and $b=2$, this value is equal to $(6 \times 10) + (2 \times 1)$. 

If we interchange the digits, $ba$ this value changes according to the places.

Using this logic we will make a pair of linear equations according to the given conditions, then solve a pair of linear equations using the elimination method to get the desired result.

Complete step-by-step answer:

Let the numbers of the two digit numbers be$x$ and $y.$

Then the two digit number is $10x+y.$

It is given that $x + y = 12$ ……….(1)

It is also given that the new number formed by reversing the digits is greater than the original number by $54.$

So,

$10y + x –(10x + y) = 54$

$\Rightarrow 9y – 9x = 54$

$\Rightarrow y – x = 6$………..(2)

On adding equation (1) and (2) we get,

$\Rightarrow x + y + y – x = 18$

$\Rightarrow$ 2y = 18

$\Rightarrow y = 9$

putting the value of $y$ in equation (1) we get,

$\Rightarrow x + 9 = 12$

$\Rightarrow x = 3$

We have assumed the number is $10x +y.$

On putting the value of $x$ and $y$ we get $10(3) + 9 =39.$

Hence the number is $39.$

We can check it by doing the sum, we get $3 + 9 = 12.$

And also by reversing the number we get $93$. So, the difference between $93$ and $39$  is $54.$

Therefore, the number 39 is the right number.

Note: While solving the problem, there should be no confusion in assuming the value of a two digit number in variables. There might be a chance of thinking the two digit number (xy) value is equal to the $x \times y$ which is incorrect. It always has to be taken according to the place values.