Question

the sum of the digit of a two digit number is 12 the number obtained by interchanging the digits exceeds the original number by 54 find the original number

Answer 1

a +b = 12
10a +b - a - 10b = 54
9a-9b= 54
a-b = 54÷9 = 6 , ( a = 6+b)

6+b+b=12. , 2b = 6. , b=.3
a+3= 12
a=.9
the numbers are 39 and 93

Answer 2

Answer:

The orignal number is 39 or 93.

Step-by-step explanation:

Given :

The sum of the digit of a two digit number is 12.

The number obtained by interchanging the digits exceeds the original number by 54.

To Find :

The original number.

Solution :

Consider the :

The digit at ones place as - x

So,

As given,

The sum of digits of number is 12.

Hence,

The digit at tens place = 12 - x

Now,

$\sf\\ \\\implies 10 \times (12 - x) + x\\ \\ \\ \implies 120 - 10x + x\\ \\ \\\implies 120 - 9x$

Now,

On interchanging the digits of given number :

• The digit at ones place become  (12 - x).
• The digits at tens place become x.

So,

New Number :

$\sf\\ \\\implies 10x + (12 - x)\\ \\ \\ \implies \: 9 x+ 12$

As given,

The new number exceeds the original number by 54.

So,

$\bigstar\;\boxed{\sf New\;Number - Original\;Number = 54}}$

So,

$\sf\\ \\\implies (9x + 12) - (120 - 9x) = 54\\ \\ \\ \implies 9x + 12 - 120 + 9x = 54\\ \\ \\ \implies 18x - 108 = 54\\ \\ \\ \implies 18x = 54 + 108\\ \\ \\ \implies 18x = 162\\ \\ \\ \implies x = \dfrac{162}{18}\\ \\ \\ \implies x = 9$

Now we get,

The digit at ones place = 9

The digit at tens place = 3

Therefore,

The orignal number is 39 or 93.