Question

3. The sum of a number of two digits and the number obtained on interchanging the digits is 99. If 5 is
added to the number, then it is 4 less than 6 times the sum of digits. find the number ​

Answer 1

Answer :-

The number is 45.

Given :-

• The sum of a number of two digits and the number obtained on interchanging the digits = 99

To FinD :-

• What is the number ?

Solution :-

Let the number is 10x + y

According to the first condition →

$\longrightarrow10x + y + 10y + x = 99 \\ \\ \longrightarrow11x + 11y = 99 \\ \\ \longrightarrow11(x + y) = 99 \\ \\ \longrightarrow \: x + y = \frac{99}{11} \\ \\ \longrightarrow \: x + y = 9 \\ \\ \longrightarrow \: y = 9 - x - - - - - (i)$

Now, According to the second condition :-

$\longrightarrow10x + y + 5 = 6(x + y) - 4 \\ \\ \longrightarrow10x + y + 5 = 6x + 6y - 4 \\ \\ \longrightarrow10x + y - 6x - 6y = - 4 - 5 \\ \\ \longrightarrow4x - 5y = - 9$

Now putting the value of y from the equation (i) →

$\longrightarrow4x - 5y = - 9 \\ \\\longrightarrow \: 4x - 5(9 - x) = - 9 \\ \\\longrightarrow 4x - 45 + 5x = - 9 \\ \\ \longrightarrow \: 9x = - 9 + 45 \\ \\ \longrightarrow \: 9x = 36 \\ \\ \longrightarrow \: x = \frac{36}{9} \\ \\ \longrightarrow \: x = 4$

Now put the value of x into the equation (i) →

$\longrightarrow \: y = 9 - x \\ \\ \longrightarrow \: y = 9 - 4 \\ \\ \longrightarrow \: y = 5$

So, the number is -

$\longrightarrow10x +y \\ \\ \longrightarrow10 \times 4 + 5 \\ \\ \longrightarrow40 + 5 \\ \\ \longrightarrow45$