Question

The sum of the digits of a two digit num​

Answer 1

Given :-

• Sum of digits is 12.
• The no. obtained by Interchanging exceed the original number by 18.

Have to Find :

• The original number.

Solution:

Let us we consider

• The digit at tens be " a " y

• The unit place digit be " b " x

According to case 1

$\implies$ y + x = 12

→ y = 12 - x ......(1)eqn

According to Case 2

(10x + y) - (10y + x) = 18

$\implies$ 9x - 9y = 18

$\implies$ x - y = 2

$\implies$ x - ( 12 - x ) = 2

$\implies$ 2x - 12 = 2

$\implies$ 2x = 14

$\implies$ x = 7

Now,

$\implies$ y = 12 - y

$\implies$ y = 12 - 7

$\implies$ y = 5

Putting x = 7 and y = 5

Hence,

The original Number = 10 (5) + 7

→ 57

Answer 2

★Solution :-

• Let x and y be the digits of two digit number such that x is placed at the units place and y is placed at Tens place.

$\sf \: let \: the \: number \: be \: N _o \\ = 10 \times y + x \\ \implies N_o \: = 10y + x \: \: (1)$

$\sf \: Let \: N_r \: be \: the \: number \: obtained \\ \sf \: \: by \: interchanging \: the \: two \: digit \\ \sf \: \: number \: of \: N_o$

$\: Then \implies\: N_r \: = 10 \times x \times y \\ \implies N_r \: = 10x + y \: \: \: (2)$

According To Question,

$x + y = 12 \\ \implies \: y = 12 - x \: \: \: (3)$

Also,

$N_r \: = N_o \: + 18$

$\implies10x + y = 10y + x + 18 \\ [Using \: (1) \: \: and \: \: (2)] \\ \implies \: 10x + y - 10y - x = 18 \\ \implies \: 9x - 9y = 18 \\ \implies \: 9(x - y) = 9 \times 2 \\ \implies \: x - y = 2 \: \: (4)$

Putting (3) and (4) ,We Get,

$x -( 12 - x) = 2 \\ \implies \: 2x = 2 + 12 \\ \implies \: 2x = 14 \\ \implies \: x = 7$

Putting x =7 in (3) , we get

$y = 12 - 7\\ \implies \: y = 5 \\ So, \: N_o = (10 \times 5) + 7 \: \\ [Using \: (1)] \\ N_o = 50 + 7 \\ \color{teal} N_o \: = 57$

Hence, The Answer is 57.