Question

A two digit number is such that sum of digit is 11 .When number with the same digit is subtracted from this number,difference is 9.What is the number?

Hint:: 9(x-y) =difference​

Answer 1

$\huge{\bf{\green{\mathfrak{\dag{\underline{\underline{Question:-}}}}}}}$

A two digit number is such that sum of digit is 11 .When number with the same digit is reversed is subtracted from this number,difference is 9. What is the number?(There is a mistake in question.)

$\huge{\bf{\green{\mathfrak{\dag{\underline{\underline{Solution:}}}}}}}$

Let tens digit of the number be x and ones digit number be y.

Then the number is 10x + y(if the number is 23 it will be expanded as 10*2 + 3, here 2 is x and 3 is y).

According to the question,

sum of the digits is 11 i.e.,

x + y = 11________(1)

and difference between the original number(10x + y) and reversed number(10y + x) is 9 i.e.,

10x + y - (10y + x) = 9

10x + y - 10y - x = 9

10x - x - 10y + y = 9

9x - 9y = 9

÷ 9

x - y = 1________(2)

(1) + (2)

$(1) \: x + \cancel{ y} = 11 \\ (2) \: \underline{ x - \cancel{ y }= 01} \\ \: \: \: \: \: \: \:\underline { 2x + 0 = 12}$

$2x = 12$

$\large{x = \frac{ \cancel{12} \: {}^{6} }{\cancel{2}} }$

$x = 6$

Apply x = 6 (1)

x + y = 11

6 + y = 11

y = 11 - 6

y = 5

$\therefore \: the \: number \: is \: 65.$