Question

sum of a two digit number and number obtained by interchanging digits is 66 . if the digits of number differ by 2 , find the number​

Answer 1

Step-by-step explanation:

hope it's helpfull

I think this is correct

Answer 2

SOLUTION :-

Let,

Digit in ten's place = x

Digit in one's place = y

Two digit number = 10x + y

According to the first condition,

$\longrightarrow \sf (10x+y)+(10y+x) = 66 \\\\\longrightarrow 10x+x+10y+y= 66 \\\\\longrightarrow 11x+11y = 66 \\\\\longrightarrow x+y = 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(1)$

According to the second condition,

$\longrightarrow\sf x-y = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(2)$

Eq (2) + Eq (1),

$\longrightarrow\sf 2x = 8 \\\\\longrightarrow \bf x = 4$

Substitute the value of x in eq (1),

$\longrightarrow\sf 4+y= 6 \\\\\longrightarrow \bf y = 2$

Two digit number = 24