Question

.) The sum of the digits of a two-digit number is 11. The number got by
interchanging the digits is 27 more than the original number. What is the
number?​

Answer 1

$\huge\boxed{\underline{\bf { \red S \green O \pink L \blue U \orange T \purple I \red O \pink N \green{..}}}}$

Let the one digit number be "R" and tenth digit number be "S".

According to first statement

$\purple\bigstar$Sum of the digits is 11.

↪ R + S = 11 _____________eqn. (1)

According to second statement

$\green\bigstar$ The number formed by reversing the digits is 27 more than the original number.

Original number = 10R + S

Reversed number = 10S + R

↪ 10S + R = 10R + S + 27

↪ 10S + R - 10R - S = 27

↪ 9R - 9S = 27

↪ R - S = 3 _____________eqn. (2)

From eqn. (1) & (2),

$\sf \: \: \: R + S = 11 \\ \sf \: R - S = 3 \\ - - - - - - \\ \sf \: 2R = 14$

↪ R = 7 $\green\bigstar$

Put R =7 in eqn. (1)

↪ 7 + S = 11

↪ S = 11 - 7

↪ S = 4 $\blue\bigstar$

So,

The number = 10S + R

↪ 10(4) + 7

↪ 40 + 7

↪ 47 $\orange\bigstar$

Answer 2

Step-by-step explanation:

Let a number be x and y

10x+y=11

10y+x=27+(10x+y)

therefore

10y+x=27+11

10y+x=38

10y=38-x

y=38-x/10. (I)

Replacing the value found in eqn 1

10x+38-x/10

100x-x+38/10

99x+38/10x =38/99