Question

the sum of the digit of a two digits of two digit number is 11 the number got by interchanging the digits is 27 rules than the original number what is the number​

Answer 1

Correct 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 this number ?

Answer:-

Let the digit in one's place be y and digit at ten's place be x.

Given:

Sum of the digits = 11

⟶ x + y = 11

⟶ x = 11 - y -- equation (1).

And,

The number formed by reversing the digits is 27 more than the original number.

⟶ Reversed number = Original number + 27

• reversed number = 10y + x.

⟶ 10y + x = 10x + y + 27

⟶ 10y + x - 10x - y = 27

⟶ 9y - 9x = 27

substitute the value x from equation (1).

⟶ 9y - 9(11 - y) = 27

⟶ 9y - 99 + 9y = 27

⟶ 18y = 27 + 99

⟶ 18y = 126

⟶ y = 126/18

⟶ y = 7

Substitute the value of y in equation (1).

⟶ x = 11 - 7

⟶ x = 4

• The number = 10(4) + 7 = 40 + 7 = 47.

∴ The required two digit number is 47.

Answer 2

Answer:

• The required two digit number is 47.

Step-by-step explanation:

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 $\orange\bigstar$