Question

The sum of the digits of a 2-digit number is 11. The number obtained interchanging the digits exceeds the original number by 27. Find the number​

Answer 1

Answer:

58

Step-by-step explanation:

Considering the provided information in the given question, we have :

• The sum of the digits of a 2-digit number is 11.
• The number obtained interchanging the digits exceeds the original number by 27.

We are asked to calculate the original number.

Let us assume the two digit number as 10x + y. Here, x and y are its digits.

― According to the question, the sum of the digits of a 2-digit number is 11. Writing this statement in the form of am equation,

$\longrightarrow \sf{\quad {x + y = 13 \quad \dots \bf {(1)} }}$

From this equation,

$\longrightarrow \sf{\quad {x = 13 - y }}$

― Also, the number obtained interchanging the digits exceeds the original number by 27. Writing this statement in the form of am equation,

$\longrightarrow \sf{\quad {10y + x = 27 + 10x + y \quad \dots \bf {(2)} }}$

Transposing the like terms.

$\longrightarrow \sf{\quad {10y - y = 27 + 10x - x }}$

Performing subtraction in L.H.S and R.H.S.

$\longrightarrow \sf{\quad {9y = 27 + 9x }}$

Substitute the value of x from equation ( 1 ) into the equation ( 2 ).

$\longrightarrow \sf{\quad {9y = 27 + 9(13 - y) }}$

Performing multiplication in R.H.S.

$\longrightarrow \sf{\quad {9y = 27 + 117 - 9y }}$

Performing addition in R.H.S, and transposing -9y from R.H.S to L.H.S.

$\longrightarrow \sf{\quad {9y + 9y = 144 }}$

Performing addition in L.H.S.

$\longrightarrow \sf{\quad {18y = 144 }}$

Transposing 18 from L.H.S to R.H.S, its arithmetic operator will get changed.

$\longrightarrow \sf{\quad {y = \cancel{\dfrac{144}{18}} }}$

Dividing 144 by 18.

$\longrightarrow \quad { \textbf{\textsf{y = 8}}}$

$\underline{ \qquad\qquad\qquad\qquad \qquad\qquad\qquad\qquad}$

Substitute the value of y in the equation ( 1 ) to find the value of x.

$\longrightarrow \sf{\quad {x = 13 - y }}$

Substituting the value of y.

$\longrightarrow \sf{\quad {x = 13 - 8 }}$

Performing subtraction.

$\longrightarrow \quad { \textbf{\textsf{x = 5}}}$

$\underline{ \qquad\qquad\qquad\qquad \qquad\qquad\qquad\qquad}$

$\longrightarrow \sf{\quad {Original \; Number = 10x + y }}$

Substitute the value of x and y.

$\longrightarrow \sf{\quad {Original \; Number = 10(5) + 8 }}$

Performing multiplication.

$\longrightarrow \sf{\quad {Original \; Number = 50 + 8 }}$

Performing addition.

$\longrightarrow \quad \underline{\boxed { \textbf{\textsf{Original \; Number = 58}}}}$

Therefore,

⠀⠀★ Required Number = 58