Question

the sum of the digits of a two-digit numbers is 11 the number formed by reversing the digits, exceeds the given number by 9. find the given number. ​

Answer 1

Answer:

let the unit digit be x

ten digit = 11 - x

two digit no. = 10× (11 - x) + ×

110 - 10x + x

if we reverse the digits

so unit digit no.= (11- x)

ten digit no.= x

so, 10× x + (11- x)

10x + 11- x = 110 - 10x + x +9

18x = 108

x= 108

x= 6

so original no.= 44

Answer 2

$\huge{\bold{\underline{Solution:}}}$

Let the digit of one's place be x and that of ten's digit be y.

A.T.Q,

• Case I

The sum of the digits of the number is 11.

This can be represented in the form of linear equation as:

$\Rightarrow x + y = 11 \qquad \bold{(1)}$

• Case II

The digits of the number (10x + y) is reversed and the number formed exceeds the original number by 9.

This can be represented in the form of linear equation as:

$\Rightarrow 10y + x = 10x + y + 9 \\ \Rightarrow 9x - 9y = -9 \qquad \bold{(2)}$

Multiply (1) by 9 to make the coefficient of y same.

On Solving the equations, We get

x = 5

Putting x = 5 in (1):

$\Rightarrow x + y = 11 \\ \Rightarrow 5 + y = 11 \\ \Rightarrow y = 11 - 5 \\ \Rightarrow y = 6$

The number is 10x + y.

=> 10 × 5 + 6

=> 50 + 6

=> 56

$\huge{\bold{Answer: \: 56}}$