Question

A number consists of two digits whose sum is 9 .if the 9 is subtracted from the the number the digits the inter change their places .find the number

Answer 1

Let the ten's digit and the one's digit be x and y respectively.

Given

Sum of digits = 9

$\implies x+y=9--(1)$

If 9 is subtracted from the the number the digits they interchange their places.

$\implies 10x+y-9=10y+x\\\implies 10x-x+y-10y=9\\\implies 9x-9y=9\\\implies x-y=1--(2)$

Solving (1) and (2), we get,

$x+y=9\\\underline{x-y=1}\\\underline{\underline{2x=10}}\\\implies x=5\\\implies y = 4$

$\boxed{\boxed{\bold{Therefore, \ the \ number \ is \ 54}}}}}$

                                                       

Answer 2

$\rule{200}{2}$

♦ GiveN InformatioN :-

• A number consists of two digits whose sum is 9.

• If the 9 is subtracted from the the number the digits the inter change their places.

♦ TO FinD :-

The original number.

♦ PrE-RequisitE :-

Let the ten's digit and the one's digit be x and y respectively.

♦ SolutioN :-

According to the given information,

x + y = 9              ...(1)

It is given that, if 9 is subtracted from the the number the digits they interchange their places.

Let the original number be 10x + y.

Then,

10x + y - 9 = 10y + x

⇒9x - 9y = 9

⇒9(x - y) = 9

⇒x - y = 1              ...(2)  

Solving eq.(1) and eq.(2) :--

In eq.(2), x = 1 + y

Putting this value in eq.(1) :-

x + y = 9

⇒1 + y + y = 9

⇒2y + 1 = 9

⇒2y = 8

⇒y = 4

Now, x = 1 + y

⇒x = 1 + 4

⇒x = 5

$\boxed{\therefore \text{The\ original\ number\ = 10x + y = 10(5) + 4 = \bold{54}}}$

$\rule{200}{2}$

Regards :)