Question

The sum of the digits of a two-digit number is 9. Also, nine times this number is
twice the number obtained by reversing the order of the digits. Find the number.​

Answer 1

Answer:

The number is 18.

Step-by-step explanation:

hope it helps.

Answer 2

ANSWER :

$\bf{\Large{\underline{\sf{Given\::}}}}$

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits.

$\bf{\Large{\underline{\sf{To\:find\::}}}}$

The number.

$\bf{\Large{\underline{\tt{\red{Explanation\::}}}}}$

Let the one's place be R.

Let the ten's place be M.

A/q

$\implies\sf{R+M\:=\:9...................(1)}$

Original Number = 10M + R.

Reversed Number = 10R + M.

So,

$\implies\sf{9(10M+R)=2(10R+M)}\\\\\\\implies\sf{90M+9R=20R+2M}\\\\\\\implies\sf{90M-2M\:=\:9R-20R}\\\\\\\implies\sf{\cancel{88}M=\cancel{11}R}\\\\\\\implies\sf{\red{8M\:=\:R................(2)}}$

Putting the value of R in equation (1), we get;

$\implies\sf{8M+M\:=\:9}\\\\\\\implies\sf{9M\:=\:9}\\\\\\\implies\sf{M\:=\:\cancel{\dfrac{9}{9} }}\\\\\\\implies\sf{\red{M\:=\:1}}$

Putting the value of M in equation (2), we get;

$\implies\sf{R\:=\:8(1)}\\\\\\\implies\sf{\red{R\:=\:8}}$

Thus,

The number is 10(1) + 8

The number is 10 + 8 = 18.