Question

The sum of the digits of a two 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:

18

Step-by-step explanation:

Let the units digit be x.

Tens digit = 9-x

Original number = 10(9-x)+x = 90-9x

After reversing digits,

Units digit = 9-x

Tens digit = x

New number = 10x+9-x = 9x+9

According to Question,

9(90-9x) = 2(9x+9)

=> 810-81x = 18x+18

=> 810-18 = 18x+81x

=> 99x = 792

=> x = 792/99

=> x = 8

Hence, units digit = 8

Tens digit = 9-8 = 1

Therefore, original number = 18

Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

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

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The number.

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

Let the tens place digit be r

Let the ones place digit be m

$\underline{\sf{The\:Original\:number=10r+m}}}}\\\underline{\sf{The\:Reversed\:number=10m+r}}}}$

A/q

$\leadsto\tt{r+m=9}\\\\\leadsto\tt{r=9-m....................(1)}$

&

$\longrightarrow\tt{9(10r+m)=2(10m+r)}\\\\\\\longrightarrow\tt{90r+9m=20m+2r}\\\\\\\longrightarrow\tt{90r-2r=20m-9m}\\\\\\\longrightarrow\tt{\cancel{88}r=\cancel{11}m}\\\\\\\longrightarrow\tt{8r=m}\\\\\\\longrightarrow\tt{8(9-m)=m\:\:\:[from(1)]}\\\\\\\longrightarrow\tt{72-8m=m}\\\\\\\longrightarrow\tt{72=m+8m}\\\\\\\longrightarrow\tt{72=9m}\\\\\\\longrightarrow\tt{m=\cancel{\dfrac{72}{9} }}\\\\\\\longrightarrow\tt{\green{m=8}}$

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

$\longrightarrow\tt{r=9-8}\\\\\longrightarrow\tt{\green{r=1}}$

Thus;

$\underbrace{\bf{The\:original\:number=(10r+m)=10(1)+8=10+8=\boxed{18}}}}$