Question

The sum of the digit of a 2 digit no. is 9.also 9 times this no. is twice the no. obtained a by reversing the digits .

Answer 1

Solution -  ​​

Let the digit in ones place be x  

So, the digit in tens place be 9 - x  

Original no. = 10(9 - x) + 1(x)  

= 90 - 10x + x  

= 90 - 9x  

New no. = 10(x) + 1(9 - x)  

= 10x + 9 - x  

= 9x + 9  

According to Question,    

9[ Original no. ] = 2[ New no. ]  

⇒ 9[90 - 9x] = 2[9x + 9]  

⇒ 810 - 81x = 18x + 18  

⇒ 810 - 18 = 18x + 81x  

⇒ 792 = 99x    ​

⇒ 792/99 = x    

⇒ 8 = x  

Original no. = 90 - 9x  

= 90 - 9(8)  

= 90 - 72  

= 18  

∴ The number is 18 or 81

#Be Brainly ^ ^

Answer 2

Given :-

The sum of digits f a two digit number is 9.

Also nine times this number is twice the number obtained by reversing the order of digit.

To Find :-

The Number

Solution :-

Let the unit digit and tens digits of the number be x and y

Number = 10y + x

Number after reversing the digits = 10x + y

According to the question,

⇒ x + y = 9 ... (i)

⇒ 9(10y + x) = 2(10x + y)

⇒ 88y - 11x = 0

⇒ -x + 8y =0 ... (ii)

Adding equation (i) and (ii), we get

⇒ 9y = 9

⇒ y = 1 ... (iii)

Putting the value in equation (i), we get

⇒ x = 8

Hence, the number is 10y + x = 10 × 1 + 8 = 18.