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:

18

Step-by-step explanation:

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.

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.

Answer 2

Answer:

Let the unit digit be x and tens digit be y.

So, the number will be = 10y + x

After interchanging the digits the number becomes = 10x + y

It is given that, The sum of the digits of two digit number is 9. Therefore we get :]

x + y = 9.......(Equation i)

It is also given that, Nine times this number is twice the number obtained by reversing the order of the digits :]

9(10y + x) = 2(10x + y)

90y + 9x = 20x + 2y

90y - 2y = 20x - 9x

188y = 11x

x = 8y.......(Equation ii)

____________________

Now,Putting the value of x = 8y in equation (i) we get :]

x + y = 9

8y + y = 9

9y = 9

y = 1

_____________________

Now, substitute the value of y in equation (ii) we get :]

x = 8y

x = 8(1)

x = 8

Therefore

The original number will be 10y + x = 10(1) + 8 = 18