Question

the sum of the digits of a two digit number is 9 also nine times the number is twice the number obtained by reversing the order of digit find the numbers​

Answer 1

Step-by-step explanation:

LET THE DIGITS BE x AND y

THEREFORE, THE NO. WILL BE 10x+y

NOW, ATQ

x+y = 9

x = 9-y

ALSO,

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

90x+9y = 20y+ 2x

88x = 11y

8x = y

8(9-y)=y

72 - 8y = y

72 = 9y

y = 8

NOW, AS 8x=y

8x = 8

x = 1

therefore, the no is 10 x 1+8

18

Answer 2

$\huge\sf{\underline{\underline{Solution:}}}$

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

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

Number = 10y + x

Number after reversing the digits = 10x + y

A.T.Q

⇒ 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.