Question

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

Answer 1

Let unit digit  = x

Tens digit  = y

Number will 10 times the tens digit + unit times the unit digit

Hence number will 10 y + x


Sum of digits are 9

So that

X + y  = 9        ………….(1)

 nine times this number is twice the number obtained by reversing the order of the digits


9 (10 y + x )  =  2 (10 x + y )

90 y + 9 x  = 20 x + 2y

88 y – 11 x  = 0

Divide by 11 we get

8 y  - x  = 0     …………..(2)

X + y  = 9        ………….(1)


Adding both equations we get

9 y = 9

Y = 9/9 = 1

Plug this value in equation first we get

X+ y = 9

X + 1 = 9

X = 8

So our original number is 10 y  + x    = 10*1 + 8 = 18

Answer 2

Answer :

☛ Required Number = 18.

Step By Step Explanation :

Let ten's place Digit Number and one's place Digit Number be x and y respectively,

•°• Number = 10x + y

°•° Reversed Number = 10y + x

Now, According to the Question!

☛ x + y = 9 ---- (I)

Again, [ nine times this number and the twice the number obtained by reversing the order of the digits. ]

☛ 9 ( Number ) = 2 ( Interchanged Number)

☛ 9 ( 10x + y ) = 2 ( 10y + x )

☛ 90x + 9y = 20y +2x

☛ 90x - 2x = 20y - 9y

☛ 88x = 11y

☛ 88x - 11y = 0

☛ 11(8x - y ) = 0

•°• 8x - y = 0 -----(II)

Now, Solving Equation 1 and 2 :

x + y = 9 × 8

8x - y = 0

_________
☛
8x + 8y = 72

8x - y = 0

_________
9y = 72

☛ y = 8

Putting Value of y in Equation (1)

☛ x + y = 9

☛ x + 8 = 9

☛ x = 9 - 8

☛ x = 1

•°• Number = 10x + y => 10 (1) + 8

Number = 10 + 8

Number = 18

Therefore, Required Number is 18.