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

Given :-

• 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.

To find :-

• Required numbers

Solution :-

Let the tens digit be x then ones digit be y

• Original number = 10x + y

Sum of the digits of a two digit number is 9.

• x + y = 9 ----(i)

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

• Reversed number = 10y + x

→ 9(10x + y) = 2(10y + x)

→ 90x + 9y = 20y + 2x

→ 90x - 2x + 9y - 20 = 0

→ 88x - 11y = 0

Take 11 as a common

→ 11(8x - y) = 0

→ 8x - y = 0 ---(ii)

Add both the equations

→ x + y + 8x - y = 0 + 9

→ 9x = 0 + 9

→ 9x = 9

→ x = 1

Put the value of x in equation (i)

→ x + y = 9

→ 1 + y = 9

→ y = 9 - 1

→ y = 8

Therefore,

• Tens digit = x = 1
• Ones digit = y = 8

Hence,

• Original number = 10x + y = 18
• Reversed number = 10y + x = 81

Answer 2

Answer:

The number is 18.

Step-by-step explanation:

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