Question

The sum of the digits of a two digit numbers in 9. Also, nine time this number is twice the number obtained by reversing the order of the digits . Find the number

Answer 1

☘️ Solution:

Let the first number be x

And the second number be y.

Case I:

Given that their sum is 9

So the equation we'll get is:

x + y = 9 ———(1)

Case II:

According to the question:

• The original number = 10x + y
• On reversing number = x + 10y

Given that, nine time this number is twice the number obtained by reversing the order of the digits.

So, the equation formed with be:

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

Now let's find the number.

→ 9(10x + y) = 2(x + 10y)

→ 90x + 9y = 2x + 20y

→ 90x + 9y - 2x - 20y = 0

→ 88x - 11y = 0

→ 8x - y = 0 ———(2)

Using elimination method:

x + y = 9 ×8

8x - y = 0 ×1

8x + 8y = 72

8x - y = 0

- + -

9y = 72

y = 72/9

y = 8

Substituting value of y in eq.1:

→ x + y = 9

→ x + 8 = 9

→ x = 9 - 8

→ x = 1

Here, x is the tens digit and y is the unit digit. So, the number obtained will be 18.

☘️ Answer:

• Number obtained = 18.