Question

The sum of the digits of a two digit number 9 .Also , nine times this number is twice the number obtained by reversing the order of two digits . find the number.​

Answer 1

Step-by-step explanation:

Let the number be 10a+b

Reverse digit number is 10b+a

Given the sum of digits is 9

a+b=9______(1)

Also given that 9 times the original number is twice the reverse digit number

So, 9(10a+b)=2(10b+a)

90a+9b=20b+2a

88a=11b

8a=b

Substitute b=8a in eqtn 1

a+b=9

a+8a=9

9a=9

a=1

b=8a=8

The original number is 18

Answer 2

Given:

The sum of digits of 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

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..