Question

The sum of 2digit no.s is 3 also nine times this no. is twice the no. obtained by reversing the order of the digit. find the no.​

Answer 1

Answer:

Step-by-step explanation:

Please note that your question seems to be incorrect as per the solution i am getting but the following is the way it is solved.

Let x and y be the tens and units digits of the 2 digit number.

Thus the number is 10x + y

Given, Sum of digits is 3

=> x  + y = 3 ----  [1]

Given, Nine times this number is twice the no. obtained by reversing the order of the digit.

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

=> 90x + 9y = 20y + 2x

=> 88x  =  11y

=> y = 8x    ----------------- [2]

Substitute value of [2] in [1]

x + y = 3

x + 8x = 3

=> 9x = 3

=> x = 1/3

=> y = 8/3

Substitute x and y in the equation 10x + y to get the number.

Answer 2

SOLUTION:-

Given:

The sum of two digit number is 3 & also 9 times the number obtained by reversing the order of the digit.

To find:

The number.

Explanation:

Let the digit at tens place be R.

Let the digit at ones place be M.

Therefore,

Original number =10R + M.

Reversing number=10M+R

According to the question:

R+M= 3.............(1)

=) 9(10R+M)= 2(10M+R)

=) 90R + 9M= 20M +2R

=) 90R -2R= 20M -9M

=) 88R = 11M

=) 8R =M.............(2)

Using substitution Method:

From equation (1) & (2), we get;

=) R + M= 3

=) R+ 8R= 3

=) 9R = 3

=) R= 3/9

=) R= 1/3

Putting the value of R in equation (1),we get;

=) R + M= 3

=) 1/3 + M= 3

=) 1+ 3M= 9

=) 3M= 9 -1

=) 3M= 8

=) M= 8/3

Now,

The original number:

$10r + m \\ \\ Original \: number = 10 \times \frac{1}{3} + \frac{8}{3} \\ \\ Original \: number = \frac{10}{3} + \frac{8}{3} \\ \\ Original \: number = \frac{10 + 8}{ 3} \\ \\ Original \: number = \frac{18}{3}$