Question

sum of the digit of a two digit number is 11 the given number is less than the number obtained by interchanging the digits by 9 find the number

Answer 1

Let the tens place digit be a

And Unit place digit be b

According to first condition,

a + b = 11 --------(1)

According to second condition

( 10a + b) - (10b +a) = 9

=> 10a + b - 10b - a = 9

=> 9a - 9b = 9

=> a - b = 1 ------(2)

On adding equation 1 and 2, we get

2a = 12

=> a = 6

Now,

On Substituting the value a in equation 1, we get

6 + b = 11

=> b = 5

Required number = 10 × 6 + 5

= 65

Answer 2

Hi.

Here is your answer---

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Let the once digit in a two-digit number be y and tens digit be x.
Thus, the Numbers = 10x + y.

According to the question,

  x + y = 11
    x = 11 - y ------------------------------eq(i)

Also,
 10x + y = 10y + x + 9
  9x - 9y =9
  x -y =1

Putting eq(i) in this,

 (11 -y) -y =1
 11 -2y = 1
  2y =10
  y =5

y = 5 in eq(i),
 
 x =11 - 5
 x = 6

Thus, Number = 10x + y
                        = 10(6) + 5
                        = 65

The two-digit number is 65.

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Hope it helps.

Have a nice day.