When you reverse the digits in a certain two digit number you decrease its value by 9. Find the number if the sum of it digits is 11.

Question

It is basically asking for a value I do not know how to find. 

Answer 1

Remember that decimal is a positional number system.

 

Let the digits of the two-digit number be "ab"     [this is a*10+b]

 

Translate:

"reverse the digits in a certain two digit number you decrease its value by 9"    means

       ba                                                               =   ab                 -    9

 

 

10b + a = 10a + b - 9

  9b - 9a  = -9

   a - b = 1                [eq1]

 

"the sum of it digits is 11"    means

   a + b = 11              [eq2]

 

Using elimination:

    a + b = 11                        [eq2]

    a  - b =  1                        [eq1]

  ----------------    [elimination; add equations]

   2a      =  12

    a      =   6          [eq3]

 

Now, use either equation [eq1] or [eq2] to find the value of b:

   a  +  b = 11         [eq2]

   a         =   6         [eq3]

 ---------------------------------    [elimination; subtract equations]

           b =   5

 

Check:

              56 = 65 -9

          6+5=11

Answer 2

Suppose the 2-digit number consists of t as the tens position, u as the units position.

Then the value of this number is 10t+u.

 

If the digits are reversed, a different number is formed & its value must be 10u+t.

 

Now you have two equations: 10t+u = 10u+t + 9 (because "When you reverse the digits in a certain two digit number you decrease its value by 9").

 

Also, t+u = 11.

 

So you have a system of two equations and two unknowns, which you can solve.

Answer 3

10a + b = 10b + a - 9

a + b = 11