Let x be the number of the adult tickets and

let y be the number of the child tickets.


Then you have these two equations

 x +  y =  500,           (1)     (counting tickets)
5x + 3y = 2300  ghc       (2)     (counting revenue in ghc)


They are your basic equations, and as soon as you got this system, the setup is completed.


There are different methods of solving such systems (Substitution, Elimination, using determinants). 

I will use the Substitution method here. 
From equation (1) express y = 500-x and substitute it into equation (2). You will get a single equation for the unknown x

5x + 3(500-x) = 2300.     (3)


5x + 3*500 - 3x = 1142  ====>  (5-3)x = 2300 - 3*500  ====>  x =   = 400.    (4)


Thus, 400 adult tickets were sold.


Hence the number of the child tickets was 500 - 400 = 100.


Check.  5*400 + 3*100 = 2300 ghc.   ! Correct. !


Answer.  400 adult tickets.

Let x be the number of adult tickets  (the unknown value under the question)

Then the number of the child tickets is (500-x),  according to the condition.


The adult tickets cost 5x ghc.
The child tickets cost 3*(500-x) ghc.

Summing up these money, you get the revenue equation

5x + 3*(500-x) = 2300.     (5)


It is your basic equation in the frame of this approach,
and as soon as you got this equation, the setup is completed.


Notice that this equation (5) coincides exactly with the equation (3) of the solution 1 above.

Solve equation (5) by the same method as it was done in the Solution 1.


Surely, you will get the same answer:  the number of adult tickets is 400.

Let assume for a minute that all 500 tickets were child tickets.

Then the total revenue would be 3*500 = 1500 ghc.

It is by 2300 - 1500 = 800 ghc less than the given revenue value.


Why did we get the difference ?  - But of course, because we counted all tickets as the child tickets.

To compensate the difference, we must replace some number of 500 child tickets by the adult tickets in our analysis.

At each replacement, we diminish the difference of 800 ghc by 2 ghc (which is the difference 
5 ghc - 3 ghc between the adult and the child tickets price).

Then it is clear that the number of adult tickets is   = 400.


Thus logical solution is completed, and you obtain the same answer.



    Notice, that this logic works EXACTLY in accordance with the formula

        number of adult tickets =     (6)

    which is EXACTLY THE SAME as the formula (4).

     The typical ticket problem is THIS :

         - You are given the total number of cheap and expensive tickets;
         - you are given the prices for tickets of these two categories;
         - you are given the total revenue;
         - you are asked to find the number of tickets in each of the two categories.


     When you solve such a problem, you may use ANY of the three methods of this presentation.


     But you also can write the solution for the number of more expensive tickets in one line as a fraction whose numerator 

     is the difference between the given revenue and the hypothetical revenue assuming that all tickets are for cheap price, 

     and whose denominator is the difference (positive difference) between expensive and cheap prices.


    The logic explained in the Solution 3 will help you to go in this way and will prevent you of making errors or steps aside.


    And when the formula is implemented and all the given numbers are plug-in, all you need to do is to copy and paste the formula
    into MS Excel in your computer (or into other similar software) and press the "Enter" button to get the final number (solution; answer).

    And it is, if you want, the fourth method of solving such problems - the shortest and the quickest.