SOLUTION: A booster club sells 180 tickets for a pancake breakfast. A child’s plate cost $4.50, and an adult plate costs $7.00. A total of $980. How many of each plate was served

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A booster club sells 180 tickets for a pancake breakfast. A child’s plate cost $4.50, and an adult plate costs $7.00. A total of $980. How many of each plate was served       Log On


   



Question 1125119: A booster club sells 180 tickets for a pancake breakfast. A child’s plate cost $4.50, and an adult plate costs $7.00. A total of $980. How many of each plate was served
Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.

        There are different methods to solve the problem.
        I will show you 3 (three) basic methods in this post.


Solution  1   (the system of  2  equations approach)

Let x be the number of the adult tickets sold and

let y be the number of the children tickets sold.


Then you have these two equations

 x +    y = 180,           (1)     (counting tickets)
7x + 4.5y = 980  dollars   (2)     (counting dollars)


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


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

I will use the Substitution method here. 
From equation (1) express y = 180-x and substitute it into equation (2). You will get a single equation for the unknown x

7x + 4.5(180-x) = 980.     (3)


7x + 4.5*180 - 4.5x = 980  ====>  (7-4.5)x = 980 - 4.5*180  ====>  x =  %28980-4.5%2A180%29%2F%287-4.5%29 = 68.    (4)


Thus, 68 adult tickets were sold.


Hence the number of children tickets was 180 - 68 = 112.


Check.  7*68 + 4.5*112 = 980 dollars.   ! Correct. !


Answer.  68 adult and 112 children tickets.


Solution  2   (one equation approach)

Let x be the number of adult tickets sold.

Then the number of children tickets sold is (180-x),  according to the condition.


The adults   tickets cost 7x        dollars.
The children tickets cost 6*(159-x) dollars.

Summing up these money, you get the revenue equation

7x + 4.5(180-x) = 980.     (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.


Solution  3   (Logical analysis)

Let assume for a minute that all 180 tickets were children.

Then the total revenue would be 4.5*180 = 810 dollars.

It is by 980 - 810 = 170 dollars less than the given revenue value.


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

To compensate the difference, we must replace some number of 180 children tickets by the adult tickets.

At each replacement, we diminish the difference of $170 by $2.50 (which is the difference $7-$4.50 between the adult and the children tickets price).

Then it is clear that the number of adult tickets is  170%2F2.50 = 68.


Hence, the number of children is  180 - 68 = 112,  and you got the same answer as in solutions 1 and 2 above.



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

        adults = %28980-4.5%2A180%29%2F%287-4.50%29    (6)

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

--------------

Congratulations !   You are now familiar with 3 methods for ticket problems solution.

I suggest that algebraic methods will be your basic methods for such problems,
and the logical analysis method will allow you to solve the problems  QUICKLY  without using equations.
I will be happy if it will make your horizon wider.

--------------
To see other similar solved problems on tickets,  look into the lessons
    - Using systems of equations to solve problems on tickets
    - Three methods for solving standard (typical) problems on tickets
in this site.

To see how the logical method works for other similar problems, look into the lessons
    - Problem on two-wheel and three-wheel bicycles
    - Problem on animals at a farm
    - Problem on pills in containers
    - What type of problems are these?
in this site.


================

The lesson to learn from my presentation :

     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 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) in one click.

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