How do I work out this question about different values?

Question

A group of 42 people go to an amusement park. The admission fee for adults is $16. The admission fee for children is $12 . The group spent $568 to get into the park . How many adults and children were in the group?

Answer 1

Let C = number of children

Let A = number of adults

 

Using these variables, we can set up two equations.  One to represent the total number of people in that group.  The other to represent the total cost.

 

A + C = 42                          eq1

16A + 12C = 568                 eq2

 

Solve the system of equations using substitution/elimination methods to find A and C.

Answer 2

For Honors Algebra, you should be capable of both elimination and substitution.

 

Let A = number of adults

     C = number of children

 

Note that dollars for each group = (dollars per ticket)*(number of tickets)

                                            [that is, 16A and 12C]

 

"42 people"    means     A+C=42  [eq1]

"adults is $16 ... children is $12 ... spent $568"   means  16A + 12C = 568   [eq2]

 

Elimination:

  16A + 12C = 568    [eq2]

  12A + 12C = 504    [12*eq1]

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

    4A           =   64

      A           =  16

      A  +  C  =  42

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

              -C = -26

               C = 26

 

Substitution:

  C = 42 - A      [from eq1]

  16A + 12(42 - A) = 568        [substitute for C in eq2]

  16A - 504 - 12A = 568

     4A = 64

       A = 16

  Substitute that into either equation [one is often easier than the other]

   C = 42 - A            [from eq1]

   C = 26

Answer 3

Let A = # of adults

Let C = # of children

 

There is a total of 42 people in the group

 

That would translate to:

 

A+C=42

 

We also know they spent a total of $568

Each adult had to pay $16 and each child had to pay $12

 

That would translate to:

 

16A+12C=568

 

This is system of 2 equations with 2 unknowns. 

 

A+C=42

16A+12C=568

 

We will use a method called elimination.

Multiply the top equation by 12

 

12(A+C=42)=12A+12C=504

 

12A+12C = 504

16A+12C=568

 

Now if we subtract these equations we will "eliminate" the C variable.

 

4A = 64

 

Now we can solve for A by dividing by 4

A =64/4 = 16

 

Therefore there are 16 Adults.

We can plug this value into the equation

A+C=42 to solve for C

16+C=42

C = 26

There are 26 children