how many people may have attended the performance?

Question

at a small theater, tickets for adults cost $12 and tickets for children at $8. at one performance sales were $480. how many people may have attended the performance?

Answer 1

Rose,

The words “may have” and “small theater” usually make most mathematicians nervous because they know how to calculate but not how to estimate.  Estimation is not taught in many math classes, but should be.  Good mathematicians become accountants, engineers, computer scientists … who also know how to interpolate, extrapolate, etc.

The problem has enough information to create a formula.  Since the total dollars is $480, and since there were adult tickets and child tickets sold, and since the number of tickets times the price per ticket (note: ticket cancels out, leaving dollars) for each group, the formula is:

      12A + 8C = 480        [note: A=number of adult tickets;  C=number of child tickets]

Now (as Robert F. showed in a table), values of A range from 0 to 40; values of C range from 0 to 60.  With this formula, the total number of people (A+C) that the problem asks us to find ranges from 40 to 60.  There aren’t any other possibilities (no half-people, etc.).

O.K., if the theater is “small,” it “may” have, for example, 38 adults and 3 children.  Or, it may have 36 adults and 6 children.  Or, …, I start hesitating because I think that 60 people in a theater is not exactly “small;”  what do you think?

Have I answered the question?  Yes !!    Adequately?  Yes !!

Now, if I had said the theater may have held 500 adults and 30 children, you would call the Fire Marshall and you would say that I’m keeping some of the ticket sales money somehow.

Life is full of estimating problems and we need to understand ranges and how to refine estimates as more information becomes available.

Best wishes!

 

Answer 2

Let a = the number of adults who attended

and c = the number of children who attended.

 

These are related as follows:  12a + 8c = 480

 

(With all the units written in, this would be ($12/adult)(a adults) + ($8/child)(c children) = $480.)

 

This is not enough information to solve for a and c uniquely, but all we need is a range of values for a+c, the total number of attendees.

 

I would suggest making a table of whole-number solutions to find this range.

 

a        0      2      ...   

c       60    57      ...   

a+c   60    59      ...   

 

Or you could graph the line in the ac-plane.

Answer 3

This is a least common multiple problem.

The total amount of sales, 480 is divisible by both 12 and 8.

One possibility is that there were 480/12=40 adults.

Another possibility is that there were 480/8=60 children.

Subtract The least common multiple of 12 and 8 is 24.

Begin with 40 adults and 0 children. Subtract 2 adults. That frees up $24 to add 3 children. So another possibility is 38 adults and 3 children. You can continue this process to find all the possibilities.


Adults      Children
40           0
38           3
36           6
34           9
32          12
30          15
. .
. .
2            57
0            60