Solving for a, b, and c in this equation for $ revenue for a circus

[PhysicsInept]

Homework Statement

Not homework but a puzzle I’m working on.

It costs money to enter the circus.
Adults are $5
Seniors are $2
Children are 10¢

100 people entered the circus and the total amounted to $100

Relevant Equations

So given this, I created the equation…..
5a + 2b + .1c = 100

Is this correct?

I’ve been able to determine the appropriate equation for calculating each of a, b, and c but of course one needs to know the values of the other 2 variables.

Is there a formula to determine this?

I haven’t yet found a method to solve my equation, assuming it can be. So far I’ve just attempted trial and error, using variations of values. I figured either 60 or 80 children in order to get the high number but just haven’t been able to tweak it yet.

Thanks

 

[.Scott]

Normally, you would want 3 equations for three unknowns.
But in this case you have one more equation (what is it) and something else about a, b, c.

Each of a b and c represent a count of people. What another way of saying that?That said, even with that other equation, you will need a bit of trial and error.
But you can put bounds on the problem. For example, does it make sense for there to be more than 100 children? So what would be the minimum dollar amount that needs to be covered by adults and seniors?

 

[FactChecker]

This is an example of an integer linear programming problem. They are NP-hard and can grow explosively. I am not an expert on them, but a small problem like this can be done easily by brute force with a small computer program running through all possible combinations.
Mentor note: solution deleted as OP had not found it yet.
There are a few methods to reduce the explosive nature of these problems, but these problems remain difficult to solve. (See this.)

 

[Frabjous]

Since the solution was given.
Mentor note: previously given answer is now deleted, so part of what was written below has been deleted.
1) you wrote down the money equation correctly.
2) <snip>
3) the third thing is that a,b and c need to be non-negative integers.

Since you only have two equations, you will only be able to eliminate 2 of the variables (i.e. two of the variables will be expressed terms of the third). You will use the fact that the variables need to be integers to figure out the third variable.

A couple of hints:
Looking at the money equation, I noticed that only one group had cents, so I chose to write c=10d so that the solution would only have dollar terms.
Think about the value of d in reference to the two equations, you will see that it needs to be 6, 7 or 8.

 

[.Scott]

So far I’ve just attempted trial and error, using variations of values. I figured either 60 or 80 children in order to get the high number but just haven’t been able to tweak it yet.

So let's work with the numbers you have.
Clearly you have discovered that choosing a children count "c" is a good starting point - because there are so many values of c that can't work. For example, any c=3 will give a total revenue that includes an integer dollar amount plus 30 cents - and thus can never work.

For c=60
a+b must be 40
5a + 2b must be 94
That's 2 equations and 2 unknown. Solve it.

For c=80
a+b must be 20
5a + 2b must be 92
That's 2 equations and 2 unknown. Solve it.