Math Problem
 

An audience for a concert totals 100 people consisting of adults, students, and children. Ticket prices are $10 for adults, $3 for students, and fifty cents for children. The total amount of money taken in for ticket sales is $100. How many adults, how many students, and how many children were in attendance?

Re: Math Problem
 

5 1 94

Re: Math Problem
 

20 students and 80 children 0 adults.

Re: Math Problem
 

There are several solutions. The equation 10a + 3s + 0.5c = 100 where a, s, c > 0 represents a plane "segment" (dunno the correct term, but this should be clear). If a, s, and c have to be integral (which they hopefully do in this case.. hehe) then there are a finite number of solutions. If they didn't have to be integral, there would be an infinite number of solutions on the plane segment.

I hope that it is something like 9 adults, 3 students, and 2 children, though... so that all those children don't run rampant...

Oops.. missed a constraint... sigh.. hehe
 

100 people, eh? Well, there are still multiple solutions because there are three variables and only two constraints.

The solutions all lie along the line segment created where the two plane segments intersect...

10a + 3s + 0.5c = 100
a + s + c = 100
a, s, c > 0

Looks like we are gonna have lots of unsupervised children... wouldn't want to be at that show.

Solution
 

Let X = # adults, Y = # students, and Z = #children. We have the following:

(1) X + Y + Z = 100

(2) 10X + 3Y + 0.5Z = $100

Re-Write (2) gives:

(3) 20X + 6Y + Z = 200

Now (3) - (1) yields:

(4) 19X + 5Y = 100

Since both 5Y and 100 are multiples of 5, then 19X must be a multiple of 5. So let 5n = X.

Then (19)(5n) = 100 - 5Y. Divide both sides by 5 yields:

(5) 19n = 20 - Y. Since Y and n must be postive integers then n = 1 and Y = 1. If n = 1 then X = 5. Then Z = 94. The answer is 5 adults, 1 student, and 94 children.

Re: Solution
 

n and Y don't have to be positive.. they just have to be non-negative. 0 adults, 20 students, and 80 children is also an answer.

Not true
 

The problem specifically states that the audience consists of adults, students, and children. Therefore, there must be at least one adult.

Fair enough *NT*
 

Re: Not true
 

Jim,

If you want to be that specific about the wording of the problem then there is no correct answer. Last time I looked 1 is a student not students. Not trying to pick any nits here just wondering why MajorKongs soultion is incorrect if yours is correct.

Jimbo