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Jim Brier
11-09-2002, 04:19 PM
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?

11-09-2002, 05:46 PM
5 1 94

11-09-2002, 08:06 PM
20 students and 80 children 0 adults.

Ed Miller
11-09-2002, 09:39 PM
There are several solutions. The equation 10a + 3s + 0.5c = 100 where a, s, c &gt; 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...

Ed Miller
11-09-2002, 09:45 PM
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 &gt; 0

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

Jim Brier
11-10-2002, 06:36 AM
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.

Ed Miller
11-10-2002, 07:03 AM
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.

Jim Brier
11-10-2002, 07:15 AM
The problem specifically states that the audience consists of adults, students, and children. Therefore, there must be at least one adult.

Ed Miller
11-10-2002, 07:22 AM

Jimbo
11-10-2002, 01:20 PM
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

Jim Brier
11-10-2002, 03:29 PM
The problem with pluralizing students is that you give away the solution to the problem if you state: "the audience consists of adults, a student, and children." Perhaps I should have put on an additional constraint with something like: "(Note: There must be at least one adult)".

The reason this problem is an important one is that algebra students are conditioned to believe that when you have N unknowns, you must have N independent equations. But this is not true. You can solve a wide variety of important problems by having fewer equations than unknowns when you have constraints.

BruceZ
11-11-2002, 08:12 PM
The reason this is solvable is because the solutions to the 2 equations in 3 unknowns must be positive integers. This is the additional constraint. Equations in more than one variable are called "diophantine equations" and are an important part of number theory.