Can someone write this out for me?

[Lestat]

It took me a while to get this, because I have no formal math skills whatsoever. I ended up just using common sense and the process of elimination, but it was frustrating not knowing how to put this into the proper mathematical equation form. Can someone on here show me the correct way to write this problem out, along with the answers? There seems to be too many variable for a simple algebraic equation. But there must be a way to put this into a proper math equation. Thanks.



500 people go to see a movie. The theater takes in a total of $500. The admission is as follows:

Men $5, Women $1, Children a penny.

How many in attendance were men? How many were Woman? And how many were Children?

I was tempted to just answer this right out, but Im assuming that this is some sort of psychological test... which tells you something about yourself based on how you answer??

I would say that 100 men went and saw it.

[Borodog]

First notice that there are several cases that must be investigated. An even multiple of 100 children must have gone, since the total box office take is an even number of dollars. Cases that must be investigated are 0, 100, 200, 300, and 400. 500 is not possible because the total take is not enough, only $5. For each case, let the number of children in attendence be denoted C.

For each of these cases you will have two equations and two unknowns. Call the number of men M and the number of women W. The first equation is then:

Total revenue = $5xM + $1xW + $0.01xC = $500

The second equation is that the total number in attendance must be equal to 500:

M + W + C = 500.

We can solve this for, say, W:

W = 500 - C - M

We can then substitute this expression into our first equation:

$5xM + $1x(500 - C - M) + $0.01xC = $500,

$5xM + $500 - $1xC -$1xM + $0.01xC = $500,

$4xM -$0.99xC = 0,

$4xM = $0.99xC

or M = 0.99xC/4.

If C = 0, then M=0 and we have a solution. 500 women go to the theater at $1 apiece for a total of $500.

C cannot be 100, 200, or 300, since M would not be a whole number.

If C is 400, M = 99, leaving 1 woman. $5x99 + $1 + $4 = $500.

So you have two solutions: 500 women, or 400 children, 99 men, and 1 woman.

okay huh?

my sarcasm detector must have been way off, but there isnt any info given to suggest that different types of people had to go!

100 men, 500 women, or 50,000 kids are all possible.. and any mix therein.

this is making me feel really stupid, someone explain why all this is needed?

[Borodog]

[ QUOTE ]
500 people go to see a movie.

[/ QUOTE ]

[Siegmund]

[ QUOTE ]

500 people go to see a movie. The theater takes in a total of $500. The admission is as follows:

Men $5, Women $1, Children a penny.

How many in attendance were men? How many were Woman? And how many were Children?

[/ QUOTE ]

To set it up:

since 500 people saw it, M + W + C = 500.
sincd $500 were collected, 5M + W + .01C = 500.

With two constraints and three unknowns, there are an infinite number of solutions to the equations. Common sense imposes an additional constraint: M, W, C all whole numbers.

Since this last constraint gives you at most six cases to check, it's probably simplest to just test them one at a time rather than fussing too much further with the equations.

[Lestat]

Wow! You're good! I forget to mention that at least one of each group (men, women, and children), were in attendance. Thanks you!

[Lestat]

My fault for not mentioning that at least one of each group attended.

[Borodog]

No problem.

[ QUOTE ]
[ QUOTE ]
500 people go to see a movie.

[/ QUOTE ]

[/ QUOTE ]

this is not a direct stipulation, and therefore I was confused... since implications dont usually play in math problems.

for future reference, should I have taken that to mean one from each?