Question 1099298: There are 100 seats for 100 people at the movie theater. The total price for the 100 people is $1.00. Men will be charged $.05 each, women $.02 and children 10 for $.01. How many men, women, and children will there be?
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
Let m be the number of men, w the number of women, and c be the number of children. Then
 [the total number of people is 100]
 [the total cost of the tickets is $1.00]
This is a problem with Diophantine equations. Diophantine equations are systems of equations in which there are fewer equations than variables, but in which it is possible to find the solution (or a family of solutions) using the requirement that all the variables must have values that are integers (usually, as in this case, positive integers).
With two variables and three equations, the procedure is to use the two equations to eliminate one of the variables, leaving one equation with two variables.
Then solve that one equation for one variable in terms of the other, and use the requirement that both variables have positive integer values to find solution(s).
With the two equations in this problem, we can eliminate variable c by multiplying the second equation by 1000 and subtracting the two equations:
With the equation in this form, we know that, because w has to be an integer, (900-49m) must be divisible by 19.
There are various ways to find what value(s) of m make this an integer. With pencil and paper, the best way is to rewrite the numerator (900-49m) in two parts, the first of which is exactly divisible by 19.
Proceeding with that idea...
With the equation in this form, we see that (2m-3) must be divisible by 19; and some quick calculations show that m must be 11. Then
And then since there are 11 men and 19 women, the number of children must be 70.
Now here is a different approach to the problem which uses more standard algebraic techniques.
Since the price for children is 10 for a penny, the number of children must be a multiple of 10.
Furthermore, a bit of logical analysis tells us that, since that average cost of the tickets is a penny, most of the people must be children.
So see if you can find a solution if the number of children is 90; if not, then try 80; then 70; and so on, until you find a solution.
If c = 90...
Multiply the first equation by 5 and then subtract the two equations:
 obviously no good...
If c = 80...
 no; 8 is not divisible by 3
If c = 70...
This works. c = 70; w = 19; that makes m = 11.
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