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.
|