AIME II Problem 4 2007

[zefuri]

The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, 100 workers can produce 300 widgets and 200 whoosits. In two hours, 60 workers can produce 240 widgets and 300 whoosits. In three hours, 50 workers can produce 150 widgets and m whoosits. Find m.

[stupidityismygam]

[Quelloquialism]

My solution was this...
1) The amount of work done depends equally on exactly two factors, the number of workers and the number of hours, so you don't need to worry about each factor individually--only them together. Multiply each number of workers with the corresponding number of hours.
Situation 1: 1 hr * 100 wrk = 100 wrk*hr
Situation 2: 2 hr * 60 wrk = 120 wrk*hr
Situation 3: 3 hr * 50 wrk = 150 wrk*hr

2) Go through your test and scribble out all instances of "widget" and replace them with "a". Then go through your test and scribble out all instances of "whoosit" and replace them with "b". I hate their names.

3) Set the problem up as a three-variable system of equations.
Situation 1: 100 = 300a + 200b
Situation 2: 120 = 240a + 300b
Situation 3: 150 = 150a + xb

4) Divide off greatest common divisors for ease of calculation.
Situation 1: 1 = 3a + 2b
Situation 2: 2 = 4a + 5b
Situation 3: 150 = 150a + xb

5) Solve the first pair of equations for a and b, using whatever method you prefer. I used elimination.
5 = 15a + 10b
4 = 8a + 10b
1 = 7a
a = 1/7
1 = 3/7 + 2b
4/7 = 2b
b = 2/7

6) Plug in the solutions for a and b into the final situation and solve for x.
150 = 150(1/7) + x(2/7)
1050 = 150 + 2x
900 = 2x
x = 450

In other words, the exact same thing as Stephen.