2007 AIME II Problems/Problem 4

Problem

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 $m$.

Solutions

Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit.

Therefore, we can write that (note that two hours is similar to having twice the number of workers, and so on):

$100 = 300x + 200y$

$2(60) = 240x + 300y$

$3(50) = 150x + my$

Solve the system of equations with the first two equations to find that $(x,y) = \left(\frac{1}{7}, \frac{2}{7}\right)$. Substitute this into the third equation to find that $1050 = 150 + 2m$, so $m = \boxed{450}$.

Video Solution by OmegaLearn

https://youtu.be/00Ngozqw2d0?t=542

~ pi_is_3.14

See also

2007 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png