2007 AIME II Problem 4

Attempt Problem 4 of the 2007 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AIME II solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

4.

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, 100100 workers can produce 300300 widgets and 200200 whoosits. In two hours, 6060 workers can produce 240240 widgets and 300300 whoosits. In three hours, 5050 workers can produce 150150 widgets and mm whoosits. Find m.m.

Answer: 450
Concepts:ratesystem of equations
Difficulty rating: 2020
Solution:

Let aa and bb be the worker-hours required to make one widget and one whoosit. The three scenarios supply 100,100, 120,120, and 150150 worker-hours, so 300a+200b=100,240a+300b=120,150a+mb=150. \begin{aligned} 300a + 200b &= 100, \\ 240a + 300b &= 120, \\ 150a + mb &= 150. \end{aligned}

The first two equations simplify to 3a+2b=13a + 2b = 1 and 4a+5b=2,4a + 5b = 2, giving a=17a = \frac{1}{7} and b=27.b = \frac{2}{7}. Substituting into the third, 1507+2m7=150,\frac{150}{7} + \frac{2m}{7} = 150, so 150+2m=1050150 + 2m = 1050 and m=450.m = 450.

← Problem 3#3
Full Exam

Problem 4 in Other Years