Difference between revisions of "2007 AIME II Problems/Problem 4"
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Solve the system of equations with the first two equations to find that <math>(x,y) = \left(\frac{1}{7}, \frac{2}{7}\right)</math>. Substitute this into the third equation to find that <math>1050 = 150 + 2m</math>, so <math>m = \boxed{450}</math>. | Solve the system of equations with the first two equations to find that <math>(x,y) = \left(\frac{1}{7}, \frac{2}{7}\right)</math>. Substitute this into the third equation to find that <math>1050 = 150 + 2m</math>, so <math>m = \boxed{450}</math>. | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/00Ngozqw2d0?t=542 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
{{AIME box|year=2007|n=II|num-b=3|num-a=5}} | {{AIME box|year=2007|n=II|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 07:04, 4 November 2022
Problems
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, workers can produce widgets and whoosits. In two hours, workers can produce widgets and whoosits. In three hours, workers can produce widgets and whoosits. Find .
Solutions
Suppose that it takes hours for one worker to create one widget, and 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):
Solve the system of equations with the first two equations to find that . Substitute this into the third equation to find that , so .
Video Solution by OmegaLearn
https://youtu.be/00Ngozqw2d0?t=542
~ pi_is_3.14
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
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.