Difference between revisions of "2007 AIME II Problems/Problem 4"
(→Solutions) |
|||
| Line 15: | Line 15: | ||
</div> | </div> | ||
| − | 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 = 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>. |
== 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 04:14, 27 November 2016
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 
.
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. 
