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 ==
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https://youtu.be/00Ngozqw2d0?t=542
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~ 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, $100$ 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

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