Difference between revisions of "2007 AIME II Problems/Problem 4"

(add solution)
 
m (Problems)
 
(4 intermediate revisions by 4 users not shown)
Line 1: Line 1:
== Problems ==
+
== 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, <math>100</math> workers can produce <math>300</math> widgets and <math>200</math> whoosits. In two hours, <math>60</math> workers can produce <math>240</math> widgets and <math>300</math> whoosits. In three hours, <math>50</math> workers can produce <math>150</math> widgets and <math>m</math> whoosits. Find <math>\displaystyle m</math>.
+
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, <math>100</math> workers can produce <math>300</math> widgets and <math>200</math> whoosits. In two hours, <math>60</math> workers can produce <math>240</math> widgets and <math>300</math> whoosits. In three hours, <math>50</math> workers can produce <math>150</math> widgets and <math>m</math> whoosits. Find <math>m</math>.
  
 
== Solutions ==
 
== 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>.
 +
 
 +
== 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}}

Latest revision as of 10:12, 20 April 2024

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