Difference between revisions of "2000 AMC 12 Problems/Problem 13"
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<cmath>\left(\frac{c}{6} + \frac{m}{4}\right)p = c + m</cmath> | <cmath>\left(\frac{c}{6} + \frac{m}{4}\right)p = c + m</cmath> | ||
Regrouping, we get <math>2c(6-p)=3m(p-4)</math>. Since both <math>c,m</math> are positive, it follows that <math>6-p,p-4</math> are also positive, which is only possible when <math>p = 5\ \mathrm{(C)}</math>. | Regrouping, we get <math>2c(6-p)=3m(p-4)</math>. Since both <math>c,m</math> are positive, it follows that <math>6-p,p-4</math> are also positive, which is only possible when <math>p = 5\ \mathrm{(C)}</math>. | ||
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| + | Alternatively, one could notice that since there are only two components to the mixture Angela must have more than her "fair share" of one and less then her "fair share" of the other in order to ensure that everyone has 8 ounces. The "fair share" is 1/p. So, | ||
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| + | <cmath>\frac{1}{6} < \frac{1}{p}<\frac{1}{4}</cmath> | ||
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| + | Which requires that p be 5, since p is a whole number. | ||
== See also == | == See also == | ||
Revision as of 12:00, 15 July 2013
- The following problem is from both the 2000 AMC 12 #13 and 2000 AMC 10 #22, so both problems redirect to this page.
Problem
One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
Solution
Let 
be the total amount of coffee, 
of milk, and 
the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so
![\[\left(\frac{c}{6} + \frac{m}{4}\right)p = c + m\]](http://latex.artofproblemsolving.com/a/0/1/a018ae1614ac885a782067b85a8b65ff51f73dca.png)
Regrouping, we get 
. Since both 
are positive, it follows that 
are also positive, which is only possible when 
.
Alternatively, one could notice that since there are only two components to the mixture Angela must have more than her "fair share" of one and less then her "fair share" of the other in order to ensure that everyone has 8 ounces. The "fair share" is 1/p. So,
![\[\frac{1}{6} < \frac{1}{p}<\frac{1}{4}\]](http://latex.artofproblemsolving.com/e/4/1/e415a2efc92b64fac3e3726552212381b837a91d.png)
Which requires that p be 5, since p is a whole number.
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
