Difference between revisions of "2000 AMC 12 Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | Let <math>a</math> be the total amount of coffee, <math>b</math> of milk, and <math>n</math> the number of people in the family. Then each person drinks the same amount of coffee and milk, so < | + | Let <math>a</math> be the total amount of coffee, <math>b</math> of milk, and <math>n</math> the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so |
| + | <cmath>\left(\frac{a}{6} + \frac{b}{4}\right)n = a + b</cmath> | ||
| + | Regrouping, we get <math>2a(6-n)=3b(n-4)</math>. Since both <math>a,b</math> are positive, it follows that <math>6-n,n-4</math> are also positive, which is only possible when <math>n = 5\ \mathrm{(C)}</math>. | ||
== See also == | == See also == | ||
Revision as of 19:38, 9 February 2008
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
![\[\left(\frac{a}{6} + \frac{b}{4}\right)n = a + b\]](http://latex.artofproblemsolving.com/2/d/e/2def09d0fb42ffc9d48d1dee708d49e68be77dbb.png)
Regrouping, we get 
. Since both 
are positive, it follows that 
are also positive, which is only possible when 
.
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 | |
