Difference between revisions of "2000 AMC 12 Problems/Problem 13"
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== Solution == | == Solution == | ||
− | Let <math> | + | Let <math>c</math> be the total amount of coffee, <math>m</math> of milk, and <math>p</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{ | + | <cmath>\left(\frac{c}{6} + \frac{m}{4}\right)p = c + m</cmath> |
− | Regrouping, we get <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>. |
== See also == | == See also == |
Revision as of 15:59, 14 June 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
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 |
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 |