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Difference between revisions of "2000 AMC 12 Problems/Problem 13"

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== Solution ==
 
== Solution ==
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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  
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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{a}{6} + \frac{b}{4}\right)n = a + b</cmath>
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<cmath>\left(\frac{c}{6} + \frac{m}{4}\right)p = c + m</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>.  
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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 14:59, 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?

$\text {(A)}\ 3 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 6 \qquad \text {(E)}\ 7$

Solution

Let $c$ be the total amount of coffee, $m$ of milk, and $p$ the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so \[\left(\frac{c}{6} + \frac{m}{4}\right)p = c + m\] Regrouping, we get $2c(6-p)=3m(p-4)$. Since both $c,m$ are positive, it follows that $6-p,p-4$ are also positive, which is only possible when $p = 5\ \mathrm{(C)}$.

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
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