2000 AMC 12 Problems/Problem 13

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

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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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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2000 AMC 12 Problems/Problem 13

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