2000 AMC 10 Problems/Problem 22

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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?

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

Solution

The exact value "8 ounces" is not important. We will only use the fact that each member of the family drank the same amount.

Let $m$ be the total number of ounces of milk drank by the family and $c$ the total number of ounces of coffee. Thus the whole family drank a total of $m+c$ ounces of fluids.

Let $n$ be the number of family members. Then each family member drank $\frac {m+c}n$ ounces of fluids.

We know that Angela drank $\frac m4 + \frac c6$ ounces of fluids.

As Angela is a family member, we have $\frac m4 + \frac c6 = \frac {m+c}n$ .

Multiply both sides by $n$ to get $n\cdot\left( \frac m4 + \frac c6 \right) = m+c$ .

If $n\leq 4$ , we have $n\cdot\left( \frac m4 + \frac c6 \right) \leq 4\cdot \left( \frac m4 + \frac c6 \right) = m + \frac{2c}3 < m+c$ .

If $n\geq 6$ , we have $n\cdot\left( \frac m4 + \frac c6 \right) \geq 6\cdot \left( \frac m4 + \frac c6 \right) = \frac{3m}2 + c > m+c$ .

Therefore the only remaining option is $n=\boxed{5}$ .

2000 AMC 10 (Problems • 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

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2000 AMC 10 Problems/Problem 22

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Problem

Solution

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