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| − | ==Problem==
| + | #REDIRECT [[2000 AMC 12 Problems/Problem 13]] |
| − | 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?
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| − | <math>\mathrm{(A)}\ 3 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7</math>
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| − | ==Solution==
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| − | The exact value "8 ounces" is not important. We will only use the fact that each member of the family drank the same amount.
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| − | Let <math>m</math> be the total number of ounces of milk drank by the family and <math>c</math> the total number of ounces of coffee. Thus the whole family drank a total of <math>m+c</math> ounces of fluids.
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| − | Let <math>n</math> be the number of family members. Then each family member drank <math>\frac {m+c}n</math> ounces of fluids.
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| − | We know that Angela drank <math>\frac m4 + \frac c6</math> ounces of fluids.
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| − | As Angela is a family member, we have <math>\frac m4 + \frac c6 = \frac {m+c}n</math>.
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| − | Multiply both sides by <math>n</math> to get <math>n\cdot\left( \frac m4 + \frac c6 \right) = m+c</math>.
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| − | If <math>n\leq 4</math>, we have <math>n\cdot\left( \frac m4 + \frac c6 \right) \leq 4\cdot \left( \frac m4 + \frac c6 \right) = m + \frac{2c}3 < m+c</math>.
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| − | If <math>n\geq 6</math>, we have <math>n\cdot\left( \frac m4 + \frac c6 \right) \geq 6\cdot \left( \frac m4 + \frac c6 \right) = \frac{3m}2 + c > m+c</math>.
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| − | Therefore the only remaining option is <math>n=\boxed{5}</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2000|num-b=21|num-a=23}}
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