Problem: $$2000-AMC12-13$$  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?   一天早上，$$Angela$$ 的每个家庭成员都喝了 8 美元一盎司的咖啡加牛奶混合物。 咖啡和牛奶的量因杯子而异，但从未为零。 安吉拉喝了牛奶总量的四分之一，咖啡总量的六分之一。 家里有多少人？
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: C
Analysis: 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 $$6-p$$ and $$p-4$$ are also positive, which is only possible when $$p=5$$ ($$\text{C}$$).  One could notice that (since there are only two components to the mixture) Angela must have more than her "fair share" of milk and less then her "fair share" of coffee in order to ensure that everyone has $$8$$ ounces. The "fair share" is $$\frac{1}{p}$$. So,  $$\frac{1}{6}\textless\frac{1}{p}\textless\frac{1}{4}$$  Which requires that $$p$$ be $$p=5$$ ($$\text{C}$$), since $$p$$ is a whole number.  Again, let $$c$$, $$m$$, sand $$p$$ be the total amount of coffee, total amount of milk, and number of people in the family, respectively. Then the total amount that is drunk is $$8p$$, and also $$c+m$$. Thus, $$c+m=8p$$, so $$m=8p-c$$ and $$c=8p-m$$.  We also know that the amount Angela drank, which is is $$\frac{c}{6}+\frac{m}{4}$$, equal to $$8$$ ounces,  thus $$\frac{c}{6}+\frac{m}{4}=8$$. Rearranging gives $$24p-c=96$$. Now notice that $$c\textgreater0$$ (by the problem statement).  In addition, $$m\textgreater0$$ so $$c=8p-m\textless8p$$. Therefore, $$0\textless c\textless8p$$ ,and so We know that so $$24p\textgreater24p-c\textgreater16p$$. We know that $$24p-c=96$$, so $$24p\textgreater96\textgreater16p$$. From the leftmost inequality, we get $$p\textgreater4$$, and from the rightmost inequality, we get $$p\textless6$$. The only possible value of $$p$$ is $$p=5$$ ($$\text{C}$$).  Let $$c$$, $$m$$, and $$p$$ be the total amount of coffee, total amount of milk, and number of people in the family, respectively. $$c$$ and $$m$$ obviously can\textquotesingle t be $$0$$. We know $$\frac{c}{6}+\frac{m}{4}=8$$ or $$3c+2m=96$$ and $$c+m=8p$$ or $$2c+2m=16p$$. Then, $$(2c+2m)+c=16p+c=96$$. Because $$16p$$ and $$96$$ are both divisible by $$16$$, $$c$$ must also be divisible by $$16$$. Let $$c=16k$$. Now, $$3(16k)+2m=2m=48k+2m=96k$$ can\textquotesingle t be $$0$$, otherwise $$c$$ is $$0$$, and $$k$$ can\textquotesingle t be $$2$$, otherwise $$m$$ is $$0$$. Therefore $$k$$ must be $$1$$, $$c=16$$ and $$m=24$$. $$c+m=16+24=40=8p$$. Therefore, $$p=5$$ ($$\text{C}$$).  Let $$m$$,$$c$$ be the total amounts of milk and coffee, respectively. In order to know the number of people, we first need to find the total amount of mixture $$t=m+c$$. We are given that $$\frac{m}{4}+\frac{c}{6}=8$$. Multiplying the equation by $$4$$ to get $$m+\frac{2}{3}x=(m+c)-\frac{1}{3}c=t-\frac{1}{3}c=32$$. Since $$\frac{1}{3}m\textgreater0$$, we have $$t\textgreater32$$.~ Now multiplying the equation by $$6$$ to get $$\frac{3}{2}m+c=(m+c)+\frac{1}{3}m=t+\frac{1}{3}m=48$$. Since, $$\frac{1}{3}m\textgreater0$$, we have $$t\textless48$$. Thus, $$32\textless t\textless48$$.  Since $$t$$ is a multiple of $$8$$, the only possible value for in that range is $$40$$. Therefore, there are $$\frac{40}{8}=5$$ people in Angela\textquotesingle s family. ($$\text{C}$$). 
Knowledge points: 课内体系->知识点->等式与不等式->等式->方程组的解集
美国AMC10/12->Knowledge Point->Algebra->Application->Linear Function Word Problems
Source: 2000年AMC12竞赛第13题