AMC8 2014

AMC8 2014 · Q24

AMC8 2014 · Q24. It mainly tests Averages (mean), Optimization (basic).

One day the Beverage Barn sold $252$ cans of soda to $100$ customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?

一天，饮料店卖出了 $252$ 罐苏打水，共给 $100$ 位顾客，每位顾客至少买了一罐苏打水。当天每位顾客购买苏打水罐数的中位数最大可能是多少？

(A) 2.5 2.5

(B) 3.0 3.0

(D) 4.0 4.0

(E) 4.5 4.5

Answer

Correct choice: (C)

正确答案：（C）

Solution

In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are $100$ people, the median will be the average of the $50\text{th}$ and $51\text{st}$ largest amount of cans per person. To minimize the first $49$, they would each have one can. Subtracting these $49$ cans from the $252$ cans gives us $203$ cans left to divide among $51$ people. Taking $\frac{203}{51}$ gives us $3$ and a remainder of $50$. Seeing this, the largest number of cans the $50$th person could have is $3$, which leaves $4$ to the rest of the people. The average of $3$ and $4$ is $3.5$. Thus our answer is $\boxed{\text{(C) }3.5}$.

为了使中位数最大化，我们需要使前半部分的数字尽可能小。因为有 $100$ 位顾客，中位数将是第 $50$ 和第 $51$ 大的购买罐数的平均值。为了让前 $49$ 个尽可能小，它们各买一罐。从 $252$ 罐中减去这 $49$ 罐，剩下 $203$ 罐分给剩下的 $51$ 个人。计算 $\frac{203}{51}$ 得到 3 余 50。由此，第 $50$ 个人最多可买 3 罐，其余人买 4 罐。第 $50$ 和第 $51$ 个人的平均数是 $3$ 和 $4$ 的平均值，即 $3.5$。因此，答案是 $\boxed{\text{(C) }3.5}$。

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