AMC12 2004 B

AMC12 2004 B · Q11

AMC12 2004 B · Q11. It mainly tests Linear equations, Averages (mean).

All the students in an algebra class took a $100$-point test. Five students scored $100$, each student scored at least $60$, and the mean score was $76$. What is the smallest possible number of students in the class?

代数班的所有学生参加了一场$100$分的测试。有五名学生得了$100$分，每位学生得分至少$60$分，平均分为$76$分。班级中最少可能有多少名学生？

(A) 10 10

(B) 11 11

(D) 13 13

(E) 14 14

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let the number of students be $n\geq 5$. Then the sum of their scores is at least $5\cdot 100 + (n-5)\cdot 60$. At the same time, we need to achieve the mean $76$, which is equivalent to achieving the sum $76n$. Hence we get a necessary condition on $n$: we must have $5\cdot 100 + (n-5)\cdot 60 \leq 76n$. This can be simplified to $200 \leq 16n$. The smallest integer $n$ for which this is true is $n=13$. To finish our solution, we now need to find one way how $13$ students could have scored on the test. We have $13\cdot 76 = 988$ points to divide among them. The five $100$s make $500$, hence we must divide the remaining $488$ points among the other $8$ students. This can be done e.g. by giving $61$ points to each of them. Hence the smallest possible number of students is $\boxed{\mathrm{(D)}\ 13}$.

设学生人数为$n\geq 5$。则他们的总分至少为$5\cdot 100 + (n-5)\cdot 60$。同时，要使平均分为$76$，等价于总分为$76n$。因此$n$必须满足必要条件：$5\cdot 100 + (n-5)\cdot 60 \leq 76n$。化简得$200 \leq 16n$。满足该不等式的最小整数$n$为$n=13$。为完成解答，我们还需给出一种$13$名学生可能的得分方式。总分为$13\cdot 76 = 988$。其中五个$100$分共$500$分，因此剩余$488$分需由另外$8$名学生分配。例如，每人$61$分即可。因此，最少可能的学生人数是$\boxed{\mathrm{(D)}\ 13}$。

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