AMC12 2021 B

AMC12 2021 B · Q12

AMC12 2021 B · Q12. It mainly tests Linear equations, Averages (mean).

Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$, then the average value (arithmetic mean) of the integers remaining is $32$. If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$. If the greatest integer is then returned to the set, the average value of the integers rises to $40$. The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$. What is the average value of all the integers in the set $S$?

假设 $S$ 是一个有限的正整数集。如果从 $S$ 中移除 $S$ 中的最大整数，则剩余整数的平均值（算术平均）为 $32$。如果再移除 $S$ 中的最小整数，则剩余整数的平均值为 $35$。如果将最大整数放回集合，则整数的平均值上升到 $40$。原集合 $S$ 中的最大整数比最小整数大 $72$。集合 $S$ 中所有整数的平均值是多少？

(A) 36.2 36.2

(B) 36.4 36.4

(D) 36.8 36.8

(E) 37 37

Answer

Correct choice: (D)

正确答案：（D）

Solution

We can then say that A+S(n)n+1=32, S(n)n=35, and B+S(n)n+1=40. Expanding gives us A+S(n)=32n+32, S(n)=35n, and B+S(n)=40n+40. Substituting S(n)=35n to all gives us A+35n=32n+32 and B+35n=40n+40. Solving for A and B gives A=−3n+32 and B=5n+40. We now need to find S(n)+A+Bn+2. We substitute everything to get 35n+(−3n+32)+(5n+40)n+2, or 37n+72n+2. Say that the answer to this is Z. Then, Z needs to be a number that makes n a positive integer. The only options that work is $\boxed{\textbf{(C) }36.6}$ and $\boxed{\textbf{(D) }36.8}$. However, if 36.6 is an option, we get n=3. So that means that A is 23 and B is 55, and S(n)=105. But if there is 3 terms, then the middle number is 105, but we said that B is the largest number in the set, so therefore our answer cannot be $\boxed{\textbf{(C) }36.6}$ and is instead $\boxed{\textbf{(D) }36.8}$ and now, we're finished! If A is smaller than B by 72 therefore from the equation on the top you can find out that N=8 using substitution the plug it in to the equation 37n+72n+2 then you will get that Z = $\boxed{\textbf{(D) }36.8}$.

我们可以表示为 $A+S(n)n+1=32, S(n)n=35,$ 和 $B+S(n)n+1=40$。展开得 $A+S(n)=32n+32, S(n)=35n,$ 和 $B+S(n)=40n+40$。代入 $S(n)=35n$ 得 $A+35n=32n+32$ 和 $B+35n=40n+40$。解得 $A=-3n+32$ 和 $B=5n+40$。现在需要找到 $S(n)+A+Bn+2$。代入得 $35n+(-3n+32)+(5n+40)n+2$，即 $37n+72n+2$。假设答案为 $Z$。则 $Z$ 需使 $n$ 为正整数。唯一可行的选项是 $\boxed{\textbf{(C) }36.6}$ 和 $\boxed{\textbf{(D) }36.8}$。但是，若选 $36.6$，则 $n=3$。这意味着 $A=23$，$B=55$，$S(n)=105$。但有 $3$ 个项，则中间数为 $105$，但我们说 $B$ 是集合中的最大数，因此答案不能是 $\boxed{\textbf{(C) }36.6}$，而是 $\boxed{\textbf{(D) }36.8}$，现在完成了！由于 $A$ 比 $B$ 小 $72$，因此从上方方程可知 $N=8$，代入 $37n+72n+2$ 得 $Z = \boxed{\textbf{(D) }36.8}$。

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