AMC10 2021 B

AMC10 2021 B · Q19

AMC10 2021 B · Q19. It mainly tests Linear equations, Systems of equations.

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合。如果从 $S$ 中移除集合中的最大整数，则剩余整数的平均值（算术平均）是 $32$。如果再移除集合中的最小整数，则剩余整数的平均值是 $35$。如果将最大整数放回集合，则平均值上升到 $40$。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 = $\boxed{\textbf{(D) }36.8}$.

我们可以设 $A+S(n)n+1=32(n+1)$，更准确地说，让 $n$ 是原始集合的大小，$A$ 是最小，$B$ 是最大，$S(n)$ 是其余 $n$ 个数的和，则： $A + S(n) = 32(n+1)$， $S(n) = 35n$， $B + S(n) = 40(n+1)$。代入 $S(n)=35n$ 得 $A + 35n = 32n + 32$ 所以 $A = -3n + 32$， $B + 35n = 40n + 40$ 所以 $B = 5n + 40$。原始总和是 $S(n) + A + B = 35n + (-3n + 32) + (5n + 40) = 37n + 72$，平均值 $Z = \frac{37n + 72}{n+2}$。此外 $B - A = 72$，即 $(5n + 40) - (-3n + 32) = 8n + 8 = 72$，所以 $8n = 64$，$n=8$。则 $Z = \frac{37\times8 + 72}{10} = \frac{296 + 72}{10} = \frac{368}{10} = 36.8$。所以答案是 $\boxed{\textbf{(D) }36.8}$。

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