AMC8 2019

AMC8 2019 · Q7

AMC8 2019 · Q7. It mainly tests Linear equations, Averages (mean).

Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ , $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?

Shauna 参加了五次考试，每次满分 $100$ 分。前三次考试成绩分别是 $76$ 、 $94$ 和 $87$ 。为了五次考试平均分达到 $81$ 分，她在剩下两次考试中最低可能得分是多少？

(A) 48 48

(B) 52 52

(D) 70 70

(E) 74 74

Answer

Correct choice: (A)

正确答案：（A）

Solution

We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] We can now cross multiply to get rid of the denominator. \[257+x+y = 405,\] \[x+y = 148.\] Now that we have this equation, we will assign $y$ as the lowest score of the two other tests, and so: \[x = 100,\] \[y=48.\] Now we know that the lowest score on the two other tests is $\boxed{\textbf{(A)}\ 48}$.

我们可以将已知信息转化为一次方程。用变量 $x$ 和 $y$ 表示最后两次考试的得分。\[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] 交叉相乘去掉分母：\[257+x+y = 405,\] \[x+y = 148.\] 设 $y$ 为两次考试中最低得分，令 $x=100$，则 $y=48$。因此剩下两次考试的最低得分是 $\boxed{\textbf{(A)}\ 48}$。

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