Question

在至少选修一门课程的 $28$ 名学生中，只选数学和英语的人数等于只选数学的人数。没有学生只选英语或只选历史，有六名学生选数学和历史但不选英语。只选英语和历史的人数是三门都选人数的五倍。若三门都选的人数为非零偶数，则只选英语和数学的人数为：

Of $28$ students taking at least one subject the number taking Mathematics and English only equals the number taking Mathematics only. No student takes English only or History only, and six students take Mathematics and History, but no English. The number taking English and History only is five times the number taking all three subjects. If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:

$5$

$6$

$7$

$8$

$9$

答案：A

知识点：韦恩图基本计数丢番图方程

难度评级：1830

小提示：

设三门都选的人数为 $t$ ，只选数学和只选数学英语的人数均为 $x$

Let $t$ be the all-three count and $x$ both the Math-only and Math-English-only count

大提示：

总人数可写成 $2x+6+6t=28$

The total becomes $2x+6+6t=28$

解答：

设三门都选的人数为 $t$ ，只选数学和英语的人数为 $x$ ，它也等于只选数学的人数。各不相交区域的人数之和为 $x+x+6+5t+t=28\text{，}$ 所以 $x=11-3t$ 。因为 $t$ 为正偶数，所以只有 $t=2$ 能使所列人数非负，此时 $x=5$ 。

因此，正确答案是 A。

Let $t$ be the number taking all three and $x$ the number taking Mathematics and English only, which also equals the Mathematics-only count. The disjoint-region total is $x+x+6+5t+t=28,$ so $x=11-3t.$ Since $t$ is positive and even, $t=2$ is the only value giving a nonnegative listed count, and $x=5.$

Therefore, the correct answer is A.

1965 AMC 12 第 29 题

答案：A

难度评级：1830

小提示：

大提示：

解答：

其他年份的第 29 题