AMC12 2011 A

AMC12 2011 A · Q4

AMC12 2011 A · Q4. It mainly tests Linear equations, Weighted average.

At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of $12$, $15$, and $10$ minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?

在一所小学，三年级、四年级和五年级学生每天平均跑步时间分别为$12$分钟、$15$分钟和$10$分钟。三年级学生人数是四年级学生人数的两倍，四年级学生人数是五年级学生人数的两倍。这些学生每天平均跑步多少分钟？

(A) 12 12

(B) \frac{37}{3} \frac{37}{3}

(D) 13 13

(E) 14 14

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let us say that there are $f$ fifth graders. According to the given information, there must be $2f$ fourth graders and $4f$ third graders. The average time run by each student is equal to the total amount of time run divided by the number of students. This gives us $\frac{12\cdot 4f + 15\cdot 2f + 10\cdot f}{4f + 2f + f} = \frac{88f}{7f} = \frac{88}{7} \Rightarrow \boxed{C}$ If you want to simplify the problem even more, just imagine/assume that only $1$ fifth grader existed. Then you can simply get rid of the variables.

设五年级学生有$f$人。根据题意，四年级学生有$2f$人，三年级学生有$4f$人。每位学生的平均跑步时间等于总跑步时间除以学生总人数，因此 $\frac{12\cdot 4f + 15\cdot 2f + 10\cdot f}{4f + 2f + f} = \frac{88f}{7f} = \frac{88}{7} \Rightarrow \boxed{C}$ 如果想进一步简化，可以不妨设只有$1$个五年级学生，这样就可以直接去掉变量。

Topics