The question is:
The average number of students per class at school X is 25 and the average number of students per class at school Y is 33. Given that there are more classes in school X than in school Y, is the average number of students per class for both schools combined less than 29?
My attempt:
Let $S_x$ denote the total students at school X and let $n_x$ denote the number of classes in school X. Then we have the average number of students per class at school X to be $A_x=S_x/n_x$, similarly, $A_y = S_y/n_y$. We know that $n_x>n_y$. We wish to find out whether $\frac{S_x+S_y}{n_x + n_y}<29$, i.e., $\frac{25n_x + 33n_y}{n_x+n_y} < 29$. Now it just so happens that 29 is the average of 25 and 33, i.e., $\frac{25+33}{2} = 29$, so essentially, we need to confirm whether \begin{align} 25\frac{n_x}{n_x+n_y} + 33\frac{n_y}{n_x+n_y} < 25\left(\frac{1}{2}\right) + 33\left(\frac{1}{2} \right) \end{align}
I have a feeling that the answer is "YES" since more "weight" is placed on 25 since $n_x > n_y$, so the weighted average on the LHS above is always smaller than the "normal" average. However, how can I "prove" this mathematically?
Arising from this question and more broadly speaking:
1) Under what conditions is the weighted average $x_1\frac{w_1}{w_1+w_2} + x_2\frac{w_2}{w_1+w_2} < \frac{x_1+x_2}{2}$?
2) What about for the more general expression, under what conditions is $$x_1\frac{w_1}{\sum_{i=1}^n w_i} + x_2\frac{w_2}{\sum_{i=1}^n w_i} + \cdots + x_n\frac{w_n}{\sum_{i=1}^n w_i} < \frac{x_1+x_2+\cdots+x_n}{n}$$?
3) Is it a coincidence that the question gave the number 29? Would the answer be "NO" had the question given any number less than 29?
Best Answer
Let N1, N2 be number of classes in the first and the second schools respectively. Then number of students in the first school is 25 N1 and in the second school is 33 N2, so the average students per class is (25 N1 + 33 N2)/ (N1 + N2) Let's see when it is < 29:
(25 N1 + 33 N2)/ (N1 + N2) < 29
25 N1 + 33 N2 < 29 (N1 + N2) = 29 N1 + 29 N2
4N2 < 4 N1
N2 < N1
As per more general question, a "simple" average implies equal weights,if the weight of a "lower" part is higher, the weighted average will be less than the simple average.