Question

The average of 17 numbers is 10.9. If the average of the first nine numbers is 10.5 and that of the last nine numbers is 11.4, the middle number is.

11.8
11.4
10.9
11.7

Kindly, tell me how this question will be solved, please.

Thank you.

Answer 1

Assume that we have a set of real positive numbers:

$$a_1, a_2,a_3,...,a_9,...,a_{17}.$$

Of course, $a_9$ is the middle number.

The average of all 17 is

$$10.9=\frac{\sum_{i=1}^{17}a_i}{17}\leftrightarrow\sum_{i=1}^{17}a_i=17\cdot10.9=185.3.$$

The average of the first nine numbers is

$$10.5=\frac{\sum_{i=1}^{9}a_i}{9}=\frac{\sum_{i=1}^{8}a_i+a_9}{9}.$$

The average of the last nine numbers is 1/9th of the sum of all 17 minus the sum of the first 8:

$$11.4=\frac{\big(\sum_{i=1}^{17}a_i-\sum_{i=1}^{8}a_i\big)}{9}.$$

Look at the last two equations: we have all we need. The sum of all 17 is 185.3, now put

$$\sum_{i=1}^{8}a_i=x,\\ a_9=y.$$

The two equations become a system with two undefined variables:

$$11.4=\frac{\big(185.3-x\big)}{9},\\\space\\ 10.5=\frac{x+y}{9}.$$

The solution for this simple system is

$$x=82.7,\\ y=11.8.$$

Thus, $a_9=y=11.8.$