Equal Means
What value of x makes the mean of the first three numbers in this list equal to the mean of the other four?
What value of $x$ makes the mean of the first three numbers in this list equal to the mean of the last four?$$15\qquad 5 \qquad x \qquad 7 \qquad 9 \qquad 17 \qquad$$
Using trial and improvement:
The first two numbers, 15 and 5, add to 20.
The last three numbers, 17, 7 and 9, add to 33.
This means that $x$ added onto 20 and 33 makes two different totals that have the same mean.
If you assume that the mean will be a whole number you can try numbers, that when added to 20 give a total that is divisible by 3 (e.g. 1, 4, 7, 10...) and when added to 33 give a total that is divisible by 4 (e.g. 3, 7, 11, 15, ...)
After some trials you may find that the number you are looking for is 19.
Noticing difference:
The first three numbers must add up to 3 times the mean, and the last four numbers must add up to 4 times the mean, so the difference between their sums will be the mean.
5 + 15 + x = 20 + x
7 + 9 + 17 + x = 33 + x
(33 + x) - (20 + x) = 13
Therefore 13 is the mean of both groups of numbers.
3 x 13 = 39
4 x 13 = 52
Therefore the first three numbers must add to 39, and the last four numbers to 52.
39 - 20 = 19
52 - 33 = 19
Therefore 19 is the missing number.
Using algebra:
The mean of the first three numbers is $\dfrac{20 + x}{3}$
and the mean of the last four numbers is $\dfrac{33 + x}{4}$
Both means are equal, so:
$$\frac{20 + x}{3}=\frac{33+ x}{4}$$
Multiplying both sides by $12$:
$$80 + 4x = 99 + 3x$$
Subtracting $80$ from both sides:
$$4x = 19 + 3x$$
Subtracting $3x$ from both sides:
$$x = 19$$
Therefore, the missing number is $19$.
When we plug it in into the original question, we find that the mean of the first three numbers is $13$ and so is the mean of the last four numbers.