An average weight is calculated after each player is weighed. Of the average weight increases by 2 pounds after each player is weighed, how much is the last player than the first?
My answer is 8 pounds. I first mark the five members with letters a to e. Since the difference in weight between a and c is 4 and the difference between c and e is also 4, what I have to do is just add the two 4s up, and get 8 as mu final solution.but I know the answer isn't 8. Could you show me a full algebraic work.
On
[Edit:] No, ignore this. ah11950's correct. I didn't read the question properly.
No, that's it.
The average weight change is the total of the weight changes divided by the number of weight changes. So the total of the weight changes is the number of weight changes multiplied by the average weight change.
${\bar {w} = \dfrac{\sum\limits_{\forall i} w_i}{n} \iff \sum\limits_{\forall i} w_i = n \bar {w}}$
The only trick here is recognising that there are but four weight changes between the first and fifth player.$$\text{Answer} = 4 \times 2 = 8$$
Let's number the players 1 through to 5, and for $1\leq i \leq 5$, let $m_i$ be the weight of player $i$.
We know that $\frac{m_1+m_2}{2} = m_1 + 2$ (the average weight of just one player is just his weight) and hence $m_1+m_2 = 2m_1+4$ so $m_2 = m_1+4$.
Similarly, $\frac{2m_1+4+m_3}{3} = \frac{m_1+m_2+m_3}{3} = \frac{m_1+m_2}{2}+ 2 = m_1 + 4$ so $m_3 = m_1+8$.
Continuing this way and substituting $m_i = m_1 + k$ once we know how much heavier player $i$ is than player 1, we find:
$m_4 = m_1+12$ and $m_5=m_1+16$, so the last player is 16 lbs heavier than the first.
There's a more elegant way to do this, but the above is very explicit and clear.