There are four hundred students in a class and their total age is 8000 years. Show that the total age of the oldest hundred is at least 2000 years.

Question

I'm pretty sure I'm supposed to use the pigeonhole principle, but not exactly sure how. It's evident that since 2000 is the average, it must be greater but I'm not sure how to show it.

Answer 1

Split the $400$ students into $4$ groups of $100$. By the pigeonhole principle, at least one of them must have a total age of at least $8000/4 = 2000$. But the oldest $100$ will have an age that is greater than or equal to any group of $100$, and you are done.

Answer 2

A sketch of a proof that $T_N,$ the total age of the $N$ oldest students is at least $20N.$

Consider the age of the $N$th oldest student, $A_N.$ There are two cases:

$A_N \geq 20.$ Conclude that $T_N \geq 20N.$

$A_N < 20.$ Conclude that $8000 - T_N < 20(400 - N)$ and therefore $T_N > 20N.$

Set $N=100$ for the proof required in the question.

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Note that this is a proof sketch. I do not claim it to be a complete proof. But if the idea is fundamentally unworkable (that is, if there is something that prevents doing a complete proof along these lines), I would appreciate if someone would point out specifically why it cannot work.

Answer 3

Let the number representing sum of ages of 100 youngest members = $n_1$.

Sum of next 100 hundred (101- 200= $n_2$

Sum of next hundred (201-300) =$n_3$

Sum of oldest membera = $n_4$

We know $n_4>n_3>n_2>n_1$

Therefore if we assume all $n_4>n_3>n_2>n_1$ <2000 , implies $n_4+n_3+n_2+n_1 < 8000$ (a contradiction) which means atleast one of them $\ge$ 2000. As $n_4$ is greatest so , it is $\ge$ 2000