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Answer: $7\frac12$


Comparing Harry and Christine with and without Betty
Harry and Betty together pack $12$ boxes per hour.
Christine and Betty together pack $9$ boxes per hour.

Therefore, in one hour, Harry packs $3$ more boxes than Christine.

In one hour, Harry and Christine together pack $18$ boxes.

Since Harry packs 3 more than Christine, Harry packs $10\frac{1}{2}$ boxes and Christine packs $7\frac{1}{2}$ boxes in an hour.


Using algebra
Say that Harry packs $H$ boxes per hour, Christine packs $C$ boxes per hour and Betty packs $B$ boxes per hour.

$2H+2C=36$ so $H+C=18$.
$3H+3B=36$ so $H+B=12$.
$4C+4B=36$ so $C+B=9$.

Adding all three equations together gives $2H+2C+2B=39$,
so $H+C+B=19 \frac12$.

Since we know that $H+B=12$,
$C=19\frac12 - 12 = 7\frac12$.


This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.