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# Packing Boxes

**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 $H+B=12$,

$C=19\frac12 - 12 = 7\frac12$.

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Age 14 to 16

ShortChallenge Level

- Problem
- Solutions

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.

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.