Average Discovery

Find the missing number if the mean, median and mode are all the same.

Algebraic Average

The mean of three numbers x, y and z is x. What is the mean of y and z?

For Richer for Poorer

Charlie has moved between countries and the average income of both has increased. How can this be so?

Jewellery Boxes

Age 14 to 16 Short Challenge Level:

Suppose that initially the jewels in $P$ are worth $(a + b + 5000)$,
and those in $Q$ are worth $(c + d + e)$.

The average value in $P$ at the start is $\frac {a + b + 5000}3$

After the jewel has been moved it is $\frac{a + b}2$.
Therefore:

\frac {a + b + 5000}3 = \frac{a + b }{2} - 1000

Multiplying by $6$ and collecting like terms gives $a + b = 16000$

The average value in $Q$ at the start is $\frac {c + d + e}3$

After the jewel has been moved, the average value is $\frac{c + d + e + 5000}4$
Therefore:

\frac {c + d + e}3 = \frac{c + d + e + 5000}{4} - 1000

Multiplying by $12$ and collecting like terms gives $c + d + e = 3000$.

Therefore the total value is
$$a + b + 5000 + c + d + e = 16000 + 5000 + 3000 = 24000$$

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