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The sum of all $64$ numbers is their mean times the number of numbers, which is $$64 \times 64 = 4096$$
Similarly the sum of the first $36$ numbers is $$36 \times 36 = 1296$$
Therefore the sum of the last $28$ numbers is $$4096 - 1296 = 2800$$
Therefore the mean of the last $28$ numbers is $$\frac{2800}{28}=100$$
Partial Means
Age 14 to 16
ShortChallenge Level 
The sum of all $64$ numbers is their mean times the number of numbers, which is $$64 \times 64 = 4096$$
Similarly the sum of the first $36$ numbers is $$36 \times 36 = 1296$$
Therefore the sum of the last $28$ numbers is $$4096 - 1296 = 2800$$
Therefore the mean of the last $28$ numbers is $$\frac{2800}{28}=100$$
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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