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Problem 164

Algebra Difficulty 1.5 Multiple choice CEMC Cayley · Canada · 2024

Ten numbers have an average (mean) of $87$ . Two of those numbers are $51$ and $99$ . The average of the other eight
numbers is

Pick one

Official solution

Since $10$ numbers have an
average of $87$ , their sum is $10 \times 87 = 870$ .

When the numbers $51$ and $99$ are removed, the sum of the remaining
$8$ numbers is $870 - 51 - 99$ or $720$ .

The average of these $8$ numbers is
$\dfrac{720}{8} = 90$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.