Maths Olympiad Prep

Library / /54 of 144

Algebra Difficulty 2.1 Junior Find the answer

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

A number or a short expression. Spacing and $ signs are ignored.

Solution

Since 10 numbers have an average of 87, their sum is 10×87=87010 \times 87 = 870. When the numbers 51 and 99 are removed, the sum of the remaining 8 numbers is 870−51−99870 - 51 - 99 or 720. The average of these 8 numbers is 7208=90\frac{720}{8} = 90.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.