2004 AMC 8 Problem 9

Attempt Problem 9 of the 2004 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

9.

The average of the five numbers in a list is 54.54. The average of the first two numbers is 48.48. What is the average of the last three numbers?
5555
5656
5757
5858
5959
Answer: D
Concepts:mean
Small hint:
Turn each average into a sum.
Big hint:
Subtract the sum of the first two numbers from the sum of all five.
Solution:
The sum of all 55 numbers is 54⋅5=270.54 \cdot 5 = 270. The sum of the first 22 numbers is 48⋅2=96.48 \cdot 2 = 96.
The sum of the last 33 numbers is 270−96=174.270 - 96 = 174. The average is therefore 174÷3=58.174 \div 3 = 58.
Thus, D is the correct answer.
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