2008 AMC 8 Problem 10 — Solution | LIVE by Po-Shen Loh

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2008 AMC 8 Problem 10

Attempt Problem 10 of the 2008 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 2008 AMC 8 solutions, or check the answer key.

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

10.

The average age of the $6$ people in Room A is $40.$ The average age of the $4$ people in Room B is $25.$ If the two groups are combined, what is the average age of all the people?

$\ 32.5$

$\ 33$

$\ 33.5$

$\ 34$

$\ 35$

Answer: D

Concepts:mean

Difficulty rating: 960

Solution:

The sum of the ages in Room A is $6\cdot 40 = 240.$

The sum of the ages in Room B is $4\cdot 25 = 100.$

The total sum is $240+100 = 340.$

The average is therefore $\dfrac{340}{10} =34.$

Thus, the answer is D .

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