2010 AMC 10B Problem 4 — Solution | LIVE by Po-Shen Loh

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2010 AMC 10B Problem 4

Attempt Problem 4 of the 2010 AMC 10B 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 2010 AMC 10B solutions, or check the answer key.

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

4.

For a real number $x,$ define $\heartsuit(x)$ to be the average of $x$ and $x^2.$ What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)?$

$3$

$6$

$10$

$12$

$20$

Answer: C

Concepts:custom operation mean

Difficulty rating: 870

Solution:

We have $\heartsuit(1) = \dfrac{1 + 1^2}{2} = 1,$ $\heartsuit(2) = \dfrac{2 + 2^2}{2} = 3,$ and $\heartsuit(3) = \dfrac{3 + 3^2}{2} = 6.$

Then $1 + 3 + 6 = 10.$

Thus, C is the correct answer.

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