2010 AMC 12B Problem 10 — Solution | LIVE by Po-Shen Loh

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2010 AMC 12B Problem 10

Attempt Problem 10 of the 2010 AMC 12B 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 12B solutions, or check the answer key.

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

10.

The average of the numbers $1, 2, 3, \ldots, 98, 99,$ and $x$ is $100x.$ What is $x?$

$\dfrac{49}{101}$

$\dfrac{50}{101}$

$\dfrac{1}{2}$

$\dfrac{51}{101}$

$\dfrac{50}{99}$

Answer: B

Concepts:mean summation linear equation

Difficulty rating: 1410

Solution:

The numbers $1$ through $99$ sum to $\dfrac{99\cdot100}{2}=4950.$

The average condition is $\frac{4950+x}{100}=100x,$ so $4950+x=10000x$ and $9999x=4950.$

Thus $x=\dfrac{4950}{9999}=\dfrac{50}{101}.$

Thus, the correct answer is B.

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