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

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2010 AMC 12B Problem 10

Below is the professionally curated solution for Problem 10 of the 2010 AMC 12B, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:mean summation linear equation

Difficulty rating: 1410

10.

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

$\dfrac{49}{101}$

$\dfrac{50}{101}$

$\dfrac{1}{2}$

$\dfrac{51}{101}$

$\dfrac{50}{99}$

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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