2010 AMC 10B Problem 14

Below is the professionally curated solution for Problem 14 of the 2010 AMC 10B, 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.

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Concepts:meanarithmetic sequencelinear equation

Difficulty rating: 1370

14.

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

49101\dfrac{49}{101}

50101\dfrac{50}{101}

12\dfrac{1}{2}

51101\dfrac{51}{101}

5099\dfrac{50}{99}

Solution:

Recall that the sum of the first nn integers is n(n+1)2.\dfrac{n(n + 1)}{2}.

Then, we have that 991002+x100=100x, \dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x, which simplifies to 9950=(10021)x 99 \cdot 50 = (100^2 - 1)x=10199x, = 101 \cdot 99x, by difference of squares. Dividing gives us x=50101.x = \dfrac{50}{101}.

Thus, B is the correct answer.

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