2010 AMC 10B Problem 14

Attempt Problem 14 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).

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}

Answer: B
Concepts:meanarithmetic sequencelinear equation
Difficulty rating: 1370
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.

← Problem 13#13
Full Exam

Problem 14 in Other Years