2014 AMC 10A Problem 10

Below is the professionally curated solution for Problem 10 of the 2014 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 10A solutions, or check the answer key.

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Concepts:arithmetic sequencemean

Difficulty rating: 900

10.

Five positive consecutive integers starting with aa have average b.b. What is the average of 55 consecutive integers that start with b?b?

a+3a+3

a+4a+4

a+5a+5

a+6a+6

a+7a+7

Solution:

Note that the average of 55 consecutive numbers starting with xx is 5x+1+2+3+45=x+2. \dfrac{5x + 1 + 2 + 3 + 4}{5} = x + 2.

This means that the average of 55 consecutive integers starting with aa is a+2,a + 2, which we know is b.b.

Furthermore, the average of 55 consecutive numbers starting with bb is b+2=a+4.b + 2 = a + 4.

Thus, B is the correct answer.

Problem 10 in Other Years