2013 AMC 10B Problem 3

Below is the professionally curated solution for Problem 3 of the 2013 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 10B solutions, or check the answer key.

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Concepts:mean linear equation

Difficulty rating: 560

On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ degrees. What was the low temperature in Lincoln that day (in degrees)?

$-13$

$-8$

$-5$

$-3$

$11$

Solution:

Let the low temperature be represented by $l.$ Then, the high temperature is $l+16.$

This makes the average $\dfrac{l+l+16}2 = l+8 = 3.$ Therefore, $l = -5.$

Thus, the correct answer is C .

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