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2013 AMC 10B Problem 3

Problem 3 of 25EasierAlgebra

On a particular January day, the high temperature in Lincoln, Nebraska, was 1616 degrees higher than the low temperature, and the average of the high and low temperatures was 33 degrees. What was the low temperature in Lincoln that day (in degrees)?

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Solution

Let ll represent the low temperature. Then the high temperature is l+16.l+16. The average satisfies l+l+162=l+8=3. \dfrac{l+l+16}2 = l+8 = 3. Therefore, l=−5.l = -5. Thus, the correct answer is C.

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Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.