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| − | ==Problem==
| + | #REDIRECT [[2013 AMC 12B Problems/Problem 1]] |
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| − | On a particular January day, the high temperature in Lincoln, Nebraska, was <math>16</math> degrees higher than the low temperature, and the average of the high and the low temperatures was <math>3\,^\circ</math>. In degrees, what was the low temperature in Lincoln that day?
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| − | <math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ -5\qquad\textbf{(D)}\ -3\qquad\textbf{(E)}\ 11</math>
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| − | ==Solution==
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| − | Let <math>L</math> be the low temperature. The high temperature is <math>L+16</math>. The average is <math>\frac{L+(L+16)}{2}=3</math>. Solving for <math>L</math>, we get <math>L=\boxed{\textbf{(C) } -5}</math>
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| − | == See also ==
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| − | {{AMC10 box|year=2013|ab=B|num-b=2|num-a=4}}
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