Difference between revisions of "2013 AMC 10B Problems/Problem 3"

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==Problem==
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#REDIRECT [[2013 AMC 12B Problems/Problem 1]]
 
 
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?
 
 
 
<math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ -5\qquad\textbf{(D)}\ -3\qquad\textbf{(E)}\ 11</math>
 
 
 
==Solution==
 
 
 
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>
 

Latest revision as of 11:08, 7 April 2013