Difference between revisions of "2013 AMC 10B Problems/Problem 3"
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| − | Let <math>L</math> be the low temperature. 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> | + | 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> |
Revision as of 15:10, 21 February 2013
Problem
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 the low temperatures was 
. In degrees, what was the low temperature in Lincoln that day?
$\textbf{(A)}\ -13 \qquad\textbf{(B)}\ \-8 \qquad\textbf{(C)}\ \-5 \qquad\textbf{(D)}\ \-3 \qquad\textbf{(E)}\ \111$ (Error compiling LaTeX. Unknown error_msg)
Solution
Let 
be the low temperature. The high temperature is 
. The average is 
. Solving for , we get 
