Difference between revisions of "2013 AMC 12B Problems/Problem 1"
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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> | <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== | ||
| + | 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 16:33, 22 February 2013
Problem
On a particular January day, the high temperature in Lincoln, Nebraska, was 
degrees higher than the low temperature, and the average of the high and low temperatures was 
. In degrees, what was the low temperature in Lincoln that day?
Solution
Let 
be the low temperature. The high temperature is 
. The average is 
. Solving for , we get 
