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

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<math>\textbf{(A)}\ -13 \qquad\textbf{(B)}\ \-8  \qquad\textbf{(C)}\ \-5 \qquad\textbf{(D)}\ \-3 \qquad\textbf{(E)}\ \111</math>
 
<math>\textbf{(A)}\ -13 \qquad\textbf{(B)}\ \-8  \qquad\textbf{(C)}\ \-5 \qquad\textbf{(D)}\ \-3 \qquad\textbf{(E)}\ \111</math>
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==Solution==
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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>

Revision as of 15:09, 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 $3\,^{\circ}\mathrm{}$. 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 $L$ be the low temperature. The average is $\frac{L+L+16}{2}=-3$. Solving for $L$, we get $L=\boxed{\textbf{(C) } -5}$