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

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==Problem==
 
==Problem==
  
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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 <math>3\,^{\circ}\mathrm{}</math>. In degrees, what was the low temperature in Lincoln that day?
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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 <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)}\ \111</math>
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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>
  
 
==Solution==
 
==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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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 16:06, 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 $3^\circ$. In degrees, what was the low temperature in Lincoln that day?

$\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ -5\qquad\textbf{(D)}\ -3\qquad\textbf{(E)}\ \ 11$

Solution

Let $L$ be the low temperature. The high temperature is $L+16$. The average is $\frac{L+(L+16)}{2}=3$. Solving for $L$, we get $L=\boxed{\textbf{(C) } -5}$

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