Difference between revisions of "2013 AMC 10B Problems/Problem 3"
m (→Problem) |
|||
| Line 1: | Line 1: | ||
==Problem== | ==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 <math>3^\circ</math>. In degrees, what was the low temperature in Lincoln that day? | + | 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? |
| − | <math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ -5\qquad\textbf{(D)}\ -3\qquad\textbf{(E)} | + | <math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ -5\qquad\textbf{(D)}\ -3\qquad\textbf{(E)}\ 11</math> |
==Solution== | ==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> | 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 16:22, 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?
Solution
Let 
be the low temperature. The high temperature is 
. The average is 
. Solving for , we get 
