Difference between revisions of "2013 AMC 12B Problems/Problem 1"

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==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>
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==Video Solution==
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https://youtu.be/OsnepwGjcw8
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 +
~savannahsolver
  
 
== See also ==
 
== See also ==

Latest revision as of 22:10, 21 September 2020

The following problem is from both the 2013 AMC 12B #1 and 2013 AMC 10B #3, so both problems redirect to this page.

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 low temperatures was $3\textdegree$. 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 $L=\boxed{\textbf{(C)} \  -5}$

Video Solution

https://youtu.be/OsnepwGjcw8

~savannahsolver

See also

2013 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2013 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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