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

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==Problem==
 
==Problem==
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On a particular January day, the high temperature in Lincoln, Nebraska, was <math>16</math> degrees higher than the low temperature, and the average of the high and low temperatures was <math>3\textdegree</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 <math>16</math> degrees higher than the low temperature, and the average of the high and low temperatures was <math>3</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)}\ 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==
 
==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 22:30, 15 April 2022

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 $3$. 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 (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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