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> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/OsnepwGjcw8 | ||
| + | |||
| + | ~savannahsolver | ||
== See also == | == See also == | ||
Revision as of 21:10, 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.
Contents
Problem
On a particular January day, the high temperature in Lincoln, Nebraska, was 
degrees higher than the low temperature, and the average of the high and 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 
Video Solution
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
