Difference between revisions of "2012 AMC 10A Problems/Problem 3"
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Crawling from <math>-2</math> to <math>-6</math> takes it a distance of <math>4</math> units. Crawling from <math>-6</math> to <math>5</math> takes it a distance of <math>11</math> units. Add <math>4</math> and <math>11</math> to get <math>\boxed{\textbf{(E)}\ 15}</math> | Crawling from <math>-2</math> to <math>-6</math> takes it a distance of <math>4</math> units. Crawling from <math>-6</math> to <math>5</math> takes it a distance of <math>11</math> units. Add <math>4</math> and <math>11</math> to get <math>\boxed{\textbf{(E)}\ 15}</math> | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/VKaj_HDCUuQ | ||
| + | |||
| + | ~Education, the Study of Everything | ||
== See Also == | == See Also == | ||
Latest revision as of 12:57, 1 July 2023
- The following problem is from both the 2012 AMC 12A #1 and 2012 AMC 10A #3, so both problems redirect to this page.
Problem
A bug crawls along a number line, starting at 
. It crawls to 
, then turns around and crawls to 
. How many units does the bug crawl altogether?
Solution
![[asy] draw((-2,1)--(-6,1),red+dashed,EndArrow); draw((-6,2)--(5,2),blue+dashed,EndArrow); dot((-2,0)); dot((-6,0)); dot((5,0)); label("$-2$",(-2,0),dir(270)); label("$-6$",(-6,0),dir(270)); label("$5$",(5,0),dir(270)); label("$4$",(-4,0.9),dir(270)); label("$11$",(-1.5,2.5),dir(90)); [/asy]](http://latex.artofproblemsolving.com/a/9/2/a927a82769c8fbe2aaae6d6429953fbf8777298e.png)
Crawling from to takes it a distance of 
units. Crawling from to takes it a distance of 
units. Add and to get 
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
See Also
| 2012 AMC 10A (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 | ||
| 2012 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
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. 
