Difference between revisions of "2012 AMC 10A Problems/Problem 3"

 
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== Problem ==
 
== Problem ==
  
A bug crawls along a number line, starting at -2. It crawls to -6, then turns around and crawls to 5. How many units does the bug crawl altogether?
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A bug crawls along a number line, starting at <math>-2</math>. It crawls to <math>-6</math>, then turns around and crawls to <math>5</math>. How many units does the bug crawl altogether?
  
 
<math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math>
 
<math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math>
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== Solution ==
 
== Solution ==
  
Crawling from -2 to -6 takes it a distance of 4 units. Crawling from -6 to 5 takes it a distance of 11 units. Add 4 and 11 to get <math>\boxed{\textbf{(E)}\ 15}</math>
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<asy>
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draw((-2,1)--(-6,1),red+dashed,EndArrow);
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draw((-6,2)--(5,2),blue+dashed,EndArrow);
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dot((-2,0));
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dot((-6,0));
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dot((5,0));
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label("$-2$",(-2,0),dir(270));
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label("$-6$",(-6,0),dir(270));
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label("$5$",(5,0),dir(270));
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label("$4$",(-4,0.9),dir(270));
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label("$11$",(-1.5,2.5),dir(90));
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</asy>
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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>
  
 
== See Also ==
 
== See Also ==
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{{AMC10 box|year=2012|ab=A|num-b=2|num-a=4}}
 
{{AMC10 box|year=2012|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2012|ab=A|before=First Problem|num-a=2}}
 
{{AMC12 box|year=2012|ab=A|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 09:32, 6 September 2014

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 $-2$. It crawls to $-6$, then turns around and crawls to $5$. How many units does the bug crawl altogether?

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$

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]

Crawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\boxed{\textbf{(E)}\ 15}$

See Also

2012 AMC 10A (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
2012 AMC 12A (ProblemsAnswer KeyResources)
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

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