Art of Problem Solving

2012 AMC 10A Problems/Problem 3: Difference between revisions

Flamewire (talk | contribs)
Wisewigglyjaguar (talk | contribs)
 
(6 intermediate revisions by 5 users not shown)
Line 1: Line 1:
== Problem 3 ==
{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #1]] and [[2012 AMC 10A Problems|2012 AMC 10A #3]]}}


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?
== Problem ==
 
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>
Line 7: Line 9:
== 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>\qquad\textbf{(E)}\ 15</math>
<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 <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>
 
==Video Solution (CREATIVE THINKING)==
https://youtu.be/VKaj_HDCUuQ
 
~Education, the Study of Everything
 
== See Also ==
 
{{AMC10 box|year=2012|ab=A|num-b=2|num-a=4}}
{{AMC12 box|year=2012|ab=A|before=First Problem|num-a=2}}
{{MAA Notice}}

Latest revision as of 21:11, 20 July 2025

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 $\boxed{\textbf{(E)}\ 15}$

Video Solution (CREATIVE THINKING)

https://youtu.be/VKaj_HDCUuQ

~Education, the Study of Everything

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

These problems are copyrighted © by the Mathematical Association of America.