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

(See Also)
Line 15: Line 15:
 
{{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}}
 +
{{MAA Notice}}

Revision as of 23:00, 3 July 2013

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png