Difference between revisions of "2006 AMC 8 Problems/Problem 3"

(Solution)
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== Solution ==
 
== Solution ==
 
When Elisa started, she finished a lap in <math> \frac{25}{10}=2.5 </math> minutes. Now, she finishes a lap is <math> \frac{24}{12}= 2 </math> minutes. The difference is <math> 2.5-2=\boxed{\textbf{(A)}\ \frac{1}{2}} </math>.
 
When Elisa started, she finished a lap in <math> \frac{25}{10}=2.5 </math> minutes. Now, she finishes a lap is <math> \frac{24}{12}= 2 </math> minutes. The difference is <math> 2.5-2=\boxed{\textbf{(A)}\ \frac{1}{2}} </math>.
 
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==See Also==
 
{{AMC8 box|year=2006|num-b=2|num-a=4}}
 
{{AMC8 box|year=2006|num-b=2|num-a=4}}

Revision as of 23:51, 12 November 2012

Problem

Elisa swims laps in the pool. When she first started, she completed 10 laps in 25 minutes. Now, she can finish 12 laps in 24 minutes. By how many minutes has she improved her lap time?

$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{3}{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$

Solution

When Elisa started, she finished a lap in $\frac{25}{10}=2.5$ minutes. Now, she finishes a lap is $\frac{24}{12}= 2$ minutes. The difference is $2.5-2=\boxed{\textbf{(A)}\ \frac{1}{2}}$.

See Also

2006 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions
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