Difference between revisions of "2012 AMC 8 Problems/Problem 2"

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<math> \textbf{(A)}\hspace{.05in}600\qquad\textbf{(B)}\hspace{.05in}700\qquad\textbf{(C)}\hspace{.05in}800\qquad\textbf{(D)}\hspace{.05in}900\qquad\textbf{(E)}\hspace{.05in}1000 </math>
 
<math> \textbf{(A)}\hspace{.05in}600\qquad\textbf{(B)}\hspace{.05in}700\qquad\textbf{(C)}\hspace{.05in}800\qquad\textbf{(D)}\hspace{.05in}900\qquad\textbf{(E)}\hspace{.05in}1000 </math>
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==Solution==
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There are 3 births and one death everyday in East Westmore. Therefore, the population increases by 2 people everyday. Thus, there are 2 X 365 = 730 people added to the population every year. Rounding, we find the answer is <math>\boxed{\textbf{(B)}\ 9}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=1|num-a=3}}
 
{{AMC8 box|year=2012|num-b=1|num-a=3}}

Revision as of 10:42, 24 November 2012

In the country of East Westmore, statisticians estimate there is a baby born every $8$ hours and a death every day. To the nearest hundred, how many people are added to the population of East Westmore each year?

$\textbf{(A)}\hspace{.05in}600\qquad\textbf{(B)}\hspace{.05in}700\qquad\textbf{(C)}\hspace{.05in}800\qquad\textbf{(D)}\hspace{.05in}900\qquad\textbf{(E)}\hspace{.05in}1000$

Solution

There are 3 births and one death everyday in East Westmore. Therefore, the population increases by 2 people everyday. Thus, there are 2 X 365 = 730 people added to the population every year. Rounding, we find the answer is $\boxed{\textbf{(B)}\ 9}$.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AJHSME/AMC 8 Problems and Solutions