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== | ||
| + | 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 09:42, 24 November 2012
In the country of East Westmore, statisticians estimate there is a baby born every 
hours and a death every day. To the nearest hundred, how many people are added to the population of East Westmore each year?
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 
.
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
