Difference between revisions of "2012 AMC 8 Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | There are <math>3</math> births and one death everyday in East Westmore. Therefore, the population increases by <math>2</math> people everyday. Thus, there are <math>2 \times 365 = 730</math> people added to the population every year. Rounding, we find the answer is <math>\boxed{\textbf{(B)}\ 700}</math>. | + | There are <math>24\text{ hours}\div8\text{ hours} = 3</math> births and one death everyday in East Westmore. Therefore, the population increases by <math>3</math> - <math>1</math> = <math>2</math> people everyday. Thus, there are <math>2 \times 365 = 730</math> people added to the population every year. Rounding, we find the answer is <math>\boxed{\textbf{(B)}\ 700}</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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:44, 14 August 2017
Problem
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 births and one death everyday in East Westmore. Therefore, the population increases by - = people everyday. Thus, there are 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 | |
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All AJHSME/AMC 8 Problems and Solutions |
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