Difference between revisions of "2012 AMC 8 Problems/Problem 2"
m (→Solution) |
(→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | There are <math>24\text{ hours}\div8\text{ hours} = 3</math> births and one death everyday in East Westmore. Therefore, the population increases by 3-1 = <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}} | {{MAA Notice}} | ||
Revision as of 19:44, 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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
