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

(Solution)
Line 4: Line 4:
  
 
==Solution==
 
==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>.
+
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>.
  
 
==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 09:46, 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 $2 \times 365 = 730$ people added to the population every year. Rounding, we find the answer is $\boxed{\textbf{(B)}\ 700}$.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS