Difference between revisions of "2011 AMC 10A Problems/Problem 8"

(Problem 8)
(Solution)
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<math> \textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math>
 
<math> \textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math>
  
== Solution ==
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== Solution 1 ==
  
 
75% of the total birds were not swans. Out of that 75%, there was <math>30\% \cdot 75\% = \boxed{40\%}</math> of the birds that were not swans that were geese. (C)
 
75% of the total birds were not swans. Out of that 75%, there was <math>30\% \cdot 75\% = \boxed{40\%}</math> of the birds that were not swans that were geese. (C)
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== Solution 2 ==
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Suppose there were 100 birds in total living on Town Lake, then 30 were geese, 25 were swans, 10 were herons, and 35 were ducks. <math>100-25 = 75</math> of the birds are not swans and 30 of these are geese, so the answer is <math>\frac{30}{75} \times 100 = \boxed{40 \%}</math>.

Revision as of 20:25, 22 February 2011

Problem 8

Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?

$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60$

Solution 1

75% of the total birds were not swans. Out of that 75%, there was $30\% \cdot 75\% = \boxed{40\%}$ of the birds that were not swans that were geese. (C)

Solution 2

Suppose there were 100 birds in total living on Town Lake, then 30 were geese, 25 were swans, 10 were herons, and 35 were ducks. $100-25 = 75$ of the birds are not swans and 30 of these are geese, so the answer is $\frac{30}{75} \times 100 = \boxed{40 \%}$.