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

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== Solution 1 ==
 
== 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)
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75% of the total birds were not swans. Out of that 75%, there was <math>30\% / 75\% = \boxed{40\%}</math> of the birds that were not swans that were geese. (C)
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 00:53, 5 January 2013

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 $30\% / 75\% = \boxed{40\%}$ 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 $\frac{30}{75} \times 100 = \boxed{40 \%}$.

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 AMC 10 Problems and Solutions