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

Revision as of 21:50, 6 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?

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

Solutions

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

2011 AMC 10A (Problems • Answer Key • Resources)
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

@@ Line 5: / Line 5: @@
 <math> \textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math>
+==Solutions==
 === Solution 1 ===
 % 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 ===
 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>.

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