Difference between revisions of "2011 AMC 12A Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | To simplify the problem, let us say that there were a total of <math>100</math> birds. The number of birds that are not swans is <math>75</math>. The number of geese is <math>30</math>. Therefore the percentage is just <math>\frac{30}{75} \times 100 = 40 \Rightarrow \boxed{C}</math> | + | To simplify the problem, WLOG, let us say that there were a total of <math>100</math> birds. The number of birds that are not swans is <math>75</math>. The number of geese is <math>30</math>. Therefore the percentage is just <math>\frac{30}{75} \times 100 = 40 \Rightarrow \boxed{C}</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=4|num-a=6|ab=A}} | {{AMC12 box|year=2011|num-b=4|num-a=6|ab=A}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 22:45, 11 February 2018
Problem
Last summer 
of the birds living on Town Lake were geese, 
were swans, 
were herons, and 
were ducks. What percent of the birds that were not swans were geese?
Solution
To simplify the problem, WLOG, let us say that there were a total of 
birds. The number of birds that are not swans is 
. The number of geese is 
. Therefore the percentage is just 
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 12 Problems and Solutions | |
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