2011 AMC 10A Problem 8

P1-10~1 minPrint

Ratios, Percents & Averages Set up the equation/formula and compute; no special trick needed Test small/specific values or special cases to find or verify the answer

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?

Show solution without answering

Key insight

Pretend there are 100 birds. Leaving out the 25 swans leaves 75 birds, and 30 of them are geese, so the answer is 30 out of 75, which is 40%.

Solution

Percents are easiest to handle when the total is $100$ , so pretend there were exactly $100$ birds.

Then there were $30$ geese, $25$ swans, $10$ herons and $35$ ducks.
The question is only about the birds that are not swans. There are $100 - 25 = 75$ of those.
All $30$ geese are in that group of $75$ .
So the geese are $\dfrac{30}{75}$ of the non-swans. Simplify: $\dfrac{30}{75} = \dfrac{2}{5} = \dfrac{40}{100}$ , which is $40\%$ .

The answer is $\boxed{\textbf{(C)}\ 40}$ .

Why this works

A percent always means "out of some whole". Here the whole changes from "all the birds" to "the birds that are not swans". The number of geese stays at $30$ , but the number you divide by shrinks to $75$ . Pretending there are $100$ birds turns every percent into a plain count, so you can see exactly what to divide by what.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Suppose there were $100$ birds. Then there were $30$ geese, $25$ swans, $10$ herons and $35$ ducks.

The birds that were not swans number $100 - 25 = 75$ , and all $30$ geese are among them. So the required percent is
$\frac{30}{75} = \frac{2}{5} = 40\%.$

The answer is $\boxed{\textbf{(C)}\ 40}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
Prealgebra, Chapter 13: Data and Statistics (§13.1)