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2011 AMC 10A Problem 8

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2011 AMC 10A Problem 8AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

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?

  1. A)

    2020

  2. B)

    3030

  3. C)

    4040

  4. D)

    5050

  5. E)

    6060

Answer

C

Key insight

Removing the swans leaves 75% of the birds, and the geese are 30% of the original total, so the fraction is 30/75 = 40%.

Solution

Suppose there were 100100 birds. Then there were 3030 geese, 2525 swans, 1010 herons and 3535 ducks.

The birds that were not swans number 100−25=75100 - 25 = 75, and all 3030 geese are among them. So the required percent is
3075=25=40%.\frac{30}{75} = \frac{2}{5} = 40\%.

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

Why this works

A percent is always relative to some whole. The question changes the whole from "all birds" to "birds that are not swans," so the numerator stays at 3030 but the denominator drops to 7575. Assuming 100100 birds makes percents into counts and keeps the arithmetic concrete.

The trap

Answering 30% because geese remain 30% of all birds, ignoring that the denominator shrinks to the non-swans.

Common mistakes

  • Answering 30% because geese remain 30% of all birds, ignoring that the denominator shrinks to the non-swans.
  • Dividing 3030 by 7070 or 6565 by subtracting the wrong category (herons or ducks) instead of the swans.

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer