1971 AMC 12 Problem 10

Attempt Problem 10 of the 1971 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1971 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

Each of a group of

50

girls is blonde or brunette and is blue-eyed or brown-eyed. If

14

are blue-eyed blondes,

31

are brunettes, and

18

are brown-eyed, then the number of brown-eyed brunettes is:

5

7

9

11

13

First find the total number of blondes

Subtract the brown-eyed blondes from all brown-eyed girls

There are

50-31=19

blondes, of whom

19-14=5

are brown-eyed. Hence the number of brown-eyed brunettes is

18-5=13.

Therefore, the correct answer is E.

Problem 10 in other years