Difference between revisions of "1971 AHSME Problems/Problem 10"
Coolmath34 (talk | contribs) (Created page with "== Problem == Each of a group of <math>50</math> girls is blonde or brunette and is blue eyed of brown eyed. If <math>14</math> are blue-eyed blondes, <math>31</math> are br...") |
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Next, we know that <math>18</math> people are brown-eyed, so there are <math>18-5=13</math> brown-eyed brunettes. | Next, we know that <math>18</math> people are brown-eyed, so there are <math>18-5=13</math> brown-eyed brunettes. | ||
| − | The answer is <math>\textbf{(E)} | + | The answer is <math>\boxed{\textbf{(E) }13}.</math> |
-edited by coolmath34 | -edited by coolmath34 | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 35p box|year=1971|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:53, 1 August 2024
Problem
Each of a group of 
girls is blonde or brunette and is blue eyed of brown eyed. If 
are blue-eyed blondes,
are brunettes, and 
are brown-eyed, then the number of brown-eyed brunettes is
Solution
There are brunettes, so there are 
blondes. We know there are blue-eyed blondes, so there are 
brown-eyed blondes.
Next, we know that people are brown-eyed, so there are 
brown-eyed brunettes.
The answer is 
-edited by coolmath34
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
