Difference between revisions of "1991 AJHSME Problems/Problem 23"
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<cmath>80+100-x=110\Rightarrow x=70.</cmath> | <cmath>80+100-x=110\Rightarrow x=70.</cmath> | ||
Thus, the number of males in band but not orchestra is <math>80-70=10\rightarrow \boxed{\text{A}}</math>. | Thus, the number of males in band but not orchestra is <math>80-70=10\rightarrow \boxed{\text{A}}</math>. | ||
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==See Also== | ==See Also== | ||
Revision as of 18:39, 7 April 2021
Problem
The Pythagoras High School band has 
female and 
male members. The Pythagoras High School orchestra has female and male members. There are 
females who are members in both band and orchestra. Altogether, there are 
students who are in either band or orchestra or both. The number of males in the band who are NOT in the orchestra is
Solution
There are 
females in either band or orchestra, so there are 
males in either band or orchestra. Suppose 
males are in both band and orchestra. By PIE,
![\[80+100-x=110\Rightarrow x=70.\]](http://latex.artofproblemsolving.com/f/5/3/f53f18c7da0d7073e5c17fa4ac18de60b7b93648.png)
Thus, the number of males in band but not orchestra is 
.
See Also
| 1991 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
