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Difference between revisions of "1991 AJHSME Problems/Problem 23"
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==Solution==
 
==Solution==
  
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There are <math>100+80-60=120</math> females in either band or orchestra, so there are <math>230-120=110</math> males in either band or orchestra.  Suppose <math>x</math> males are in both band and orchestra. By PIE,
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There are <math>100+80-60=120</math> females in either band or orchestra, so there are <math>230-120=110</math> males in either band or orchestra.  Suppose <math>x</math> males are in both band and orchestra.
 
<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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PIE
  
 
==See Also==
 
==See Also==

Latest revision as of 17:23, 29 December 2022

Problem

The Pythagoras High School band has $100$ female and $80$ male members. The Pythagoras High School orchestra has $80$ female and $100$ male members. There are $60$ females who are members in both band and orchestra. Altogether, there are $230$ students who are in either band or orchestra or both. The number of males in the band who are NOT in the orchestra is

$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 70$

Solution

There are $100+80-60=120$ females in either band or orchestra, so there are $230-120=110$ males in either band or orchestra. Suppose $x$ males are in both band and orchestra. \[80+100-x=110\Rightarrow x=70.\] Thus, the number of males in band but not orchestra is $80-70=10\rightarrow \boxed{\text{A}}$.

PIE

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

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