1. At Cornthwaite H.S., many students enroll in an after-school arts program. The program offers a drama class and a music class. Each student enrolled in the program is in one class or both classes.
(a) This year, 41 students are in the drama class and 28 students are in the music class. If 15 students are in both classes, how many students are enrolled in the program?
(b) In 2014, a total of 80 students enrolled in the program. If $3 x-5$ students were in the drama class, $6 x+13$ students were in the music class, and $x$ students were in both classes, determine the value of $x$.
(c) In 2013, half of the students in the drama class were in both classes and onequarter of the students in the music class were in both classes. A total of $N$ students enrolled in the program in 2013. If $N$ is between 91 and 99 , inclusive, determine the value of $N$.
1. (a) Solution 1
Since 41 students are in the drama class and 15 students are in both drama and music, then $41-15=26$ students are in the drama class but not in the music class.
Since 28 students are in the music class and 15 students are in both drama and music, then $28-15=13$ students are in the music class but not in the drama class.
Therefore, there are $26+13=39$ students in exactly one class and 15 students in two classes.
Thus, there are $39+15=54$ students enrolled in the program.
Solution 2
If we add the number of students in the drama class and the number of students in the music class, we count the students in both classes twice.
Therefore, to obtain the total number of students enrolled in the program, we add the number of students in the drama class and the number of students in the music class, and subtract the number of students in both classes. (This has the effect of counting the students in the intersection only once.)
Thus, the number of students in the program is $41+28-15=54$.
(b) Solution 1
As in (a), there are $(3 x-5)-x=2 x-5$ students enrolled in the drama class but not in the music class.
Similarly, there are $(6 x+13)-x=5 x+13$ students enrolled in the music class but not in the drama class.
Therefore, there are $(2 x-5)+(5 x+13)=7 x+8$ students enrolled in exactly one class and $x$ students in two classes.
Since there were a total of 80 students enrolled in the program, then $(7 x+8)+x=80$ or $8 x=72$ and so $x=9$.
Solution 2
Using the method from Solution 2 to part (a), we obtain the equation $(3 x-5)+(6 x+$ 13) $-x=80$.
Simplifying, we obtain $8 x+8=80$ or $8 x=72$, and so $x=9$.
(c) Suppose that the number of students that were in both classes was $a$.
Since half of the students in the drama class were in both classes, there were also $a$ students in the drama class who were not in the music class.
Since one-quarter of the students in the music class were in both classes, then threequarters of the students in the music class were in music only. Thus, the number of students in music only was three times the number of students in both classes, or $3 a$.
Therefore, there were $a+3 a=4 a$ students in exactly one class and $a$ students in two classes.
Since a total of $N$ students were in the program, then $4 a+a=N$ or $N=5 a$.
This tells us that $N$ is a multiple of 5 , since $a$ must be an integer.
Since $N$ is between 91 and 99 and is a multiple of 5 , then $N=95$.