Question

A certain high school has $5000$ students. Of these students, $x$ are taking music, $y$ are taking art and $z$ are taking both music and art. How many students are taking neither music nor art?

• A. $5000-z$
• B. $5000-x-y$
• C. $5000-x+z$
• D. $5000-x-y-z$
• E. $5000-x-y+z$

Answer 1

Answer: (E)

$5000-x-y+z$

Let $S\to$students in a high school .

$A\to$students who take music.

$B\to$students who take arts.

$C\to$Students who take both arts and music.

$n\left(S\right)=5000$

$n\left(A\right)=x$

$n\left(B\right)=y$

$n\left(C\right)=n(A\cap B)=Z$

Number of students who take neither music nor arts $=n\left(S\right)-n\left(A\right)-n\left(B\right)+n\left(C\right)$

$=n\left(S\right)-n\left(A\right)-n\left(B\right)+n(A\cap B)$

$=5000-x-y+z$