In a class of 300 students, 242 take math, 208 take science, and 183 take both math and science. How many students take neither math nor science?

Aug 11, 2016

$33$ students take neither of the subjects.

Explanation:

We will use the formulas $\left(1\right) : n \left(A \cup B\right) = n \left(A\right) + n \left(B\right) - n \left(A \cap B\right)$,

$\left(2\right) : n \left(A - A \cap B\right) = n \left(A\right) - n \left(A \cap B\right)$.

$\left(3\right) : \left(A \cup B\right) ' = A ' \cap B '$...............[De'Morgan's law].

where, A & B sub U and, $n \left(A\right)$ denotes the Number of Elements in a Set $A \subset U$, the Universal Set .

Let $M =$ The Set of students taking Maths. , and, $S$ that of

students taking Sc.. Hence, $M \cap S$ is the set of students taking both the subjects, whereas, $M ' \cap S '$ is the set of students opting neither of the subjects.

Our goal is to find $n \left(M ' \cap S '\right) = n \left(\left(M \cup S\right) '\right) ,$ because of $\left(3\right)$.

Now, let us observe that, $\left(M - M \cap S\right) \cup \left(M \cap S\right) = M$ and, their intersection is $\phi$, so, by $\left(1\right)$, we get,

$n \left(M\right) = n \left(M - M \cap S\right) + n \left(M \cap S\right)$, i.e.,

$242 = n \left(M - M \cap S\right) + 183 \Rightarrow n \left(M - M \cap S\right) = 59$.

$\left(2\right) \Rightarrow n \left(M\right) - n \left(M \cap S\right) = 59$,

From $\left(1\right)$, then, we have,

$n \left(M \cup S\right) = n \left(M\right) + n \left(S\right) - n \left(M \cap S\right) = 59 + 208 = 267$

Therefore, $n \left(M ' \cap S '\right) = n \left(\left(M \cup S\right) '\right) = n \left(U - M \cup S\right)$

$= n \left(U\right) - n \left(M \cup S\right) = 300 - 267 = 33$.

Enjoy Maths.!