Question

In a B school there are $15$ teachers who teach marketing or finance. Of these $8$ teach finance and $4$teach both marketing and finance. How many teach marketing but not finance?
A. \[15\]
B. \[20\]
C. \[11\]
D. None of these.

Answer 1

Hint: To solve this question we will use the formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. We will consider two sets M and N as number of teachers who teaches marketing and finance. The total number of teachers in the school will be the union of both sets that is $n(F\cup M)$and the total number of teachers who teaches both marketing and finance will be the intersection of both sets that is $n(F\cap M)$. We will then substitute all the given values in the formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$ and find the number of teachers who teaches marketing only.



Formula Used:$n(A\cup B)=n(A)+n(B)-n(A\cap B)$



Complete step by step solution:We are given that in a school B there are $15$ teachers who teach marketing or finance in which $8$ teaches finance and we have to determine how many teach marketing but not finance.
Let us assume that teacher who teaches finance is F and who teaches marketing is M.
Now as given,
The number of teachers who teaches finance or marketing $=n(F\cup M)$
                                       $=15$
The number of teachers who teaches finance and marketing both $=n(F\cap M)$
   $=4$
Number of teachers who only teaches finance only $=n(F)$
      $=8$
To find the number of teachers who teaches marketing but not finance we have to find the value of $n(M)$.
We will now use the formula \[n(A\cup B)=n(A)+n(B)-n(A\cap B)\],
$\begin{align}
  & n(F\cup M)=n(F)+n(M)-n(F\cap M) \\
 & 15=8+n(M)-4 \\
 & n(M)=11
\end{align}$



Option ‘C’ is correct



Note: The union of two sets contains all the elements of both the sets while intersection of two sets contain all the elements which are common on both the sets. That is why total number of teachers in the school was taken as union of sets M and N while number of teachers which teaches both the subjects finance and marketing is taken as intersection of both sets M and N.