An investigator interviewed 100 students to determine the performance of three drinks: milk, coffee and tea.
The investigator reported that
\[ n(M\cap C\cap T)=10 \]
\[ n(M\cap C)=20 \]
\[ n(M\cap T)=25 \]
\[ n(C\cap T)=20 \]
Milk only \(=12\), Coffee only \(=5\), Tea only \(=8\)
Solution
Students taking only two drinks:
\[ (M\cap C)\text{ only}=20-10=10 \]
\[ (M\cap T)\text{ only}=25-10=15 \]
\[ (C\cap T)\text{ only}=20-10=10 \]
Total students taking at least one drink:
\[ 12+5+8+10+15+10+10 \]
\[ =70 \]
Therefore, students taking none:
\[ 100-70=30 \]
Answer
\[ \boxed{30} \]
Correct option: (d)