Question

In a class of 55 students, the number of students studying different subjects are 23 in Mathematics, 24 in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9 in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects. The number of students who have taken exactly one subject is

• A. 6
• B. 9
• C. 7
• D. All of these

Answer 1

Answer: (D)

All of these

Answer 2

$n(M) = 23, n(P) = 24, n(C) =19$ $n(M \cap P) =12,n(M \cap C) = 9,n(P \cap C) = 7$ $n(M \cap P \cap C) = 4$ We have to find $n(M \cap P' \cap C'),n(P \cap M ' \cap C')$, $n(C \cap M \cap P)$ Now $n(M \cap P'\cap C')$ = $n[M \cap(P \cup C) ']$ $= n(M) - n[(M \cap (P\cup C)]$ $= n(M) - n[(M \cap P)\cup (M \cap C)]$ $= n(M) - n(M \cap P) - n(M \cap C) + n(M \cap P \cap C)$ $= 23 - 12 - 9 + 4 = 27 - 21 = 6$ $n(P\cap M ' \cap C') = n[P \cap (M \cup C) ']$ $= n(P) - n[P \cap (M \cup C)]$ $= n(P) - n[(P \cap M)\cup (P \cap C)]$ $= n(P) - n(P \cap M) - n(P \cap C) + n(P \cap M \cap C)$ $= 24 - 12 - 7 + 4 = 9$ $n(C \cap M ' \cap P')$ $= n(C) - n(C \cap P) - n(C \cap M) + n(C \cap P \cap M)$ $= 19 - 7 - 9 + 4 = 23 - 16 = 7$