In a class of 30 pupils,12 take needle work,1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $N$,$P$,and $H$ represent the sets of students taking needle work,physics,and history,respectively.Given: $n(N) = 12$,$n(P) = 16$,$n(H) = 18$,

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) Let $N$,$P$,and $H$ represent the sets of students taking needle work,physics,and history,respectively.Given: $n(N) = 12$,$n(P) = 16$,$n(H) = 18$,and $n(N \cup P \cup H) = 30$.Since no one takes all three subjects,$n(N \cap P \cap H) = 0$.Using the principle of inclusion-exclusion for three sets:$n(N \cup P \cup H) = n(N) + n(P) + n(H) - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + n(N \cap P \cap H)$.Substituting the values:$30 = 12 + 16 + 18 - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + 0$.$30 = 46 - [n(N \cap P) + n(P \cap H) + n(N \cap H)]$.Therefore,$n(N \cap P) + n(P \cap H) + n(N \cap H) = 46 - 30 = 16$.The number of students taking exactly two subjects is given by $[n(N \cap P) + n(P \cap H) + n(N \cap H)] - 3n(N \cap P \cap H)$.Since $n(N \cap P \cap H) = 0$,the number of students taking exactly two subjects is $16 - 3(0) = 16$.

In a class of $30$ pupils,$12$ take needle work,$16$ take physics,and $18$ take history. If all the $30$ students take at least one subject and no one takes all three,then the number of pupils taking $2$ subjects is:

Similar Questions

If $A = \{ 1, 2, 3, 4, 5 \}$,then the number of proper subsets of $A$ is

The domain of the function $\sqrt{\log \left( \frac{5x - x^2}{6} \right)}$ is

If $n(A) = 3$ and $n(B) = 6$ and $A \subseteq B$,then the number of elements in $A \cap B$ is equal to

For which of the following cases are sets $A$ and $B$ equivalent?

The number of elements in the set $\{ (a, b) : 2a^2 + 3b^2 = 35, a, b \in \mathbb{Z} \}$,where $\mathbb{Z}$ is the set of all integers,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module