In a class of 100 students,55 students have p | vedclass

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(D) Let $n(M)$ be the number of students who passed in Mathematics and $n(P)$ be the number of students who passed in Physics.Given: $n(M) = 55$,$n(P)

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) Let $n(M)$ be the number of students who passed in Mathematics and $n(P)$ be the number of students who passed in Physics.Given: $n(M) = 55$,$n(P) = 67$,and the total number of students $n(M \cup P) = 100$.Using the formula for the union of two sets: $n(M \cup P) = n(M) + n(P) - n(M \cap P)$.Substituting the values: $100 = 55 + 67 - n(M \cap P)$.$100 = 122 - n(M \cap P)$.Therefore,$n(M \cap P) = 122 - 100 = 22$.This represents the number of students who passed in both Mathematics and Physics.The number of students who passed in Physics only is given by $n(P 	ext{ only}) = n(P) - n(M \cap P)$.$n(P 	ext{ only}) = 67 - 22 = 45$.

In a class of $100$ students,$55$ students have passed in Mathematics and $67$ students have passed in Physics. Then the number of students who have passed in Physics only is

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