In a certain class,72 % of the students took | vedclass

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(C) Let $n(A)$ be the percentage of students who took Biology and $n(B)$ be the percentage of students who took Mathematics.Given: $n(A) = 72 \%$,$n(B

Vedclass · Accepted Answer

$250$

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(C) Let $n(A)$ be the percentage of students who took Biology and $n(B)$ be the percentage of students who took Mathematics.Given: $n(A) = 72 \%$,$n(B) = 44 \%$.Since every student took at least one subject,the union of both sets is $n(A \cup B) = 100 \%$.Using the formula $n(A \cup B) = n(A) + n(B) - n(A \cap B)$:$100 \% = 72 \% + 44 \% - n(A \cap B)$$100 \% = 116 \% - n(A \cap B)$$n(A \cap B) = 116 \% - 100 \% = 16 \%$.It is given that $40$ students took both subjects,so $16 \%$ of the total students $x$ is $40$.$\frac{16}{100} 	imes x = 40$$x = \frac{40 	imes 100}{16} = 250$.Therefore,the total number of students in the class is $250$.

In a certain class,$72 \%$ of the students took Biology and $44 \%$ took Mathematics. If each student took at least one of Biology or Mathematics and $40$ students took both of these subjects,the total number of students in the class is:

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