Set Theory Questions in English

(B) Let $R, T,$ and $V$ represent the sets of families owning Radios,$TVs,$ and $VCRs$ respectively.
Given:
Total families $= 100$
$n(R) = 50, n(T) = 75, n(V) = 25$
$n(R \cap T \cap V) = 10$
Since every $VCR$ owner also has a $TV$,$V \subset T$,which implies $n(V \cap T) = n(V) = 25$.
Families having only $VCR$ and $TV$ (but not Radio) $= n(V \cap T) - n(R \cap T \cap V) = 25 - 10 = 15$.
Let $x$ be the number of families having both Radio and $TV$ but not $VCR$.
From the Venn diagram,the total number of families is the sum of all disjoint regions:
$(40 - x) + x + 10 + 15 + (50 - x) = 100$
$115 - x = 100$
$x = 15$
Now,the number of families having only $TV$ is $n(T) - [n(T \cap R \text{ only}) + n(T \cap V \text{ only}) + n(R \cap T \cap V)]$
$= 75 - [15 + 15 + 10] = 75 - 40 = 35$.

(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 $40$.
$\frac{16}{100} \times x = 40$
$x = \frac{40 \times 100}{16} = 250$.
Therefore,the total number of students in the class is $250$.