This problem involves using percentages and the principle of inclusion-exclusion to find the total number of students.
If $15\%$ failed both subjects, the remaining percentage must have passed at least one subject (Physics or Chemistry or Both).
Percentage Passed at least one subject $= 100\% - (\text{Percentage Failed Both})$
Percentage Passed at least one subject $= 100\% - 15\% = 85\%$
This represents the union of the sets of students who passed Physics and Chemistry, i.e., $P \cup C$. So, $n(P \cup C) = 85\%$.
We use the principle of inclusion-exclusion for two sets: $n(P \cup C) = n(P) + n(C) - n(P \cap C)$
Substituting the known values: $85\% = 80\% + 70\% - n(P \cap C)$ $85\% = 150\% - n(P \cap C)$
Rearranging the equation to find the percentage who passed both: $n(P \cap C) = 150\% - 85\%$ $n(P \cap C) = 65\%$
We know that $65\%$ of the total students passed both subjects, and this number is equal to $325$. Let $T$ be the total number of students who appeared for the examination.
$65\% \text{ of } T = 325$
$\frac{65}{100} \times T = 325$
Solving for $T$: $T = \frac{325 \times 100}{65}$ $T = \frac{32500}{65}$ $T = 500$
The total number of students who appeared in the examination is $500$.
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