Key information provided:
Candidates passing at least one section are those not failing both.
Percentage Passed (At Least One) = 100% - Percentage Failed Both
Percentage Passed (At Least One) = 100% - 18% = 82%
Let $P(A)$ denote the percentage passed in Section A, $P(B)$ the percentage passed in Section B, and $P(A \cap B)$ the percentage passed in both. The percentage passed in at least one section is $P(A \cup B)$.
Using the inclusion-exclusion principle:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$Substitute known values:
$82\% = 80\% + 77\% - P(A \cap B)$
$82\% = 157\% - P(A \cap B)$
Solve for the percentage passed in both sections:
$P(A \cap B) = 157\% - 82\% = 75\%$
We established that 75% of total candidates passed both sections. This corresponds to 1425 students.
Let $T$ represent the total number of candidates.
$75\% \times T = 1425$
Convert percentage to decimal for calculation:
$0.75 \times T = 1425$
Solve for $T$:
$T = \frac{1425}{0.75}$
$T = \frac{1425}{3/4}$
$T = 1425 \times \frac{4}{3}$
$T = 475 \times 4$
$T = 1900$
Thus, 1900 candidates appeared for the examination.
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