In an examination,52 % of the candidates fail | vedclass

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(A) Let $E$ be the set of candidates who failed in English and $M$ be the set of candidates who failed in Mathematics.Given: $n(E) = 52 \%$,$n(M) = 42

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$23$

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(A) Let $E$ be the set of candidates who failed in English and $M$ be the set of candidates who failed in Mathematics.Given: $n(E) = 52 \%$,$n(M) = 42 \%$,and $n(E \cap M) = 17 \%$.The percentage of candidates who failed in at least one subject is given by the formula: $n(E \cup M) = n(E) + n(M) - n(E \cap M)$.$n(E \cup M) = 52 + 42 - 17 = 77 \%$.The percentage of candidates who passed in both subjects is the complement of those who failed in at least one subject.Percentage of passed candidates $= 100 \% - n(E \cup M) = 100 - 77 = 23 \%$.

In an examination,$52 \%$ of the candidates failed in English and $42 \%$ failed in Mathematics. If $17 \%$ failed in both the subjects,then the percentage of candidates,who passed in both the subjects,was

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