Question

In an examination, 52% candidates failed in English and 42% failed in Mathematics. If 17% candidates failed in both English and Mathematics, what percentage of the candidates passed in both the subjects?

Answer 1

Hint: Candidates passing in both the exams means the intersection of the two events has to be evaluated, if one event is passing in mathematics and the other one is passing in English. Use the direct formula for set and relations which give the relation between such events but remember that % for fail is given, so in the final answer subtract with 100 to get the required passing %

Complete step-by-step answer:

Let the total number of candidates = 100%
Now it is given 52% candidates failed in English (E) and 42 % candidates failed in Mathematics (M)
So let n(E) = 52%, n(M) = 42%
Now if 17% candidates failed in both English and Mathematics.
Therefore $n\left( {E \cap M} \right) = 17$ %
So first find out total candidates failed in both English and Mathematics i.e. $n\left( {E \cup M} \right)$.
As we know $n\left( {E \cup M} \right) = n\left( E \right) + n\left( M \right) - n\left( {E \cap M} \right)$ so use this property of set we have,
$ \Rightarrow n\left( {E \cup M} \right) = 52 + 42 - 17 = 77$ %
So total candidates failed in both English and Mathematics is 77%.
So the total candidates passed in both English and Mathematics is the subtraction of total number of candidates and the total candidates failed in both English and Mathematics.
Therefore total candidates passed in both English and Mathematics is
= (100 - 77) %
= 23%.
So this is the required answer.
Hence option (c) is correct.

Note: In this question since things were given in percentage so on direct substitution of values back into the main formula $n\left( {E \cup M} \right) = n\left( E \right) + n\left( M \right) - n\left( {E \cap M} \right)$ will eventually give out the required quantity in percentage only. No need to evaluate the percentage differently.