Finding the proportion of candidates that passed both Math and English.

Question

In an examination, $35$% of the candidates failed in Mathematics and $25$% in English. If $10$% failed in both Math and English then how many percent passed in both the subjects?

ATTEMPT Let $x$ be total candidates. Then $\frac{35x}{100}$ failed in Math, $\frac{x}{4}$ failed in English, $\frac{x}{10}$ failed in both.

So now i look for candidates who failed in either Math or English which is the union of the above two.

Candidates who failed in either Math or English = Candidates who failed in Maths + Candidates who failed in English - Candidates who failed in both.

SO it becomes $7x/20 + x/4 - x/10$

Now percent of candidates who passed in both are $1 - \{ 7x/20 + x/4 - x/10 \}$.

But i am kind of stuck here. Thanks.

Answer 1

$7/20 = 0.35$

$1/4 = 0.20$

$1/10 = 0.10$

now, $1 - { 7/20 + 1/4 - 1/10 } = 0.5$

Thus, ($1 - 0.50)*100$ That is, 50% students passed.

Answer 2

Stop using x.

Let $P(M)$ be the percent of students who failed math.

Let $P(E)$ be the percent of students who failed English.

The percent of students who failed one or the other (or both) $ = P(M \lor E)$.
$$P(M \lor E) = P(M) + P(E) - P(M \land E)$$ $$P(M \lor E) = .35 + .25 - .10 = .50$$

The percent of students who PASSED both is the complement of $P(M \lor E)$.