Three categories of candidates appear for an admission test: diligent (10%), lazy (30%) and confused (60%). A diligent candidate is 10 times more likely to clear the admission test compared to a lazy candidate.
If 40% of the candidates clearing the admission test are confused, what is the MAXIMUM possible value of the probability of a confused candidate clearing the test?

Let us assume that the total number of candidates who appeared for the admission test$$=100k$$

So, as per the data, the number of diligent candidates$$=10k$$

The number of lazy candidates$$=30k$$

and, the number of lazy candidates$$=60k$$

Now, let us assume that the probability of a lazy candidate to clear the test $$=p$$

$$\therefore$$ The probability of a lazy candidate to clear the test $$=10p$$

Let us also assume that the probability of a confused candidate to pass the exam $$=q$$

Now, we know that 

Probability of a deligent candidate passing the test,$$10p=\dfrac{\text{No of deligent candidates who passed}}{\text{total deligent candidates who appeared}}$$

or, $$p=\dfrac{\text{No of deligent candidates who passed}}{10k}$$

or, number of the diligent candidates who passed the test $$=10k*10p=100kp$$

Similarly, the number of lazy candidates who passed the test $$=30k*p=30kp$$

and, the number of confused candidates who passed the test $$=60k*q=60kq$$

Now, we are told that 40% of the candidates who clear the admission test are confused.

$$\therefore \dfrac{60kq}{100kp+30kp+60kq}=\dfrac{40}{100}$$

On simplifying this equation, we get:

$$q=\dfrac{13p}{9}$$

Now, in order to maximise $$q$$, we need to maximise $$p$$.

Now we know that the probability of a diligent candidate passing the test, $$10p\leq1$$

or, $$p\leq\dfrac{1}{10}$$

So the maximum value of $$p=\dfrac{1}{10}$$

Now, calculating the maximum value of $$q$$.

Hence, the MAXIMUM possible value of the probability of a confused candidate clearing the test, $$q=\dfrac{13}{90}$$

Hence, option B is correct.