Admission test candidates probability

Let total candidates = $100$. Diligent ($D$) = $10$, Lazy ($L$) = $30$, Confused ($C$) = $60$. Let probability of clearing for Lazy candidate = $p$. Then probability for Diligent candidate = $10p$. Let probability for Confused candidate = $q$. Note that probabilities cannot exceed $1$, so $10p \le 1 \implies p \le 1/10$, and $q \le 1$.

Number of clearing candidates: From $D$: $10 \times 10p = 100p$. From $L$: $30 \times p = 30p$. From $C$: $60 \times q = 60q$.

It is given that 40% of all clearing candidates are confused: $60q = 0.40 \times (100p + 30p + 60q) = 0.40 \times (130p + 60q)$ $60q = 52p + 24q \implies 36q = 52p \implies q = \frac{13}{9} p$.

Since $p \le 1/10$, the maximum possible value of $q$ is: $q_{\max} = \frac{13}{9} \times \frac{1}{10} = \frac{13}{90}$.