[Two-stage mixing]

Alcohol fractions: $A = 5/8$, $B = 7/12$. Target for $C$: $8/13$.

Stage 1. If $A : B = p : q$, then

$

\frac{\tfrac{5p}{8} + \tfrac{7q}{12}}{p+q} = \tfrac{8}{13} \Longrightarrow 13(15p + 14q) = 192 \cdot 24 \cdot \tfrac{p+q}{24} \cdot \tfrac{24}{1}.

$

Working it cleanly: $13(15p + 14q) = 192(p+q) \Rightarrow 3p = 10q$, i.e.\ $p:q = 10:3$. (This step is only to confirm $C$ is achievable; the alcohol fraction of $C$ is fixed at $8/13$.)

Stage 2. Take 13 units of $C$: alcohol $= 8$, water $= 5$. Add $k$ units of water. Final alcohol fraction $= 40$:

$

\frac{8}{13 + k} = \frac{2}{5} ;\Longrightarrow; 13 + k = 20 ;\Longrightarrow; k = 7.

$

Answer: (C) 7.

Why CAT-level: Two linked mixings. Starting with alligation on $C$ + water is the elegant path; mixing $A$ + $B$ explicitly is the false-start consumer of time.