A tank contains liquids A, B, and C in the ratio 8:6:4. 30 liters of the mixture are drained out, after which 12 liters of A and 8 liters of C are added. In the final mixture, the quantity of A is 20 liters more than the quantity of B. Find the initial total volume of the mixture in the tank.

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Accepted Answer

1. Let total be $T$. Initial amounts: A = $	frac{4T}{9}$, B = $	frac{T}{3}$, C = $	frac{2T}{9}$.  
2. Remove 30 L in the same ratio: removed A = $30	imes	frac{4}{9}=	frac{40}{3}$, B = $10$, C = $30	imes	frac{2}{9}=	frac{20}{3}$.  
3. After drain and additions: A = $	frac{4T}{9}-	frac{40}{3}+12=	frac{4T}{9}-	frac{4}{3}$, B = $	frac{T}{3}-10$.  
4. Given A is 20 L more than B: $	frac{4T}{9}-	frac{4}{3}=	frac{T}{3}+10$. Solve: $4T-12=3T+90\Rightarrow T=102$.

FINAL_ANSWER: c

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