There are two mixtures of alcohol and water. In 48 L of first mixture 32 L is alcohol while in 32 L of second mixture 20 L is alcohol. If these mixtures are mixed in a large container in such a way that per cent of water in final mixture becomes 36.8%, then find that in what ratio these two mixtures are mixed to form final mixture?

Mixture 1: 48 L with 32 L alcohol (16 L water) = 33.33% water. Mixture 2: 32 L with 20 L alcohol (12 L water) = 37.5% water. Let ratio be m1:m2. Water in final = (16m1 + 12m2)/(48m1 + 32m2) = 36.8%. Solving: 16m1 + 12m2 = 0.368(48m1 + 32m2). This gives m1/m2 ≈ 21:104, confirming option B. (The calculation involves solving linear equations for mixture ratios.)