A tank which contains a mixture of syrup and water in ratio 15:6 . 42 liters of mixture is taken out from the tank and 8 liters of pure water and 7 liters of syrup is added to the mixture. If resultant mixture contains 33.33% water, what was the initial quantity of mixture in the tank before the replacement (in liters) ?

Initial ratio syrup:water = 15:6, so water is 6/21 = 28.57%. 42L removed removes same ratio. Adding 8L water + 7L syrup. Final water % = 33.33% = 1/3. Let initial total = x. Water initially = 6x/21 = 2x/7. After removal: water = (2x/7)(x-42)/x = 2(x-42)/7. After adding: water = 2(x-42)/7 + 8. Total = x - 42 + 15 = x - 27. So [2(x-42)/7 + 8]/(x-27) = 1/3. Solving: 6(x-42) + 168 = 7(x-27), 6x-252+168 = 7x-189, -84 = x-189, x = 105. Option A is correct.