Vessels A and B contains mixture of copper and zinc.Copper content is 80% and 62.5% in vessel A and B respectively.25 kg of mixture is taken out from vessel A and poured into vessel B so that copper content becomes 100% more than that of zinc in vessel B. If new quantity of mixture in vessel B is 16% less than initial quantity of mixture in vessel A. Find initial quantity of copper in vessel A?

Let initial quantities in A and B be a and b. A has 80% copper = 0.8a, B has 62.5% copper = 0.625b. After transferring 25 kg from A to B, B's new copper = 0.625b + 0.8*25 = 0.625b + 20. B's zinc becomes 0.375b + 5. New copper in B is 100% more than zinc means: 0.625b + 20 = 2 * (0.375b + 5). Solving: 0.625b + 20 = 0.75b + 10 gives 0.125b = 10, so b = 80. New quantity in B = b + 25 = 105, which is 16% less than a. So 105 = 0.84a, giving a = 125. Initial copper in A = 0.8 * 125 = 100 kg. Option E is correct.