Alligation or Mixture problems
4 kg of a metal contains 1/5 copper and rest in Zinc. Another 5 kg of metal contains 1/6 copper and rest in Zinc.The ratio of Copper and Zinc into the mixture of these two metals:
Copper in 4 kg = 4/5 kg and Zinc in 4 kg = 4 x (4/5) kg Copper in 5 kg = 5/6 kg and Zinc in 5 kg = 5 x (5/6) kg Therefore, Copper in mixture = 4 5 + 5 6 = 49 30 kg and Zinc in the mixture = 16 5 + 25 6 = 221 30 kg Therefore the required ratio = 49 : 221
Quantity of alohol in the mixture = 40 x 5/8 = 25 lit Quantity of water = 40 - 25 = 15 lit According to question, Required ratio = 20 - 40 x 20 100 x 5 8 15 - 40 x 20 100 x 3 8 + 40 x 20 100 = 20 15 - 3 + 8 = 1 : 1
By the rule of alligation, we have: Strength of first jar Strength of 2nd jar 40% MeanStrength 26% 19% 7 14 So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2 ∴ Required quantity replaced = 2 3