Alligation or Mixture problems
The diluted wine contains only 8 liters of wine and the rest is water. A new mixture whose concentration is 30%, is to be formed by replacing wine. How many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture?
Wine Water 8L 32L 1 : 4 20 % 80% (original ratio) 30 % 70% (required ratio) In ths case, the percentage of water being reduced when the mixture is being replaced with wine. so the ratio of left quantity to the initial quantity is 7:8 Therefore , 7 8 = 1 - K 40 => K = 5 Lit
Water in 60 gm mixture=60 x 75/100 = 45 gm. and Milk = 15 gm. After adding 15 gm. of water in mixture, total water = 45 + 15 = 60 gm and weight of a mixture = 60 + 15 = 75 gm. So % of water = 100 x 60/75 = 80%.
Ratio of milk and water = 2 : 1Quantity of milk = 60 X 2/3 = 40 litreQuantity of water = 20 litreTo make ratio, 1: 2, we have to double the water that of milkSo, water should be 80 litre.That means 80 ? 20 = 60 litre water to be added.
Given, Manideep purchases 30kg of barley at the rate of 11.50/kg nad 20kg at the rate of 14.25/kg. Total cost of the mixture of barley = (30 x 11.50) + (20 x 14.25) => Total cost of the mixture = Rs. 630 Total kgs of the mixture = 30 + 20 = 50kg Cost of mixture/kg = 630/50 = 12.6/kg To make 30% of profit => Selling price for manideep = 12.6 + 30% x 12.6 => Selling price for manideep = 12.6 + 3.78 = 16.38/kg.