Alligation or Mixture problems
If a dairy mixes cow's milk which contains 10% fat with buffalo's milk which contains 20% fat, then the resulting mixture has fat (120/7)% of fat. What ratio was the cow's milk mixed with buffalo's milk?
% of milk in first bottle = 64% % of milk in second bottle = 100 - 26 = 74% Now, ATQ 64% 74% 68% 6 4 Hence, by using allegation method, Required ratio = 3 : 2
Customer ratio of Milk and Water is given by Milk :: Water 6.4 0 8 1 + 3 8 = 64 11 64 11 64 10 - 64 11 => Milk : Water = 110 : 11 = 10 : 1 Therefore, the proportionate of Water to Milk for Customer is 1 : 10
Let he mixes the oils in the ratio = x : y Then, the cost price of the oils = 60x + 65y Given selling price = Rs. 68.20 => Selling price = 68.20(x+y) Given profit = 10% = SP - CP => 10/100 (60x + 65y) = 68.20(x+y)-(60x + 65y) => 6x + 6.5y = 8.20x + 3.20y =>2.2x = 3.3y => x : y = 3 : 2
If the two alloys are mixed, the mixture would contain 15 gms of each metal and it would cost Rs. (150 + 120) = Rs. 270.
Cost of (15 gms of metal A + 15 gms of metal B) = Rs. 270
Cost of (1 gm of metal A + 1 gm of metal B) = Rs. (270 / 15) = Rs. 18
Cost of 1 gm of metal B = Rs. (18 ? 6) = Rs. 12
Average cost of original piece of alloy = (150 / 15) = Rs. 10 per gm.
Quantity of metal / A Quantity of metal B = (2 / 4) = (1 / 2)
Quantity of metal B = 2 (1 + 2) × 15 = 10 gms.