Alligation or Mixture problems
Two bottles contains mixture of milk and water. First bottle contains 64% milk and second bottle contains 26% water. In what ratio these two mixtures are mixed so that new mixture contains 68% 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
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.
Initial quantity of copper = 80 100 x 50 = 40 g And that of Bronze = 50 - 40 = 10 g Let 'p' gm of copper is added to the mixture => 50 + p x 90 100 = 40 + p => 45 + 0.9p = 40 + p => p = 50 g Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.
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.
EXAMIANS