Alligation or Mixture problems
In a 48 ltr mixture, the ratio of milk and water is 5:3. How much water should be added in the mixture so as the ratio will become 3:5?
Given mixture = 48 lit Milk in it = 48 x 5/8 = 30 lit => Water in it = 48 - 30 = 18 lit Let 'L' lit of water is added to make the ratio as 3:5 => 30/(18+L) = 3/5 => 150 = 54 + 3L => L = 32 lit.
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.
% 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