Alligation or Mixture problems
A milk man sells the milk at the cost price but he mixes the water in it and thus he gains 9.09%. The quantity of water in the mixture of 1 liter is :
Profit (%) = 9.09 % = 1/11 Since the ratio of water and milk is 1 : 11, Therefore the ratio of water is to mixture = 1:12 Thus the quantity of water in mixture of 1 liter = 1000 x (1/12) = 83.33 ml
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
EXAMIANS