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.
Let C.P. of 1 liter milk be Re. 1, Gain = 16 2/3 % = 50/3 %and S.P. of 1 liter mixture = Re. 1 then C.P. of 1 liter mixture = (1 x (100 x 3) / 350) = Re. (6 / 7) By the rule of alligation,Hence, required ratio = (1/ 7) : (6 / 7) = 1 : 6
Number of liters of water in 150 liters of the mixture = 20% of 150 = 20/100 x 150 = 30 liters.P liters of water added to the mixture to make water 25% of the new mixture.Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).(30 + P) = 25/100 x (150 + P)120 + 4P = 150 + P => P = 10 liters.
According to figure we find that the ratio will be 3 : 1.Quantity sold at 20% profit = 3 / (3 + 1) × 50 = 37.5 kgs. Quantity sold at 40% profit = (50 ? 37.5) = 12.5 kgs.
EXAMIANS