Alligation or Mixture problems
A piece of an alloy of two metals (A and B) weighs 15 gms and costs Rs. 150. If the weights of the two metals be interchanged, the new alloy would be worth Rs. 120. If the price of metal A is Rs. 6 per gm, find the weight of the other metal in the original piece of alloy.
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.
By the rule of alligation: C.P. of 1 kg sugar of 1st kind C.P. of 1 kg sugar of 2nd kind Therefore, Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3. Let x kg of sugar of 1st kind be mixed with 27 kg of 2nd kind. Then, 7 : 3 = x : 27 or x = (7 x 27 / 3) = 63 kg.
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
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.
Given, Manideep purchases 30kg of barley at the rate of 11.50/kg nad 20kg at the rate of 14.25/kg. Total cost of the mixture of barley = (30 x 11.50) + (20 x 14.25) => Total cost of the mixture = Rs. 630 Total kgs of the mixture = 30 + 20 = 50kg Cost of mixture/kg = 630/50 = 12.6/kg To make 30% of profit => Selling price for manideep = 12.6 + 30% x 12.6 => Selling price for manideep = 12.6 + 3.78 = 16.38/kg.
EXAMIANS