Alligation or Mixture problems
8 litres are drawn from a flask containing milk and then filledwith water. The operation is performed 3 more times. Theratio of the quantity of milk left and total solution is 81/625.How much milk the flask initially holds?
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.
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.
Let the price per kg of mixed variety be Rs. P; then By the rule of alligation,Now, (20 - P) / (P - 15) = (2 / 3) ? 60 ? 3P = 2P ? 30 ? 5P = 90 ? P = Rs. 18
EXAMIANS