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.
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 initially milk and water in container B is 3x liter and x liter respectively
Now, 3x + (8/9) × 18 ? x ? (1/9) × 18 = 30
3x + 16 ? x ? 2 = 30
x = 8
Initial quantity is container B = 8 (3 + 1) = 32 Liter.
Here total parts of milk and water in the solution is 6+2 = 8 parts. 1part = 640/8 = 80 old mixture contains 6parts of milk and 2 parts of water(6:2). To get new mixture containing half milk and half water, add 4parts of water to the old mixture then 6:(2+4) to make the ratio same. i.e, 4 x 80 = 320 ml.
EXAMIANS