Alligation or Mixture problems
2/3 of a milk-water mixture was milk. There was 21 lit of the mixture. If 4 lit of water is added to it, the % of milk in the new mixture will be:
Let the amount of juice and water in original mixture '4x' litre and '3x' litre respectively. According to given data, 4x/3x+6 =8/7 28x=24x+48 28x?24x=48 4x = 48 x = 12 Amount of juice = 4x = 4×12 = 48 litre.
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.
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.
As per the given question , We are concerned with solid part of the fruit (pure portion). Assume Q kg of dry fruit is obtained.? Solid part in fresh fruit = Solid part in dry fruit? 0·28 × 100 = 0·8 × Q? Q = 35 kg? 35 kg of dry fruit can be obtained from 100 kg fresh fruit.
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 the price of the mixed variety be Rs. x per kg. By the rule of alligation, we have : Cost of 1 kg of type 1 rice Cost of 1 kg of type 2 rice ? (20-x)/(x-15) = 2/3 ? 60 - 3x = 2x - 30 ? x = 18.
EXAMIANS