Alligation or Mixture problems
A can contains a mixture of 2 liquids A and B in proportion 7 : 5 when 9 liters of mixture are drawn off and the can is filled with B, the proportion of A and B becomes 7 : 9. How many liters of liquid A was contained by the can initially?
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.
Quantity of milk @ Rs.10 per liter / Quantity of milk @ Rs. 16 per liter = 1 / 2 So, quantity of milk @ Rs. 10 per liter = 26 / 2 = 13 liter.2nd Method Let us assume shopkeeper buy P liter milk of price @ Rs. 10 per liter.Buy price of 26 liter of milk @ Rs. 16 per liter = 26 x 16Buy price of P liter of milk @ Rs. 10 per liter = P x 10Sell price of total milk ( P + 26 ) @ Rs. 14 per liter = 14 x ( P + 26 ) According to question there is no loss or no profit.Then Buy Price = Sell Price 26 x 16 + P x 10 = 14 x ( P + 26 ) ? 26 x 16 + 10P = 14P + 14 x 26 ? 26 x 16 - 26 x 14 = 14P - 10P? 2 x 26 = 4P? 4P = 2 x 26? P = 2 x 26 / 4 = 13So, quantity of milk @ Rs. 10 per liter = 13 liter.
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.
EXAMIANS