Alligation or Mixture problems
A shopkeeper buys 26 liter of milk @ Rs. 16 per liter. He also buys from another source an inferior quality of milk @ Rs. 10 per liter. How much quantity of the later should he buy to mix it with the former so that he can sell the mixture @ Rs. 14 per liter without making any loss?
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.
Let C.P. of 1 liter milk be Re. 1, Gain = 16 2/3 % = 50/3 %and S.P. of 1 liter mixture = Re. 1 then C.P. of 1 liter mixture = (1 x (100 x 3) / 350) = Re. (6 / 7) By the rule of alligation,Hence, required ratio = (1/ 7) : (6 / 7) = 1 : 6
Let he mixes the oils in the ratio = x : y Then, the cost price of the oils = 60x + 65y Given selling price = Rs. 68.20 => Selling price = 68.20(x+y) Given profit = 10% = SP - CP => 10/100 (60x + 65y) = 68.20(x+y)-(60x + 65y) => 6x + 6.5y = 8.20x + 3.20y =>2.2x = 3.3y => x : y = 3 : 2