Alligation or Mixture problems
8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine the cask hold originally?
Let the quantity of the wine in the cask originally be x litres Then, quantity of wine left in cask after 4 operations = x 1 - 8 x 4 litres ? x 1 - 8 x 4 x = 16 81 ? 1 - 8 x 4 = 2 3 4 ? x = 24
Let us assume the number of boys = B and number of girls = G.According to question,B + G = 30Lets us assume total weight of boys = W1 and total weight of girls = W2average weight of boys = total weight of boys/number of boystotal weight of boys/number of boys = 20W1/B = 20W1 = 20Baverage weight of girls = total weight of girls/number of girls25 = W2/GW2 = 25GData is not sufficient to solve the equation.since we do not know either the average weight of the whole class or the ratio of no. of boys to girls.
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
EXAMIANS