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 water is 16 : 65. How much wine did 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 ❩ 4 ] litres. x ∴ ❨ x(1 - (8/x))4 ❩ = 16 x 81 ⟹ ❨ 1 - 8 ❩ 4 = ❨ 2 ❩ 4 x 3 ⟹ ❨ x - 8 ❩ = 2 x 3 ⟹ 3x - 24 = 2x ⟹ x = 24.
W 1 : A 1 W 2 : A 2 . . . . . W N : A N 2 : 3 4 : 5 5 : 7 W 1 W 1 + A 1 = 2 5 W 2 W 2 + A 2 = 4 9 W N W N + A N = 5 12 = 72/180 = 80/180 = 75/180 => 5 : 3 Therefore, the ratio is 5: 3
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.
Initial quantity of copper = 80 100 x 50 = 40 g And that of Bronze = 50 - 40 = 10 g Let 'p' gm of copper is added to the mixture => 50 + p x 90 100 = 40 + p => 45 + 0.9p = 40 + p => p = 50 g Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.
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.
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