Alligation or Mixture problems
A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:
By the rule of alligation, we have: Strength of first jar Strength of 2nd jar 40% MeanStrength 26% 19% 7 14 So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2 ∴ Required quantity replaced = 2 3
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.
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
Ratio of Milk and water in a vessel A is 4 : 1 Ratio of Milk and water in a vessel B is 3 : 2 Ratio of only milk in vessel A = 4 : 5 Ratio of only milk in vessel B = 3 : 5 Let 'x' be the quantity of milk in vessel C Now as equal quantities are taken out from both vessels A & B => 4/5 : 3/5 x 3/5-x x - 4/5 => 3 5 - x x - 4 5 = 1 1 (equal quantities) => x = 7/10 Therefore, quantity of milk in vessel C = 7 => Water quantity = 10 - 7 = 3 Hence the ratio of milk & water in vessel 3 is 7 : 3
EXAMIANS