Alligation or Mixture problems
A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:
By the rule of alligation, we have: Profit on 1st part Profit on 2nd part 8% Mean Profit 14% 18% 4 6 Ration of 1st and 2nd parts = 4 : 6 = 2 : 3 ∴ Quantity of 2nd kind = ❨ 3 x 1000 ❩kg = 600 kg.
% of milk in first bottle = 64% % of milk in second bottle = 100 - 26 = 74% Now, ATQ 64% 74% 68% 6 4 Hence, by using allegation method, Required ratio = 3 : 2
Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4 => Quantity of milk in Jar A = 5/9 x 36 = 20 litres Quantity of water in Jar A = 36 - 20 = 16 litres Let quantity of water in Jar B = x litres => Quantity of milk in Jar B = (20 - x) litres Acc. to ques, =>[20 + (20-x)]/(16+x) = 5/3 => 120?3x = 80+5x => 5x +3x = 120?80 => 8x = 40 => 5 litres.
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