Alligation or Mixture problems
The amount of water (in ml) that should be added to reduce 9 ml lotion, containing 50% alcohol, to a lotion containing 30% alcohol is?
Let us assume that the lotion has 50% alcohol and 50% water.ratio = 1:1As the total solution is 9mlalcohol = water = 4.5mlNow if we want the quantity of alcohol = 30%The quantity of water = 70%The new ratio = 3:7Let x ml of water be addedWe get, 4 . 5 4 . 5 + x = 3 7 => x=6Hence 6ml of water is added.
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.
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