The ratio of spirit and water in a mixture is 1:3. If the volume of the solution is increased by 25% by adding spirit only, what is the resultant ratio of spirit and water?
Answer: A Assuming that there is 1L of spirit and 3L of water. The total volume of the mixture is 4L. After increasing its volume by 25%, volume of the mixture = 4+ 25% of 4 = 5L So the ratio of volumes of spirit to water = 2:3
Q. No. 14:
A solution contains 20L of pure milk. 20% of this is taken out. Another 20% of the remaining solution is taken out and finally 20% of the contents left is taken out. The total content taken out is replaced with water. Find the ratio of milk to water in the final solution.
Answer: D Three successive decrease of 20% in the solution would mean that 51.2% of original amount of the contents is left. Therefore ratio is 512:488
Q. No. 15:
A can is full of paint out of which 5L is removed and it is substituted by a thinning liquid. The process is repeated once more. Now the ratio of volume of paint to the volume of thinner is 49:15. What is the capacity of the can?
Answer: C Let volume of can be V litres. Amount of paint left / Amount of paint originally = {(Volume of can - volume replaced) /Volume of can}2 => (49/64 *V)/V = [(V-5)/V]2 => V= 40 Litres.
Q. No. 16:
50 litres of a cocktail of vodka and "Lime cordial" contains 32 litres of vodka. In the first round 10 litres of the cocktail is taken out and is replaced by water. In the second round, 20 litres of cocktail is taken out and is replaced by lime cordial. The percentage of vodka in the final concentration is
Answer: D The important thing to understand here is that it was need to focus on the concentration of vodka, it does not matter that the replaced fluid is water or lime cordial or any other fluid as that does not affect the concentration of vodka. What matters is the quantity of mixture replaced only. Initial concentration of vodka = 32/50 = 16/25 After the first operation, it becomes (16/25)* ((50 - 10)/50) = 64/125 After second round, it becomes (64/125)*((50-20)/50) = 192/625 = 30.72%
Q. No. 17:
A vessel has 10ml of a solution of milk and water containing 20% milk. x ml of milk was added to the vessel to reverse this ratio. y ml of water was then added to the vessel to reverse the ratio once again. Find x+y.
Answer: B Initial quantity of milk in the vessel = 0.2(10) = 2ml Initial quantity o water in the vessel = 0.8(10)= 8ml After x ml of milk was added, the ratio of milk and water would become 4:1. x= (4*8) - 2 = 30 ml After y ml of water was added, ratio of milk and water would again become 1:4 y = (4*4*8) - 8 = 120 Thus, x+y = 120 + 30 = 150 ml.
Q. No. 18:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
Answer: A Let the capacity of vessel be x ml Amount of milk originally in the vessel after replacement by water is 6x/100 ml. Amount of milk in the 9 L withdrawn = (9*9)/100 = 81/100 ml Hence, 9x/100 - 6x/100 = 81/100 => 3x =81 => x= 27 L