A 20 litre mixture of milk and water contains milk and water

x/8=(1-10/20)^2 => x=amount of water left after 2 removements. its 9:1...

Chris I dont really understand your reasoning, care to explain a bit where the numbers come from. Thanks.

usually i use this formula for these kind of questions...
amount of x after removal/total amount of x=(1-(amount of y that is replaced)/total volume)^number of replacements...not the best but the fastest way unless its trickier...

Where did you get it from? Might be useful to memorize a few of those formulas for the actual exam.

he 20 litre mixture contains milk and water in the ratio of 3 : 2. Therefore, there will be 12 litres of milk in the mixture and 8 litres of water in the mixture.

Step 1.
When 10 litres of the mixture is removed, 6 litres of milk is removed and 4 litres of water is removed. Therefore, there will be 6 litres of milk and 4 litres of water left in the container. It is then replaced with pure milk of 10 litres. Now the container will have 16 litres of milk and 4 litres of water.

Step 2.
When 10 litres of the new mixture is removed, 8 litres of milk and 2 litres of water is removed. The container will have 8 litres of milk and 2 litres of water in it. Now 10 litres of pure milk is added. Therefore, the container will have 18 litres of milk and 2 litres of water in it at the end of the second step. Therefore, the ratio of milk and water is 18 : 2 or 9 : 1.

Shortcut.
We are essentially replacing water in the mixture with pure milk.
Let W_o be the amount of water in the mixture originally = 8 litres.
Let W_r be the amount of water in the mixture after the replacements have taken place.
Then,{W_r}/{W_o}= (1-R/M)^n
where R is the amount of the mixture replaced by milk in each of the steps, M is the total volume of the mixture and n is the number of times the cycle is repeated.

Hence, {W_r}/{W_o} =(1/2)^2  =1/4
Therefore, "W_r ={W_o}/4= 8/4 = 2 litres

Reference: lofoya. com/Aptitude-Questions-And-Answers/Alligation-or-Mixture/l3p1 .htm

Hope it helps

I just learned the new method to solve this problem as discussed by Karishma from veritas link, Trying to solve this question using that method

For future reference

total solution : 20 liters
milk : water 3:2

milk = 12 liters
water = 8 liters

Amount = Concentratio * volumn
therefore concentration of water in the solution is

8 = Cw * 20
Cw = 2/5

Step 1: 10 liters from 20 liters solution is removed - in the leftover solution the concentration of water remains the same. i.e. 2/5. Intial volume of solution was 20, new volumn is 10 (after 10 liter is removed)

Step 2: 10 liters of pure milk is added to the solution, therefore new solution is again 20 liters

Initial COncentration (Ci) * Initial Volumn (Vi) = Final Contentration (Cf) * final Volumn (Vf)
(2/5) * 10 = Cf * 20
Cf = 2/5 (1/2) (the amount of water will remain the same when as initial in (10 liter solution) but concentration of water will changes when milk is added to intial 10 liter water)


Step 3 now again from the Step 2 solution we remove 10 liter solution and added 10 liter pure milk

cf = (2/5) (1/2) (1/2)
cf = 1/10 (this is a new concentration of Water in the solution)

there fore concentration of milk is 9/10

therefore ratio of milk : water is 9:1

The attachment is written with invisible ink
i cant see anything..pls can you upload it again thanks

This attachment is lost.

Great question, but it may be 600 level.

I would not recommend this method, but if you want to save time you can plug in the formula. The solution is given below. Chris gave the original solution, however I think he is lacking a bit of explanation. This is my take on his original solution.

The formula is -

\(\frac{QuantityOfAleft}{QuantityOfAoriginallyPresent} = (1-\frac{R}{T})^n\)
R = Replacing quantity of B
T = Total quantity of mixture
n = number of operations of replacement

We are replacing water in the mixture with pure milk.
Quantity of A originally present = 8 litres (water)
Then, \(\frac{Quantity of A left}{8} = (1−\frac{10}{20})^n\), where R is the amount of the mixture replaced by milk in each of the steps = 10 litres, T is the total volume of the mixture = 20 litres, and n is the number of times the cycle is repeated.

Therefore, Quantity of A (water) left = 2 litres.
This is the quantity of water in the mixture after the process is done twice.
So, the final quantity of water in the 20-litre mixtures is 2 litres.
Hence, the mixture will have 18 litres of milk and 2 litres of water.

Ratio of milk to water = 18:2 = 9:1

based on the forumula,

Final Volume of Milk /initial volume of milk = (1- quantity replaced/total initial quantity)^n

total 20 ltrs, ratio = 3:2 that is 12 ltrs milk and 8 ltrs water.

x/12 = (1-10/20)^2

x/12 = 1/4

x= 9/1 = 9:1

New Concentration of Water = Old concentration of Water * (V1/V2)^n

New Concentration of Water = 2/5 * (1/2)^2 = 1/10

Ratio of milk:water = 9:1

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