Let the capacity of the vessel be \(x\).

After the first removal there would be \(x-8\) liters of water left in the vessel. Note that the share of the water would be \(\frac{x-8}{x}\);

After the second removal, the removed mixture of 8 liters will contain \(8*\frac{x-8}{x}\) liters of water, so there will be \((x-8)-8*\frac{x-8}{x}=\frac{(x-8)^2}{x}\) liters of water left.

As the ratio of water to milk after that is \(\frac{9}{40}\), then the ratio of water to the capacity of the vessel would be \(\frac{9}{40+9}=\frac{9}{49}\).

So \(\frac{\frac{(x-8)^2}{x}}{x}=\frac{9}{49}\) --> \(\frac{(x-8)^2}{x^2}=\frac{9}{49}\) --> \(\frac{x-8}{x}=\frac{3}{7}\) --> \(x=14\).

Answer: D.
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the equation you'd get once 8 lts are drawn the second time is: (w-16)/16 = 9/40 where w is the capacity of the vessel.
Solving for w, w = 19.6. So, its C

Oops, should've read the question better..sorry abt that....thanks Bunuel...

Step1 :-

Water in vessel = x

8 liters is removed and replaced with milk

so now,

Water = x-8
Milk = 8

Step 2 :-

Now again 8 liters of mixture is removes , so water in that mixture is w = (x-8/x) * 8

So Total water removed = x-8-w = x-8-(x-8/x)*8 = (x^2-8x-8x-64)/x = (x^2-16x-64)/x

one more way:

Lets start with option:

Vessel capacity be 20litre:

8litre water removed and milk added: W=12litre, M=8litre

8litre mixture removed, (removed proportionately) : W= 12-4.8 = 7.2litre, M= 8-3.2 = 4.8 litre

8litre milk added ; W=7.2 litre, M=12.8 litre

Ratio will be 72/128 = 18/32 = 9/16

As we can see that milk is less so we have to select the ratio even less than 20,

Vessel capcity be 14 litre

8litre water removed and milk added: W=6 litre, M=8 litre

8litre mixture removed, (removed proportionately) : W= 6-3.4 = 2.6 litre, M= 8-4.6 = 3.4 litre

8litre milk added ; W=2.6 litre, M=11.4 litre

Ratio will be 26/114 = 8.66/38 (approx) very close to 9/40

So correct option is D

Bunuel, could you please pin point to a mistake in my reasoning?

Let T = total capacity of vessel

1.) the concentration of milk after the first round is 8/T
2.) After the second round the concentration of milk is \(40/49\)

\(8/T\) * (T-8) + 8 = \(40/49\) T

Now, I can't solve this equation. And even substituting answer choices won't help.

Where's the mistake?

Thank you very much.

Bunuel, thanks a lot. I must have made some miscalculation.

Conceptually the question is interesting. However the numbers are quite "difficult" to work with. Do you think that a real GMAT question would have better numbers?

Actually, the numbers are quite suitable for a very efficient, quick and oral solution. This is what I thought of when I came up with the answer in 20 secs. Mind you, you need to go through the link provided below to understand this theory. Else the 20 sec solution will probably not make sense to you.

We are substituting milk so we should work with water.

Final concentration of water = 9/49
There were two iterations. So, 9/49 = (100%)*(Vi/Vf)^2
Vi/Vf = 3/7
Since we are putting 8 liters of water but difference between Vi and Vf is 4, final volume (which is also equal to volume of the vessel) must be twice too i.e. 7*2 = 14 liters.

http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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if we are removing 8 litres of mixture,does that mean we are removing water share ?? because in mixture,how can we remove water now,as it contains milk too now. why do we use 8*water share??Pls clarify..

Hi Karishma , Could you please elaborate the highlighted portion ? I am unable to grasp the last part of the solution . Thanks in advance .