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PS Mixtures: 30 ml solution replaced with water 3 times [#permalink]
08 Oct 2011, 16:42

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Question Stats:

0% (00:00) correct
100% (02:32) wrong based on 2 sessions

A vessel has 300 ml of pure milk. 30 ml of milk is removed and 30 ml of water is added into the vessel. If this operation is repeated another 2 times, what is the % of milk in the vessel at the end?. _________________

Re: PS Mixtures: 30 ml solution replaced with water 3 times [#permalink]
09 Oct 2011, 01:48

2

This post received KUDOS

maheshsrini wrote:

A vessel has 300 ml of pure milk. 30 ml of milk is removed and 30 ml of water is added into the vessel. If this operation is repeated another 2 times, what is the % of milk in the vessel at the end?.

You can solve these kind of questions with a standard formula..

If there is P volume of pure liquid initially and in each operation ,Q volume is take out and replaced by Q volume of water ,then at the end of n such operations ,the concentration (k) of the liquid in the solution is given by

{ (P-Q)/P }^n = k

This gives the concentration (k) of the liquid as a PROPORTION of the total volume of the solution. If concentration has to be expressed as percentage ,then it will be 100k.

Here , we have P= 300 ; Q = 30 and n =3(2+1)

{(300-30)/300}^3 = (9/10)^3 = 0.729

Therefore the percentage of the alcohol in the vessel is 72.9 % . _________________

Re: PS Mixtures: 30 ml solution replaced with water 3 times [#permalink]
09 Oct 2011, 09:09

gmatcracker24 seems like a big fan of alcohol. Its milk in here dude

gmatcracker24 wrote:

maheshsrini wrote:

A vessel has 300 ml of pure milk. 30 ml of milk is removed and 30 ml of water is added into the vessel. If this operation is repeated another 2 times, what is the % of milk in the vessel at the end?.

You can solve these kind of questions with a standard formula..

If there is P volume of pure liquid initially and in each operation ,Q volume is take out and replaced by Q volume of water ,then at the end of n such operations ,the concentration (k) of the liquid in the solution is given by

{ (P-Q)/P }^n = k

This gives the concentration (k) of the liquid as a PROPORTION of the total volume of the solution. If concentration has to be expressed as percentage ,then it will be 100k.

Here , we have P= 300 ; Q = 30 and n =3(2+1)

{(300-30)/300}^3 = (9/10)^3 = 0.729

Therefore the percentage of the alcohol in the vessel is 72.9 % .

_________________

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ MGMAT 6 650 (51,31) on 31/8/11 MGMAT 1 670 (48,33) on 04/9/11 MGMAT 2 670 (47,34) on 07/9/11 MGMAT 3 680 (47,35) on 18/9/11 GMAT Prep1 680 ( 50, 31) on 10/11/11

Re: PS Mixtures: 30 ml solution replaced with water 3 times [#permalink]
09 Oct 2011, 09:14

The formula { ( p-q)/p } ^n gives the %age of pure liquid left where q is volume of liquid taken out of p and replaced by another liquid (same quantity) n is the number of times this process is repeated.

eg in our case pure MILK = { (300-30)/300 }^3 * 300 = 729/1000 * 300 = 218.7 %age of pure milk is { (300-30)/300 }^3 = 72.9 % _________________

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ MGMAT 6 650 (51,31) on 31/8/11 MGMAT 1 670 (48,33) on 04/9/11 MGMAT 2 670 (47,34) on 07/9/11 MGMAT 3 680 (47,35) on 18/9/11 GMAT Prep1 680 ( 50, 31) on 10/11/11

Re: PS Mixtures: 30 ml solution replaced with water 3 times [#permalink]
09 Oct 2011, 09:16

catfreak wrote:

gmatcracker24 seems like a big fan of alcohol. Its milk in here dude

gmatcracker24 wrote:

maheshsrini wrote:

A vessel has 300 ml of pure milk. 30 ml of milk is removed and 30 ml of water is added into the vessel. If this operation is repeated another 2 times, what is the % of milk in the vessel at the end?.

You can solve these kind of questions with a standard formula..

If there is P volume of pure liquid initially and in each operation ,Q volume is take out and replaced by Q volume of water ,then at the end of n such operations ,the concentration (k) of the liquid in the solution is given by

{ (P-Q)/P }^n = k

This gives the concentration (k) of the liquid as a PROPORTION of the total volume of the solution. If concentration has to be expressed as percentage ,then it will be 100k.

Here , we have P= 300 ; Q = 30 and n =3(2+1)

{(300-30)/300}^3 = (9/10)^3 = 0.729

Therefore the percentage of the alcohol in the vessel is 72.9 % .

lol I am not a big fan of alcohol bro...neither fan of milk but unfortunately these kind of questions always have either alcohol or milk ...in it _________________

Re: PS Mixtures: 30 ml solution replaced with water 3 times [#permalink]
10 Oct 2011, 21:18

maheshsrini wrote:

But why is there a difference of 2.9% between the formula and the trial version?. Can somebody pls explain?

There should not be any difference between both.The difference is because in the trial versions they have removed the milk 30+30+30 which is not correct because after the first operation 30 litre of water is mixed to the solution ,and next operation onwards 30 litre removed contains water + milk. The approach should have been different.

The formula takes care of these conditions and hence accurate. _________________

Re: PS Mixtures: 30 ml solution replaced with water 3 times [#permalink]
10 Oct 2011, 22:38

I am not sure if the question is flawed or not. However, if we are NOT losing any solution after each operation the Final Solution should be 300 ml. In that case, I think, we must have 210 (300-90) ml Milk in the 300 ml Solution.

However, If we are taking out solution only to have 30 ml Milk: in the second operation we must take out 33.33 ml solution to have 30 ml of milk of which 3.33 ml will be water, because after the 1st operation 90% of the solution is Milk. Therefore, after the 2nd operation we will have 296.67 ml of solution. If we continue the process: after the 3rd operation we have 289.59 ml of the solution of which 210 ml is milk. The left over milk is 72.5%+ of the solution.

Here we are assuming that we must take out as much solution as to have 30 ml of milk. Hence, we are losing approximately 10.41 ml of the water.

Re: PS Mixtures: 30 ml solution replaced with water 3 times [#permalink]
12 Oct 2011, 20:42

Nice observation there.

Most of the solutions including mine assumes that we are removing 30ml solution and not 30ml Milk but the question says otherwise. Having said that, one can't just remove Milk from the mixture it has to be a solution ( with some water and milk in some %ages). Generally in such type of questions its the solution that is removed. So i think the question should clarify this more. Someone with better explanation, please clarify.

BDSunDevil wrote:

I am not sure if the question is flawed or not. However, if we are NOT losing any solution after each operation the Final Solution should be 300 ml. In that case, I think, we must have 210 (300-90) ml Milk in the 300 ml Solution.

However, If we are taking out solution only to have 30 ml Milk: in the second operation we must take out 33.33 ml solution to have 30 ml of milk of which 3.33 ml will be water, because after the 1st operation 90% of the solution is Milk. Therefore, after the 2nd operation we will have 296.67 ml of solution. If we continue the process: after the 3rd operation we have 289.59 ml of the solution of which 210 ml is milk. The left over milk is 72.5%+ of the solution.

Here we are assuming that we must take out as much solution as to have 30 ml of milk. Hence, we are losing approximately 10.41 ml of the water.

Can someone please verify?

_________________

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ MGMAT 6 650 (51,31) on 31/8/11 MGMAT 1 670 (48,33) on 04/9/11 MGMAT 2 670 (47,34) on 07/9/11 MGMAT 3 680 (47,35) on 18/9/11 GMAT Prep1 680 ( 50, 31) on 10/11/11