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cfc198
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 04 Apr 2019, 10:59
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To a solution having milk and water in the ratio 2 : 3, 15 lts of water is added and then 15 lts of the solution is drawn out. This operation is done a total of thrice. If the ratio of milk and water now is 25 : 231, find the volume of original solution.

(1) 25 lts
2) 30 lts
(3) 40 lts
(4) 45 lts
(5) 60 lts
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KarishmaB
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 12 Jul 2019, 05:59
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cfc198
To a solution having milk and water in the ratio 2 : 3, 15 lts of water is added and then 15 lts of the solution is drawn out. This operation is done a total of thrice. If the ratio of milk and water now is 25 : 231, find the volume of original solution.

(1) 25 lts
2) 30 lts
(3) 40 lts
(4) 45 lts
(5) 60 lts
Show more


The removal/replacement method is explained in detail here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... -mixtures/

Using this, let initial volume be V. After the addition step, the volume becomes V+15

\(C2 = C1 * (\frac{V1}{V2}) = (\frac{2}{5})*\frac{V}{(V+15)}\)

This operation is done a total of 3 times so

\(\frac{25}{256} = (\frac{2}{5})*(\frac{V}{(V+15)})^3\)
V = 25

Answer (A)

Note that in this question addition happens first and removal later. It doesn't bother us. We just need to consider the concentration and volume before and after addition.
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 12 Jul 2019, 01:47
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IMO-A

Initial Vol= x , M:W=2:3=> Milk:Solution= 2:(2+3)=2:5
x*2/5=(x+15)*C (Final)
C(Final)= {x/(x+15)} * 2/5
Similarly for successive operation, C= {x/(x+15)}^n *2/5
n=3
{x/(x+15)}^3 *2/5 = 25/(25+231)
=> x=25L
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 05 Apr 2019, 11:34
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Provide solution for the question please

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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 06 Apr 2019, 07:01
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AjjayKannan
Provide solution for the question please

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I hope this is clear enough. Sorry, I'm not an expert at explaining things, but here's how I did it, and no, I do not know how I would do this quickly without a calculator.

First, you need to see how the two ratios relate to each other, so they need to be similar. Change 2:3 to 50:75. Change 25:231 to 50:462.
Now you can clearly see how the water is affecting the solution. So, the difference in water "concentration" from beginning to end is 462-75=387.
Since the process is performed 3 times, divide 387 by 3 to get 129, which is how much is changing per addition of 15lts of water.
Now you are ready to set up the equation. Because water is being added, all ratios in the equation should be in terms of water.

The original mixture is 2:3=4:6, which is 60% water, so the volume of water is 6/10*x where x is the total volume.
The equation becomes 6/10*x + 15lts = 129/179*(x+15) ; You are adding the original mixture and the 15lts of pure water, and setting it equal to resultant mixture.
Now you solve for x.
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate Updated on: 12 Jul 2019, 01:15
Originally posted by effatara on 07 Apr 2019, 08:23.
Last edited by effatara on 12 Jul 2019, 01:15, edited 1 time in total.
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Solution is given in the attachment.
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Process.docx [17.06 KiB]
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 04 Jul 2019, 03:59
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drakell
AjjayKannan
Provide solution for the question please

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I hope this is clear enough. Sorry, I'm not an expert at explaining things, but here's how I did it, and no, I do not know how I would do this quickly without a calculator.

First, you need to see how the two ratios relate to each other, so they need to be similar. Change 2:3 to 50:75. Change 25:231 to 50:462.
Now you can clearly see how the water is affecting the solution. So, the difference in water "concentration" from beginning to end is 462-75=387.
Since the process is performed 3 times, divide 387 by 3 to get 129, which is how much is changing per addition of 15lts of water.
Now you are ready to set up the equation. Because water is being added, all ratios in the equation should be in terms of water.

The original mixture is 2:3=4:6, which is 60% water, so the volume of water is 6/10*x where x is the total volume.
The equation becomes 6/10*x + 15lts = 129/179*(x+15) ; You are adding the original mixture and the 15lts of pure water, and setting it equal to resultant mixture.
Now you solve for x.
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Hi :) where did you get 179 from? Thank you
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate Updated on: 08 Aug 2019, 21:55
Originally posted by effatara on 12 Jul 2019, 01:05.
Last edited by effatara on 08 Aug 2019, 21:55, edited 1 time in total.
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After several attempts, I have been able to devise a quick and simple way to solve this problem by applying the following formula which can be very useful in solving problems where we start with an unknown quantity of a two-component solution A and B (milk and water in this case) in a given ratio (2:3 in this case) and then a certain quantity (15 liters in this case) of B is added and then exactly the same quantity (15 liters) is drawn out from the resultant solution. This process is repeated a given number of times (thrice in this case) at the end of which the ratio of the components (25/231 in this case) is given and you are required to find the quantity of the original solution.

FORMULA (Explanation is given in attachment): x{T/(T+y)}^n = Quantity of A in the Final Solution(FS)
(where 'x' is the quantity of A in the Original Solution(OS), T is the quantity of OS (which is the same as FS because the quantity of solution remains unchanged since the same amount is added and removed every time), 'y' is the quantity added and drawn out each time and 'n' is the number of times the process is repeated)

Quantity of Milk in FS (expressed as a fraction of the total quantity of FS/OS)=[x{T/(T+y)}^n]/T=[(2/5)T{T/(T+15)}^3]/T=25/(231+25)=25/256. Therefore:

[2{T/(T+15)}^3]/5=25/256...> [T/(T+15)]^3=5*25/2*256...> [T/(T+15)]^3=(5*5*5)/(8*8*8)=(5/8)^3. Extracting the cube-root from both sides of the equation:
T/(T+15)=5/8...> 8T=5T+75...> T=25
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FORMULA.docx [16.19 KiB]
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 31 Jul 2019, 18:09
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Why did we consider the new volume v+15 and not v+45 if the process is repeated thrice ?
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 08 Aug 2019, 06:22
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VeritasKarishma
cfc198
To a solution having milk and water in the ratio 2 : 3, 15 lts of water is added and then 15 lts of the solution is drawn out. This operation is done a total of thrice. If the ratio of milk and water now is 25 : 231, find the volume of original solution.

(1) 25 lts
2) 30 lts
(3) 40 lts
(4) 45 lts
(5) 60 lts


The removal/replacement method is explained in detail here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... -mixtures/

Using this, let initial volume be V. After the addition step, the volume becomes V+15

\(C2 = C1 * (\frac{V1}{V2}) = (\frac{2}{5})*\frac{V}{(V+15)}\)

This operation is done a total of 3 times so

\(\frac{25}{256} = (\frac{2}{5})*(\frac{V}{(V+15)})^3\)
V = 25

Answer (A)

Note that in this question addition happens first and removal later. It doesn't bother us. We just need to consider the concentration and volume before and after addition.
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Hey, I really think that something is wrong here either in the question or in the answer. Bcuz the formula for this type of ques is:

New proportion (of milk)= old proportion * (initial operational volume/post operation volume)^n

n= no. Of times it is done.

Now, the initial operational volume is the one that we get after the solution is touched for the very first time and post operational value is the one that we get when we get the original volume back

Here, especially, in this question amount is added first to the original solution and then it is removed. So, i think that volume1 should be "v+15" and volume2 should ve "v" i.e. original volume.

Please let me know where am i wrong. Cheers.

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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 26 Jan 2021, 04:15
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Hi VeritasKarishma,

Not clear why we don't have to worry about the removal process? Could you help me understand?

Thank you
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 27 Jan 2021, 00:56
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Hi VeritasKarishma,

Not clear why we don't have to worry about the removal process? Could you help me understand?

Thank you
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We are dealing with homogenous mixtures. Say a jug of milk and water mixed and stirred such that you have 50% milk and 50% water. Now if you take 1 cup of mixture out, the concentration will still be 50% milk and 50% water in the jug as well as the cup.
Removal only reduces volume of the mixture in the jug, not the concentration.

So at removal concentration does not change. When we add back, concentration changes and hence we get C1 and C2. So the step to worry about is when we add back.

Check the link: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... -mixtures/
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To a solution having milk and water in the ratio 2 : 3, 15 lts of wate 17 Jun 2025, 03:12
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