From a 3 : 5 solution of milk and water, 20% is taken out and replaced

An alternate approach that does not deal with the formula.

Let us take 800 units of mixture as the ratio is 3:5.
So water is 500 and milk is 300.

Now, whenever we take out of 20% of the mix, then 20% of each of water and milk goes down. And when we increase it by that amount, the amount of milk increases by that.
For example...
When we take out 20% for the first time, the amount of M and W becomes 80% of initial, so M and W become 0.8*300 and 0.8*500 respectively. That is M=240 and W=400.
Then we add this amount 20% of 800 back as only milk, so milk becomes 240+160 =400 and W remains 400. Total after first replacement = 400+400=800.

Now the entire solution remains the same, so if the final ratio of M:W=17:8, amount of water = \(\frac{8}{(8+17)}*800=\frac{8}{25}*800=256\).
Thus, we carry on with replacement until water reduces to 256 from 500.

As the water is not added and just decreases by 20% every time, let us concentrate only on water.
1) First replacement.. Water = \(500*0.8=400\)
2) Second replacement.. Water = \(400*0.8=320\)
3) Third replacement .. Water = \(320*0.8=256\), the answer we are looking for.

So three replacements

C
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Archit3110

Please check the concept video to understand how to approach any question with successive replacements.

The video explanation of this question is attached as well.

Video Explanation:




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You are taking out the solution and putting back milk. So let's work with the concentration of water.

Cf = Ci * (Vi/Vf)^n

8/25 = 5/8 * (8/10)^n

64/125 = (4/5)^n

n = 3

Answer (C)

Check out this post for the details of this method:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2012/0 ... -mixtures/
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First we know that the TOTAL mixture remains the same, so let us make both mixs equal..

Now...Initial weight of water=5/8, and final = 8/25....
Now \(\frac{5}{8}=\frac{5*25}{8*25}=\frac{125}{200}\), and \(\frac{8}{25}=\frac{8*8}{25*8}=\frac{64}{200}\)

So when the mix is 200, water changes from 125 to 64..
So \(\frac{W_F}{W_I}=\frac{64}{125}=(1-\frac{20}{100})^n=(\frac{4}{5})^n\)
Now, \(\frac{64}{125}=\frac{4^3}{5^3}=(\frac{4}{5})^3=(\frac{4}{5})^n....n=3\)

C
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VeritasKarishma GMATinsight ; hello could you please advise on how to solve this mixture question?

Do not fully understand how the total mixture remains the same. Can someone post an alternate method or approach to solve this problem?

The amount of the mix remains the same because you are replacing 20% of mixture with milk.
So some amount taken out and SAME amount of milk substituted. The total amount will remain the same.
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Why conceptual thinking is more important than formula? Let's analyze we know that for the case as above removal and replacement in the mixture (M) containing A and B elements is given by= Amount of A or B left out /Amount of A/B originally present = [1-x/M]^n where M is mixture weight, x is weight removed, n is a number of times the removal/replacement is done.

In Problem we have a scenario where water from 62.5 percentage is dropping to 32 percent in volume.
Well, I tried to use the above formula and reach a conclusion but the ratios didn't make sense to me, So quickly shifted to reasoning.

Let's assume we have 32 liters of Mixture (because we have 8 parts keeping multiples of 8 is good practice).
Deductions Milk is 3*4=12 liters, Water =20 liters
The concentration of Milk is 37.5% Objective is to reach 68 percent, so let's begin.
First removal- 20 percent of the mixture then supplant it with Milk of the same amount, 20% of 32 is 6.4 Remaining=32-6.4=25.6 (Key concept is concentration remains same-->Deduction Milk qty is 37.5% of 25.6=9.6 liters of milk, and Water qty is 16 liters )
First replacement = 6.4 liters of milk is added to so in end we have 9.6+6.4 liters of milk=16 liters of milk over the 16 liters of water so the ratio is 1/1 <17/8 so keep going.
Second removal and replacement=16/16 again 6.4 is removed we have 25.6 remaining now we look for 50/50 concentration, therefore Milk 12.8 liters and water is 12.8 liters then add 6.4 liters milk 19.2/12.8 still less than 17/8.
Do once more 3rd removal and replacement 32-6.4=25.6 (how de split this? look for earlier concentration milk is 60 percent so 60% of 25.6 is 15.36 liters of Milk add 6.4 again to it, 10.24 liters is water =21.76/10.24=17/8
BOOM. hence C.

i am not applying any hardcore formulae but simple logic

3x +5x is the content distribution of the solution
and the content taken out distribution = 1/5 * 8x *y (where y represents the no of time it's taken out )

the ratio being [3x- 1/5(3x) *y] / [5x- (1/5 *5y) = 17/8

The x can be taken out from denominator and numerator
and the only variable left will be y

Which when solved y= 305 / 108 approximately 3

Therefore IMO C