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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 14 May 2020, 05:00
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From a 3 : 5 solution of milk and water, 20% is taken out and replaced by milk. How many times should this process be done to make the ratio of milk to water as 17 : 8?
(a) Once
(b) Twice
(c) Thrice
(d) Four times
(e) Five times
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C
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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 15 May 2020, 22:40
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preetamsaha
From a 3 : 5 solution of milk and water, 20% is taken out and replaced by milk. How many times should this process be done to make the ratio of milk to water as 17 : 8?
(a) Once
(b) Twice
(c) Thrice
(d) Four times
(e) Five times
Show more


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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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 14 May 2020, 05:39
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preetamsaha
From a 3 : 5 solution of milk and water, 20% is taken out and replaced by milk. How many times should this process be done to make the ratio of milk to water as 17 : 8?
(a) Once
(b) Twice
(c) Thrice
(d) Four times
(e) Five times
Show more

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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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 15 May 2020, 16:40
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chetan2u
preetamsaha
From a 3 : 5 solution of milk and water, 20% is taken out and replaced by milk. How many times should this process be done to make the ratio of milk to water as 17 : 8?
(a) Once
(b) Twice
(c) Thrice
(d) Four times
(e) Five times

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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Do not fully understand how the total mixture remains the same. Can someone post an alternate method or approach to solve this problem?
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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 15 May 2020, 22:28
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saipradyoth
chetan2u
preetamsaha
From a 3 : 5 solution of milk and water, 20% is taken out and replaced by milk. How many times should this process be done to make the ratio of milk to water as 17 : 8?
(a) Once
(b) Twice
(c) Thrice
(d) Four times
(e) Five times

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


Do not fully understand how the total mixture remains the same. Can someone post an alternate method or approach to solve this problem?
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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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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 16 May 2020, 01:58
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preetamsaha
From a 3 : 5 solution of milk and water, 20% is taken out and replaced by milk. How many times should this process be done to make the ratio of milk to water as 17 : 8?
(a) Once
(b) Twice
(c) Thrice
(d) Four times
(e) Five times

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



CONCEPT VIDEO:
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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 18 May 2020, 22:19
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preetamsaha
From a 3 : 5 solution of milk and water, 20% is taken out and replaced by milk. How many times should this process be done to make the ratio of milk to water as 17 : 8?
(a) Once
(b) Twice
(c) Thrice
(d) Four times
(e) Five times
Show more

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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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 29 Jul 2021, 06:48
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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.
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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 22 Aug 2021, 00:25
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preetamsaha
From a 3 : 5 solution of milk and water, 20% is taken out and replaced by milk. How many times should this process be done to make the ratio of milk to water as 17 : 8?
(a) Once
(b) Twice
(c) Thrice
(d) Four times
(e) Five times
Show more

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
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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 08 Oct 2024, 06:20
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For the curious souls wondering why the formula requires us to consider water concentration (0% milk) instead of milk concentration (100% milk) :
If we consider milk concentration, the process becomes lengthy and we just can't take the nth power directly.
That's why water concentration is used, though counter intuitive.


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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 28 Oct 2025, 19:26
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Given:
Initial ratio of milk : water = 3 : 5
→ So milk fraction = 38\frac{3}{8}83
→ Water fraction = 58\frac{5}{8}85
Let total volume = 111 unit (for simplicity).
So initially:
  • Milk = 38\frac{3}{8}83
  • Water = 58\frac{5}{8}85
[hr]
Process:
Each time, 20% of the total mixture is removed and replaced with pure milk.
That means:
  • 20% mixture removed → 80% remains.
  • Since the removed part has the same composition, 80% of milk and water remain.
  • Then we add 20% milk.
[hr]
After one operation:
Remaining milk = 0.8×38+0.2×10.8 \times \frac{3}{8} + 0.2 \times 10.8×83+0.2×1
=0.8×3+0.2×88=2.4+1.68=48=12= \frac{0.8 \times 3 + 0.2 \times 8}{8} = \frac{2.4 + 1.6}{8} = \frac{4}{8} = \frac{1}{2}=80.8×3+0.2×8=82.4+1.6=84=21
So after one time:
  • Milk = 1⁄2
  • Water = 1⁄2
    → Ratio = 1 : 1
[hr]
We need final ratio = 17 : 8
Milk fraction = 1717+8=1725=0.68\frac{17}{17+8} = \frac{17}{25} = 0.6817+817=2517=0.68
We’ll find how many times this process is done (n).
[hr]
General formula:
If fff fraction is replaced each time by pure milk,
and initial milk fraction = M0M_0M0,
then after nnn operations:
Mn=1−(1−M0)(1−f)nM_n = 1 - (1 - M_0)(1 - f)^nMn=1−(1−M0)(1−f)n
[hr]
Here,
M0=38=0.375M_0 = \frac{3}{8} = 0.375M0=83=0.375
f=0.2f = 0.2f=0.2
We need Mn=0.68M_n = 0.68Mn=0.68
0.68=1−(1−0.375)(0.8)n0.68 = 1 - (1 - 0.375)(0.8)^n0.68=1−(1−0.375)(0.8)n 0.32=0.625(0.8)n0.32 = 0.625(0.8)^n0.32=0.625(0.8)n (0.8)n=0.320.625=0.512(0.8)^n = \frac{0.32}{0.625} = 0.512(0.8)n=0.6250.32=0.512
Now 0.83=0.5120.8^3 = 0.5120.83=0.512
So n=3n = 3n=3
[hr]
✅ Answer: (c) Thrice
[hr]
Reason total mixture remains same:
Each time you remove 20% of the mixture and replace it with the same quantity (20%) of milk — so total volume never changes, only composition does.



saipradyoth



Do not fully understand how the total mixture remains the same. Can someone post an alternate method or approach to solve this problem?
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From a 3 : 5 solution of milk and water, 20% is taken out and replaced 14 Nov 2025, 03:53
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8/25=5/8 (1-0.2)^n
0.8^n=64/125
(4/5)^3=64/125
n=3
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