A 100-litre mixture of milk and water contains 36 litres of

I just found a direct way of doing this.

For replacement problems

C(Final)/C(Initial) = \((1-x/y)^n\)

Where C is the concentration
x is the amount replaced
y is the total amount
and
n is the number of replacement

Therefore for the given problem
(25/36) = \((1- x/100 )^2\)

Take Square root on both sides
5/6 = (1- x/100)

x= 16.67 litres

My guess is that the question is incorrect because if 16.67 L is the answer, then the amount of milk removed would be 36% of 16.67 L = 6 L. So milk percentage should be reduced to 30% and not 25%.

Please correct me if I am wrong.

Oh, I forgot to read that the process is repeated once.

In which case,

36 - 25 = 36%x + (36 - 36%x)%x
11 = 36% (x + x - x^2/100)
110000/36 = -x^2 + 200x
x = 16.67


The above approach involved wolfram alpha, so not useful on the gmat.

Simplest approach is:

To get from 36 to 25 in two independent iterations, we can spot that there is only one way by removing a constant 1/6th is removed at each stage: 6 + 5.

(1/6)*100 = 16.67 L
Ans.

Milk ................ Water .................. Total

36 .................... 64 ......................... 100

X litres of mixture is removed; so water & milk gets removed proportionally

\(36 - \frac{36x}{100}\) ............. \(64 - \frac{64x}{100}\) ............... 100 - x

Now x litres of water is added

\(36 - \frac{36x}{100}\) ............. \(64 - \frac{64x}{100} + x\) ............... 100

Process is repeated; Again x litres of mixture is removed

Now, milk in the new solution \(= (36-\frac{9x}{25})%\)

So, milk is x litres of mixture \(= (36-\frac{9x}{25}) * \frac{x}{100}\)

Milk content in new mixture\(= 36 - \frac{36x}{100} - (36-\frac{9x}{25}) * \frac{x}{100} = 25\)

This is a complex quadratic equation..... this question is good indeed, seems some flaw in the numeric....

Bunuel, your kind comments please..........

Responding to a pm:

First check this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... -mixtures/

You are removing a part of the solution and adding water. When you add water, the amount of milk remains the same.

C1*V1 = C2*V2
This is repeated once more.

C2 = C1 * (V1/V2)^2

\(.25 = .36 * [\frac{(100 - x)}{100}]^2\)

\(\frac{(100 - x)}{100} = \frac{5}{6}\)

x = 100/6 = 16.67 litres

36(1-x)^2=25

Solve for x: 36x^2-72x+11=0 (-66x-6x)

Factors: x=1/6, x=11/6

1/6*100=16.67

11/6*100=183 (not possible)

Hence, x=16.67

After first removal, the mixture would be \(100-x\) and will contain \(36*\frac{(100-x)}{100}\) liter milk.
After second removal, the mixture would be \(100-x\) and will contain \(36*\frac{(100-x)}{100}*\frac{(100-x)}{100}\)liter milk

∴ \(36*\frac{(100-x)}{100}*\frac{(100-x)}{100}=25\)
\(=>\)\((\frac{(100-x)}{100})^2\)=\(\frac{25}{36}\)
\(=>(\frac{(100-x)}{100})\)=\(\frac{5}{6}\)
\(=>600-6x=500\)
\(=> -6x=-100\)
\(=>x=16.67\)
_________________

That's fine as long as you get the logic behind it is similar to decline in population.

Final = Original * ( 1 - rate)^n

Well This was my approach.
In 100L mixture we have 36 L milk and 64 L water.
Hence M and W are in the ratio of 9:16.
x L of solution has been removed.
Hence we have 36-9/25x of milk and 64- 16/25x of water.
FOR CALCULATION SIMPLICITY multiply and divide 9/25x by 4 WE GET 36/100x
Since this procedure is repeated 2 times.
This was my equation
25=36*(36-36/100x)^2)/(36)^2

By solving it.
We get x=100/6 OR 50/3 OR 16.67

FOCUS ON CONCEPT RATHER THAN MEMORISING THE FORMULAS

can anyone explain me, I did

0.36(100-x)+0x=0.27(100)
36-0.36x=27
x=25

and what is wrong?

36- 36x/100 ...
Why 36x/100

x% of 36 = 36% of x of milk is removed in the first iteration.

I agree with this... The question should say "the process is repeated AGAIN" .. once seems to suggest that the process happened once.

This took me around 4 minutes to solve..

I was thinking this:

Current mixture: 36% or 36 liters of milk. Since my total liters don't change my new milk content is 25 liters. Thus 36-(36/100)=25. Not sure where I'm going wrong either.

The iteration is twice, kindly refer my post above......

For 100 lts of mixture, we have 36 lts of milk.(36%)
after removing some amount of mixture and adding the same quantity of water the milk content is 25lts.(25%)

Difference in milk = 9lts

100 lts of mixture has 36 lts of milk, so how much mixture will have 9lts of milk
100 - 36
? - 9

=>(9*100)/36 = 25

So 25 lts of mixture is replaced by 25lts of water.

Per my understanding there is something missing in the question???

No, process is repeated, so twice totally.

Please refer above post.

Posted from my mobile device

OK, Thanks mate. I get it now.
I think the question would be more clear if it problem states "the process is repeated once again".

I believe I would not be the first person to say that this is a very BAD question for GMAT practice.

Water = 64/100
Milk = 36/100
so when we remove x, we remove 0.36x of milk, and we are left with 36-0.36x of milk.
then we add x of water, and we have 64+0.36x water.

since we need to repeat this..
we need to find new ratio of milk and water:
milk = (36-0.36x)/100
and we remove again (36-0.36x)x/100

we then are told that
36-0.36x - (36-0.36x)x/100 = 25
11=0.36x + (36-0.36x)x/100
multiply by 100
1100 = 36x+36x-0.36x^2
1100=72x-0.36x^2

to solve this..is a nightmare...to be honest..I spent almost 1 hour to try to figure out maybe there's a shortcut..but there is not

YOU WILL NEVER SEE SUCH A COMPLEX QUADRATIC EQUATION on GMAT!

Working formula ...
Initial Concentration*Initial Volume=Final Concentration*Final Volume.
Let X is the part removed from 100 lts.
36%(1-X/100)^2 = 25% * 100%
(1-x/100)^2=25/36------>(1-x/100)^2=(5/6)^2
100-X=500/6
x=100/6------>16.67...
Ans B