A 20 litre mixture of milk and water contains milk and water in the ra

Question

A 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removals and replacement, what is the ratio of milk and water in the resultant mixture?

A. 9 : 1
B. 17 : 3
C. 5 : 3
D. 3 : 17
E. 1 : 17

A. 9 : 1
B. 17 : 3
C. 5 : 3
D. 3 : 17
E. 1 : 17

CrackverbalGMAT · Accepted Answer

We have a rule for iterations which really cuts down a lot of time

\frac{Final}{Initial} = (1 - \frac{b}{a})^n

Final = The Final Quantity of that component who's   concentration is being reduced  .

Initial = The Initial Quantity of that component who's   concentration is being reduced  .

b = Amount of liquid replaced

a = Final Volume in the container   after the replacement   of quantity b.

n = number of times the operation is done

Here since milk is being added, the concentration of water is reduced.

Initial Quantity of water = (2/5) * 20 = 8 liters, b = 10 (amount of milk replaced), a = 20 (final amount of solution), n = 2

Therefore 
 \frac{Final}{8} = (1 - \frac{10}{20})^2

Final amount of water = 8 * (1/2) * (1/2) = 2 L

Final Amount of milk = 20 - 2 = 18 L

Ratio of Milk and Water = 18 : 2 = 9 : 1

Option A

Arun Kumar

SangamJena · Answer

Ratio of milk to water is 
3:2 => 3*4 :2*4 => 12:8 (Gives our initial 20lts)

1st Operation :: 
10lts Removed (Half from each)
6parts from 12lts & 4parts 8lts

Hence the ratio becomes  6:4

Now,Adding 10lts milk 
Ratio becomes  16 : 4

2nd Operation 
Remove half again from 16:4
Ratio becomes  8:2

Adding 10lts milk 
Ratio now is 18:2 (Which is the required ratio)

Required Ratio is  9:1

Option A

ScottTargetTestPrep · Answer

Solution:
 
We see that, in the original 20-liter mixture, 12 liters are milk and 8 liters are water (notice that 
12 : 8 = 3 : 2). Therefore, when 10 liters (i.e., half of the mixture) are removed, 6 liters of milk and 4 liters of water have been removed. After the replacement by 10 liters of milk, we have 12 - 6 + 10 = 16 liters of milk and 8 - 4 = 4 liters of water. Now, if this process is repeated again, 8 liters of milk and 2 liters of water are removed (recall that again half of the mixture is removed). Finally, after the replacement by 10 liters of milk, we have 16 - 8 + 10 = 18 liters of milk and 4 - 2 = 2 liters of water in the final mixture. Therefore, the ratio of milk to water is 18 : 2 = 9 : 1.

Answer: A

CEdward · Answer

Finally understand these questions. My recommendation is to avoid using formulas, at least until you really understand what's happening conceptually. So the first thing to understand is that in general, for these questions to work, the mixtures must be homogenous. What that means is when you remove x volume from the solution, the amount of each constituent in the mix that gets removed is proportional to how much of it actually exists.

We are given 20L of the solution which has a milk:water ratio of 3:2 (12:8 split in terms of litres). When we remove 10L, that effectively means getting rid of 6L milk and 4L water. Then we replace the lost 10L with pure milk (for a net gain of 4).

Iteration 1: 12 - 6 + 10 / 20 = 16 / 20 <---- Ratio of milk:water is now 4:1 (16/20 L )
Iteration 2: 16 - 8 + 10 / 20 = 18 / 20

Therefore, the final ratio of milk to water is 9:1.

Answer is A.

kkannan2 · Answer

Hi @CrackverbalGMAT-I like the simplicity of having a rule for such questions, so thank you for your post. I'm curious how a question like this one can be solved in a quick and efficient way using the same formula: https://gmatclub.com/forum/a-container- ... 86312.html

Is it possible to simplify the algebra in this case to avoid tedious calculations? Trying to understand how we could approach such a question when n is much greater than 2 (in this case, n=19 iterations).

Thanks in advance, Kaushik

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A 20 litre mixture of milk and water contains milk and water in the ra

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