How many litres of a 12 litre mixture containing milk and water in the

I have one solution for this question, please correct me if I am wrong.

here it is saying that Total =12 litres ,
milk:water = 2:3
let milk=2x and water = 3x
therefore, 2x+3x=12.
Now let z litres of milk be replaced.
so final mixture contains milk:water=1:1
therefore 2x+z/3x=1
which yields x=z

Now, 2x+z+3x=12-z(here I am subtracting z from 12 since z has to be replaced)
so, 2x+2z+3x=12.
since x=z(from previous equation)
6x=12
implies x=2
Therefore z=2.
answer B is correct that is 2 litres have to be replaced.
Kindly correct me if i am wrong

Let the quantity of the solution removed and subsequently replaced by milk be 'x' liters. In order for the final solution to contain milk and water in the ratio of 1:1 the quantity of milk needs to be 6 liters since the solution quantity is unchanged at 12 liters. ( We don't need to bother about the water quantity because,if the milk qty is 6L, water qty will automatically be 6L because the solution contains only milk and water, nothing else).

Initial qty of milk = (2/5)*12 = (24/5)
Qty of milk remaining after removing 'x' liters of solution = 24/5 - {(24/5)/12}*x = 24/5 - 2x/5
Qty of milk in the final solution (after 'x' liters of milk is added) = (24/5 - 2x/5) + x = 24/5 - 3x/5
Therefore, 24/5 - 3x/5 = 6.....> x=2. ANS: B

Given: How many litres of a 12 litre mixture containing milk and water in the ratio of 2:3 be replaced with milk so that the resultant mixture contains milk and water in equal proportion?

Milk in 12 litre mixture = .4 * 12 = 4.8 litres
Milk in final solution = 6 litres
Milk to be added = 1.2 litres

Let us assume x litres of mixture is to be replaced with x litres of milk
In x litres of mixture, water = .6x
In x litres of mixture, milk = .4x
Addition of milk = x - .4x = .6x litres = 1.2
x =2

IMO B

Desired final concentration is 50%, or 1/2. Initial concentration of milk in solution is 3/5:

\(\frac{1}{2}=\frac{3}{5}*\frac{x}{12}\)

where x represent the volume of the solution after you have removed some of the original solution but before adding pure milk, and 12 is the total volume of the solution

\(x=10\) so we have to remove 2 liters of the original. Answer is B

con.final=con.initial(volume-removed/volume-removed+pure)
water.initial=3/5, water.final=1/2, volume=12, removed=x, pure=x
\(0.5=3/5(12-x/12-x+x)…30-36=-3x…x=2\)

Answer (B)

Tough question.

12 litres that has a milk:water = 2:3
milk = 4.8L
water = 7.2L

7.2 - 0.6x / 12 = 1/2
x = 2

Answer is B.

12 liter in the ratio of 2:3 = 4.8:7.2
We want to make it 1:1 or 6:6
So, 7.2-6 = 1.2

60% of 1.2 = 2l -- ans

I used the replacement theory described here
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/01 ... -mixtures/

Note that in the entire replacement process,
a) Mixture is homogenous (as its all liquid)
b) Milk is \(\frac{2}{5}\) parts and Water is \(\frac{3}{5}\) parts
c) Quantity of water is untouched (constant) =>\(\frac{ 3}{5}\) in 12 ltr of mixture will be changed to \(\frac{ 1}{2}\) after x ltr of solution is replaced by pure milk
d) quantity of milk is varying (milk is being added to make a new concentration)

Hence going by the replacement formula,
we have to go by water (constant)

x is the quantity of mixture removed (in ltr)

Ci*Vi = Cf*Vf
\(\frac{3}{5}(12-x) = \frac{1}{2}* 12\)
\(=> 12-x = 10\)
\(=> x = 2 ltr\)

Hence answer is B

I feel the question begs for you to assume the resultant mixture has to be 12 litres. If that is the case, then you can solve by this method as well:

..........ratios-> 1 50% 5
Diluted mixture ---I-------------- Pure milk
......40% milk ......................... 100% milk

since the separation has to be 6x apart, and the total has to be 12 litres, then 5x2= will be the old mixture (implying 2 litres removed), and 2 litres added with pure milk

Amount of milk before removing = (2/5) * 12
let Amount of solution removed = X
Amount of milk removed = (2/5)*X
Amount of milk added 100% * X

Final amount of milk = 6 (since ratio has to be same)
12*(2/5) - (2/5)*X + X = 6

X= 2 litres.

could you explain the logic behind this ?

Check this post:
https://gmatclub.com/forum/weighted-ave ... l#p1586082

Posted from my mobile device

Volume of water in original mixture= (3/5)*12

Volume of water in 'x' amount of mixture= (3/5)*x

Volume of water in mixture after replacement= 12/2


\((\frac{3}{5})*12 - (\frac{3}{5})*x = \frac{12}{2}\)

\((\frac{3}{5})*12 -(\frac{3}{5})*(\frac{12}{12})*x = 6\)

\((\frac{3}{5})*12[1-\frac{x}{12}] =6\)

\(7.2[1-\frac{x}{12}] =6\)

Basically we're just subtracting the amount of water that is present in the mixture that we replaced from the amount of water present in the mixture initially.

I think your ratios are wrong. Initial milk:water= 2:3 hence milk cant be 3/5.

step 1 : Original quantities of mixture

Milk = 2x and Water = 3x
Total = 12 litres
So x= 12/5 ---------(1)

Step 2: Remove mixture
Lets say we remove y litres , milk and water will get removed in same proportion as above.
So, milk = 2x- (2/5)y and water = 3x-(3/5)y
Total = 12-y

Step 3 : Add y litres of milk to the mixture
So milk = 2x-(2/5)y + y and water = 3x-(3/5)y
Total = 12-y+y=> 12 litres

As milk and water are in equal proportion ,
2x-(2/5)y + y = 3x-(3/5)y
-> y=(5/6)x -----------(2)


Using 1 and 2 , we know
y=2 litres

Hence option B