A certain quantity of 40% solution is replaced with 25% solu

Or use our standard mixtures formula for replacements too. In a certain quantity of 40% solution, 25% solution is added to give 35% solution.
w1/w2 = (A2 - Aavg)/(Aavg - A1) = (40 - 35)/(35 - 25) = 1/2
So quantity of 40% sol:25% solution = 2:1
This means the initial total solution was 3 and the fraction of 25% now in the mixture is 1. Therefore, 1/3 of the 40% solution was removed and replaced with 25% solution.

Yet another way:


Initial solution = x
concentration of solvent = .4x

Lets remove 'y' from the total solution
Solvent in the removed solution = .4y

We add back 'y' into the solution
Solvent in the added solution = .25y
___________________________________________________
Adding,
Total solution = x-y+y = x
Solvent = .4x - .4y + .25y = .4x - .15y

Now,
.4x - .15y = .35x (new concentration)

Solve for y = 1/3 of x.

In this kinds of problems, we should always try to apply the concept of weighted average.

(strength of one solution) (quantity of that solution) + (strength of another solution) (quantity of that solution) = (strength of resultant solution) (quantity of the resultant solution)

(0.40) (1-q) + (0.25)q = (0.35) (1)

Solving for q = 1/3

Responding to a pm:



From your calculations, you get that when you mix 2 parts of 40% solution with 1 part of 25% solution, you get resultant 35% solution.

Initially, you had only 40% solution. You removed say x of it and put x of 25% solution in its place. This x was 1 part and you had 2 parts of 40% solution left. So initially, you must have had 3 parts of 40% solution. You then must have removed 1 part and replaced it with 25% solution. That is how you would have ended up mixing 2 parts of 40% with 1 part of 25% to get 35% solution.

So we must have replaced 1 part out of 3 (i.e. 1/3) of the original 40% solution.

Let actual solution be "T"
Replaced solution be "R"

(T-R)-> 40%
R->25%
Average->35%

(T-R)0.4+R*0.25=0.35T
0.4T-0.4R+0.25R=0.35T
0.05T=0.15R
R/T=0.05/0.15=1/3

Ans: "B"

OR using other form of Weighted Average:

\(\frac{T-R}{R}=\frac{35-25}{40-35}\)

\(\frac{T-R}{R}=\frac{10}{5}=2\)

\(\frac{T}{R}-1=\frac{10}{5}=2\)

\(\frac{T}{R}=3\)

Invertendo:
\(\frac{R}{T}=\frac{1}{3}\)

oringal quantity of solution=1
solution replaced = x

0.4 *(1-x)+0.25x= 0.35 *1

0.05= 0.15 x--> x= 1/3

let x=fraction of solution replaced
.4-.4x+.25x=.35
x=1/3

When solving mixture questions, we can often get a much clearer picture by sketching the solutions with the components separated

Since the question asks us to find a certain FRACTION, let's assign a nice value to the starting volume.

So let's start with 100 ml of a solution that's 40% salt (Why salt? Why not?).
If we separate the two components, we get the following:



Now let's REMOVE x ml of solution.
This means 100 - x = the resulting volume of the entire solution
Since 40% of this volume is salt, the resulting volume of salt = 40% of (100 - x) ml
In other words: the resulting volume of salt = 0.4(100 - x) ml
Expand to get: the resulting volume of salt = 40 - 0.4x ml



Now let's ADD x ml of the 25% solution.
The volume of salt in this solution = 25% of x
In other words: the volume of salt in this solution = 0.25x ml




To determine how much salt there is in the resulting mixture, simply add the volume of salt from the two mixtures we're combining
In other words, the TOTAL amount of salt = (40 - 0.4x ml) + 0.25x ml
= 40 - 0.15x ml
We get:



Since we are told the resulting mixture is 35% salt, we can write: (40 - 0.15x)/100 = 35/100
This means that: 40 - 0.15x = 35
Rearrange to get: 5 = 0.15x
Solve: x = 5/0.15 = 500/15 = 100/3 = 33 1/3

In other words, 33 1/3 mls were removed from the original 100 ml of solution.
In other words 1/3 of the solution was removed

Answer: B

Cheers,
Brent

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Ok Shabang, I'll try to explain this to you since it appears most people are away celebrating x-mas. Studying for the GMAT is how I'm celebrating X-mas this year.

I would pick numbers here and scan the answer choices (also think logically - the difference in the percentage of the solution declines by only 5% when added with the diluted solution - thus I would get rid of answer choices c,d,e - so I'm left with a and b) I chose b off the bat:

Pick 60 (ml, oz, whatever) as the total mixture - it works well with 3, 4, and 5.

You have a mixture that is 40% solution: 2:5=x:60 thus x = 24 solution : 60 total mixture

Using answer choice B 1/3 - plug it in. 1/3 of 60 is 20 so you're left with 40 oz of the solution. Thus the new solution is 2:5=x:40 x=16 solution: 40 total mixture. You're adding 20 oz of a diluted mixture. thus 1/4 = x/20 = 5 solution: 20 total mixture. Add them together you have: 21 solution : 60 total mixture or 21/60 = 35%.

I'm a little buzzed - I hoep ti amkes sense.

Let me try out, the logic here is

You removed x quantity of 40% concentration solution from 1 and added same x quantity of 25% concentration solution, which total to original quantity 1 of solution with 35% concentration. Hence

Remaining quantity of 40% solution + added quantity of 25% solution = Total solution with 35% concentration.
0.4(1-x) + 0.25x = .35

1. (Quantity of 40%solution- quantity removed) *(concentration of the solution) + (Quantity of 25% solution added) *(concentration of the solution)/ (Initial Quantity) = 35/100
2. Let x be the initial quantity of 40% solution. Let y of it be removed and y of 25% solution added.
3. (x -y)*0.4 + y*0.25/x = 0.35
4. y/x=1/3 or in other words 1/3 of the solution was replaced.

Original = 100mL
100(0.4) - x (0.4) + x (0.25) = 100 (0.35)
x = 33.3333
33.3333/100 = 1/3

HereтАЩs my solution:


Original quantity = A
Substituted quantity = B

Then:

(A*0.4 + 0.25*B тАУ 0.4*B ) / A = 0.35

0.4 + (B/A)*(-0.15)=0.35

B/A=-0.05/-0.15=1/3

This is how I did it. But it took some time for me to come up with a solution.

I like beer so I will go with this example.
The beer contained 40% alcohol 60% water. from this x amount was taken out. This x amount will carry same amount of alcohol with it so we have

0.4a + 0.6w - ( 0.4ax + 0.6wx )

then we add same x with 25% alcohol
so we have

0.4a+0.6w - ( 0.4ax + 0.6wx ) + ( 0.25ax + 0.75 wx )
= 0.4a+0.6w-(0.15ax - 0.15wx )
this equals beer with 35% alcohol

0.4a+0.6w-(0.15ax-0.15wx) = 0.35a+0.65w

0.15ax-0.15wx = 0.05a-0.05w
so
x = 0.05(a-w) / 0.15(a-w) = 5/15 = 1/3

Can u please the below eqn ?

i belive here 1-x is the remaining soln. Now can you why are we taking 40 % of 1-x. I do not get it. Appreciate ur help.

0.4 *(1-x)+0.25x= 0.35 *1

I back solved it:
Since it deals with percent, lets take 100 as the base.

When 1/4 - total mixrure is 40 + 25*4 / 500 = 140/500 ; this is not eq 35.
When 1/3 - total mixture is 40 + 25*3/ 400 = 115/400 = 35% - this is the ans.
When 1/2 - total mixture is 40 + 20*2/ 300 = 80/300 ; not eq 35%
Similarly for D & E.

Thus B is correct. IMO

Geethu - how did u convert ratio 2:1 to 1/3..?