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

Question

A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?

(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 3/4

Geethu · Accepted Answer

I find it difficult to explain it in words. So, I am pasting the way i did it.

Paul · Answer

Indeed, not easy to explain with words. My explanation is:

Let's say that the total original mixture  A  is 100ml

The original mixture  A  thus has 40ml of alcohol out of 100ml of solution
You want to replace some of that original mixture  A  with another mixture  B  that contains 25ml of alcohol per 100ml. Thus, the difference between 40ml and 25ml is 15ml per 100ml of mixture. This means that everytime you replace 100ml of the original mixture  A  by 100ml of mixture  B , the original alcohol concentration will decrease by 15%. The question says that the new mixture, let's call it  C , must be 35% alcohol, a decrease of only 5%. Therefore, 5 out of 15 is 1/3 and B is the answer. Was that clear?

Titleist · Answer

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.

Ro · Answer

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.

gmatblast · Answer

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

x2suresh · Answer

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

tkarthi4u · Answer

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

humans · Answer

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

fluke · Answer

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}

KarishmaB · Answer

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.

KarishmaB · Answer

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.

gracie · Answer

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

SVaidyaraman · Answer

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.

BrentGMATPrepNow · Answer

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: 
 https://i.imgur.com/vjTkmxQ.png

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  
 https://i.imgur.com/zW4vAkr.png

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    
  https://i.imgur.com/YvjD2wH.png

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: 
 https://i.imgur.com/f7G24ZJ.png

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

RELATED VIDEO 
 https://www.youtube.com/watch?v=7Or9gzROsjQ

Malpora · Answer

Let total quantity be 100 mL,
Let quantity replaced be 'y' mL,

By mass balance,

0.25 * y + 0.40 * (100-y) = 35
0.15 * y = 5
y = 100 /3

fraction = 1/3

maelstrom93 · Answer

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

iamsujatha · Answer

let 100ml be the total amount of liquid, let's call the solution "A"
Therefore, it has 40% of the solution i.e. 40ml
Now, let " x " ml be the amount of solution that was replaced
 Tip:  just remember that in solution replacement related problems,  the total amount of liquid replaced = the total amout of liquid that is added
Hence, the overall quantity of the liquid will not change, i.e. it will remain 100 
Which means, if "x" amount of liquid was replaced, within x, % of the solution A is 40% i.e. 0.4x
similarly, the new liquid "x" added has 25% of that solution i.e. 0.25x

So our equations comes to -

(40-0.4x+0.25x) / 100 = 35/100 -
when you solve this you get x = 33.3333

now we need the fraction of the solution that was replaced -
which is -
Amount replaced / original amount

33.3333/ 100 which is equivalent to 1/3

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

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