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

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A certain quantity of 40% solution is replaced with 25% solu [#permalink] New post 24 Dec 2003, 20:28
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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
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 [#permalink] New post 25 Dec 2003, 13:51
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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. :P

Last edited by Titleist on 25 Dec 2003, 16:04, edited 1 time in total.
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 [#permalink] New post 25 Dec 2003, 14:07
I find it difficult to explain it in words. So, I am pasting the way i did it.
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 [#permalink] New post 25 Dec 2003, 14:43
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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?[/b]
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 [#permalink] New post 26 Dec 2003, 11:43
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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
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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.
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 [#permalink] New post 26 Dec 2003, 12:05
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
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 [#permalink] New post 31 Dec 2003, 08:21
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
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 [#permalink] New post 10 Jan 2004, 17:53
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
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Re: Mixture problem [#permalink] New post 08 May 2009, 10:41
gmatprep09 wrote:
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


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
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Re: Mixture problem [#permalink] New post 10 May 2009, 03:29
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
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Re: Mixture problem [#permalink] New post 19 May 2009, 23:23
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
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Re: Mixture problem [#permalink] New post 20 May 2009, 02:13
tkarthi4u wrote:
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


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
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Re: Mixture problem [#permalink] New post 21 Jul 2011, 20:09
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gmatprep09 wrote:
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


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}
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Re: Mixture problem [#permalink] New post 22 Jul 2011, 02:30
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gmatprep09 wrote:
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


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.

For explanation of the formula, see:
http://www.veritasprep.com/blog/2011/03 ... -averages/
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Re: A certain quantity of 40% solution is replaced with 25% [#permalink] New post 30 May 2014, 11:34
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Re: A certain quantity of 40% solution is replaced with 25%   [#permalink] 30 May 2014, 11:34
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