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

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Senior Manager
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A certain quantity of 40% solution is replaced with 25% [#permalink]  24 Dec 2003, 20:28
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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shubhangi

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Just out of curiosity - is this a Kaplan problem?
Senior Manager
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No this is not from Kaplan.. and answer is B ......how did u get it Geetu.Plz explain.
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shubhangi

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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.

Last edited by Titleist on 25 Dec 2003, 16:04, edited 1 time in total.
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I find it difficult to explain it in words. So, I am pasting the way i did it.
Attachments

mixture problem.JPG [ 26.98 KiB | Viewed 435 times ]

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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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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
___________________________________________________
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.
Senior Manager
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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
Senior Manager
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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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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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Hi gmatblast,

Your solution to the problem is the best. I will remember the formula.

Anand.
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