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Some part of the 50% solution of acid was replaced with the

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Re: PS: Mixture Problem [#permalink] New post 07 Nov 2010, 09:10
If you notice average of 50 and 30 = 40
That means, taking the weighted average, we know 50% and 30% solution should be of the same quantity.

That mean 1/2 part of 50% solution was replaced by 30% solution
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Re: PS: Mixture Problem [#permalink] New post 08 Nov 2010, 07:10
Thanks a lot guys for your help. :)
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Re: PS: Mixture Problem [#permalink] New post 11 Feb 2011, 14:11
Wonderful stuff this!!

If anyone is looking for a tabular approach, here is a link you may find useful:

http://www.onlinemathlearning.com/mixtu ... s.html#add
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Re: PS: Mixture Problem [#permalink] New post 11 Feb 2011, 23:55
One may solve it by alligation as well.

50 30
\ /
40
/ \
10 10

50 - Percent of acid in first solution
30 - Percent of acid in second solution
40 - Percent of acid in mixture of both
LHS 10 -> Percent of acid in mixture of both - Percent of acid in second solution = 40-30 -> Portion of the second in the mixture
RHS 10 -> Percent of acid in first solution - Percent of acid in mixture of both = 50-40 ->Portion of the first in the mixture

\frac{Portion \quad of \quad second}{Portion \quad of \quad first}=\frac{10}{10}=\frac{1}{1}
\frac{Portion \quad of \quad first}{Total \quad mixture}=\frac{1}{1+1}=\frac{1}{2}

If the new mixture contains 1/2 of the first solution; then half of the first solution must have been replaced with second solution.

Ans: "C"
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Re: PS: Mixture Problem [#permalink] New post 12 Feb 2011, 11:16
Karishmas system her works great.

the 50% "pulls" as hard as the 30% so if its going to 40% it means both "pulls" exactly the same.

so it means we have the same amount of 30% and 50% solution. so we have 1/2 replaced.
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Re: PS: Mixture Problem [#permalink] New post 12 Mar 2011, 22:27
Answer is 50%

for replacement problems use the following formula.

Whole - part removed + part added = New total.
Assume 100 is the total here
Therefore, 50% (100) - 50% (x) + 30% (X) = 40% (100)

Solve...
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Re: PS: Mixture Problem [#permalink] New post 26 Jun 2011, 16:48
Good problem this...took more more than 2 mins -- but C it is. Enjoyed reading the explanations -- which seem to use much more efficient methods.
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Re: PS: Mixture Problem [#permalink] New post 28 Jun 2011, 11:19
Seems like it's easier just to solve it straight.
(30%+50%)/2=40%. So the amount replaced was 50-30 or 20%.
Or am I missing something?
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Re: PS: Mixture Problem [#permalink] New post 28 Jun 2011, 13:05
imerial wrote:
Seems like it's easier just to solve it straight.
(30%+50%)/2=40%. So the amount replaced was 50-30 or 20%.
Or am I missing something?



I don't think this is correct.

You have a 50% solution and a 30% solution. You are replacing part of the 50% solution with the 30% solution to make a 40% solution.

There are lots of different approaches to solve the problem.

Intuitively -- you know that 40% is in between 30% and 50% -- so 1/2 of the 50% solution will need to be replaced.

You can use algebra and solve this or use the tables. You can also use the "allegation" method -- which is quite useful in these problems. I don't quite follow how you got 20%.
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Re: PS: Mixture Problem [#permalink] New post 28 Jun 2011, 19:25
Sorry, the 20% was a typo. My thought process was:

1) We have a 50% solution.
2) We want a 40% solution.
3) We have at our disposal a 30% solution.
4) 40 is the average of 50 and 30, so the solution needs to be half 50 and half 30.
5) Therefore, we replace half the 50% solution with 30% solution.
6) Now instead of being all 50 we're half 50 half 30. That means we replaced half the solution.

Does that work or have I misunderstood? I saw all the complex math in the answers and got confused as to why it was necessary.
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Re: PS: Mixture Problem [#permalink] New post 28 Jun 2011, 20:13
imerial wrote:
Sorry, the 20% was a typo. My thought process was:

1) We have a 50% solution.
2) We want a 40% solution.
3) We have at our disposal a 30% solution.
4) 40 is the average of 50 and 30, so the solution needs to be half 50 and half 30.
5) Therefore, we replace half the 50% solution with 30% solution.
6) Now instead of being all 50 we're half 50 half 30. That means we replaced half the solution.

Does that work or have I misunderstood? I saw all the complex math in the answers and got confused as to why it was necessary.



Yeah..that's perfect.
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Re: PS: Mixture Problem [#permalink] New post 07 Nov 2011, 02:24
lonewolf wrote:
Hermione wrote:
Does anyone have a quick and easy way of solving this problem? I had to pick from the answer choices but I want to know how to solve it outright.

Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

1/5
1/4
1/2
3/4
4/5


Let X = Part of acid solution replaced, then 1-X will be the parts not replaced.

(1-X)*0.5 + 0.3*x = 0.4
0.5 - 0.5X +0.3X = 0.4
-0.2X = -0.1
X=1/2

So the answer is C


Good explanation
Thank you
Re: PS: Mixture Problem   [#permalink] 07 Nov 2011, 02:24
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