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Re: Some part of the 50% solution of acid was replaced with the [#permalink]
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.
Re: Some part of the 50% solution of acid was replaced with the [#permalink]
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%.
Re: Some part of the 50% solution of acid was replaced with the [#permalink]
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.
Re: Some part of the 50% solution of acid was replaced with the [#permalink]
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.
Re: Some part of the 50% solution of acid was replaced with the [#permalink]
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.
Re: Some part of the 50% solution of acid was replaced with the [#permalink]
06 Sep 2014, 18:17
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Re: Some part of the 50% solution of acid was replaced with the [#permalink]
06 Sep 2014, 18:54
Expert's post
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{2}\) D. \(\frac{3}{4}\) E. \(\frac{4}{5}\)
Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.
Answer: C.
Alternately you can do the following, say \(x\) part of the original solution was replaced then: \((1-x)*0.5+x*0.3=1*0.4\) --> \(x=\frac{1}{2}\).
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