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M25-30

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M25-30  [#permalink]

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New post 16 Sep 2014, 00:24
1
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A
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C
D
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  25% (medium)

Question Stats:

77% (01:13) correct 23% (01:45) wrong based on 402 sessions

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Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as the result, a 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. \(\frac{4}{5}\)

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Re M25-30  [#permalink]

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New post 16 Sep 2014, 00:24
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Official Solution:

Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as the result, a 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.

Alternatively, you can do the following, say \(x\) part of the original solution was replaced then: \((1-x)*0.5+x*0.3=1*0.4\), which gives \(x=\frac{1}{2}\).


Answer: C
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Re: M25-30  [#permalink]

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New post 28 Feb 2015, 12:41
Bunuel wrote:
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. \(\frac{4}{5}\)


Hi,

Can I got the answer right but I hope my approach is correct in tackling this question. Let x equal to the original solution, now if we remove 50% from the original we get:

100%x - 50%x = 50%x remaining solution

from the remaining solution, 30% was added. Therefore we get 50%x+30%x = 80%x now we know the remaining is 40% solution. So we can create an equation of 80%x=40% and now if we isolate x we get 1/2 -> Answer choice C
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Re: M25-30  [#permalink]

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New post 20 Feb 2016, 20:45
At first I thought that this was too easy so maybe I fell for a trick. But after triple checking, yeah its just half.
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Re: M25-30  [#permalink]

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New post 07 Mar 2016, 18:49
Another way to solve this: When I add 30% acid I end up with 40% acid so the "x" amount i subtracted at the beginning leaves a 10% acid of the original solution and a 90% water of the original solution.
let's say I had 10 acid and 10 water>I extract> I now have 1 acid and and 9 water. I have 10 units vs the 20 I had to start off with, so I extracted 50%.
This would also answer the question, what % of water and acid did I extract? 40% acid and 10% water,
hope I was helpful :)
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Re: M25-30  [#permalink]

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New post 24 Jan 2017, 10:04
why are we doing (1−x)∗0.5? why it shouldn't be x*0.5?
pls explain
thnx
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New post 25 Jan 2017, 05:23
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Re: M25-30  [#permalink]

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New post 27 Oct 2017, 18:21
I have a slightly extended version for a question like this:
1/2L of 30% acid solution was replaced with 1/4L of 50% solution. If the original solution had 40% solution of acid, what was is the amount of acid in the new solution?

Can someone solve this, please?
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Re M25-30  [#permalink]

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New post 22 Nov 2017, 00:38
I agree with explanation. other method can be alligation:

since 50% and 30% solutions giving 40%. therefore, new ratio is 1:1 (meaning both original solutions formed equal part). therefore, 50% solution's only half part was replaced (1/2. as other half formed by 30% solution)
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Re: M25-30  [#permalink]

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New post 10 Jul 2018, 15:55
I like to do the "see-saw" approach for this.
You started at 50, replaced with 30. So 50-30 = 20. 20 is the total of the potential change.
The result was 40. So 50-40 = 10. 10 is the change in the composition of the new mixture vs. the composition of the old mixture.
If we take the actual change over the potential change, or 10/20, we get 1/2, which is the answer.
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Re: M25-30 &nbs [#permalink] 10 Jul 2018, 15:55
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