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A vessel is filled with liquid, 3 parts of which are

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A vessel is filled with liquid, 3 parts of which are  [#permalink]

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New post 28 Aug 2018, 04:04
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A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/9
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A vessel is filled with liquid, 3 parts of which are  [#permalink]

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New post 28 Aug 2018, 08:42
1
shard87 wrote:
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/9



Adding two more ways..

(I) Straight answer would be the fraction that is removed of total will fraction of syrup that is removed, so 3:5 becomes 4:4, so change in syrup is (5-4) in 5 units so \(\frac{5-4}{5}\)=1/5


Other ways to think of is..

(II) Let total be 80 so water : syrup = 30:50
This becomes 40:40 after transfer
So 10 of syrup moves from syrup to water, but it is a combination of syrup and water that is removed...
So if in 50, we remove 10,
therefore in total (50+30=80), we remove \(\frac{10*80}{50}\)=16

So as a fraction, 16 out of 80 is 16/80=1/5

(III) Ratio \(= 3:5 = 3y:5y\)
say we take out x units this will contain \(\frac{3x}{3+5}\) water and \(\frac{5x}{3+5}\) syrup..
Overall effect on water = \(\frac{x-3x}{8}=\frac{5x}{8}\)
So new ratio = \(3y+\frac{5x}{8}:5y-\frac{5x}{8}=1:1.........(3y+\frac{5x}{8})/(\frac{5y-5x}{8})=\frac{1}{1}\)......
\(3y+\frac{5x}{8}=\frac{5y-5x}{8}...........2y=\frac{5x}{4}.....y=\frac{5x}{8}.\)..

Initial = \(3y+5y=8y=8*\frac{5x}{8}=5x\)
So x is \(\frac{x}{5x}\) of 5x....or 1/5

C
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Re: A vessel is filled with liquid, 3 parts of which are  [#permalink]

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New post 28 Aug 2018, 10:39
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shard87 wrote:
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/9
Answer is C
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Re: A vessel is filled with liquid, 3 parts of which are  [#permalink]

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New post 28 Aug 2018, 21:05
sumit411

1) Why did you take initial concentation 5/8
2) how could the final volume be 8

Nice approach but I guess I am missing some concept on my side.
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Re: A vessel is filled with liquid, 3 parts of which are  [#permalink]

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New post 28 Aug 2018, 22:28
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1) Initial concentration is 5 volumes of syrup in 8 units of total solution. Now this is the basic definition of concentration

Eg what you mean by 10% alcohol solution?
10 units of alcohol in 100 units of solution

2) 3+5=8
So I used 8 units as total volume.

https://gmatclub.com/forum/weighted-ave ... l#p1586082

Check out removal/replacement in mixture topic.

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Re: A vessel is filled with liquid, 3 parts of which are  [#permalink]

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New post 29 Aug 2018, 04:01
shard87 wrote:
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/9


VeritasKarishma and Bunuel May I request you to solve this with an easy approach.
I tried to understand the explanations given in the thread, but couldn't.
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A vessel is filled with liquid, 3 parts of which are  [#permalink]

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New post 29 Aug 2018, 04:23
1
siddreal wrote:
shard87 wrote:
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/9


VeritasKarishma and Bunuel May I request you to solve this with an easy approach.
I tried to understand the explanations given in the thread, but couldn't.


Hi..

Straight answer would be the fraction that is removed of total will fraction of syrup that is removed, so 3:5 becomes 4:4, so change in syrup is (5-4) in 5 units so \(\frac{5-4}{5}\)=1/5


Other way....

Let total be 80 so water : syrup = 30:50
This becomes 40:40
So 10 of syrup moves from syrup to water, but it is a combination of syrup and water that is removed...
So if in 50, we remove 10,
therefore in total (50+30=80), we remove \(\frac{10*80}{50}\)=16

So as a fraction 16 out of 80 is 16/80=1/5
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Re: A vessel is filled with liquid, 3 parts of which are  [#permalink]

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New post 29 Aug 2018, 05:38
1
siddreal wrote:
shard87 wrote:
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/9


VeritasKarishma and Bunuel May I request you to solve this with an easy approach.
I tried to understand the explanations given in the thread, but couldn't.



You can just use the weighted avg formula here:

w1/w2 = (A2 - Aavg) / (Aavg - A1)
The concentration of water in original solution is A1 = 3/8, the concentration of water in second solution is 1 (it is pure water). The concentration of water in final solution is A2 = 1/2

w1/w2 = (1 - 1/2) / (1/2 - 3/8) = 4/1

The original solution must be 4 parts and the water added (by removing some solution) must be 1 part. So 1/5 of the solution must have been replaced with water.

Answer (C)

Check this post for another such question: https://www.veritasprep.com/blog/2012/0 ... -mixtures/
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Re: A vessel is filled with liquid, 3 parts of which are  [#permalink]

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Re: A vessel is filled with liquid, 3 parts of which are   [#permalink] 08 Sep 2019, 00:21
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