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# mixture problem

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Manager
Joined: 31 Aug 2009
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mixture problem [#permalink]  07 Sep 2009, 18:40
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(N/A)

Question Stats:

100% (01:47) correct 0% (00:00) wrong based on 8 sessions
I actually don't even understand the explanation. Can someone elaborate? Thanks.

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?

1/5
1/4
1/2
3/4
4/5
Senior Manager
Joined: 08 Jan 2009
Posts: 329
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Kudos [?]: 88 [0], given: 5

Re: mixture problem [#permalink]  07 Sep 2009, 23:46
50% Acid soln A : W = 1 : 1
30 % Acid son A :W = 3 :7

so quantity x is removed from and replaced with 30 %

x liters is taken out from 50% Acid soln

the remaining is A: W = 1 -x/2 : 1 -x/2

when x liters is taken from 30% Acid soln they come out as = 3x/10 and W = 7x/10

and it is added to 50% acid soln = 1-x/2 + 3x/10 : 1-x/2 + 7x/10

so (1-x/2 + 3x/10 ) / (1-x/2 + 3x/10 + 1-x/2 + 7x/10) = 40 /100

so i got x = 1

so what part of the original was replaced i got 1/2.

Let me know if i am wrong
Manager
Joined: 29 Jul 2009
Posts: 123
Location: France
GPA: 3.95
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 96 [1] , given: 15

Re: mixture problem [#permalink]  08 Sep 2009, 03:27
1
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Ratio 1 : 1
i.e. 1 +1 = 2 Total

= 1 /2
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Manager
Joined: 10 Jul 2009
Posts: 129
Location: Ukraine, Kyiv
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Re: mixture problem [#permalink]  17 Sep 2009, 10:29
post OA, please. Got 1/2 too.
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Never, never, never give up

Manager
Joined: 27 Oct 2008
Posts: 185
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Kudos [?]: 102 [1] , given: 3

Re: mixture problem [#permalink]  19 Sep 2009, 05:37
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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 a result, 40% solution of acid was obtained, what part of the original solution was replaced?

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

Soln: Lets take the quantity of 50% solution of acid to be 1 liter
Now let x liter of 50% solution be removed and an equal amount of 30% solution be added to make a liter of 40% solution of acid.
initial quantity of acid in 1 liter of 50% solution is .5
quantity of acid that will be removed when x liter of 50% solution is removed is .5x
quantity of acid that will be added when x liter of 30% solution is added is .3x
so the new ratio of acid in solution is,
(.5 -.5x + .3x)/1 = 40%
.5 - .2x = .4
.2x = .1
x = 1/2

Therefore the quantity that will get replaced is = 1/2 liter
Re: mixture problem   [#permalink] 19 Sep 2009, 05:37
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# mixture problem

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