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# 30%-solution of alcohol was mixed with a 50%-solution of

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VP
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30%-solution of alcohol was mixed with a 50%-solution of [#permalink]

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08 Mar 2008, 05:47
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30%-solution of alcohol was mixed with a 50%-solution of alcohol to obtain 10 liters of a 45%-solution of alcohol. How much of the 30%-solution was used?
 2.0 liters
 2.5 liters
 2.7 liters
 3.0 liters
 3.2 liters
Director
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08 Mar 2008, 09:26

Using the mixtures principles

30 50
45
5 15

So the new solution will have 30% to 50% solutions in 1:3 ratio, but the total volume is 10 Ltrs

thus 30% solution will be 2.5 Ltrs
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SVP
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08 Mar 2008, 09:31
prasannar wrote:

Using the mixtures principles

30 50
45
5 15

So the new solution will have 30% to 50% solutions in 1:3 ratio, but the total volume is 10 Ltrs

thus 30% solution will be 2.5 Ltrs

could you elaborate on what the principles are, and work through this one step by step ?
VP
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08 Mar 2008, 09:45
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pmenon wrote:
prasannar wrote:

Using the mixtures principles

30 50
45
5 15

So the new solution will have 30% to 50% solutions in 1:3 ratio, but the total volume is 10 Ltrs

thus 30% solution will be 2.5 Ltrs

could you elaborate on what the principles are, and work through this one step by step ?

I'll tell you the underlying concept.

let C1 be the first concentration and C2 the second.
C1=30/100 of x, where x is the relative amount of water needed
C2=50/100 of y, where y is the relative amount of water needed

thus 30/100 x * 50/100 y= 45/100 (x+y)

and

x+y=10

solving for x and y you'll find the answer.

the mechanical formula to solve mixtures problems is found here

7-t26249

hope this helps
CEO
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08 Mar 2008, 19:05
marcodonzelli wrote:
30%-solution of alcohol was mixed with a 50%-solution of alcohol to obtain 10 liters of a 45%-solution of alcohol. How much of the 30%-solution was used?
 2.0 liters
 2.5 liters
 2.7 liters
 3.0 liters
 3.2 liters

x+y=10

.3x+.5y=.45x+.45y --> .05y=.15x y=3x

thus 4x=10 --> 10/4=x --> 5/2 thus 2.5 liters.
Re: Mixtures   [#permalink] 08 Mar 2008, 19:05
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