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A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10

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A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 15 Aug 2007, 19:06
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A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 litre mixture of 45% alcohol. How much of the 30% mixture was used?
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 15 Aug 2007, 19:10
fresinha12 wrote:
a 30% alchol mixture is added to a 50% alchol mixture to form a 10 litre mixture of 45% alchol. How much of the 30% mixture was used?


Hi Hi
x = amount of 30% solution
30x + 50(10-x) = 45*10
x = 2.5
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 15 Aug 2007, 20:10
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a 30% alchol mixture is added to a 50% alchol mixture to form a 10 litre mixture of 45% alchol. How much of the 30% mixture was used?


.30x + .50y = .45(x+y); x+y=10; y=10-x;

.30x + .5(10-x) = .45x + .45(10-x)
.30x + 5 -.5x = .45x + 4.5 - .45x
5 - .2x = 4.5
0.5 = .2x
x = 5/2 or 2.5
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 15 Aug 2007, 21:00
for some odd reason I get 7.5


here is my working


solution A=50% solution solution b=30% solution

30%-45% =15% solution B
50%-45%=5% soltuion A

so the ratio is 3:1 for 30%:50% solutions

3/4 *10 liter=7.5 for 30% solution and 2.5 for 50% solution..

what am i doing wrong??? maybe its too late (midnight nyc time) and i am as usual fried
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 15 Aug 2007, 21:31
fresinha12 wrote:
for some odd reason I get 7.5


here is my working


solution A=50% solution solution b=30% solution

30%-45% =15% solution B
50%-45%=5% soltuion A

so the ratio is 3:1 for 30%:50% solutions

3/4 *10 liter=7.5 for 30% solution and 2.5 for 50% solution..

what am i doing wrong??? maybe its too late (midnight nyc time) and i am as usual fried


Your ratios are backwards - the ratios should be inverted:

A : B (30%:50%) should be 1:3 not 3:1

here is a helpful link for good reference: http://www.gmatclub.com/phpbb/viewtopic.php?t=49897
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 16 Aug 2007, 08:14
fresinha12 wrote:
a 30% alchol mixture is added to a 50% alchol mixture to form a 10 litre mixture of 45% alchol. How much of the 30% mixture was used?


Let x be the volume of 30% mixture
Let y be the volume of 50% mixture

From the problem we can get two equations:
Equation 1 (alcohol concentration of the resulting mixture): (x·0.3+y·0.5)/10=0.45
Equation 2 (total volume of the resulting mixture): x+y = 10

from equation 2 => y= 10 -x
I substitue that in Equation 1 to get: (x·0.3 + (10-x)·0.5 / 10 = 0.45

solving this simple equation gets you to x = 2.5 L
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 25 Jan 2010, 22:41
x+y=10
15x=5y
y=3x
x+3x=10
x=2.5
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 11 Jul 2010, 06:42
Please tell me whether this method is correct or not:

Original Added Result
Concentration .5 .3 .45
Amount (10-x) x 10
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 13 Mar 2011, 01:57
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0.30 X + 0.5 (Y) = 0.45 (10)
Remember X+Y=10
Therefore 2 equations can be used to solve...
x=2.5
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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New post 13 Mar 2011, 04:45
I don't think we need 2 variables :

0.3x + (10-x)0.5 = 4.5

=> 5 - 0.2x = 4.5
=> x = 2.5
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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10 [#permalink]

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Re: A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10   [#permalink] 05 Jun 2016, 03:33
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A 30% alcohol mixture is added to a 50% alcohol mixture to form a 10

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