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# Two different solutions of alcohol

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Two different solutions of alcohol [#permalink]  13 Aug 2009, 15:12
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100% (04:11) correct 0% (00:00) wrong based on 2 sessions
Two different solutions of alcohol with respective proportions of alcohol to water of 3:1 and 2:3 were combined. What is the concentration of alcohol in the new solution if the first solution was 4 times the amount of the second solution?

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Re: Two different solutions of alcohol [#permalink]  13 Aug 2009, 15:19
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tejal777 wrote:
Two different solutions of alcohol with respective proportions of alcohol to water of 3:1 and 2:3 were combined. What is the concentration of alcohol in the new solution if the first solution was 4 times the amount of the second solution?

SOL:
Lets assume that 40L of sol1 was mixed with 10L of sol2.

Alcohol in Sol1: 40*3/4 = 30L
Alcohol in Sol2: 10*2/5 = 4L

Total Alcohol in the mixed solution of 50L = 30 + 4 = 34L

Alcohol concentration in the new sol: (34/50)*100 = 68%
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Re: Two different solutions of alcohol [#permalink]  04 Feb 2010, 18:40
1
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I have one more way of doing this with the grid

Assume 40 Liters for sol 1 and 10 Liters for sol 2

Liters.....%.......Totals
40........75%....40(.75)
10........40%......10(.40)
50.........X........

So you have 50X=40(.75)+10(.40)
50x=30+4
50x=34
x=.68 or 68%
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Re: Two different solutions of alcohol [#permalink]  11 Dec 2011, 09:57
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Second solution = 20
Thus first solution = 80
New solution (20+80) = 100

Alcohol in first solution = 80*3/4 = 60
Alcohol in second solution = 20*2/5 = 8
Total Alcohol = 68
Concentration of Alcohol 68/100*100= 68% Ans.
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Re: Two different solutions of alcohol [#permalink]  14 Aug 2009, 16:51
Thank you.It's perfect!
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Re: Two different solutions of alcohol [#permalink]  15 Aug 2009, 02:35
Good strategy by samrus. A slightly different way using variables:

Let sol1 have x liters, and sol2 have y liters.
The ratio of alcohol to water in the final solution will be :
[ (3/4)*x + (2/5)*y ] / [ (1/4)*x + (3/5)*y ]

Now, x=4y, substituting x in the above ratio, we get 17/8 as the ratio of alcohol to water in the final solution.
Hence, concentration of alcohol is (17/25)*100 % = 68%
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Re: Two different solutions of alcohol [#permalink]  16 Apr 2010, 14:30
Solution B Let it be X
Ratio - 2:3
2y+3Y=X
5y=X
so y=X/5
Now ratio converted to amount becomes 2X/5 and 3X/5

Solution A will be 4X
Ratio - 3:1
3y+1y=4X
y=X
Ratio converted to amount = 3X and 1X

Mixture (A+B) Amount = 3x+1x = 4x
Alcohol amount = 3X+ 2X/5 -----------------------FROM ABOVE
= 17X/5

Concentration = individual amount /total amount *100 = 17x/5 * 1/5x *100 = 17/25*100 = 68%
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Re: Two different solutions of alcohol [#permalink]  16 Apr 2010, 16:36
A different approach:
Alcohol in soln1 = 3/4 and in soln2 = 2/5
Using Alligation:
3/4--------?--2/5
4 : 1
So the resultant ratio (the ?) is at a 1/5th distance from 3/4 (or conversely at a 4/5th distance from 2/5.
3/4 - 2/5 = 7/20
Now, calculating 1/5th of 7/20 and subtracting it from 3/4
(3/4) - (1*5)/(7*20)
= 75/100 - 7/100
= 68/100
= 68%

This approach is not quite advisable in this particular problem unless one is instinctive wrt alligation, but definitely awfully handy in other scenarios. This is just to enlighten people towards proper understanding of Alligation
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Re: Two different solutions of alcohol [#permalink]  11 Dec 2011, 22:08
perfect explained by samrus98
Re: Two different solutions of alcohol   [#permalink] 11 Dec 2011, 22:08
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