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# Two different solutions of alcohol with respective

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Two different solutions of alcohol with respective [#permalink]  07 Jan 2008, 08:04
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: mixtures - alcohol [#permalink]  07 Jan 2008, 08:14
I get 17:8

For the amount of solution A to be 4 times that of solution B, we need a total of 20 units of A to 5 of B. 20 of A will include 15 units of alcohol, 5 of water. We already know that 5 units of B contains 2 units of alcohol and 3 of water. Add the two together and you get 17:8.
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Re: mixtures - alcohol [#permalink]  07 Jan 2008, 09:16
Let's say we have 100 liters of the new mixture
80 of solution A
20 of solution B

3:1 ratio of 80L will be 60 liters alcohol to 20 liters water
2:3 ratio of 20L will be 8 liters alcohol to 12 liters water

68/100 = concentration of alcohol in new solution = 17/25 or 68%

17:8 is the ratio of the new solution, but they're asking for the concentration of alcohol
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Re: mixtures - alcohol [#permalink]  07 Jan 2008, 10:36
i would sum the values in the following way
(3+1)*4+(2+3) to get 25 units of a comdined substance
of which (20*3/4 + 2) units is an alcohol, so
17/25 should be it.
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Re: mixtures - alcohol [#permalink]  07 Jan 2008, 13:57
17:8

First solution is 4 times the amount of second solution, so assume..

First solution is 30:10 ( same as 3:1, but total solution is 40 units..)
Second solution is 4:6 ( same as 2 : 3, but total solution is 10 units..)

Mix this.. then the ratio of alcohol to water is 34 : 16, i.e 17 : 8
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Re: mixtures - alcohol [#permalink]  07 Jan 2008, 14:43
eschn3am wrote:
Let's say we have 100 liters of the new mixture
80 of solution A
20 of solution B

3:1 ratio of 80L will be 60 liters alcohol to 20 liters water
2:3 ratio of 20L will be 8 liters alcohol to 12 liters water

68/100 = concentration of alcohol in new solution = 17/25 or 68%

17:8 is the ratio of the new solution, but they're asking for the concentration of alcohol

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Re: mixtures - alcohol [#permalink]  07 Jan 2008, 22:11
Let x be the quantity of first solution and y the quantity of second solution. The concentration of alcohol in the mix will be:
[3/4x+2/5y]/[x+y]*100. Since x/y=4 we know that [(3/4)*4+2/5]/[4+1]*100=68%
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Re: mixtures - alcohol [#permalink]  07 Jan 2008, 22:38
It should be 17/25.
Because the question is asking about the concentration of Alcohal but not the ratio of alcohal to water.

Ratio of Alcohal to Water in the solution is 17:8.

But concentration of Alcohal = Amount of alcohal / totla solution ==> it is 17/(17+8) ==17/25
Re: mixtures - alcohol   [#permalink] 07 Jan 2008, 22:38
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# Two different solutions of alcohol with respective

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