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Two different solutions of alcohol with respective proportio [#permalink]

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13 Aug 2009, 15:12

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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]

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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?

Detailed explanations please!

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]

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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%

Re: Two different solutions of alcohol [#permalink]

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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 _________________

In a Normal Distribution, only the Average'Stand Out'

Re: Two different solutions of alcohol [#permalink]

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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]

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01 Jun 2014, 23:11

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Re: Two different solutions of alcohol with respective proportio [#permalink]

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10 Sep 2014, 13:11

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?

Re: Two different solutions of alcohol with respective proportio [#permalink]

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01 Sep 2016, 04:05

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Two different solutions of alcohol with respective proportio [#permalink]

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01 Sep 2016, 09:10

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?

let x=% of alcohol in new solution 3/4*4+2/5=5x x=68%

Re: Two different solutions of alcohol with respective proportio [#permalink]

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11 Sep 2016, 22:55

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Can someone help to explain why solving the question as below didnt give OA? x: the proportion of alcohol to water of new solution (x-2/3)/(3-x)=4 15x=38 x=38/15 =>> the concentration of alcohol in the new solution is 38/(38+15)=38/53??

Thank you in advance!!!

gmatclubot

Re: Two different solutions of alcohol with respective proportio
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11 Sep 2016, 22:55

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