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Two different solutions of alcohol with respective proportions of

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tejal777
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Two different solutions of alcohol with respective proportions of [#permalink] New post   13 Aug 2009, 16:12
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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?

A. 17%
B. 25%
C. 34%
D. 50%
E. 68%
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E
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samrus98
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Re: Two different solutions of alcohol with respective proportions of [#permalink] New post   13 Aug 2009, 16:19
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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 with respective proportions of [#permalink] New post   15 Aug 2009, 03:35
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 with respective proportions of [#permalink] New post   04 Feb 2010, 19:40
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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 with respective proportions of [#permalink] New post   16 Apr 2010, 15:30
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 with respective proportions of [#permalink] New post   16 Apr 2010, 17:36
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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 with respective proportions of [#permalink] New post   11 Dec 2011, 10:57
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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 with respective proportions of [#permalink] New post   10 Sep 2014, 14:11
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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?



(x-(2/5))/((3/4)-x) = 4/1

5x=2/5+3

x=17/25

%= 68%
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Two different solutions of alcohol with respective proportions of [#permalink] New post   Updated on: 29 Jan 2018, 13:40
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?

A. 17%
B. 25%
C. 34%
D. 50%
E. 68%

let x=amount of second solution
let a=% of alcohol in combined solution
3/4*4x+2/5*x=a*5x
→3/4*4+2/5=5a
→a=.68=68%
E

Originally posted by gracie on 01 Sep 2016, 10:10.
Last edited by gracie on 29 Jan 2018, 13:40, edited 1 time in total.
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Re: Two different solutions of alcohol with respective proportions of [#permalink] New post   11 Sep 2016, 23:55
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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!!!
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Re: Two different solutions of alcohol with respective proportions of [#permalink] New post   10 Oct 2017, 08:41
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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?

Track on alcohol; use water only to determine the percentage of alcohol in each solution.

Mixture A = .75 alcohol
\(\frac{3A}{1W}\): 4 parts total
Alcohol is 3 parts of 4, or
\(\frac{3}{4}\) = .75 of Mixture A

Mixture B = .40 alcohol
\(\frac{2A}{3W}\): 5 parts total
Alcohol is 2 parts of 5, or
\(\frac{2}{5}\) = .40 of Mixture B

Percent alcohol = concentration

\((Concen_{A})(Vol_{A}) + (Concen_{B})(Vol_{B})=\)

\((Concen_{A+B})(Vol_{A+B})\)

Mixture A has 4 times the volume of Mixture B. Let volume of A = 4 and volume of B = 1, where x = percentage of alcohol in final mixture

(.75)(4) + (.40)(1) = (x)(4+1)
3 + .4 = 5x
3.4 = 5x

x = \(\frac{3.4}{5} = \frac{34}{50} = \frac{68}{100} = x\)

x = 68 percent
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Re: Two different solutions of alcohol with respective proportions of [#permalink] New post   27 Jan 2018, 22:34
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?



simplest approach


(3/4 (4) + 2/5(1) ) / (1/4 (4) + 3/5 (1)) = 17/ 8
so conc of alcohol is 17/25
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Re: Two different solutions of alcohol with respective proportions of [#permalink] New post   28 Jan 2018, 07:17
email2vm wrote:
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?



(x-(2/5))/((3/4)-x) = 4/1

5x=2/5+3

x=17/25

%= 68%


Hi generis, :)

can you help me to understand why in the picture above (alligation method) instead of 4/1 is "x" :? isnt 4/1 combined ratio ? :?

i am trying to understand alligation method and drum this concept into my head :)

is there difference between weighted average method and alligation ?

many thanks for your contribution to my understanding :)

D. :-)
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Re: Two different solutions of alcohol with respective proportions of [#permalink] New post   28 Jan 2018, 13:13
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dave13 wrote:
email2vm wrote:
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?



(x-(2/5))/((3/4)-x) = 4/1

5x=2/5+3

x=17/25

%= 68%


Hi generis, :)

can you help me to understand why in the picture above (alligation method) instead of 4/1 is "x" :? isnt 4/1 combined ratio ? :?

i am trying to understand alligation method and drum this concept into my head :)

is there difference between weighted average method and alligation ?

many thanks for your contribution to my understanding :)

D. :-)


Hi dave13

In allegation method, the center point (which is \(x\) here) usually refers to the quantity of the mixture.

So if you have two solutions A & B whose alcohol content is \(\frac{3}{4}\) & \(\frac{2}{5}\) respectively, then alcohol content of mixture will be represented by \(x\) here.

the ratio of difference of individual content with the content of the mixture gives you the ratio of quantities in the mixture.

\(=>(x-2/5)/(3/4-x)=\frac{4}{1}\) (this is similar to weighted average method)

The important thing to note in allegation method is the center point refers to the quantity of the mixture and not ratio of elements of the mixture.
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Re: Two different solutions of alcohol with respective proportions of [#permalink] New post   29 Jan 2018, 12:12
Many thanks niks18 for clear explanation!:) I bookmarked your post :) what do you think of this idea :) https://gmatclub.com/forum/bookmarks-25 ... l#p2005645
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Re: Two different solutions of alcohol with respective proportions of [#permalink] New post   19 Aug 2021, 00:25
Hi,

Can someone please explain me why are we considering Sol1 40L and Sol2 10L?. Question says Sol1 was 4 times Sol2.
So it should be Sol1 = 4 * Sol2 right?

Thanks
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