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If two different solutions of alcohol with a respective prop

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If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?

A. 30.0%
B. 36.6%
C. 42.5%
D. 44.4%
E. 60.0%

This is how i did it:
propn of alcohol in first solution : 1:4
propn of alcohol in second solution: 3:5
so, propn of alcohol in the new solution = 4:9 = 44.4%
[Reveal] Spoiler: OA

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Re: mixture [#permalink]

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New post 20 Aug 2010, 23:59
If you mix 1:4 and 3:5 in equal amounts, the result is not 4/9 but 17/40 which should be 42.5%
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bibha wrote:
If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?
• 30.0%
• 36.6%
• 42.5%
• 44.4%
• 60.0%

This is how i did it:
propn of alcohol in first solution : 1:4
propn of alcohol in second solution: 3:5
so, propn of alcohol in the new solution = 4:9 = 44.4%


In this case you are adding fractions to generate a new total.
Common denominator of the to 2 fractions is 20 (1/4=5/20 and 3/5=12/20).
Combining the mixtures mean you will be adding the numerators and denominators separately yielding 17/40.

Then to get the percentage 40 x 2.5=100 17x2.5=42.5.

There may be a quicker way of getting to this.

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Re: mixture [#permalink]

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New post 21 Aug 2010, 15:13
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Alcohol: Water

First mixture -- 3:1 -- 15:5
Second mixture -- 2:3 -- 8:12

Equal amounts of the mixture were combined, hence the final mixture will be in the ratio -- (15+8): (5+12) -- 23:17

% of alcohol -- 17/(23+17) * 100

= (17/40) * 100

= 42.5
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Re: mixture [#permalink]

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New post 05 Sep 2010, 00:35
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Lets do it in the Matrix form which is much easier to understand:

Given W1/A1 (First solution) = 3/1
so concentration of Alcohol = 1/4 (A1/A1+W1)

W2/A2 = 2/3
so concentration of Alcohol = 3/5 (A2/A2+W2)
Let concentration of Alcohol in the mixed solution be X
Matrix :

(Sol 1) (Sol 2)
1/4..................3/5
............X
3/5-X.......:.......X-1/4
1...................1 (Since both are mixed in equal amounts)

Solving the above equation we get X =85/2=42.5% - C
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New post 05 Sep 2010, 03:10
picking numbers:

let 20 l of each is mixed. [20 is divisible by both the proportions]
in solution 1 amount of alcohol = 20X(1/4) = 5 L
in solution 2 amount of alcohol = 20X(3/4)=12 L

total 17 L alcohol in 40 L of solution => (17/40)X100 = 42.5%

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take the amount of two different solutions as 20
In 1st solution concentration of alcohol=(1/4)20=5
In 2nd solution concentration of alcohol=(3/5)20=12
total concentration=40
amount of alcohol=17
% of alcohol=17/40*100=42.5%

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Re: mixture [#permalink]

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New post 11 Feb 2011, 15:19
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bibha wrote:
If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?
• 30.0%
• 36.6%
• 42.5%
• 44.4%
• 60.0%

This is how i did it:
propn of alcohol in first solution : 1:4
propn of alcohol in second solution: 3:5
so, propn of alcohol in the new solution = 4:9 = 44.4%


I think this is an averages question.

%ge of alcohol in solution 1: \(\frac{1}{4}= 25%\)
%ge of alcohol in second solution= \(\frac{3}{5}= 60%\)

Since they are mixed in equal quantities, Average alcohol %= \(\frac{(60%+25%)}{2}= 42.5%\)
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New post 13 Feb 2011, 00:40
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1st solution
3:1::Water:Alcohol
Percent of alcohol in the 1st solution= 1/(3+1) = 1/4 = 0.25

2nd Solution
2:3::Water:Alcohol
Percent of alcohol in the 2nd solution = 3/(2+3)=3/5=0.6

Lets say we mix 1 liter from each of the two solutions and form a new solution.

Alcohol in the new solution from the first solution: 0.25 litres
Alcohol in the new solution from the second solution: 0.6 litres

Total alcohol in the new solution: 0.25+0.6 = 0.85 litres
Total weight of the new solution: 1+1=2 litres

Let x be the percent of alcohol in the new solution.

0.85 = 2 * x
x = 0.85/2 = 0.425 = 42.5%

Ans: "C"
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If two different solutions of alcohol with a respective proportion of [#permalink]

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If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?

A) 30.0%
B) 36.6%
C) 42.5%
D) 44.4%
E) 60.0%

Last edited by Bunuel on 31 Mar 2015, 05:13, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.

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Re: If two different solutions of alcohol with a respective proportion of [#permalink]

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New post 01 Mar 2015, 08:03
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IMO: C

There are a few different ways to go about this problem.

Since we know that the original solutions were mixed in equal amounts, we can just take the average of the alcohol concentration in each solution to get the % alcohol in the new combined solution.

Solution 1 has W:A of 3:1. This means that there are 4 total parts to the solution, one of which is alcohol. Thus alcohol is 25% of this solution.

Solution 2 has W:A of 2:3. This means that there are 5 total parts to the solution, three of which are alcohol. Thus alcohol is 60% of this solution.

From here, we can just take the average (.25 + .60)/2 = .425.
Another way to do this last step is just multiply .25 * .5 and then add that to .60 * .5
.25 * .5 + .60 * .5 = .425
This last step is the exact same, its just a matter of whether you choose to break up the fraction into 2 separate pieces.

Main parts about this question are to realize that the solutions are mixed in equal amounts and to understand how each ratio allows you to know the alcohol concentration.

Also worth noting why choice D is incorrect. If you just add the mixtures together (3 + 2):(1 + 3) you will get 5:4 and thus assume alcohol is 4/9 of the solution. You cannot do this since this incorrectly takes the weighted average of each solution. You need to break each solution down into its own % alcohol and then take the weighted average (in this case the weights are 50-50).

Definitely check out the Quant forum for more questions like this. gmat-quantitative-section-7/?fl=menu
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Hi Anupamt,

As alexau has pointed out, this question can be solved as a straight "average" of two percents. Most questions in this category are "weighted averages" though, so you have to adjust your approach (and your "math") to whatever information you're given to work with.

If you're not great at dealing with ratios yet, then you can TEST VALUES and get the solution that way. Here's how....

We're given two solutions of alcohol:
1) The first is 3 parts water to 1 part alcohol
2) The second is 2 parts water to 3 parts alcohol

We're asked to mix EQUAL AMOUNTS of each alcohol. Since the first alcohol has "4 parts" and the second alcohol has "5 parts", we want to use a common multiple of both. The least would be 20....

So let's say that we have....
20 ounces of the first solution: 15 ounces of water and 5 ounces of alcohol
20 ounces of the second solution: 8 ounces of water and 12 ounces of alcohol

Mixing those 40 ounces together, we have (17 ounces of alcohol)/(40 ounces total)

17/40 = 42.5% alcohol

Final Answer:
[Reveal] Spoiler:
C


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If two different solutions of alcohol with a respective prop [#permalink]

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New post 23 Mar 2015, 09:33
Since,in final solution they are in equal amount we can apply a trick here.

first solution is having 25% alcohol solution
second solution is having 60% alcohol solution.

Now, upon mixing their concentration should lie in between 25<----->60
If you see difference is 35 and since volume of each solution is equal, we can divide the difference in two parts and add/subtract to/from 25/60,which will give us 42.50 %

Note: We can apply this only when both solutions are in equal volume. If not then make both of them in equal volume before you apply this logic.


Hope, this was helpful ! :)

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If two different solutions of alcohol with a respective prop [#permalink]

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New post 28 May 2016, 14:03
a=% alcohol in new solution
a=(.25+.60)/2=.425=42.5%

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Re: If two different solutions of alcohol with a respective prop [#permalink]

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Since the new mixture is a result of two mixtures in equal weights, we can simply average the ratio of alochol to total mixture.
the average of 1/4 and 3/5 = 17/40.
To quickly approximate this in decimal, we can notice that 40 * 2.5 = 100.
17*2.5 = 34+8.5 = 42.5==> answer is 42.5 %.

This was easier since the two mixtures were mixed in the same proportion.

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Re: If two different solutions of alcohol with a respective prop [#permalink]

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New post 21 Jan 2017, 10:59
If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?

A. 30.0%
B. 36.6%
C. 42.5%
D. 44.4%
E. 60.0%

let x=amount of each original solution
y=% of alcohol in new solution
x/4+3x/5=2xy
y=42.5%
C

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Re: If two different solutions of alcohol with a respective prop [#permalink]

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The approach I have taken is :

Conc1 * Quantity1+Conc2*Quantity2=Final Conc(quantity1+quantity2)

Here Quantity1 =Quantity2

Let the quantity be "x" and final solutions quantity be "c"

(1/4)*x +(3/5)*x=c(x+x)

C= 17/40
= 42.5%

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Re: If two different solutions of alcohol with a respective prop [#permalink]

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New post 17 Aug 2017, 05:52
Answer is C

ratio1 of water : alcohol = 3:1 , it means 4 Ltr Mix has 1 ltr alcohol , 1 ltr has 250 ml alcohol, 0.5 ltr = 125 ml alc
ratio2 of water : alcohol = 2:3 , it means 5 ltr mix has 3 ltr alcohol , 1 ltr has 600 ml alcohol, 0.5 ltr = 300 ml alc
totalling the mix of 0.5 ltr + 0.5 ltr gives 125 + 300 ml alcohol which is 425 ml alcohol
1 ltr mix has 425 ml alcohol , % of alcohol is 42.5%
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Re: If two different solutions of alcohol with a respective prop [#permalink]

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New post 21 Aug 2017, 07:50
bibha wrote:
If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?

A. 30.0%
B. 36.6%
C. 42.5%
D. 44.4%
E. 60.0%

This is how i did it:
propn of alcohol in first solution : 1:4
propn of alcohol in second solution: 3:5
so, propn of alcohol in the new solution = 4:9 = 44.4%


Using allegation :
(w1/w2)=(A2-Aavg)/(Aavg-A1)

w1/w2 = 1/1

A1=1/4
A2=3/5

Hence Aavg = (17/4)*100 = 42.5%

C

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Re: If two different solutions of alcohol with a respective prop [#permalink]

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New post 27 Aug 2017, 22:18
First Solution -

Water:Alcohol :: 3:1
In First solution water is 3/4
In First solution alcohol is 1/4

Similarly, in second solution -

Water : Alcohol :: 2:3
In Second solution water is 2/5
In Second solution alcohol is 3/5

In final solution -

Water: Achohol :: 3/4+2/5 : 1/4+3/5
=23:17

Percentage of alcohol in final solution =17/40*100 =42.5%

Option C

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Re: If two different solutions of alcohol with a respective prop   [#permalink] 27 Aug 2017, 22:18
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