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# M22-07

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:16
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Difficulty:

65% (hard)

Question Stats:

61% (01:10) correct 39% (01:24) wrong based on 148 sessions

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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%

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16 Sep 2014, 00:16
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Official Solution:

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%

The strength of the first solution $$= \frac{1}{3 + 1} = 25\%$$. The strength of the second solution $$= \frac{3}{2 + 3} = 60\%$$. Because the solutions were mixed in equal amounts, the strength of the new solution is $$\frac{60\% + 25\%}{2} = 42.5\%$$.

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07 Nov 2015, 17:38
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I think this is a high-quality question and I agree with explanation. I like it! Another way of viewing this. If you are going to use 5 L of the 2:3 solution, you must use 5L of the 3:1 solution. Thus you will be adding 3.75:1.25 L to 2:3 L which results in a 10 L solution that has 5.75:4.25 ratio. You want to know what 4.25L is as a % of 10L, so just times 4.25 / 10 by 10 -> 42.5 / 100
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15 May 2016, 05:24
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Bunuel's solution is pretty awesome, I didn't think one could solve it that simply. I took a different approach and worked with actual numbers:

1) We know that the ratio of water:alcohol in the first solution is $$\frac{3}{1}$$ (which give 4 parts) and we know that the ratio of water:alcohol in the second solution is $$\frac{2}{3}$$ (which give 5 parts).

2) 20 is the LCM of 4 and 5, so let's assume that both solutions have a volume of 20 units. From the first solution, we have that water is $$\frac{3}{4}$$ of the solution (which we said had a volume of 20), which gives 15 units of water and thus 5 units of alcohol. From the second solution, we have that water is $$\frac{2}{5}$$ of the solution, which gives 8 units of water and 12 units of alcohol.

3) Since we are mixing both solutions with an equal amount, 20 units, our new solution has a total volume of 40 units. From 2), combining the water and alcohol in both solutions, we get that our new solution has 17 units of alcohol (5 units from the first solution and 12 units from the second solution) and 23 units of water (15 units from the first solution and 8 units from the second solution). Thus, the amount of alcohol in the mixed solution is $$\frac{17}{40}$$ = 42.5%.

PS: Time-wise I managed to do this slightly north of 2 minutes, but Bunuel's approach is of course much faster if one can identify the validity of that approach right off the bat.

/S
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21 Nov 2017, 23:44
my method is simple alligation

25% and 60% mix leading to x but the ratio of both original solutions in mix is 1:1. therefore:

x-25=60-x means 2x=85. finally x=42.5%
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Joined: 09 Aug 2017
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23 Nov 2017, 17:01
Bunuel wrote:
Official Solution:

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%

The strength of the first solution $$= \frac{1}{3 + 1} = 25\%$$. The strength of the second solution $$= \frac{3}{2 + 3} = 60\%$$. Because the solutions were mixed in equal amounts, the strength of the new solution is $$\frac{60\% + 25\%}{2} = 42.5\%$$.

\frac{60\% + 25\%}{2} = 42.5\ how did you got equation ? Why you have divided by 2 ? @bunuel

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Joined: 02 Sep 2009
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23 Nov 2017, 20:19
Spongebob02 wrote:
Bunuel wrote:
Official Solution:

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%

The strength of the first solution $$= \frac{1}{3 + 1} = 25\%$$. The strength of the second solution $$= \frac{3}{2 + 3} = 60\%$$. Because the solutions were mixed in equal amounts, the strength of the new solution is $$\frac{60\% + 25\%}{2} = 42.5\%$$.

\frac{60\% + 25\%}{2} = 42.5\ how did you got equation ? Why you have divided by 2 ? @bunuel

Sent from my Redmi 3S using GMAT Club Forum mobile app

Because we are told that the original solutions were mixed in EQUAL amounts. So, the percentage of alcohol in the resulting mixture would be the average of the percentages in two mixtures.

Check other discussion here: https://gmatclub.com/forum/if-two-diffe ... 99529.html

Hope it helps.
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Joined: 17 Dec 2017
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02 Jan 2018, 07:08
Sallyzodiac wrote:
Bunuel's solution is pretty awesome, I didn't think one could solve it that simply. I took a different approach and worked with actual numbers:

1) We know that the ratio of water:alcohol in the first solution is $$\frac{3}{1}$$ (which give 4 parts) and we know that the ratio of water:alcohol in the second solution is $$\frac{2}{3}$$ (which give 5 parts).

2) 20 is the LCM of 4 and 5, so let's assume that both solutions have a volume of 20 units. From the first solution, we have that water is $$\frac{3}{4}$$ of the solution (which we said had a volume of 20), which gives 15 units of water and thus 5 units of alcohol. From the second solution, we have that water is $$\frac{2}{5}$$ of the solution, which gives 8 units of water and 12 units of alcohol.

3) Since we are mixing both solutions with an equal amount, 20 units, our new solution has a total volume of 40 units. From 2), combining the water and alcohol in both solutions, we get that our new solution has 17 units of alcohol (5 units from the first solution and 12 units from the second solution) and 23 units of water (15 units from the first solution and 8 units from the second solution). Thus, the amount of alcohol in the mixed solution is $$\frac{17}{40}$$ = 42.5%.

PS: Time-wise I managed to do this slightly north of 2 minutes, but Bunuel's approach is of course much faster if one can identify the validity of that approach right off the bat.

/S

Took the same approach. What if the question asked for the solutions to be mixed in different proportions instead of the same? How would you use this method with unequal proportions?
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02 Jan 2018, 08:28
The answer is option C. The first soln has water and alcohol is the ration 75:25 and the second solution in the ratio 40:60. If equal amount of the two are taken, the ratio of water and alcohol is 115:85. The percentage of alcohol is (85/200)*100 or 42.5%
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19 Jun 2018, 13:37
I think this is a high-quality question and I agree with explanation.
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Joined: 06 Dec 2018
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03 Feb 2019, 04:38
I just summed up the sums of the solutions which yielded 9 ( (1+3) + (3+2) = 9)
Afterwards I did the same with the alcohol solutions which yielded 4 (3 + 1)
THe percentage of alcohol would be 4/9 = 0.4444 , Can you tell me what is wrong?
Re: M22-07   [#permalink] 03 Feb 2019, 04:38
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# M22-07

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