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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?
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 \(\frac{60\% + 25\%}{2} = 42.5\%\).
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
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.
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 \(\frac{60\% + 25\%}{2} = 42.5\%\).
Answer: C
\frac{60\% + 25\%}{2} = 42.5\ how did you got equation ? Why you have divided by 2 ? @bunuel
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 \(\frac{60\% + 25\%}{2} = 42.5\%\).
Answer: C
\frac{60\% + 25\%}{2} = 42.5\ how did you got equation ? Why you have divided by 2 ? @bunuel
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.
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?
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%
_________________
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?