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08 Jun 2024, 07:17
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First mixture: 24
Final mixture: 40
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
43% (02:19) correct
57%
(02:33)
wrong
based on 867
sessions
History
Date
Time
Result
Not Attempted Yet
Juanita combined two mixtures of lemonade. The first had 4 parts water to 1 part lime juice, and the second had 3 parts water to 2 parts lime juice. The final mixture contained 28% lime juice.
Select for First mixture the number of liters that could be in the first mixture, and for Final mixture the number of liters that could be in the final mixture that are jointly consistent with the given information. Make only two selections, one in each column.
Select for First mixture the number of liters that could be in the first mixture, and for Final mixture the number of liters that could be in the final mixture that are jointly consistent with the given information. Make only two selections, one in each column.
| First mixture | Final mixture | |
| 4 | ||
| 12 | ||
| 16 | ||
| 24 | ||
| 30 | ||
| 40 |
ShowHide Answer
Official Answer
First mixture: 24
Final mixture: 40
29 Nov 2024, 08:37
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Bunuel
Official Solution:
Since the first mixture had 4 parts water to 1 part lime juice, it contained 1/5 or 20% lime juice.
Since the second mixture had 3 parts water to 2 parts lime juice, it contained 2/5 or 40% lime juice.
Assuming the first mixture was \(x\) liters and the second was \(y\) liters, equating the amount of lime juice gives \(0.2x + 0.4y = 0.28(x + y)\), which simplifies to \(2x = 3y\).
Thus, the first mixture was \(x\) liters, and the final mixture was \(x + y = x + \frac{2x}{3} = \frac{5x}{3}\). The ratio of these quantities is \(x:\frac{5x}{3}=3:5\). Therefore, we are looking for options in a 3:5 ratio. Only 24 and 40 work.
Correct answer:
First mixture "24"
Final mixture "40"
Attachment:
GMAT-Club-Forum-he2ql5ps.png [ 11.76 KiB | Viewed 5279 times ]
General Discussion
29 Nov 2024, 10:10
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1. Let's say there were x liters of mixture 1 and y liters of mixture 2. We're interested in finding x and x+y.
2. An equation can be written down for the percentage of lime juice: \(\frac{20\% * x + 40\% * y}{x + y} = 28\%\).
3. \(\frac{20\% * x + 40\% * y}{x + y} = 28\% \rightarrow 12\% * y = 8\% * x \rightarrow \frac{y}{x} = \frac{8\%}{12\%} = \frac{2}{3}\).
4. We are given no other information, so, let's look at the ratio of the answers: \(\frac{x+y}{x} = 1 + \frac{y}{x} = 1 + \frac{2}{3} = \frac{5}{3}\).
5. Looking at the possible answer choices, only 40 and 24 have such a ratio.
6. Our answer will be: First mixture - 24 and Final mixture - 40.
2. An equation can be written down for the percentage of lime juice: \(\frac{20\% * x + 40\% * y}{x + y} = 28\%\).
3. \(\frac{20\% * x + 40\% * y}{x + y} = 28\% \rightarrow 12\% * y = 8\% * x \rightarrow \frac{y}{x} = \frac{8\%}{12\%} = \frac{2}{3}\).
4. We are given no other information, so, let's look at the ratio of the answers: \(\frac{x+y}{x} = 1 + \frac{y}{x} = 1 + \frac{2}{3} = \frac{5}{3}\).
5. Looking at the possible answer choices, only 40 and 24 have such a ratio.
6. Our answer will be: First mixture - 24 and Final mixture - 40.
