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11 Nov 2024, 06:29
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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:
55%
(hard)
Question Stats:
47% (02:26) correct
53%
(03:46)
wrong
based on 34
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
11 Nov 2024, 06:29
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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"
Bunuel
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 |
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"
29 Nov 2024, 08:34
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