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01 Sep 2022, 06:22
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A
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Jeeves prepares a hangover cure using four identical cocktail shakers. The first shaker is 1/2 full, the second shaker is 1/3 full, the third shaker is 1/4 full and the last one is empty. If Jeeves redistributes all the content of the shakers equally into the four shakers, what fraction of each shaker will be filled?
A. 13/48
B. 4/13
C. 13/36
D. 9/13
E. 35/48
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A. 13/48
B. 4/13
C. 13/36
D. 9/13
E. 35/48
Fresh GMAT Club Tests' Question
Try harder version of the above question HERE.
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19 Feb 2023, 11:52
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Bunuel
Official Solution:
Jeeves prepares a hangover cure using four identical cocktail shakers. The first shaker is \(\frac{1}{2}\) full, the second shaker is \(\frac{1}{3}\) full, the third shaker is \(\frac{1}{4}\) full and the last one is empty. If Jeeves redistributes all the content of the shakers equally into the four shakers, what fraction of each shaker will be filled?
A. \(\frac{13}{48}\)
B. \(\frac{4}{13}\)
C. \(\frac{13}{36}\)
D. \(\frac{9}{13}\)
E. \(\frac{35}{48}\)
Let's denote the capacity of each shaker as \(x\) and calculate the total amount of liquid in them:
\(\frac{x}{2} + \frac{x}{3} + \frac{x}{4} = \frac{13x}{12}\).
Now, if we redistribute the liquid equally among the four shakers, each shaker will receive 1/4 of the total amount of liquid, which is:
\(\frac{1}{4}*\frac{13x}{12} = \frac{13x}{48}\).
Therefore, each shaker will be filled with \(\frac{13}{48}\) of its capacity.
Answer: A
General Discussion
01 Sep 2022, 09:47
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IMO A?
\(\frac{F}{2 }\)+ \(\frac{F}{3}\) +\( \frac{F}{4}\) = \(\frac{13F}{12} \) (since the last one is empty we don't add it here)
and there are 4 identical containers, so the capacity of each (F) will be the same.
\(\frac{13F}{12*4}\) - Total quantity is divided into 4 containers.
= \(\frac{13F}{48 }\)
\(\frac{F}{2 }\)+ \(\frac{F}{3}\) +\( \frac{F}{4}\) = \(\frac{13F}{12} \) (since the last one is empty we don't add it here)
and there are 4 identical containers, so the capacity of each (F) will be the same.
\(\frac{13F}{12*4}\) - Total quantity is divided into 4 containers.
= \(\frac{13F}{48 }\)
