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19 Feb 2023, 12:12
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Jeeves prepares a hangover cure using four identical cocktail shakers. The first shaker is \(\frac{1}{2}\) full, the second shaker is \(\frac{4}{5}\) full, the third shaker is \(\frac{1}{k}\) full and the last one is empty. After Jeeves redistributed all the content of the shakers equally into the four shakers, each shaker became \(\frac{31}{80}\) full. What is the value of \(k\)?
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
19 Feb 2023, 12:12
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Official Solution:
Jeeves prepares a hangover cure using four identical cocktail shakers. The first shaker is \(\frac{1}{2}\) full, the second shaker is \(\frac{4}{5}\) full, the third shaker is \(\frac{1}{k}\) full and the last one is empty. After Jeeves redistributed all the content of the shakers equally into the four shakers, each shaker became \(\frac{31}{80}\) full. What is the value of \(k\)?
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
Let's denote the capacity of each shaker by \(x\).
Then, the amount of liquid in the first shaker is \(\frac{1}{2}*x\), the amount of liquid in the second shaker is \(\frac{4}{5}*x\), the amount of liquid in the third shaker is \(\frac{1}{k}*x\), and the amount of liquid in the fourth shaker is 0.
After redistributing the liquid equally among the four shakers, each shaker will contain \(\frac{1}{4}^{th}\) of the total amount of liquid, which is given to be \(\frac{31}{80}*x\), so we have:
\(\frac{1}{4}(\frac{1}{2}*x + \frac{4}{5}*x + \frac{1}{k}*x) =\frac{31}{80}*x\);
\(\frac{1}{4}(\frac{1}{2} + \frac{4}{5} + \frac{1}{k}) =\frac{31}{80}\) (reduced by \(x\));
\(\frac{1}{2} + \frac{4}{5} + \frac{1}{k} =\frac{31}{20}\) (multiplied by 4);
\(\frac{13}{10} + \frac{1}{k} =\frac{31}{20}\);
\(\frac{1}{k} =\frac{31}{20}-\frac{13}{10} \);
\(\frac{1}{k} =\frac{1}{4}\);
\(k =4\).
Answer: C
Jeeves prepares a hangover cure using four identical cocktail shakers. The first shaker is \(\frac{1}{2}\) full, the second shaker is \(\frac{4}{5}\) full, the third shaker is \(\frac{1}{k}\) full and the last one is empty. After Jeeves redistributed all the content of the shakers equally into the four shakers, each shaker became \(\frac{31}{80}\) full. What is the value of \(k\)?
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
Let's denote the capacity of each shaker by \(x\).
Then, the amount of liquid in the first shaker is \(\frac{1}{2}*x\), the amount of liquid in the second shaker is \(\frac{4}{5}*x\), the amount of liquid in the third shaker is \(\frac{1}{k}*x\), and the amount of liquid in the fourth shaker is 0.
After redistributing the liquid equally among the four shakers, each shaker will contain \(\frac{1}{4}^{th}\) of the total amount of liquid, which is given to be \(\frac{31}{80}*x\), so we have:
\(\frac{1}{4}(\frac{1}{2}*x + \frac{4}{5}*x + \frac{1}{k}*x) =\frac{31}{80}*x\);
\(\frac{1}{4}(\frac{1}{2} + \frac{4}{5} + \frac{1}{k}) =\frac{31}{80}\) (reduced by \(x\));
\(\frac{1}{2} + \frac{4}{5} + \frac{1}{k} =\frac{31}{20}\) (multiplied by 4);
\(\frac{13}{10} + \frac{1}{k} =\frac{31}{20}\);
\(\frac{1}{k} =\frac{31}{20}-\frac{13}{10} \);
\(\frac{1}{k} =\frac{1}{4}\);
\(k =4\).
Answer: C
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05 Feb 2025, 00:24
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I did not quite understand the solution. couldn't understand after redistributing the liquid can you explain in detail
