Discussion
A cask contains 81 litres of pure wine. 9 litres are drawn and replaced with water, and the process is repeated twice more. How much wine remains?
Answer: D
Correct answer: 56.89 litres
Each operation leaves the same fraction of wine behind, so after \( n \) operations the quantity remaining is \[ Q\left(1 - \frac{x}{Q}\right)^{n} = 81\left(1 - \frac{9}{81}\right)^{3} = 81 \times \left(\frac{8}{9}\right)^{3}. \] Computing, \[ 81 \times \frac{512}{729} = \frac{41472}{729} = 56.89 \text{ litres}. \] The reason a formula is needed rather than repeated subtraction is that after the first replacement the cask holds a mixture, so each later withdrawal removes water as well as wine. Subtracting 9 litres three times would give 54, which is the distractor and the commonest error. Following it through confirms the result: 72 litres of wine remain after the first replacement, 64 after the second, and \( 64 \times \frac{8}{9} = 56.89 \) after the third.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development