Sura, a strongly distilled alcoholic drink was used in ancient India as an…
2026
Sura, a strongly distilled alcoholic drink was used in ancient India as an anesthetic by surgeons. A 15 liters cask initially contains pure Sura up to the brim. The Sura is diluted by removing 5 liters and replacing that quantity with water. If Sura is diluted twice, what is the ratio of Sura to water in the cask?
- A.
2:1
- B.
1:1
- C.
1:2
- D.
4:5
Attempted by 2 students.
Show answer & explanation
Correct answer: D
Concept: When a fraction of a mixture is repeatedly removed and replaced by one of its components (here, water), the amount of the original substance left after n such removals follows the rule: remaining original quantity = initial quantity (1 − removed fraction)n. Each removal always takes out liquid at the CURRENT concentration of the cask, never a fixed volume of either substance.
Application:
Removed fraction per step = 5 litres removed ÷ 15 litres cask = 1/3, so the retained fraction each step is 1 − 1/3 = 2/3.
After the first removal-and-replacement: Sura remaining = 15 × 2/3 = 10 litres; the cask is topped back up to 15 litres with water, so Water = 5 litres. Ratio Sura:Water = 10:5 = 2:1 at this stage.
After the second removal-and-replacement, the same 2/3 retention factor applies again to whatever Sura is present at that point: Sura remaining = 15 × (2/3)2 = 15 × 4/9 = 20/3 litres.
Water present after the second step = 15 (total cask volume) − 20/3 (Sura) = 25/3 litres.
Final ratio of Sura to water = 20/3 : 25/3 = 20:25 = 4:5.
Cross-check: Track the second removal directly instead of using the formula. At that point the cask holds Sura:Water = 2:1, so the 5 litres removed carries that same split — Sura removed = 5 × 2/3 = 10/3 litres, Water removed = 5 × 1/3 = 5/3 litres. Sura left = 10 − 10/3 = 20/3 litres; Water left = 5 − 5/3 = 10/3 litres, then 5 litres of fresh water is added, giving Water = 10/3 + 5 = 25/3 litres. This reproduces Sura:Water = 20/3 : 25/3 = 4:5, confirming the result from the formula.