From a vessel filled with alcohol, 1/5 of its content is removed, and the vessel is then filled up with water. If this be done five times in succession, what proportion of the alcohol originally contained in the vessel will have been removed from it?
Solution
Correct Answer: Option B
Let the initial volume of alcohol present be v.
After 1st time removal of 1/5 of the contents of the vessel, alcohol remaining = v – v/5 = 4v/5
After 2nd time removal of 1/5 of the contents of the vessel, alcohol remaining = 4v/5 – 1/5 × 4v/5
⇒After 2nd time removal of 1/5 of the contents of the vessel, alcohol remaining = 16v/25
After 3rdtime removal of 1/5 of the contents of the vessel, alcohol remaining = 16v/25 – 1/5 × 16v/25
⇒After 3rd time removal of 1/5 of the contents of the vessel, alcohol remaining = 16v/25 – 16v/125
⇒After 3rd time removal of 1/5 of the contents of the vessel, alcohol remaining = 64v/125
After 4thtime removal of 1/5 of the contents of the vessel, alcohol remaining = 64v/125 – 1/5 × 64v/125
⇒After 4thtime removal of 1/5 of the contents of the vessel, alcohol remaining = 64v/125 – 64v/625
⇒After 4th time removal of 1/5 of the contents of the vessel, alcohol remaining = 256v/625
After 5thtime removal of 1/5 of the contents of the vessel, alcohol remaining = 256v/625 – 1/5 × 336v/625
⇒After 5th time removal of 1/5 of the contents of the vessel, alcohol remaining = 1024v/3125
∴alcohol removed from the vessel = v – 1024v/3125
⇒alcohol removed from the vessel = 2101v/3125
Thus the volume of alcohol removed compared to initial volume of alcohol present = 2101/3125
ALTERNATE SOLUTION
Volume of Alcohol Remaining after n times = v × (1- 1/5)n= v × (4/5)n
⇒ after 5 times i.e. for n =5,
⇒ Volume of Alcohol Remaining = v × (4/5)5 = 1024v/3125
∴alcohol removed from the vessel = v – 1024v/3125
⇒alcohol removed from the vessel = 2101v/3125
Thus the volume of alcohol removed compared to initial volume of alcohol present = 2101/3125
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