Subject
Quantitative Aptitude
Topic
Miscellaneous
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?
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?
ক)
2202/3125খ)
2101/3125গ)
2101/3145ঘ)
1024 / 3125ঙ)
None of theseExplanation
Let the initial volume of alcohol present be v. After 1 st time removal of 1/5 of the contents of the vessel, alcohol remaining = v – v/5 = 4v/5 After 2 nd time removal of 1/5 of the contents of the vessel, alcohol remaining = 4v/5 – 1/5 × 4v/5 ⇒After 2 nd time removal of 1/5 of the contents of the vessel, alcohol remaining = 16v/25 After 3 rd time removal of 1/5 of the contents of the vessel, alcohol remaining = 16v/25 – 1/5 × 16v/25 ⇒After 3 rd time removal of 1/5 of the contents of the vessel, alcohol remaining = 16v/25 – 16v/125 ⇒After 3 rd time removal of 1/5 of the contents of the vessel, alcohol remaining = 64v/125 After 4 th time removal of 1/5 of the contents of the vessel, alcohol remaining = 64v/125 – 1/5 × 64v/125 ⇒After 4 th time removal of 1/5 of the contents of the vessel, alcohol remaining = 64v/125 – 64v/625 ⇒After 4 th time removal of 1/5 of the contents of the vessel, alcohol remaining = 256v/625 After 5 th time removal of 1/5 of the contents of the vessel, alcohol remaining = 256v/625 – 1/5 × 336v/625 ⇒After 5 th 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 = 2101/3125Related questions
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