Two pipes can fill a cistern in 15 hours and 12 hours respectively and a third pipe
can empty the full tank in 4 hours. If these pipes are opened respectively at 8 AM
9 A.M. and 11 A.M., then at what time the tank shall be empty ?
Answers
Answer:
DOWNLOADAPP
Home » Aptitude » Pipes and Cistern » Question
InterviewMania App

Pipes and Cistern
Easy Questions
Moderate Questions
Difficult Questions
Pipes and Cistern Tutorial
Aptitude
Number System
Simplification
Fractions
Elementary Algebra
LCM and HCF
Average
Approximation
Unitary Method
Linear Equation
Quadratic Equation
Discount
Surds and Indices
Percentage
Square root and cube root
Order of Magnitude
Profit and Loss
Odd Man Out and Series
Work and Wages
Algebra
Stocks and Shares
True Discount
Ratio, Proportion
Partnership
Alligation or Mixture
Time and Work
Pipes and Cistern
Speed, Time and Distance
Problem on Trains
Height and Distance
Banker's Discount
Boats and Streams
Races and games
Problems on Ages
Clocks and Calendars
Simple interest
Compound Interest
Sets and Functions
Area and Perimeter
Volume and Surface Area of Solid Figures
Sequences and Series
Plane Geometry
Logarithm
Probability
Permutation and Combination
Statistics
Mensuration
Trigonometry
Aptitude miscellaneous
Two pipes can fill a tank with water in 15 and 12 hours respectively and a third pipe can empty it in 4 hours. If the pipes be opened in order at 8, 9 and 11 a.m. respectively, the tank will be emptied at
11 : 40 a.m.
12 : 40 p.m.
1 : 40 p.m.
2 : 40 p.m.
Correct Option: A
On the basis of given details in question ,
Part filled by A from 8 a.m to 11 a.m. =3=1155
Part filled by B from 9 a.m. to 11 a.m. =2=1126
Total Part filled till 11 a.m. =1+1=6 + 5=11563030
At 11 a.m. pipe C is opened to empty it.
∴ Part of tank emptied in 1 hour =1-1-141512
Part of tank emptied in 1 hour =15 - 4 - 5=16010
∴11part will be emptied in11× 10 =11hours or 32303033
i.e. 3 hours 40 minutes or at 11 : 40 a.m.
Therefore , the tank will be emptied at 11 : 40 a.m.