Quantitative Aptitude Ques 2374

Question: Pipe A can fill a tank in 4h and pipe B can fill it in 6h. If they are opened on alternate hours and if pipe A is opened first, then in how many hours the tank shall be full?

Options:

A) $ 4\frac{1}{2}h $

B) $ 4\frac{2}{3}h $

C) $ 3\frac{1}{2}h $

D) $ 3\frac{1}{4}h $

Show Answer

Answer:

Correct Answer: B

Solution:

  • Part filled by pipe A in $ 1,h=\frac{1}{4} $ Part filled by pipe B in $ 1,h=\frac{1}{6} $ Part filled by $ (A+B) $ in $ 2,h=\frac{1}{4}+\frac{1}{6}=\frac{3+2}{12}=\frac{5}{12} $

$ \therefore $ Part filled by $ (A+B) $ in $ 4h=\frac{5}{12}\times 2=\frac{5}{6} $ Remaining part $ =1-\frac{5}{6}=\frac{1}{6} $ Now, it is the turn of A. Time taken by A to fill $ \frac{1}{6} $ part of the tank $ =\frac{1}{6}\times 4=\frac{2}{3}h $

$ \therefore $ Total time taken $ =( 4+\frac{2}{3} )h=4\frac{2}{3}h $