SSC Sample Paper Mock Test-2 SSC CGL Tear-II Paper-1

A cistern has 3 pipes A, B and C. A and B can fill it in 3 h and 4 h respectively while C can empty the completely filled cistern in 1 h. If the pipes are opened in order at 3, 4 and 5 pm respectively. At what time will the cistern be empty?

A) 6 : 15 pm

B) 7 : 12 pm

C) 8 : 23 pm

D) 8 : 35 pm

Correct Answer: B

Solution :

Pipe A is opened at 3 pm, pipe B at 4 pm and the pipe C at 5 pm

Part of the tank filled by pipe A in 2 h \[=\frac{2}{3}\]

Part of the tank filled by pipe B in 1 h \[=\frac{1}{4}\]

Part of the tank filled till 5 pm \[=\frac{2}{3}+\frac{1}{4}=\frac{8+3}{12}=\frac{11}{12}\]

Remaining part \[=1-\frac{11}{12}=\frac{1}{12}\]

Net part emptied when A, B and C are opened

\[=\frac{1}{3}+\frac{1}{4}-1=\frac{4+3-12}{12}=\frac{-\,\,5}{12}\]

\[\therefore \]\[\frac{5}{12}\]Part is emptied in 1 h

\[\therefore \]\[\frac{11}{12}\]is emptied m\[=\frac{12}{5}\times \frac{11}{12}=\frac{11}{5}\]h=2h 12 min

Hence, the tank will be emptied at 7 : 12 pm.