SSC CHSL 4 Aug 2021 Shift 1 Paper

Section: Quantitative Aptitude

Question : 54 of 100

Marks: +1, -0

Tap

P

can fill a cistern in

6 {h}

and

t a p Q

can empty the full cistern in

10 {h}

. If both taps

P

and

Q

are kept open simultaneously, then in how many hours will the empty cistern be completely full?

Solution:

Part filled by Tap

P

1 {h}=\;{1}/{6}

Part emptied by Tap

Q

1 {h}=\;{1}/{10}

Part filled by both Tap

P

and

Q

1 {h}

=\;{1}/{6}-\;{1}/{10}=\;{10-6}/{60}=\;{1}/{15}

Hence, both taps together will fill the empty cistern

=\;{1}/{{1}/{15}}=15 {h}

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