(d) Since, pipe A takes 3h to fill
th part of the tank.
So, pipe A can fill the full tank in
x 3 i.e. in
h.
Let C can empty the full tank in xh.
Then, according to the question,
A's (5 + 3) h work + C’s 5h work = 20%
⇒
8×−×5= ⇒
=− ⇒
= ⇒
= ∴ C can empty the tank in
h
Thus, time taken by C to empty 20% of the tank (as only 20% part of the tank is filled)
=
×=h