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Question:

A cylindrical overhead tank is filled by two pumps – P1 and P2. P1 can fill the tank in 8 hours while P2 can fill the tank in 12 hours. There is a pipe P3 which can empty the tank in 8 hours. Both the pumps are opened simultaneously. The supervisor of the tank, before going out on a work, sets a timer to open P3 when the tank is half filled so that tank is exactly filled up by the time he is back. Due to technical fault P3 opens when tank is one third filled. If the supervisor comes back as per the plan what percent of the tank is still empty?

Explanation:

Time required by P1 and P2 to fill the tank = $\frac{8\times 12}{8+12}$ = $\frac{24}{5}$ hrs.

∴ Time required by P1 and P2 to fill half the tank = $\frac{12}{5}$hrs.

The remaining half of the tank will be filled by P1 and P2 along with P3 with the rate of P2 only, because P3 will empty the tank with the rate of P1.

∴ Time required to fill the remaining half of the tank = 12/2 = 6 hrs.

∴ Time after which supervisor return as per the plan = $\frac{12}{5}$ + 6 = $\frac{42}{5}$ hrs.

Time required by P1 and P2 to fill $\frac{1}{3}$rd of the tank = $\frac{8}{5}$ hrs.

Time required to fill the remaining $\frac{2}{3}$rd of the tank (rate of P2) = 12 × $\frac{2}{3}$ = 8 hrs.

∴ Time after which tank will be full = $\frac{8}{5}$ + 8 = $\frac{48}{5}$ hrs.

∴ Difference in time = $\frac{48}{5}$ - $\frac{42}{5}$ = $\frac{6}{5}$ hrs.

With the rate of P2 (i.e. in 12 hrs.), 1 tank can be filled.

∴ In $\frac{6}{5}$ hrs. the part of the tank still remaining empty = $\frac{6}{5}$ × $\frac{1}{12}$ = $\frac{1}{10}$ = 10%

Hence, option (c).