Two inlets fill a cistern in 12 minutes and 15 minutes, respectively, and a third pipe can empty a full cistern in 6 minutes. The first two pipes are kept open for 5 minutes; then the third pipe is also opened along with them. From the instant the third pipe is opened, in what time is the cistern emptied (reduced from its then level to empty)?

Introduction / Context:This repeats the common “head start fill, then drain” scenario. We again compute the partial fill after 5 minutes of inlets, then use the net draining rate once the outlet joins to find the emptying time from that point.

Given Data / Assumptions:

• Inlet rates: 1/12 and 1/15 tank/min.
• Outlet rate: −1/6 tank/min.
• First phase (5 min): inlets only; second phase: inlets + outlet until empty.

Concept / Approach:Phase 1 fill = 5 * (1/12 + 1/15) = 3/4. Net rate with outlet = 1/12 + 1/15 − 1/6 = −1/60 tank/min. Time = (3/4) / (1/60) = 45 min.

Step-by-Step Solution:

Verification / Alternative check:If the question were asking total time from the very start to become empty, it would be 5 + 45 = 50 minutes; however, the options indicate the intended interpretation is “from when the third pipe is opened.”

Why Other Options Are Wrong:30/33/37.5 min do not remove the full 3/4 at the given net rate.

Common Pitfalls:Misreading the reference time; confirm whether the clock starts at the third pipe opening.

Final Answer:45 min