Assume the tank is full at 7 a.m. plus t hours. The combined work rate of pipes A, B, C and D is 1/15 + 1/20 + 1/30 + 1/60, which equals 12/60 or 1/5 per hour. The work done by each pipe equals its rate multiplied by its operating time: A works for (3 + t) hours, B works for (2 + t) hours, C works for (1 + t) hours, and D works for t hours. Setting the equation, (3+t)/15 + (2+t)/20 + (1+t)/30 + t/60 = 1, and multiplying by 60 yields 4(3+t) + 3(2+t) + 2(1+t) + t = 60. Simplifying gives 20 + 10t = 60, meaning t = 4. Since D opened at 7 a.m., adding 4 hours gives 11:00 a.m.