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A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 66 hours when 66 filling and 55 draining pipes are on, but this time becomes 6060 hours when 55 filling and 66 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

Entered answer:

Solution

✅ Correct Answer: 10

We have a tank with two types of pipes:

Filling pipes: All work at the same rate (we call this rate f)

Draining pipes: All work at the same rate (we call this rate d)

The key insight is that we need to find the net rate of filling by subtracting the draining rate from the filling rate.


We define:

  • f = rate of each filling pipe (fraction of tank filled per hour)
  • d = rate of each draining pipe (fraction of tank drained per hour)

Scenario 1: 6 filling + 5 draining pipes → Tank filled in 6 hours

Net filling rate = 6f−5d=166f - 5d = \frac{1}{6} tanks per hour

Scenario 2: 5 filling + 6 draining pipes → Tank filled in 60 hours

Net filling rate = 5f−6d=1605f - 6d = \frac{1}{60} tanks per hour

Why do we subtract? Draining pipes work against filling pipes, so we subtract their effect from the filling rate.


We have:

  • 6f−5d=166f - 5d = \frac{1}{6} ... (1)
  • 5f−6d=1605f - 6d = \frac{1}{60} ... (2)

Equation (1) by 6 and equation (2) by 5:

  • 36f−30d=136f - 30d = 1 ... (3)
  • 25f−30d=11225f - 30d = \frac{1}{12} ... (4)

Subtracting equation (4) from equation (3):

(36f−30d)−(25f−30d)=1−112(36f - 30d) - (25f - 30d) = 1 - \frac{1}{12}

11f=1212−112=111211f = \frac{12}{12} - \frac{1}{12} = \frac{11}{12}

f=112f = \frac{1}{12} tanks per hour


Substituting back to find d:

6(112)−5d=166(\frac{1}{12}) - 5d = \frac{1}{6}

12−5d=16\frac{1}{2} - 5d = \frac{1}{6}

5d=12−16=36−16=26=135d = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}

d=115d = \frac{1}{15} tanks per hour


Question: How long with 2 filling + 1 draining pipe?

Net rate = 2f−d=2(112)−1152f - d = 2(\frac{1}{12}) - \frac{1}{15}

2(112)=212=162(\frac{1}{12}) = \frac{2}{12} = \frac{1}{6}

Finding common denominator for 16\frac{1}{6} and 115\frac{1}{15}:

LCM of 6 and 15 = 30

16=530\frac{1}{6} = \frac{5}{30}

115=230\frac{1}{15} = \frac{2}{30}

Net rate = 530−230=330=110\frac{5}{30} - \frac{2}{30} = \frac{3}{30} = \frac{1}{10} tanks per hour

Time = 1÷(110)=101 ÷ (\frac{1}{10}) = 10 hours


Answer: 10 hours

Quick Check: With a net rate of 110\frac{1}{10} tanks per hour, it makes sense that it takes 10 hours to fill 1 complete tank!

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