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Question

Three inlet pipes A, B, and C can fill a tank in 8 hours, 12 hours, and 24 hours, respectively. First, pipe A is opened alone for 1 hour. Then pipe B is opened, and both A and B run together for the next 2 hours. After that, pipe C is also opened, and all three pipes run together until the tank is full. How much total time, from the start, will it take to fill the tank?

The correct answer is
4 hour 50 minutes

Understanding Pipe Filling Rates

First, let's determine the rate at which each pipe fills the tank. The rate is the fraction of the tank filled per hour.

  • Rate of Pipe A = $ \frac{1}{8} $ tank per hour
  • Rate of Pipe B = $ \frac{1}{12} $ tank per hour
  • Rate of Pipe C = $ \frac{1}{24} $ tank per hour

Calculating Work Done in Stages

The filling process occurs in three distinct stages.

Stage 1: Pipe A Alone (1 hour)

Pipe A works alone for the first hour.

  • Work done by A in 1 hour = Rate of A $ \times $ Time = $ \frac{1}{8} \times 1 = \frac{1}{8} $ tank.
  • Tank capacity remaining = $ 1 - \frac{1}{8} = \frac{7}{8} $ tank.

Stage 2: Pipes A and B Together (2 hours)

Next, pipes A and B work together for 2 hours.

  • Combined rate of A and B = $ \frac{1}{8} + \frac{1}{12} $.
  • To add these fractions, find a common denominator (24): $ \frac{3}{24} + \frac{2}{24} = \frac{5}{24} $ tank per hour.
  • Work done by A and B in 2 hours = Combined rate $ \times $ Time = $ \frac{5}{24} \times 2 = \frac{10}{24} = \frac{5}{12} $ tank.
  • Total work done after 2 hours = Work from Stage 1 + Work from Stage 2 = $ \frac{1}{8} + \frac{5}{12} $.
  • Common denominator (24): $ \frac{3}{24} + \frac{10}{24} = \frac{13}{24} $ tank.
  • Tank capacity remaining = $ 1 - \frac{13}{24} = \frac{11}{24} $ tank.

Stage 3: Pipes A, B, and C Together

Finally, all three pipes work together until the tank is full.

  • Combined rate of A, B, and C = $ \frac{1}{8} + \frac{1}{12} + \frac{1}{24} $.
  • Common denominator (24): $ \frac{3}{24} + \frac{2}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4} $ tank per hour.

Determining Final Filling Time

Now, calculate the time needed for Stage 3 and the total time.

  • Time required for Stage 3 = Remaining capacity / Combined rate of A, B, C
  • Time for Stage 3 = $ \frac{11/24}{1/4} = \frac{11}{24} \times 4 = \frac{44}{24} = \frac{11}{6} $ hours.
  • Total time to fill the tank = Time (Stage 1) + Time (Stage 2) + Time (Stage 3)
  • Total time = $ 1 \text{ hour} + 2 \text{ hours} + \frac{11}{6} \text{ hours} = 3 + \frac{11}{6} $ hours.
  • Total time = $ \frac{18}{6} + \frac{11}{6} = \frac{29}{6} $ hours.

Convert the total time into hours and minutes:

  • $ \frac{29}{6} $ hours = 4 whole hours and $ \frac{5}{6} $ of an hour.
  • $ \frac{5}{6} $ hour = $ \frac{5}{6} \times 60 $ minutes = 50 minutes.
  • Therefore, the total time is 4 hours and 50 minutes.
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Important Questions from Pipe and Cistern

  1. Two pipes A and B can independently fill a tank completely in 20 and 30 minutes respectively. If both the pipes are opened simultaneously, how much time will they take to fill the tank completely?

  2. The compound interest on Rs. 64,000 for 3 years, compounded annually at 7.5% p.a. is

  3. A water tank can be emptied in 40 minutes by a pipe of d 'diameter, so how long will it take for a 2d diameter pipe to be emptied?

  4. A pipe can fill a tank in 4 hours, while a leak which is at one-fourth of the height of the tank from bottom can empty upto that part in 2 hours. If both are operated simultaneously and initially the tank is full, then when it will be one-fourth full?

  5. Two pipes X and Y can fill an empty tank in 16 hours and 20 hours respectively. Pipe Z alone can empty the completely filled tank in 25 hours. Firstly both pipes X and Y are opened and after 6 hours pipe Z is also opened. What will be the total time (in hours) taken to completely fill the tank?

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