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Question

Two inlet pipes can fill the tank in 15 hours and 18 hours and one outlet pipe can empty the tank in 20 hours. If all the pipes are opened simultaneously, then in how much time the tank will be filled completely?

This question was previously asked in
ESIC UDC Mains MBT (30 Apr 2022)
The correct answer is

180/13 hours

This problem involves calculating the time taken to fill a tank when multiple pipes with different rates (inlet and outlet) are operating simultaneously. We need to find the combined rate of filling and then determine the total time.

Calculating Pipe Filling Rates

First, we determine the rate at which each pipe works. The rate is the fraction of the tank that can be filled or emptied in one hour.

  • Inlet Pipe 1: Fills the tank in 15 hours. Its filling rate is $\frac{1}{15}$ of the tank per hour.
  • Inlet Pipe 2: Fills the tank in 18 hours. Its filling rate is $\frac{1}{18}$ of the tank per hour.
  • Outlet Pipe 3: Empties the tank in 20 hours. Its emptying rate is $\frac{1}{20}$ of the tank per hour. Since it empties the tank, we consider its contribution as negative.

Finding the Combined Rate of Filling

When all pipes are opened together, the net rate of filling is the sum of the rates of the inlet pipes minus the rate of the outlet pipe.

Combined Rate = (Rate of Pipe 1) + (Rate of Pipe 2) - (Rate of Pipe 3)

Combined Rate = $\frac{1}{15} + \frac{1}{18} - \frac{1}{20}$ tank per hour.

Determining the Least Common Multiple (LCM)

To add and subtract these fractions, we need a common denominator. We find the LCM of 15, 18, and 20.

  • Prime factorization of 15: $3 \times 5$
  • Prime factorization of 18: $2 \times 3^2$
  • Prime factorization of 20: $2^2 \times 5$

The LCM is the product of the highest powers of all prime factors involved: $2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$.

Calculating the Net Filling Rate

Now we convert each fraction to have the denominator 180:

  • $\frac{1}{15} = \frac{1 \times 12}{15 \times 12} = \frac{12}{180}$
  • $\frac{1}{18} = \frac{1 \times 10}{18 \times 10} = \frac{10}{180}$
  • $\frac{1}{20} = \frac{1 \times 9}{20 \times 9} = \frac{9}{180}$

Substitute these values back into the combined rate equation:

Combined Rate = $\frac{12}{180} + \frac{10}{180} - \frac{9}{180}$

Combined Rate = $\frac{12 + 10 - 9}{180} = \frac{22 - 9}{180} = \frac{13}{180}$ tank per hour.

Time Taken to Fill the Tank

The time required to fill the tank is the reciprocal of the combined filling rate. This is because Time = Total Work / Rate, and here the Total Work is filling 1 tank.

Time = $\frac{1}{\text{Combined Rate}}$

Time = $\frac{1}{\frac{13}{180}}$ hours

Time = $\frac{180}{13}$ hours.

Therefore, if all the pipes are opened simultaneously, the tank will be filled completely in $\frac{180}{13}$ hours.

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