There are three inlet pipes X, Y and Z in a tank. X and Y can fill the tank in 45 hours. Y and Z can fill the

Question

There are three inlet pipes X, Y and Z in a tank. X and Y can fill the tank in 45 hours. Y and Z can fill the tank in 36 hours. Z and X can fill the tank in 72 hours. The number of hours in which all the three pipes can fill the tank, if operated simultaneously, is:

Accepted Answer

Let the total capacity of the tank be the

LCM

of the given times (

45, 36, 72

).

LCM (45, 36, 72) = 360 units

Now, let's find the individual combined efficiencies per hour:
*

Efficiency of (X + Y) = \frac{360}{45} = 8 units/hour

*

Efficiency of (Y + Z) = \frac{360}{36} = 10 units/hour

*

Efficiency of (Z + X) = \frac{360}{72} = 5 units/hour

Adding these three equations:

2 (Efficiency of X + Y + Z) = 8 + 10 + 5 = 23 units/hour

Efficiency of X + Y + Z = \frac{23}{2} units/hour

The time taken by all three pipes together to fill the tank is:

Time = \frac{Total Capacity}{Combined Efficiency} = \frac{360}{\frac{23}{2}} = \frac{720}{23} hours

Let's convert

\frac{720}{23}

into a mixed fraction:

720 \div 23 = 31 with a remainder of 7

Time = 31 \frac{7}{23} hours

There are three inlet pipes X, Y and Z in a tank. X and Y can fill the tank in $45$ hours. Y and Z can fill the tank in $36$ hours. Z and X can fill the tank in $72$ hours. The number of hours in which all the three pipes can fill the tank, if operated simultaneously, is:

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