Three pipes and a tank

How to avoid fractions in work problems.

Two pipes A and B can fill a tank in 60 min and 90 min respectively. A leak is present at \(\tfrac{3}{4}\) of the height. The leak alone takes 36 min to empty till \(\tfrac{3}{4}\) height of the tank. Find the time taken to fill the tank when all of the taps and the leak are opened simultaneously.

1. 39 min
2. 27 min
3. 37 min
4. 33 min
5. 45 min
   

If you calculate per-minute rates, you get fractions you’d like to add. There are two ways to avoid them.

1. Introduce a probably fictitious unit for the work to be done so that the rates become integer-valued. Here we want to divide by 60, 90, 36, and 4. A common multiple of these numbers will do the trick. Therefore, you can assume that the tank contains 180 units of water (gallons, fl oz, pints, pips, whatever).
2. You don’t have to calculate rates to compare or add productivities. It’s only important to let the pipes work the same amount of time.

So let’s solve the problem using the latter strategy.

Pipes A and B work together. To combine the two, we let them operate 180 minutes, a multiple of 60 and 90, and get \[ A +B \xrightarrow {\text{180 min}} (3 +2)\text{ tanks} = 5\text{ tanks} \] The combination AB of the two pipes fills the tank in 36 minutes.
The leak C empties 1/4 of the tank in 36 minutes, or the equivalent of 1 tank in 144 minutes. We combine the pipes C and AB by letting them work 144 minutes each, a multiple of 36: \[ AB -C \xrightarrow {\text{144 min}} (4 -1)\text{ tanks} = 3\text{ tanks} \]

The combination ABC of the three pipes fills (the equivalent) of 1 tank in 48 minutes.
Since AB supplies 3/4 of the tank and ABC the remaining 1/4, it takes \[ \tfrac{3}{4}\times 36\,\text{min} +\tfrac{1}{4} \times 48\,\text{min} = 39\,\text{min} \] to fill the tank.