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.

- 39 min
- 27 min
- 37 min
- 33 min
- 45 min

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

- 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).
- 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.