This is an interesting variant of a work / rate problem where the second "worker" does negative work. We'll use the formula Work = Rate * Time (W = RT). Here, the "work" consists of emptying the cistern. We want to know how many liters the cistern holds, which is the same as the amount of work done (the number of liters emptied). There are two scenarios:
Leak Only:The leak empties the cistern in 8 hours. If we set the work (the liters to be emptied) as w and the leak rate as r, you have:
w = r * 8
Leak and Tap:The leak still operates, but now a tap fills the cistern at 6 liters per minute, which impedes the work. Let's multiply that by 60 to get 360 liters per hour (keeping the units consistent). Together the tap and the leak take 12 hours to empty the cistern.
Remember, in work / rate problems, when multiple workers work together, you add their rates to find the combined rate. It's exactly the same here, except that the tap is actually
reversing the work that the leak does. So, to find the combined rate of the leak and tap together, you need to
subtract the tap's rate from the leak's rate.So, the combined rate is (r - 360) and the work / rate equation is:
w = (r - 360) * 12
Since the work (w) is the same in both scenarios, set the equations equal to each other and solve:
8r = 12(r - 360)
2r = 3(r - 360)
2r = 3r - 1080
r = 1080
Now plug into the first equation and solve for w:
w = 1080 * 8
w = 8640
The answer is A.On work / rate problems, remember that you can add rates (don't add the times!). And when something is actually creating more work rather than doing work, subtract its rate instead.