On Friday I posted this puzzle….
Imagine a water-cask and three different sized water-taps. The smallest tap can fill the water-cask in 20 minutes. The middle tap can fill the cask in 12 minutes. And the largest tap can fill the water-cask in 5 minutes. How long does it take to fill the water-cask with the three taps together?
If you have not tried to solve it, have a go now. For everyone else the answer is after the break.
The smallest tap fills a twentieth of the water-cask in 1 minute. The middle tap fills a twelfth of water-cask in 1 minute. The largest tap fills a fifth of water-cask in 1 minute. Therefore, together they fill a third of the water-cask in 1 minute and so the water-cask is filled in 3 minutes.
Did you solve it?
Richard Wiseman 
It’s like parallel resistors in electronics. The answer is 1 / (1/time1 + 1/time2 +1/time3)
Yes, did it like this.
Yes; what he said.
The explanation should include this to make it clearer.
1/20+1/12+1/5=3/60+5/60+12/60=20/60 which means 20/60 or 1/3 is filled in 1 minute.
Yes indeed, that’s how I solved it as well.
Same here; just simple math.
I came at it from a different direction.
The smallest tap takes 20 minutes so will fill 3 casks in an hour
The middle tap takes 12 minutes so will fill 5 casks in an hour
The largest tap takes 5 minutes so will fill 12 casks in an hour
So all 3 taps will fill 3+5+12 casks = 20 casks per hour
So to fill 1 cask takes 1/20th of an hour = 3 minutes.
Excellent analysis!
I got the right answer, but in a different way.
I chose an arbitary volume for the container, then worked out the flow rates for each tap.
For convenience, the volume I chose was derived from the time each tap took to fill said container:
20 x 12 x 5 = 1200 units
That way I could be sure there would be no decimals in my results.
Then the flow rates were calculated as volume/time:
= 60, 100, 240 units per min, for each of the 3 taps.
From this I calculated a total flow rate for all 3 taps together:
400 units per min
So to fill a container of 1200 units would take 3 mins @ 400 units/min
A longer way, but the same answer.
Get back to your pit.
How rude.
Ignore Eddie, PM, he’s not had his pills yet this morning.
yes… but in stead of 20 x 12 x 5, one can consider 20 = 2 x 2 x 5 and 12 = 2 x 2 x 3 and 5 = 5 (get it to prime factors). Then you see that you don’t need decimals when you use 2 x 2 x 3 x 5 = 60.
I did it the same way with 60 units instead of 1200.
Good puzzle. Thanks
I did it the old-fashioned way with Algebra. x/20 + x/12 +x/5 = 1; 20x = 60; x = 3.
wow
I did it lindamp’s way.
I did it a special way, using complex numbers, but which is too complicated to describe here.
I did the simple ‘how full after 1min?’ approach, then went on to develop a general theory of taps and a special theory of dripping taps both of which are too comlex to describe anywhere.
+p
What if the three taps are having a race, with someone on the outside being paid to interfere with the faster or higher-capacity ones, so that the smallest wins? (Yes, I know, I’m odd. I’ve also been called out for personifying my breakfast cereal.)
Does it involve corks? And if so where do you put them?
20*3 = 12*5 = 5*12 = 60
3+5+12=20
60/20 = 3
Another approach: common denominator of 20,12, 5 = 240.
1/20 = 12/240.
1/12 = 20/240
1/5 = 48/240
Total = 80/240 or 1/3.
damn man, that really hurt my head! I spent ages writing this out.
I I tried using simultaneuos equations, so
20a = x
12b = x
5c = x
I am still convinced that this method will work… I think I am just not maths smart enought to pull the thing off! 😉
I guess I did it very simply. Tap 1 fills 3 casks/hour, no. 2 8 casks/h, no. 3 12 casks /h, which gives a filling speed of 20 casks /hour. 1 hour divided by 20 is three minutes.
I did it by throwing dollops of treacle pudding at my next-door neighbour’s pet armadillo until I received inspiration from the background collective consciousness of the universe.
Like John, I used the old-fashioned way of working it out. 🙂
I checked out my answer with algebra afterwards, mind…
20-12=8-5=3
I liked Arneb’s short-cut which was a ‘hair’ quicker/simpler than the traditional way I used. I did it the way I learned from text books dealing with ‘work’ problems. When given the time a person, machine, etc. does something, you were taught to take the inverse for 1 minute. Then find the lowest common denominator which in this problem is 60 (why use 240 or something larger?)
If the smallest tap fills the cask in 20 minutes then it does 1/20 in 1 minute.
You get 1/12 and 1/5 for the other two (for 1 minute). The lowest common denominator for 1/20, 1/12, and 1/5 is 60 ……..= 3/60 + 5/60 + 12/60 for 1 minute = 20/60 for 1 min.
If 20/60 for 1 minute then it would take 3 minutes to fill working together
(20/60 + 20/60 + 20/60 = 60/60).