Quick and dirty: painting a fence

There are four fences, which are geometrical identical (same heigth and same length). The following four people need the specified time to paint it:

• Bob needs 2 hours.
• Simon needs 3 hours.
• Tommy needs 5 hours.
• Johnny needs 8 hours.

How long does it take, if they paint one mentioned fence together?

Bonus: what is the special electrotechnical approach to this issue?

• I'm voting to close this question as off-topic because it's a math problem and not a math puzzle.
– xnor
Commented Jan 30, 2015 at 11:52
• There is an explicit puzzle feature: recognizing the parallelism between the electrotechnical formula and the described situation. Commented Jan 30, 2015 at 11:58
• @xnor Isn't migration (to Math SE) a better option than closing? Commented Jan 30, 2015 at 14:59
• @ghosts_in_the_code You're right, but cross-site migration doesn't seem to be enabled here. I think a mod would need to move it.
– xnor
Commented Jan 30, 2015 at 22:19

Nice! The answer to the bonus question is: adding up the resistance of parallel resistors.

• Mathematically, the work paces of the four guys are $B=1/2$, $C=1/3$, $T=1/5$ and $J=1/8$.

• If we put them in parallel, the resulting work pace is $B+C+T+J=139/120$.

• The total work time is the reciprocal of this, that is $1/(B+C+T+J)=120/139\approx0.863$ hours $\approx 52$ minutes.

• This is close. However, you simply add the work paces to find the total pace. To see that I'm correct, use your method for 2xBob (or even Bob alone). Commented Jan 30, 2015 at 12:05

It helps being very explicit about the units in which the various figures are expressed:

Bob: 2 hours/fence $$\rightarrow$$ $$\frac12$$ fence/hour

Simon: 3 hours/fence $$\rightarrow$$ $$\frac13$$ fence/hour

Tommy: 5 hours/fence $$\rightarrow$$ $$\frac15$$ fence/hour

Johnny: 8 hours/fence $$\rightarrow$$ $$\frac18$$ fence/hour

Painting in parallel: $$(\frac12 + \frac13 + \frac15 + \frac18)$$ fence/hour = $$\frac{139}{120}$$ fence/hour $$\rightarrow$$ $$\frac{120}{139}$$ hour/fence.