# 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 Jan 30 '15 at 11:52
• There is an explicit puzzle feature: recognizing the parallelism between the electrotechnical formula and the described situation. – Alexis Jan 30 '15 at 11:58
• @xnor Isn't migration (to Math SE) a better option than closing? – ghosts_in_the_code Jan 30 '15 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 Jan 30 '15 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). – frodoskywalker Jan 30 '15 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.