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?

  • 2
    $\begingroup$ I'm voting to close this question as off-topic because it's a math problem and not a math puzzle. $\endgroup$
    – xnor
    Commented Jan 30, 2015 at 11:52
  • 1
    $\begingroup$ There is an explicit puzzle feature: recognizing the parallelism between the electrotechnical formula and the described situation. $\endgroup$
    – Alexis
    Commented Jan 30, 2015 at 11:58
  • $\begingroup$ @xnor Isn't migration (to Math SE) a better option than closing? $\endgroup$ Commented Jan 30, 2015 at 14:59
  • $\begingroup$ @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. $\endgroup$
    – xnor
    Commented Jan 30, 2015 at 22:19

2 Answers 2


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.

  • $\begingroup$ 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). $\endgroup$ 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.


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