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A farmer is looking to hire some hired hands. His usual three hires are named Barry, Harry, and Larry.

He knows from hiring them before that:

  • Working together, Barry and Harry can plow an acre in two hours.
  • Working together, Barry and Larry can plow an acre in three hours.
  • Working together, Harry and Larry can plow an acre in five hours.

How many hours would it take for all three of them working together to plow an acre?

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closed as off-topic by Len, Engineer Toast, Tryth, Cows quack, CAS Jul 1 '15 at 8:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Len, Engineer Toast, Tryth, Cows quack, CAS
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ So they have to all work the same ammount and they can't work alone? $\endgroup$ – martijnn2008 May 17 '14 at 18:21
  • $\begingroup$ No, they each work individually at their own rate, but the sum of their work speed is represented. $\endgroup$ – Joe Z. May 17 '14 at 18:21
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Suppose we divide an acre into 30 sections.

Then:

  • Barry and Harry can plow 15 sections an hour.
  • Barry and Larry can plow 10 sections an hour.
  • Larry and Harry can plow 6 sections an hour.

If we add all these three up and divide by 2, we get the number of sections all three of them can plow in one hour:

  • Barry, Harry, and Larry can plow 15.5 sections an hour.

(Looks like poor Larry works 10 times slower than either of the other two boys.)

Thus, it will take all three of them (30/15.5 hours) = about 1 hour and 56 minutes to plow an acre.

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  • 2
    $\begingroup$ Aw, I was about to answer! Sneaaaaky! $\endgroup$ – Aza May 17 '14 at 18:26
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    $\begingroup$ I'm just doing the grunt work of posting as many traditional riddles as I can think of during the private beta. $\endgroup$ – Joe Z. May 17 '14 at 18:26
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    $\begingroup$ To be honest, it's actually probably a very good idea to post canonical riddles. Thank you for helping! $\endgroup$ – Aza May 17 '14 at 18:27
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    $\begingroup$ There's a meta question on that subject that you might want to weigh in on. $\endgroup$ – Joe Z. May 17 '14 at 18:28
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    $\begingroup$ the point is the answer that he gave, but this is not to the place to discuss this. But to be clear there is NO REAL GAIN to answer this kind of question within an hour yourself. $\endgroup$ – martijnn2008 May 20 '14 at 22:19
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Here's another way of looking at it.

Let's imagine each of Barry, Harry, and Larry gets cloned, so there are two copies of each (both with the same working speed, naturally). In 1 hour, we know how much Barry1 and Harry1 together can get done ($\frac{1}{2}$ acre), how much Barry2 and Larry1 together can get done ($\frac{1}{3}$ acre), and how much Harry2 and Larry2 together can get done ($\frac{1}{5}$ acre). So 2 copies of everyone get $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{31}{30}$ acres ploughed in an hour.

So Barry, Harry, and Larry together get half this amount ($\frac{31}{60}$ acres) done in an hour, which means they take $\frac{60}{31}$ hours or about 1 hour 56 minutes to plough an acre.

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  • $\begingroup$ That's exactly my answer, though, just without the "dividing into sections" bit. $\endgroup$ – Joe Z. Jun 24 '15 at 19:03
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    $\begingroup$ @JoeZ. Well, of course it gives the same answer! It's just a different approach, dividing up time instead of space. You don't reckon it's worth anything? $\endgroup$ – Rand al'Thor Jun 24 '15 at 22:34
  • $\begingroup$ The difference is a bit too trivial. You're not dividing up time at all, just space, which is exactly what I did as well. (For what it's worth, I'm not the one who downvoted your answer.) $\endgroup$ – Joe Z. Jun 24 '15 at 23:02
  • $\begingroup$ @JoeZ. Ah well, maybe I'll delete my answer then. But I'm now wondering if the question is no longer on-topic since the maths problems / maths puzzles ruling. What do you reckon? $\endgroup$ – Rand al'Thor Jun 24 '15 at 23:43
  • $\begingroup$ Yeah, I think nowadays this might be classified as a math problem. But personally I'd just bury this question instead of closing it outright. $\endgroup$ – Joe Z. Jun 24 '15 at 23:53

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