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An old farmer was walking beside a tractor, which was driven by another farmer. The farmer driving the tractor was dragging a pipe behind him. The old farmer wants to figure out how long the pipe is. He starts by starting at the very end of the pipe (farthest away from the tractor) and walks forward until he reaches the other end of the pipe. This take 140 steps. He then walks back to the other end, which takes only 20 steps, as this time he is walking in the opposite direction of the tractor.

The tractor is moving constantly at a uniform speed. The farmer is also moving at a uniform speed, and each of his steps is one meter long.

How long is the pipe?

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    $\begingroup$ Ehm.. another math-only based problem. $\endgroup$ Mar 29, 2015 at 17:27
  • $\begingroup$ Problem with those problems? $\endgroup$
    – Bretsky
    Mar 29, 2015 at 17:31
  • $\begingroup$ Take a look here $\endgroup$ Mar 29, 2015 at 17:32
  • $\begingroup$ @MarcoBonelli since when you were been tracking math-only subjects , go to math.se there s way way deep problems worth to be called "mathematical dilemma", i think the op was doing well animating this website with such puzzles because , it looks like riddle and rebus tags are overcoming and crawlin over all the ground. $\endgroup$
    – Abr001am
    Mar 29, 2015 at 20:02
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    $\begingroup$ @MarcoBonelli if you want to remove this math from this place so we have to rename it riddles-and-detective-mysteries . besides . maths are needed everywhere even a nutcracker works with maths $\endgroup$
    – Abr001am
    Mar 29, 2015 at 20:11

1 Answer 1

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The pipe is

35 metres long.

Proof: let $F$ and $T$ be the speed of the old farmer and the tractor respectively, $L$ be the length of the pipe, and $t_1$, $t_2$ be the times taken by the old farmer to walk along the pipe in each direction.

The old farmer walks $140$ metres in time $t_1$; this includes both walking the length of the pipe and walking as far as the tractor moves in time $t_1$. So $140=Ft_1=L+Tt_1$.

Then he walks $20$ metres in time $t_2$; this includes walking the length of the pipe minus as far as the tractor moves in time $t_2$. So $20=Ft_2=L-Tt_2$.

We solve these two simultaneous equations by making the following deductions:

  • $140=Ft_1$ and $20=Ft_2$, so $t_1=7t_2$
  • $120=140-20=(L+Tt_1)-(L-Tt_2)=T(t_1+t_2)=8Tt_2$, so $Tt_2=15$
  • $L=20+Tt_2=35$.
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  • $\begingroup$ I don't know why people are downvoting you for answering the question. Regardless of the fact that perhaps my question doesn't belong here, you still provided the right answer. $\endgroup$
    – Bretsky
    Mar 29, 2015 at 20:06
  • $\begingroup$ @Bretsky I don't think it's anything to do with this particular answer or its quality. I've had 12 downvotes today on a wide variety of my posts. But thanks :-) $\endgroup$ Mar 29, 2015 at 20:47
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    $\begingroup$ The down-voting may be because you also voted to close - considered "bad form" by some. $\endgroup$
    – Joffan
    Mar 29, 2015 at 21:18
  • $\begingroup$ @Joffan Fair enough. It was fun to solve, and I did upvote the question ... but Marco Bonelli is right that it's off-topic for the site, so I decided after solving it that I should VTC. $\endgroup$ Mar 30, 2015 at 9:32

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