From the 2013 Pink Kangaroo:

*Yurko saw a tractor slowly pulling a long pipe down the road. Yurko walked along beside the pipe in the same direction as the tractor, and counted 140 paces to get from one end to the other. He then turned around and walked back to the other end, taking only 20 paces. The tractor and Yurko kept to a uniform speed, and Yurko’s paces were all 1 metre long. *

*How long was the pipe?*

__________

When I read this sort of problem, I think two things:

(i) The solution will be quite short, and will come from forming one or two equations and solving them.

(ii) Forming the equations could make my head hurt.

Given the nature of the problem, we will almost certainly be making use of the formula ‘distance = speed x time’.

Let’s call the length of the pipe x metres. On the first walk, Yarko covers 140m. He has to walk the length of the pipe plus the distance moved by the pipe.

Now it makes sense to label some other quantities. Call the tractor’s speed u m/s and Yarko’s speed v m/s. So Yarko covers 140m in 140/v seconds. In that time the pipe moves 140u/v metres. We now have our first equation:

x + 140u/v = 140.

On the return walk, Yarko covers 20 metres in 20/v seconds This time he has to walk the length of the pipe *minus *the distance moved by the pipe. In 20/v seconds, the pipe moves 20u/v metres. We have our second equation:

x-20u/v = 20.

Subtracting the second from the first yields

160u/v = 120.

So u/v = 3/4. Substituting back into the second equation above yields

x – 15 = 20,

thus x = 35, and the pipe is 35 metres long.

I am a huge fan of this type of puzzle. All word problems require you to turn something written in everyday language into mathematics, and distance – speed – time problems always require a bit of thought to form the correct equations.

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