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Pudcat | 08:53 Fri 22nd Sep 2017 | Quizzes & Puzzles
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We have to assume that running with the wind the dog is assisted as much as it is hindered when running against. My way (I'm going to work in mph) u = dog's speed w = speed by which the wind assists / hinders dog distance = speed x time with wind - 1 = (u + w) x (1/20) so u + w = 20 against wind - 1 = (u - w) x (1/15) so u - w = 15 adding 2u = 35 u = 17.5 mph w = 20 - 17.5 = 2.5 mph Check 1 mile at...
11:42 Fri 22nd Sep 2017
Need a question.
I assume the poster meant today's puzzle
http://www.bbc.co.uk/programmes/p057wxwl

Puzzle No. 60 - Friday September 22
A dog runs a mile in three minutes with the wind and returns in four minutes against the wind.
How long would it take the dog to run the mile if there was no wind?

Need to look at identical periods of time not distances to find average speed.
With wind - in 4mins dog would travel 1 + 1/3 miles
Against wind - in 4 mins dog would travel 1 mile.
with the wind helping exactly half the time and hindering him for the other half. On a windless day he could go 2 1/3 miles in eight minutes, or one mile in 3 and 3/7 minutes.


Answer looks okay. I'm between lessons so hard to concentrate but I would have worked it out slightly differently. I'd have said the wind-assisted journey was at 20 mph, the return journey against the wind was 15 mph. So the wind speed was +2.5 mph out and -2.5 coming back, meaning true speed without wind was 17.5 mph. So each mile takes 60/17.5 minutes= 120/35 = 24/7 minutes= 3 and 3/7 minutes.
Sotty if a bit garbled. Will look back later
I'd have said the wind made a minute's difference and gone in between for 3.5 minutes.
Unfortunately that's slightly wrong, and I agree with ff and Fibonacci.
Apparently so :-). Just not sure why.
I suppose the most natural way of trying to explain it is that whether you are with the wind or against it changes your speed directly, as ff gives, and changes the time in a different way -- through the relationship speed = distance/time. So it's not a linear relationship between speed and time, and you can't just split the difference in the times as well as the speeds.
Ok, thanks Jim xx. Not entirely sure I get the logic, but i'll go with it :-)
I suppose you can also see it with the idea that, if speed doubles, then time taken halves. And if speed doubles again, then time taken halves again. And so on. Joining the dots in a straight line (which is what would be implied by the 3.5 minutes answer) would then lead to the graph of speed against time going in a sort of jagged shape, changing direction at the corners. But that's not really reasonable at all, because there's nothing special about doubling speed.

So you have to join everything up with a smooth curve, and that means that increasing speed doesn't decrease time taken linearly.

Does that make more sense?
I think so, Jim. Thank you xx i'm sure we did this at school, but that was hundreds of years ago now :)
Glad to hear it :) Also I'm sure it was no more than 75 years ago, don't be so harsh on yourself :P
We have to assume that running with the wind the dog is assisted as much as it is hindered when running against.
My way (I'm going to work in mph)
u = dog's speed
w = speed by which the wind assists / hinders dog
distance = speed x time
with wind - 1 = (u + w) x (1/20)
so u + w = 20
against wind - 1 = (u - w) x (1/15)
so u - w = 15
adding
2u = 35
u = 17.5 mph
w = 20 - 17.5 = 2.5 mph
Check
1 mile at (17.5 + 2.5) takes 60/20 = 3 minutes
1 mile at (17.5 - 2.5) takes 60/15 = 4 minutes

I have followed the maths puzzles in this series with interest - what did others make of Puzzle No. 7, Tuesday 11 July, probability of an explorer surviving after consulting a soothsayer?

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