please empty your brain below 
Yesterday's answers
1) apricot 2) ONE 3) every 40 years 4) A→B→D, D→C, C→E in 1 hour and 9 minutes. 5) 29 (10 out of 29 = 34%) 
i took the 11:15 from Eston, changed at Bilton for the 11:44, Denge at 12:02 and then then 12:13 to Apcot at 12:39. Takes 1 hour 24 minutes. Is this ok too?
(you didn't say it had to be the fastest way) dg writes: Perfectly OK, but not what was intended, so I've now tweaked the timetable to give a unique solution. 
Five out of five. Percentage took me the longest. Figured it had to be less than 17/50 but more than 8.5 out of 25. Checked by hand lower prime numbers, and then the next one up was 29.
Planets were trial and error too, but quicker. Clearly more than 15 years, and more than 24 years, and more than 30 years, and more than 32.5 years, and more than 36 years, but less than 45 or 48 years. Obvious one between 36 and 45 worked. 
I agree with the above for question 4, nothing about doing the journey as fast as possible, which direction to make it, or allowing time for connections, anyway the A&E has Japanese levels of timekeeping and Bilton has an island platform :)

Five out of five.
@Andrew. for Q3, I worked it out mathematically. One orbit every 15 years = 24° per year. Similarly, one orbit every 24 years = 15°, hence the faster one catches the other at 9° per year. So, 360° divide by 9°/ year = 40 years. Q5: I used the bounds of 0.345 to 0.335 applied to every number starting from 1 which gives the same integer result until one reaches 29. Then it was a simple matter of testing that it worked (!) 
Thanks for a very enjoyable Boxing Day diversion. (I did use the computer for question 5 since I couldn't really be bothered with the mental arithmetic  too much like my day job!

All I got was the first one. As for the planets one, the nearest I got was a supposition that a probable reason for the different orbit times was that one was on a different, much longer, radius to the other, with the result that their relative positions to each other could actually remain pretty much constant. I'm not sure I thought it right through, but came to the idea that they don't necessarily overtake each other at all.

Bad memories of the 11 plus exam.

5/5. Interesting puzzles, thanks for setting dg!
shirokazan's way of doing 5 is the easiest, I think. To phrase it differently, you're looking for smallest n with a multiple of 100 existing between 33.5*n and 34.5*n, so you just look until you find one. @Andrew, I don't think it needed to be prime. 
Roger W : they will indeed be on orbits with different radius. But this does not affect their overtaking behaviour, just like the different lengths of the hands of a clock does not affect their overtaking behaviour!

I don't really see why the number of respondents has to be a prime number  though it does happen that this one is.
What would be really nice would be a formula to calculate the result if, in that question, 34% was replaced by any other percentage. 
Thanks, Malcolm... I think I knew there was something like that: thanks for making the explanation so obvious.

You do realise you're all mad don't you.
My brain fell out just reading the questions! 
For queston 3 where D is the number of orbits of the slower planet and T is the time in years, the following solution works algebraically.
D=T/24 D+1=T/15 T/24+1=T/15 15T+360=24T 9T=360 T=40 I believe that substituting the 1 for 2,3 etc should give the solution for subsequent overtakes. 
No I didn't think 5 was 'hard' more that it took trial and error  although the range could be narrowed by applying some logic.
Is there any algebraic way of cutting out the trial and error? 
1. Easy
2. Didn't get it after 2 mins of trying, so gave up. I never think of spelling numbers out in English, since I don't see numbers as English words  and have been caught out by this before when everyone else in the room got it quickly... Here I was trying to see if there was a relationship between the letters. 3. Easy  lowest common multiple of 15 and 24 is 120. Every 120 years, A goes round 8 times and B goes round 5 times (which I already knew from working out the LCM  it really takes longer to explain than to just do it!) Thus A overtakes B 85=3 times every 120 years, i.e. once every 40 years. 4. I got the same answer as Geoff and still anon, but missed DG's own answer. 5. I did it in just about 50 seconds, so DG's comment yesterday was right on the mark :) But it will take me much longer than 50 seconds to explain! Obviously, 100 works (as 34% is just 34/100), so the answer is either 100 or less than 100. The quickest way is thus to try every number, with a calculator for the denominators where I haven't memorised the decimal representation, or mental arithmetic is too taxing. No need for spreadsheets! It's obviously not 1, 2, 3, 4 or 5. 2/6 is 33.33%. So the numerator to try is now 3, if we want a number bigger than 33.33%, since 2/7 is smaller than 2/6. 3/7 = 42...% 3/8 = 37...% 3/9 is 33.33%. So the numerator to try is now 4. 4/10 = 40% 4/11 = 36...% 4/12 is 33.33%. So the numerator to try is now 5. 5/13 = 38...% ... ... 9/27 is 33.33%. So the numerator to try is now 10. 10/28 = 35.xx% 10/29 = 34.48% AHA! *** For the multiples of 3, it's important to realise that since we are looking at numbers below 100, there's no need to try the next numerator until the denominator is incremented (until we get closer to 100): We want a result between 33.5% and 34.5%. We don't want to miss any results 34.5%33.33%=1.17% greater than the number we are trying, if the denominator is a multiple of 3. If we went past 66 without finding an answer, then for example 22/66 is 33.33% and 23/66 is 1/66 greater. 1/66 is about 1.5%, which is getting close to 1.17%. So just to be safe I would have tried 24/69 as well, because it is faster to just type it into the calculator than to work out that 24/69 couldn't be the answer. But for the smaller numbers, I knew that I didn't need to bother trying 4/9, 5/12 ... 10/27 etc since the difference between those and 33.33% could not possibly result in those fractions being less than 34.5%. Then since the answer was found long before trying 66, I didn't need to bother with this footnote. And it took about 1000 times longer to figure out how to express these fleeting thoughts in words than just getting on with it! 
Me too @Chris!
I got 1/1! :D 
In question 5, it turns out that every other percentage below 50% can be done with fewer than 29 people.
dg writes: Actually, no it doesn't. 
@DG 4:05pm
Surely it then follows that the only percentages which require as many as 29 people are 34% and 66% 
@dg I think 1% needs 67 people, and 2% needs 41. 3% also requires 29, and 49% requires 35.
@timbo I don't think it follows  you can get 13% with 8 people, but for 87% you will need 15. 
Milo  depends if you are rounding 0.5 up to the next whole number or to the nearest even number.

Perhaps I was lucky with the primes, but I had a feeling it must be less than 17/50, and could not be a multiple of 2 or 3, and multiples of 5 did not look like they would work either. That took out most (all?) of the nonprime numbers below 25!
Knew the right way to do the planets would be with the angular velocity, but trial and error required less thinking! Haven't had the time to look at the next five, so studiously looking away from the answers for now. Thanks for the entertainment, DG. 
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