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« Reply #45 on: April 03, 2020, 07:05:31 AM » |
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I'm going with 26, but I'm now paranoid and trying to see if the flowers change, the leaf is on a different side ........ this is quite stressful.  last blue flower only has 4 petals! - should have gone to specsavers, oh wait! not essential so they are closed  I think the answer is 24 AHHHHHHHHHHHHHHHHHHHHHHH for F F F F Flowery flippin sakes, that's just cruel!
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Just call me Mart. Cardiff: 5.58kWp PV - (3.58kWp SE3500 + 2kWp SE2200 WNW)
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todthedog
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« Reply #46 on: April 03, 2020, 09:05:20 AM » |
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Good luck  A. 6 8 2 – One Number is correct and well placed B. 6 1 4 – One Number is correct but wrong place C. 2 0 6 – Two Numbers are correct but Wrong Places D. 7 3 8 – Nothing is correct E. 8 7 0 – One Number is correct but wrong place
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Kidwelly South Wales
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Antman
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« Reply #47 on: April 03, 2020, 09:21:54 AM » |
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042 ? Here's my working... Strike 7 3 & 8 A. 6 8 2 – One Number is correct and well placed B. 6 1 4 – One Number is correct but wrong place C. 2 0 6 – Two Numbers are correct but Wrong Places D. 7 3 8 – Nothing is correct E. 8 7 0 – One Number is correct but wrong place So from C and E, 0 must be first digit From C, 2 must be third digit From B, 4 must be 2nd digit I still need an explanation of the fruit trader please cos I'm dim  Antman
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20 x 47mm, 172 l cylinder, Heat Dump, 15 x Sanyo HIT-H250E, SB4000TL, Nestor Martin IQ13 WBS DIY Solar System Support at http://www.handyantman.co.uk/antman.htmlAll support is voluntary and free of charge. I'm not employed by Navitron so responses may not be same-day
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marshman
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« Reply #48 on: April 03, 2020, 09:34:33 AM » |
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Agree 042.
Every time the fruit trader goes through a checkpoint to start with he gives 3 fruits away. But he also moves 2 pieces of fruit from 1 sack into the other sacks. This keeps two sacks full and empties one sack after only 10 checkpoints. Then as he only has 2 sacks he only has to give away 2 fruits at each check point and move 1 fruit from one sack to the other to help empty the 2nd sack quicker. So after another 15 checkpoints the second sack is empty. There are only 5 checkpoints left and he now has one sack of 30 fruits left, so at the end has 25 fruits left.
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biff
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« Reply #49 on: April 03, 2020, 09:47:55 AM » |
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Good one Roger, that had me stumped also , There is no word of the Fox, The Duck and the Bag of Corn. Or the man who had to cross the river.
NB, it was that long ago , The fox had rabies and bit the man. The man had an aversion to water could not cross the river ate the duck and had pop corn for after, Not quite. Biff
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« Last Edit: April 03, 2020, 09:51:19 AM by biff »
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omega1
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« Reply #50 on: April 03, 2020, 10:10:07 AM » |
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I would agree that the combo lock question is 042. Playing devils advocate regarding the nomad question I dont think you can answer it with the info you are given. The wording states that he has three sacks which can not did hold 30 fruits each. None of the sacks can hold more than 30 fruits. Doesnt actually say how many fruits each sack actually held to start with  This is keeping us all occupied in these trying times. keep em coming Biff  Kevin
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Ted
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« Reply #51 on: April 03, 2020, 10:17:17 AM » |
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Workings for the fruit trader:
stop 1: 30 + 30 + 30 - lose 3 x 1 stop 2: 29 + 29 + 29 - lose 3 x 1 stop 3: 28 + 28 + 28 - lose 3 x 1 stop 4: 27 + 27 + 27 - lose 3 x 1 stop 5: 26 + 26 + 26 - lose 3 x 1 stop 6: 25 + 25 + 25 - lose 3 x 1 stop 7: 24 + 24 + 24 - lose 3 x 1 stop 8: 23 + 23 + 23 - lose 3 x 1 stop 9: 22 + 22 + 22 - lose 3 x 1 stop 10: 21 + 21 + 21 - lose 3 x 1 - place remaining 60 into 2 sacks stop 11: 30 + 30 - lose 2 x 1 stop 12: 29 + 29 - lose 2 x 1 stop 13: 28 + 28 - lose 2 x 1 stop 14: 27 + 27 - lose 2 x 1 stop 15: 26 + 26 - lose 2 x 1 stop 16: 25 + 25 - lose 2 x 1 stop 17: 24 + 24 - lose 2 x 1 stop 18: 23 + 23 - lose 2 x 1 stop 19: 22 + 22 - lose 2 x 1 stop 20: 21 + 21 - lose 2 x 1 stop 21: 20 + 20 - lose 2 x 1 stop 22: 19 + 19 - lose 2 x 1 stop 23: 18 + 18 - lose 2 x 1 stop 24: 17 + 17 - lose 2 x 1 stop 25: 16 + 16 - lose 2 x 1 - place remaining 30 into 1 sack stop 26: 30 - lose 1 stop 27: 29 - lose 1 stop 28: 28 - lose 1 stop 29: 27 - lose 1 stop 30: 26 - lose 1 at end: 25 remaining
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Antman
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« Reply #52 on: April 03, 2020, 10:43:38 AM » |
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Doh  Never thought of emptying sacks. I just thought logically. That's cheating  Thanks Antman
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todthedog
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« Reply #53 on: April 03, 2020, 10:46:26 AM » |
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5 pirates of different ages have a treasure of 100 gold coins.
On their ship, they decide to split the coins using this scheme:
The oldest pirate proposes how to share the coins, and ALL pirates (including the oldest) vote for or against it.
If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain.
As pirates tend to be a bloodthirsty bunch, if a pirate would get the same number of coins if he voted for or against a proposal, he will vote against so that the pirate who proposed the plan will be thrown overboard.
Assuming that all 5 pirates are intelligent, rational, greedy, and do not wish to die, (and are rather good at math for pirates) what will happen?
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Kidwelly South Wales
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todthedog
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« Reply #54 on: April 03, 2020, 10:48:46 AM » |
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I will put he solutions on this afternoon cheers tod
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Kidwelly South Wales
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Countrypaul
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« Reply #55 on: April 03, 2020, 11:07:18 AM » |
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You are getting too Good
Solution: 25
Explanation: == Equation1 == RedFlower + RedFlower + RedFlower = 60 => RedFlower = 20
== Equation2== RedFlower + 5BluePetals + 5BluePetals = 30 => 20 + 10BluePetals = 30 => 10BluePetals = 10 => 1BluePetals = 1
== Equation3 == 5bluePetals + 2yellowFlower = 3 => 5 + 2yellowFlower = 3 => 2yellowFlower = 2 => 1yellowFlower = 1
== Required == YellowFlower + RedFlower + 4BluePetals = ? => 1 + 20 + 4 = 25 ... The Answer
Nice try, but you appear to have missed a vital sign. The 5 bluepetals + 2 yellowflowers = 3 is not a "+" but a "-". At least you can have some marks for showing your working out
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« Last Edit: April 03, 2020, 11:16:14 AM by Countrypaul »
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Antman
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« Reply #56 on: April 03, 2020, 03:36:26 PM » |
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The lack of replies means either everyone has found something to do or are all as stumped as me If they are greedy then they are not likely to accept equal shares at first go. So if we take the intelligent, rational and not wishing to die traits, then maybe they all agree to invest it in their Occupational Pension scheme  Either that or they are so busy arguing that they run aground and die as shark feed  Antman
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20 x 47mm, 172 l cylinder, Heat Dump, 15 x Sanyo HIT-H250E, SB4000TL, Nestor Martin IQ13 WBS DIY Solar System Support at http://www.handyantman.co.uk/antman.htmlAll support is voluntary and free of charge. I'm not employed by Navitron so responses may not be same-day
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Countrypaul
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« Reply #57 on: April 03, 2020, 03:51:10 PM » |
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For the pirates, couldn't the olddest one propose the other 4 share it equally, that way the second oldest and third oldest would see the advantage in supporting him before they too were invited to go swimming.
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todthedog
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« Reply #58 on: April 03, 2020, 04:04:09 PM » |
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A. 6 8 2 – One Number is correct and well placed – Number cannot be 6 as that will make Statement B wrong so the number in code is 2 which is correctly placed also(on Position 3) B. 6 1 4 – One Number is the correct but wrong place – 6 cannot be that number as per Statement A, It cannot be 1 also as if the number is 1 it can be on either on 1st or 3rd place those are already taken by digits 0 and 2 so the only number left is 4 C. 2 0 6 – Two Numbers are correct but Wrong Places – One of digit which is correct is 0 as per Statement E but its place cannot be at second(as per statement C) and third position( as per Statement E) so it can be only one first place, another digit is 2 as per statement A. D. 7 3 8 – Nothing is correct – We can rule out these 3 digits from other statements E. 8 7 0 – One Number is the correct but wrong place – We know 0 is one of the digit in code but the place for it would be not 3 From statements A, B and E we got our numbers. Statement A gave a position for Number 2. – Statement A -> _ _ 2 Statement C and E gave a position for the other 2 numbers. Statement A, C and E-> 0 _ 2 Statement B -> last digit 4 which will come in the middle so code would be 0 4 2 Finally, the code is 0 4 2
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Kidwelly South Wales
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todthedog
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« Reply #59 on: April 03, 2020, 04:07:04 PM » |
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Pirates The oldest pirate will propose a 98 : 0 : 1 : 0 : 1 split, in other words the oldest pirate gets 98 coins, the middle pirate gets 1 coin and the youngest gets 1 coin.
Let us name the pirates (from oldest to youngest): Alex, Billy, Colin, Duncan and Eddie.
Working backwards:
2 Pirates: Duncan splits the coins 100 : 0 (giving himself all the gold). His vote (50%) is enough to ensure the deal.
3 Pirates: Colin splits the coins 99 : 0 : 1. Eddie will accept this deal (getting just 1 coin), because he knows that if he rejects the deal there will be only two pirates left, and he gets nothing.
4 Pirates: Billy splits the coins 99 : 0 : 1 : 0. By the same reasoning as before, Duncan will support this deal. Billy would not waste a spare coin on Colin, because Colin knows that if he rejects the proposal, he will pocket 99 coins once Billy is thrown overboard. Billy would also not give a coin to Eddie, because Eddie knows that if he rejects the proposal, he will receive a coin from Colin in the next round anyway.
5 Pirates: Alex splits the coins 98 : 0 : 1 : 0 : 1. By offering a gold coin to Colin (who would otherwise get nothing) he is assured of a deal.
(Note: In the final deal Alex would not give a coin to Billy, who knows he can pocket 99 coins if he votes against Alex's proposal and Alex goes overboard. Likewise, Alex would not give a coin to Duncan, because Duncan knows that if he votes against the proposal, Alex will be voted overboard and Billy will propose to offer Duncan the same single coin as Alex. All else equal, Duncan would rather see Alex go overboard and collect his one coin from Billy.)
I found these on the net I hope that you enjoyed Tod
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Kidwelly South Wales
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