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Logic Puzzles

  • 06-12-2009 06:53PM
    #1
    Registered Users, Registered Users 2 Posts: 489
    ✭✭


    Aine, Brid and Cliona were talking together. Each belongs either to the Firinne family, whose members always tell the truth, or to the Breag family, whose members always lie. Aine says, ‘either I belong or Brid belongs to a different family from the other two.’
    It is possible to determine the family of
    A: Aine
    B: Brid
    C: Cliona
    D: None of the girls


    The person who answers must put up a new riddle? Or if that doesn't take off, maybe I'll throw up another couple during the week.
    Tagged:


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Comments

  • Registered Users, Registered Users 2 Posts: 391 twerg_85
    ✭✭


    Spoiler
    B


  • Registered Users, Registered Users 2 Posts: 489 clartharlear
    ✭✭


    Nope.


  • Registered Users, Registered Users 2 Posts: 40,307 Mellor
    ✭✭✭✭


    Spoiler
    D


  • Registered Users, Registered Users 2 Posts: 489 clartharlear
    ✭✭


    Oh gosh, twerg, I'm so sorry. You were right! I could have sworn you said
    Spoiler
    C.
    The answer is
    Spoiler
    B.
    If you didn't get it, try thinking of the four possible cases.


    Two years ago, Andrew was one-fifth as old as Brendan, whose age was 2.5 times that of Ciara. If in 4 years’ time, Brendan is twice as old as Andrew, what is Ciara’s current age?


  • Registered Users, Registered Users 2 Posts: 391 twerg_85
    ✭✭


    Good stuff !
    Spoiler
    6


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  • Registered Users, Registered Users 2 Posts: 489 clartharlear
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    Tres bien!

    It's your turn to pose a puzzle, if you like?


  • Registered Users, Registered Users 2 Posts: 391 twerg_85
    ✭✭


    12 objects. 11 are exact same weight, 1 is different weight.
    Using a balance scales just 3 times, find the object that's a different weight than the others.

    Note : not sure if I remember the solution 100% so might have to guess it again meself.


  • Registered Users, Registered Users 2 Posts: 1,009 KarmaGarda
    ✭✭✭


    Jeez! That was a good one!
    Spoiler
    Split into 3 groups of 4 (A, B, C)

    Let's split each group into
    A1, A2, A3, A4:
    B1, B2, B3, B4:
    C1, C2, C3, C4:

    Now weigh A vs B

    Scenario 1 (easy scenario):
    They weigh the same. Item is in Group C.

    Weigh C1 & C2 against 2 random items in Group A (they will be the standard weight).

    Scenario 1a:
    If these are same weight, weigh C3 against an item in group A. If they're the same then C4 is the different weight item. If they're different then C3 is the item

    Scenario 1b:
    If these are a different weight, weigh C1 against a random in Group A. If these differ C1 is the different weight item. If they are the same C2 is the different weight.


    Next, Scenario 2 (hard one!):
    A and B are different weights. Make a note of the heavier side. For this example let's say it's A.

    Now, let's pick up A1, A2, B3, and weigh against B1, B2, A3. (Notice B3 and A3 swapped sides)

    Scenario 2a:
    If they weight the same, then weigh A4 against C1. If they also weight the same B4 is the item. If they don't then A4 is the item.

    Scenario 2b:
    If the heavy side didn't change we know that the item we're looking for has not changed sides. So it's in A1, A2 (they didn't change sides).
    Weight A1 vs C1. If they differ A1 is the item, if they don't A2 it the item.

    Scenario 2c:
    The heavy side changed sides... Therefore it's in the swapped items (B3 and A3)
    Weight A3 vs C1. If they differ A3 is the item, if they don't B3 it the item.


  • Registered Users, Registered Users 2 Posts: 1,009 KarmaGarda
    ✭✭✭


    Actually, reading over it I'm not 100% there... give me another minute!


  • Registered Users, Registered Users 2 Posts: 1,009 KarmaGarda
    ✭✭✭


    Spoiler
    Split into 3 groups of 4 (A, B, C)

    Let's split each group into
    A1, A2, A3, A4:
    B1, B2, B3, B4:
    C1, C2, C3, C4:

    Now weigh A vs B

    Scenario 1 (easy scenario):
    They weigh the same. Item is in Group C.

    Weigh C1 & C2 against 2 random items in Group A (they will be the standard weight).

    Scenario 1a:
    If these are same weight, weigh C3 against an item in group A. If they're the same then C4 is the different weight item. If they're different then C3 is the item

    Scenario 1b:
    If these are a different weight, weigh C1 against a random in Group A. If these differ C1 is the different weight item. If they are the same C2 is the different weight.


    Next, Scenario 2 (hard one!):
    A and B are different weights. Make a note of the heavier side. For this example let's say it's A.

    Now, let's pick up A1, A2, B3, and weigh against B1, C1, A3. (Notice B3 and A3 swapped sides and we've added C1 to the selection)

    Scenario 2a:
    If they weight the same, then the item we're looking for is in A4, B2, B4...
    Weigh A4 + B2 against C1 + C2. If they also weight the same B4 is the item (only 1 removed from scales!). If they differ, and A4 + B2 is heavier, A4 is the item (Group A is heavier!). If it's lighter, B2 is the item.

    Scenario 2b:
    If the heavy side didn't change we know that the item we're looking for has not changed sides. So it's in A1, A2, or B1 (they didn't change sides).
    Weight A1 + B1 vs C1 + C2. If they weight the same A2 is the item (only 1 removed from scales!). If they differ, and A4 + B2 is heavier, A1 is the item (Group A is heavier!). If it's lighter, B1 is the item.

    Scenario 2c:
    If the heavy side DID change we know that the item we're looking for also changed sides. So it's in A3 or B3.
    Weight A3 vs C1. If they weight the same B3 is the item. If they differ, A3 is the item.

    I think that covers all scenarios? Bloody better, I'm after killing a couple of hours on that blasted thing :D


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  • Registered Users, Registered Users 2 Posts: 1,009 KarmaGarda
    ✭✭✭


    Ok, well presuming that's right then I'll post another :pac:


    When Andy was twice the age that Susan was when Andy was the age that Susan is now, Susan was 1 third the age that Andy will be when Susan is as old as Andy is now. If 1 is a teenager what ages are Susan and Andy?

    (Ages don't go into fractions at any point)


  • Registered Users, Registered Users 2 Posts: 40,307 Mellor
    ✭✭✭✭


    Oh gosh, twerg, I'm so sorry. You were right! I could have sworn you said
    Spoiler
    C.
    The answer is
    Spoiler
    B.
    If you didn't get it, try thinking of the four possible cases.

    Could you explain this. As I don't think you are right. Or else I'm missing something.


  • Registered Users, Registered Users 2 Posts: 489 clartharlear
    ✭✭


    KarmaGarda wrote: »
    Ok, well presuming that's right then I'll post another :pac:


    When Andy was twice the age that Susan was when Andy was the age that Susan is now, Susan was 1 third the age that Andy will be when Susan is as old as Andy is now. If 1 is a teenager what ages are Susan and Andy?

    (Ages don't go into fractions at any point)
    This is a brilliant puzzle. It's wrecking my head. Could you give any kind of a hint that wouldn't give the whole game away? Or I'll continue to struggle...
    Mellor wrote: »
    Could you explain this. As I don't think you are right. Or else I'm missing something.
    Spoiler
    If A’s statement is true, either A is the only one who is a Firinne, or B is the only who is a Breag
    If A’s statement is false, A is NOT the only one who is a Breag, and B is NOT the only one who is a Firinne, so either C is the only one who is a Firinne or all three are Breags.
    I hope!


  • Registered Users, Registered Users 2 Posts: 40,307 Mellor
    ✭✭✭✭


    Spoiler
    If A’s statement is true, either A is the only one who is a Firinne, or B is the only who is a Breag
    If A’s statement is false, A is NOT the only one who is a Breag, and B is NOT the only one who is a Firinne, so either C is the only one who is a Firinne or all three are Breags.
    I hope!

    LOL, I thought there were 4 girls, and even after reading your post, I still didn't relise and thought you mised a load or combos


  • Registered Users, Registered Users 2 Posts: 1,009 KarmaGarda
    ✭✭✭


    This is a brilliant puzzle. It's wrecking my head. Could you give any kind of a hint that wouldn't give the whole game away? Or I'll continue to struggle...

    It's quite difficult to give a hint without giving the puzzle away! Andy is obviously older than susan. And most of the puzzle is breaking it down and understanding exactly what's being said, as opposed to straight maths. That's all I can give away for now!


  • Registered Users, Registered Users 2 Posts: 19 The Thinker


    KarmaGarda wrote: »
    Ok, well presuming that's right then I'll post another :pac:


    When Andy was twice the age that Susan was when Andy was the age that Susan is now, Susan was 1 third the age that Andy will be when Susan is as old as Andy is now. If 1 is a teenager what ages are Susan and Andy?

    (Ages don't go into fractions at any point)

    Hi,
    I've tried all ages to allow 1 of them to be a teenager and I can't quite get it.

    See trial grid below:
    Spoiler
    Andy Susan
    When 14 7
    Now 21 14
    Will Be 28 21

    "Susan was 1 third the age that Andy will be" it will work if the statement is "1 Quarter".

    Maybe I'm misunderstanding the statement!:confused:


  • Registered Users, Registered Users 2 Posts: 391 twerg_85
    ✭✭


    KarmaGarda wrote: »
    Ok, well presuming that's right then I'll post another :pac:


    When Andy was twice the age that Susan was when Andy was the age that Susan is now, Susan was 1 third the age that Andy will be when Susan is as old as Andy is now. If 1 is a teenager what ages are Susan and Andy?

    (Ages don't go into fractions at any point)
    Spoiler
    Andy 16, Susan 11


  • Registered Users, Registered Users 2 Posts: 1,009 KarmaGarda
    ✭✭✭


    Hi,
    I've tried all ages to allow 1 of them to be a teenager and I can't quite get it.

    See trial grid below:
    Spoiler
    Andy Susan
    When 14 7
    Now 21 14
    Will Be 28 21

    "Susan was 1 third the age that Andy will be" it will work if the statement is "1 Quarter".

    Maybe I'm misunderstanding the statement!:confused:

    I think you might have slightly misunderstood! It's quite a riddle along with a puzzle I have to admit.... but on saying that...
    twerg_85 wrote: »
    Spoiler
    Andy 16, Susan 11

    Spot on twerg!! You're up!


  • Registered Users, Registered Users 2 Posts: 391 twerg_85
    ✭✭


    4 dudes (Speedy,Quickly,Slowly,Crawly) need to cross a bridge.

    They have one torch and can only cross the bridge with that torch. Max of 2 people can cross at same time. They cross at the speed of the slower dude.

    Speedy takes 1 min, Quickly 2 min, Slowly 5 mins and Crawly takes 10 mins.

    What's quickest time for all 4 of them to cross the bridge ?


  • Registered Users, Registered Users 2 Posts: 556 MudSkipper
    ✭✭✭


    Spoiler
    19
    Spoiler
    speedy and crawly 10mins, speedy crosses back 1min, speedy and slowly 5mins, speedy crosses back 1min, speedy and quickly 2min

    Here's another crossing one:
    your stuck on a riverbank with a dog, a cat and a mouse. you have one boat and the boat can only carry you plus on animal. the dog or mouse cannot be left with cat. how do you get you and the 3 animals over the river?


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  • Registered Users, Registered Users 2 Posts: 391 twerg_85
    ✭✭


    Nope. Can be done faster.


  • Registered Users, Registered Users 2 Posts: 19 The Thinker


    twerg_85 wrote: »
    4 dudes (Speedy,Quickly,Slowly,Crawly) need to cross a bridge.

    They have one torch and can only cross the bridge with that torch. Max of 2 people can cross at same time. They cross at the speed of the slower dude.

    Speedy takes 1 min, Quickly 2 min, Slowly 5 mins and Crawly takes 10 mins.

    What's quickest time for all 4 of them to cross the bridge ?

    Spoiler
    Speedy and Quickly (2 mins across + 1 min back = 3 mins)
    Speedy and Slowly (5 mins across + 1 min back = 6 mins)
    Speedy and Crawly (10 mins across)

    Total of 18 mins


  • Registered Users, Registered Users 2 Posts: 19 The Thinker


    MudSkipper wrote: »
    Spoiler
    19
    Spoiler
    speedy and crawly 10mins, speedy crosses back 1min, speedy and slowly 5mins, speedy crosses back 1min, speedy and quickly 2min

    Here's another crossing one:
    your stuck on a riverbank with a dog, a cat and a mouse. you have one boat and the boat can only carry you plus on animal. the dog or mouse cannot be left with cat. how do you get you and the 3 animals over the river?
    Spoiler
    Bring the cat over 1st (Leaving Dog & Mouse) - Go back alone.
    Bring the dog over next - Bring back the cat.
    Leave the cat and bring over the mouse (Now Dog & Mouse on far side) - Go back alone.
    Finally - bring over the cat!!


  • Registered Users, Registered Users 2 Posts: 19 The Thinker


    KarmaGarda wrote: »
    I think you might have slightly misunderstood! It's quite a riddle along with a puzzle I have to admit.... but on saying that...



    Spot on twerg!! You're up!

    Still can't work it out - even with the answer. :confused:


  • Registered Users, Registered Users 2 Posts: 1,009 KarmaGarda
    ✭✭✭


    Still can't work it out - even with the answer. :confused:

    Ok, let me show you:
    Spoiler
    Firstly let's identify where their starting ages are in the statement:

    When Andy was twice the age that Susan was when Andy was the age that *Susan is now*, Susan was 1 third the age that Andy will be when Susan is as old *as Andy is now*. If 1 is a teenager what ages are Susan and Andy?

    Let's use these as our anchor points to how we use the rest of the puzzle.

    So firstly it says:
    When Andy was twice the age *that Susan was when Andy was the age that Susan is now*...

    Let's work back from Susans age to the start of the sentence...

    Let's say susan is S. and Andy is A.
    When Andy was susans age that = S.
    This was also (A - S) years ago.
    What age was susan at this point?
    Susans age was S - (A - S). Let's calculate that out:
    S - (A - S)
    If you multiple the minus into the bracket it also =
    S - A + S
    = 2S - A

    So susans age back then was (2S - A)

    Now to the part of the sentence before that... when Andy was twice THAT age (i.e. 2S - A), Susan was 1/3rd of blah blah blah (let's ignore this part for now.). So firstly, what age was susan at this point?

    Starting again, when Andy was twice the age... that age is 2 x (2S - A). Let's calculate that out:
    2 (2S-A) = 4S - 2A

    When Andy was that age what age was susan? Well, we know already that the difference in their ages is (A-S) therefore she was:

    (4S - 2A) - (A - S)
    Multiply the minus into the bracket:
    4S - 2A - A + S =
    4S + S - 2A - A
    5S - 3A.

    So... this is the age we need that is 1/3 of the "blah blah" part of the puzzle...

    Let's put that aside for a minute and move to the next part:

    Susan was 1 third the age *that Andy will be when Susan is as old as Andy is now*.

    Again, the difference between their ages is (A-S) so... the second part means when Susan is A. What age is andy at this point? A + (A - S)...
    =>
    A + A - S.
    =>
    2A - S

    From the puzzle it tells us that our first calculation is 1/3 of this calculation...

    Therefore, from our 2 calculations:
    (5S - 3A) x 3 = (2A - S)
    =>
    15S - 9A = 2A - S
    =>
    15S + S = 2A + 9A
    16S = 11A

    .... what's the easiest calculation to make these equal each other?
    S = 11
    A = 16

    Andy is a teenager... and he's 16!


    Does that make anymore sense? If not PM me what's confusing and I'll try explain it best I can :)


  • Registered Users, Registered Users 2 Posts: 1,009 KarmaGarda
    ✭✭✭


    twerg_85 wrote: »
    4 dudes (Speedy,Quickly,Slowly,Crawly) need to cross a bridge.

    They have one torch and can only cross the bridge with that torch. Max of 2 people can cross at same time. They cross at the speed of the slower dude.

    Speedy takes 1 min, Quickly 2 min, Slowly 5 mins and Crawly takes 10 mins.

    What's quickest time for all 4 of them to cross the bridge ?

    I know the answer to this, but I've heard it before! So will give everyone else a chance to get it ;)


  • Registered Users, Registered Users 2 Posts: 556 MudSkipper
    ✭✭✭


    twerg_85 wrote: »
    Nope. Can be done faster.

    my bad
    Spoiler
    17
    Spoiler
    1+2 -> 2
    <- 1 1
    10+5 -> 10
    <- 2 2
    1+2 -> 2

    Bit like the cat one :-) s
    Spoiler
    peedy and quickly will be tired of that bridge (or seasick if it was a boat)


  • Registered Users, Registered Users 2 Posts: 391 twerg_85
    ✭✭


    you have 3 cards in a bag. One is black on both sides, one is red on both sides, 3rd is black on one side, red on the other side.

    you draw a card at random and look at just one side of it. It's red.

    If you draw another card (without replacement) and look at just one side, what's the probability of it being red ?


  • Registered Users, Registered Users 2 Posts: 223 Neonjack
    ✭✭


    twerg_85 wrote: »
    4 dudes (Speedy,Quickly,Slowly,Crawly) need to cross a bridge.

    They have one torch and can only cross the bridge with that torch. Max of 2 people can cross at same time. They cross at the speed of the slower dude.

    Speedy takes 1 min, Quickly 2 min, Slowly 5 mins and Crawly takes 10 mins.

    What's quickest time for all 4 of them to cross the bridge ?
    Spoiler
    12 mins?
    Crawly and Slowly cross - 10 mins
    They throw the torch back to Speedy and Quickly
    Speedy and quickly cross - 2 mins
    Total 12


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  • Registered Users, Registered Users 2 Posts: 391 Naz_st
    ✭✭


    twerg_85 wrote: »
    you have 3 cards in a bag. One is black on both sides, one is red on both sides, 3rd is black on one side, red on the other side.

    you draw a card at random and look at just one side of it. It's red.

    If you draw another card (without replacement) and look at just one side, what's the probability of it being red ?
    Spoiler
    1 / 6 ?


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