Logic Puzzles

clartharlear

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
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.

twerg_85

B

clartharlear

Nope.

Mellor

D

clartharlear

Oh gosh, twerg, I'm so sorry. You were right! I could have sworn you said

C.

The answer is

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?

twerg_85

Good stuff !

6

clartharlear

Tres bien!

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

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.

KarmaGarda

Jeez! That was a good one!

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.

KarmaGarda

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

KarmaGarda

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

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)

Mellor

clartharlear wrote: »

Oh gosh, twerg, I'm so sorry. You were right! I could have sworn you said

C.

The answer is

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.

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.

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!

Mellor

clartharlear wrote: »

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

KarmaGarda

clartharlear wrote: »

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!

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:

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!

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)

Andy 16, Susan 11

KarmaGarda

The Thinker wrote: »

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:

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!

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: »

Andy 16, Susan 11

Spot on twerg!! You're up!

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 ?

MudSkipper

19

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?

twerg_85

Nope. Can be done faster.

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 ?

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

The Thinker

MudSkipper wrote: »

19

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?

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!!

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.

KarmaGarda

The Thinker wrote: »

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

Ok, let me show you:

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

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

MudSkipper

twerg_85 wrote: »

Nope. Can be done faster.

my bad

17

1+2 -> 2
<- 1 1
10+5 -> 10
<- 2 2
1+2 -> 2

Bit like the cat one :-) s

peedy and quickly will be tired of that bridge (or seasick if it was a boat)

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 ?

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 ?

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

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 ?

1 / 6 ?

Naz_st

Assuming I was right with the above answer:

A king was looking for a knight for a dangerous quest. He has 3 equally brave, excellent and clever knights at his disposal, but he needs the smartest for this quest. So he blindfolds the three knights and brings them in to his audience chamber. He then places a hat on each of their heads and tells them: "I have placed either a white hat or a black hat on each of your heads. When your blindfolds are removed, I want you to raise your hand if you see a black hat, and lower your hand when you know the colour of your own hat". Now the wily king had placed a black hat on each knight’s head, and as soon as the blindfolds were removed, all 3 knights raised their hands. After a minute or so, one of the knights lowered his hand and said "My hat is black". How did he know? (And no, there are no mirrors etc in the room!)

twerg_85

Neonjack wrote: »

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

Mudskipper got the right answer, but I should have specified that torch had to be carried across.

twerg_85

Naz_st wrote: »

1 / 6 ?

Nope.

twerg_85

Naz_st wrote: »

Assuming I was right with the above answer:

A king was looking for a knight for a dangerous quest. He has 3 equally brave, excellent and clever knights at his disposal, but he needs the smartest for this quest. So he blindfolds the three knights and brings them in to his audience chamber. He then places a hat on each of their heads and tells them: "I have placed either a white hat or a black hat on each of your heads. When your blindfolds are removed, I want you to raise your hand if you see a black hat, and lower your hand when you know the colour of your own hat". Now the wily king had placed a black hat on each knight’s head, and as soon as the blindfolds were removed, all 3 knights raised their hands. After a minute or so, one of the knights lowered his hand and said "My hat is black". How did he know? (And no, there are no mirrors etc in the room!)

Actually, I think they should all have lowered their hand at the same time if they're equally clever ......
Each one knows that if they have a white hat, then their 2 opponents are each seeing one white and one black hat. Then your opponents would know that their hat is black so they would lower their hand. Since they didn't lower their hand, your hat is not white, i.e. black.

Naz_st

twerg_85 wrote: »

Nope.

Was a bit hasty there! A little trickier than I thought...

1 / 3 :

P(RR|R) = P(R|RR).P(RR) / P(R)
P(RR|R) = (1).(1/3) / (1/2) = 2/3

P(RB|R) = 1 - 2/3 = 1/3

x = (P(RR|R) * P(RB) * P(R)) + (P(RB|R) * P(RR) * P(R))

x = (2/3 * 1/2 * 1/2) + (1/3 * 1/2 * 1)
x = 2/12 + 1/6
x = 2/6 = 1/3

twerg_85

Naz_st wrote: »

Was a bit hasty there! A little trickier than I thought...

1 / 3 :

P(RR|R) = P(R|RR).P(RR) / P(R)
P(RR|R) = (1).(1/3) / (1/2) = 2/3

P(RB|R) = 1 - 2/3 = 1/3

x = (P(RR|R) * P(RB) * P(R)) + (P(RB|R) * P(RR) * P(R))

x = (2/3 * 1/2 * 1/2) + (1/3 * 1/2 * 1)
x = 2/12 + 1/6
x = 2/6 = 1/3

Spot on.