*OFFICAL* Weekend Brainteaser Thread!! (wear hard hats)

[Placer14]

Okay...so you have 12 marbles. 1 Marble has a different wieght from the other 11. (It could be heavier or lighter.) You also have a scale that is perfectly balenced. In 4 different wieghing attempts, you have to decipher which marble is the oddball and if it's heavier or lighter than the others. In order to answer this riddle, you have to list EVERY single possible outcome for every attempt and it should make sense in all cases.

First person to get it right, wins a cookie. Enjoy.

And it doesn't need to be said, but NO CHEATING! Or you get an ookie cookie instead. 😉

 

[Whitecloak]

this is too easy and can be done in 3 attempts. I am not gonna answer it though and will simply nef here. 😛

 

[Placer14]

Heh...i might've misquoted the riddle, if you can do it in 3 attempts, you get bonus points.

 

[LordSnailz]

hehe - this is an easy one ... someone make an OFFICIAL Friday BrainTeaser thread!

 

[SagaLore]

Start with 12 marbles.

A. SPLIT into two groups. (#1) Put on scale. 6x6

B. Whichever is lighter side, take off scale and put aside. SPLIT remainder into two groups. (#2) Put on scale. 3x3

C. If scale remains balanced, proceed to D. Otherwise proceed to I.

D. Remove current group, SPLIT previously put aside group. (#3) Put on scale. 3x3

E. Whichever is heavier side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#4) 1x1

F. If scale remains balanced, proceed to G. If one side is lighter, proceed to H.

G. REMAINING MARBLE, HELD IN HAND, IS THE LIGHTEST MARBLE. done

H. THE MARBLE ON THE SIDE THAT RAISED IS THE LIGHTEST MARBLE. done

I. Whichever is lighter side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#3) 1x1

J. If scale remains balanced, proceed to K. If one side is heavier, proceed to L.

K. THE MARBLE, HELD IN HAND, IS THE HEAVIEST MARBLE. done

L. THE MARBLE ON THE SIDE THAT LOWERED IS THE HEAVIEST MARBLE. done



I think I got that right.

 

[Placer14]

Originally posted by: SagaLore
Start with 12 marbles.

A. SPLIT into two groups. (#1) Put on scale. 6x6

B. Whichever is lighter side, take off scale and put aside. SPLIT remainder into two groups. (#2) Put on scale. 3x3

C. If scale remains balanced, proceed to D. Otherwise proceed to I.

D. Remove current group, SPLIT previously put aside group. (#3) Put on scale. 3x3

E. Whichever is heavier side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#4) 1x1

F. If scale remains balanced, proceed to G. If one side is lighter, proceed to H.

G. REMAINING MARBLE, HELD IN HAND, IS THE LIGHTEST MARBLE. done

H. THE MARBLE ON THE SIDE THAT RAISED IS THE LIGHTEST MARBLE. done

I. Whichever is lighter side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#3) 1x1

J. If scale remains balanced, proceed to K. If one side is heavier, proceed to L.

K. THE MARBLE, HELD IN HAND, IS THE HEAVIEST MARBLE. done

L. THE MARBLE ON THE SIDE THAT LOWERED IS THE HEAVIEST MARBLE. done



I think I got that right.

Short answer, no. How do you the marble you're wieghing is a normal wieghted marble or a differently wieghted marble? The left side could go down and the right side could go up and you have two possibilities. The right side marble is the odd marble and is lighten than the rest or the left side marble is odd marble and heavier than the rest.

There's a better solution. Good try.

 

[LordSnailz]

someone put up another one!

 

[Placer14]

Originally posted by: LordSnailz
someone put up another one!

I don't see another thread if that's what you mean.

 

[Keego]

Step 1: throw all 12 marbles at Placer14
Step 2: ask him which one hurt more
Step 3: throw scale at him too 😛

 

[SagaLore]

Originally posted by: Placer14

Originally posted by: SagaLore
Start with 12 marbles.

A. SPLIT into two groups. (#1) Put on scale. 6x6

B. Whichever is lighter side, take off scale and put aside. SPLIT remainder into two groups. (#2) Put on scale. 3x3

C. If scale remains balanced, proceed to D. Otherwise proceed to I.

D. Remove current group, SPLIT previously put aside group. (#3) Put on scale. 3x3

E. Whichever is heavier side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#4) 1x1

F. If scale remains balanced, proceed to G. If one side is lighter, proceed to H.

G. REMAINING MARBLE, HELD IN HAND, IS THE LIGHTEST MARBLE. done

H. THE MARBLE ON THE SIDE THAT RAISED IS THE LIGHTEST MARBLE. done

I. Whichever is lighter side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#3) 1x1

J. If scale remains balanced, proceed to K. If one side is heavier, proceed to L.

K. THE MARBLE, HELD IN HAND, IS THE HEAVIEST MARBLE. done

L. THE MARBLE ON THE SIDE THAT LOWERED IS THE HEAVIEST MARBLE. done



I think I got that right.

Short answer, no. How do you the marble you're wieghing is a normal wieghted marble or a differently wieghted marble? The left side could go down and the right side could go up and you have two possibilities. The right side marble is the odd marble and is lighten than the rest or the left side marble is odd marble and heavier than the rest.

There's a better solution. Good try.

It is determined by the intial weigh-in. Earlier on you find out if the marble is light or heavy, and that dictates the logical outcome later.

 

[Kev]

Originally posted by: SagaLore

Originally posted by: Placer14

Originally posted by: SagaLore
Start with 12 marbles.

A. SPLIT into two groups. (#1) Put on scale. 6x6

B. Whichever is lighter side, take off scale and put aside. SPLIT remainder into two groups. (#2) Put on scale. 3x3

C. If scale remains balanced, proceed to D. Otherwise proceed to I.

D. Remove current group, SPLIT previously put aside group. (#3) Put on scale. 3x3

E. Whichever is heavier side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#4) 1x1

F. If scale remains balanced, proceed to G. If one side is lighter, proceed to H.

G. REMAINING MARBLE, HELD IN HAND, IS THE LIGHTEST MARBLE. done

H. THE MARBLE ON THE SIDE THAT RAISED IS THE LIGHTEST MARBLE. done

I. Whichever is lighter side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#3) 1x1

J. If scale remains balanced, proceed to K. If one side is heavier, proceed to L.

K. THE MARBLE, HELD IN HAND, IS THE HEAVIEST MARBLE. done

L. THE MARBLE ON THE SIDE THAT LOWERED IS THE HEAVIEST MARBLE. done



I think I got that right.

Short answer, no. How do you the marble you're wieghing is a normal wieghted marble or a differently wieghted marble? The left side could go down and the right side could go up and you have two possibilities. The right side marble is the odd marble and is lighten than the rest or the left side marble is odd marble and heavier than the rest.

There's a better solution. Good try.

It is determined by the intial weigh-in. Earlier on you find out if the marble is light or heavy, and that dictates the logical outcome later.

I concur.

 

[Placer14]

Originally posted by: maladroit

Originally posted by: SagaLore

Originally posted by: Placer14

Originally posted by: SagaLore
Start with 12 marbles.

A. SPLIT into two groups. (#1) Put on scale. 6x6

B. Whichever is lighter side, take off scale and put aside. SPLIT remainder into two groups. (#2) Put on scale. 3x3

C. If scale remains balanced, proceed to D. Otherwise proceed to I.

D. Remove current group, SPLIT previously put aside group. (#3) Put on scale. 3x3

E. Whichever is heavier side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#4) 1x1

F. If scale remains balanced, proceed to G. If one side is lighter, proceed to H.

G. REMAINING MARBLE, HELD IN HAND, IS THE LIGHTEST MARBLE. done

H. THE MARBLE ON THE SIDE THAT RAISED IS THE LIGHTEST MARBLE. done

I. Whichever is lighter side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#3) 1x1

J. If scale remains balanced, proceed to K. If one side is heavier, proceed to L.

K. THE MARBLE, HELD IN HAND, IS THE HEAVIEST MARBLE. done

L. THE MARBLE ON THE SIDE THAT LOWERED IS THE HEAVIEST MARBLE. done



I think I got that right.

Short answer, no. How do you the marble you're wieghing is a normal wieghted marble or a differently wieghted marble? The left side could go down and the right side could go up and you have two possibilities. The right side marble is the odd marble and is lighten than the rest or the left side marble is odd marble and heavier than the rest.

There's a better solution. Good try.

It is determined by the intial weigh-in. Earlier on you find out if the marble is light or heavy, and that dictates the logical outcome later.

I concur.

Touche....good job. Pardon my logic. :: hides face in humilty...and to protect himself from flying marbles ::

Now I tried to see if I could remember how to do it in 3 tries and couldn't do it. Any takers?

 

[SagaLore]

Originally posted by: Placer14

Originally posted by: maladroit

Originally posted by: SagaLore

Originally posted by: Placer14

Originally posted by: SagaLore
Start with 12 marbles.

A. SPLIT into two groups. (#1) Put on scale. 6x6

B. Whichever is lighter side, take off scale and put aside. SPLIT remainder into two groups. (#2) Put on scale. 3x3

C. If scale remains balanced, proceed to D. Otherwise proceed to I.

D. Remove current group, SPLIT previously put aside group. (#3) Put on scale. 3x3

E. Whichever is heavier side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#4) 1x1

F. If scale remains balanced, proceed to G. If one side is lighter, proceed to H.

G. REMAINING MARBLE, HELD IN HAND, IS THE LIGHTEST MARBLE. done

H. THE MARBLE ON THE SIDE THAT RAISED IS THE LIGHTEST MARBLE. done

I. Whichever is lighter side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#3) 1x1

J. If scale remains balanced, proceed to K. If one side is heavier, proceed to L.

K. THE MARBLE, HELD IN HAND, IS THE HEAVIEST MARBLE. done

L. THE MARBLE ON THE SIDE THAT LOWERED IS THE HEAVIEST MARBLE. done



I think I got that right.

Short answer, no. How do you the marble you're wieghing is a normal wieghted marble or a differently wieghted marble? The left side could go down and the right side could go up and you have two possibilities. The right side marble is the odd marble and is lighten than the rest or the left side marble is odd marble and heavier than the rest.

There's a better solution. Good try.

It is determined by the intial weigh-in. Earlier on you find out if the marble is light or heavy, and that dictates the logical outcome later.

I concur.

Touche....good job. Pardon my logic. :: hides face in humilty...and to protect himself from flying marbles ::

Now I tried to see if I could remember how to do it in 3 tries and couldn't do it. Any takers?

It seems that theoretically, you need 4 weigh-in's to cover ALL possible scenarios. But it is possible to do it in 3, determined by whether the marble is light or heavy and how you start the process. But the riddle calls for a complete logical process which needs 4.

 

[Placer14]

I tried again, (attempting to use my best logic) and I could get close, but I was still missing 2 marbles unaccounted for. Whitecloak, can you help us out?

 

[Lounatik]

This is a brainteaser for all the old farts out there. Lets see if we have any braincells left!

On the Beatles "White Album" (Technically called "The Beatles") there are 29 songs. Supposing these songs are on, what us oldtimers call "vinyl", you would have two albums/four sides. Okay,on side one you have as follows:

1. Back in the USSR
2. Dear Prudence
3. Glass onion
4. Ob la di ob la da
5. Wild honey pie
6. Continuing story of bungalow Bill
7. While my guitar gently weeps
8. Happiness is a warm gun

On side two:

9. Martha my dear
10. I'm so tired
11. Blackbird
12. Piggies
13. Rocky raccoon
14. Don't pass me by
15. Why don't we do it in the road

on side three:

16. I will
17. Julia
18. Yer blues
19. Mother nature's son
20. Everybody's got something to hide except me and my monkey
21. Sexy Sadie
22. Helter skelter
23. Long long long

and finally side four:

24. Revolution
25. Honey pie
26. Savoy truffle
27. Cry baby cry
28. Revolution 9
29. Goodnight

Now the riddle is this: The total album time is 93 minutes. Your job is to tell me how many grooves are on each side of each record. Side one has x grooves,side two has x etc.

 

[Broncho]

The record one is too easy. There is one groove on each side of a vinyl record (aka LP) for a total of four groves on a 2 LP set. I'm 23 and I still know that. 😀

As far as the marble one goes, I don't feel like thinking that hard today. 😛

edit: spelling

 

[atrowe]

Originally posted by: Lounatik
This is a brainteaser for all the old farts out there. Lets see if we have any braincells left!

On the Beatles "White Album" (Technically called "The Beatles") there are 29 songs. Supposing these songs are on, what us oldtimers call "vinyl", you would have two albums/four sides. Okay,on side one you have as follows:

1. Back in the USSR
2. Dear Prudence
3. Glass onion
4. Ob la di ob la da
5. Wild honey pie
6. Continuing story of bungalow Bill
7. While my guitar gently weeps
8. Happiness is a warm gun

On side two:

9. Martha my dear
10. I'm so tired
11. Blackbird
12. Piggies
13. Rocky raccoon
14. Don't pass me by
15. Why don't we do it in the road

on side three:

16. I will
17. Julia
18. Yer blues
19. Mother nature's son
20. Everybody's got something to hide except me and my monkey
21. Sexy Sadie
22. Helter skelter
23. Long long long

and finally side four:

24. Revolution
25. Honey pie
26. Savoy truffle
27. Cry baby cry
28. Revolution 9
29. Goodnight

Now the riddle is this: The total album time is 93 minutes. Your job is to tell me how many grooves are on each side of each record. Side one has x grooves,side two has x etc.

There is exactly one long spiraled groove on each side of each record. Next one:

you are a prisoner in a foreign land. your fate will be determined by a little game. there are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)?

 

[HomeBrewerDude]

W1) m1-4 vs m5-8

if W1 balanced, m1-8 are all standard.

::W2) m1-2 vs m9-10

::if W2 balanced then ODD marble is m11 or m12

::::W3)m1 vs m11 <--reveals weight
::::W4)m1 vs m12 <--reveals weight

::if W2 unbalanced then ODD marble is m9 or m10

::::W3)m1 v m9 <--reveals weight
::::W4)m1 v m10 <-reveals weight

if W1 is UNbalanced, m9-12 are all standard

::W2) m1-4 v m9-12

::if W2 is balanced, then you know that m1-4 and m9-12 are all standard. m5-8 has ODD marble, and now result of W1 reveals relavtive weight )

::::W3) m5,6 v m8,9

::::if W3 balanced, then ODD marble is m7 or m8

::::::W4) m7 v m9

::::::if W4 equal, then m8 is ODD and relative wieght known from 1)
::::::if W4 unequal, then m7 is ODD and relative weight is know from 1) and 3)

::::if W3 is unbalanced, then ODD marble is m5 or m6

::::::W4) m5 v m8

::::::if W4 is balanced then m6 is ODD and relative weight is known from W3
::::::if W4 is unbalanced then W5 is ODD and relative weight is known from W4

::if W2 is unbalanced, then you know that ODD marble is in m1-4 and the relative weight from W1

::::W3) m1, m2 v m9, m10

::::if W3 is balanced then ODD marble is m3 or m4

::::::W4) m3 v m9

::::::if W4 is balanced then m4 is ODD and weight is known from W1
::::::if W4 is unbalanced then m3 is ODD and weight is known from W3

::::if W3 is unbalanced then ODD marble is m1 or m2

::::::W4) m1 v m9

::::::if W4) is balanced then m2 is ODD and weight is known from W1
::::::if W4 is unbalanced then m1 is ODD and weight is known from W4

 

[atrowe]

Nobody's interested? I'll post another:


you have two ropes, each of which takes one hour to burn completely. both of these ropes are nonhomogeneous in thickness, meaning that some parts of the ropes are chunkier than other parts of the rope. using just these nonhomogeneous ropes and a lighter, accurately time 45 minutes.

Note: Some clarification on what is meant by nonhomogeneous. For instance, maybe a particular section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. then it would take 10 minutes to burn off the remaning 7/8, since we know that the whole rope takes an hour to burn off. that's just an example; we don't know any such ratios beforehand. The point is, if you look at one of your ropes and cut it into pieces, you have no clue how long any individual piece will take to burn off.

 

[Chubs]

Originally posted by: Placer14

Originally posted by: SagaLore
Start with 12 marbles.

A. SPLIT into two groups. (#1) Put on scale. 6x6

B. Whichever is lighter side, take off scale and put aside. SPLIT remainder into two groups. (#2) Put on scale. 3x3

C. If scale remains balanced, proceed to D. Otherwise proceed to I.

D. Remove current group, SPLIT previously put aside group. (#3) Put on scale. 3x3

E. Whichever is heavier side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#4) 1x1

F. If scale remains balanced, proceed to G. If one side is lighter, proceed to H.

G. REMAINING MARBLE, HELD IN HAND, IS THE LIGHTEST MARBLE. done

H. THE MARBLE ON THE SIDE THAT RAISED IS THE LIGHTEST MARBLE. done

I. Whichever is lighter side, take off scale and put aside. Out of 3 remaining marbles, place 1 on each side, holding 3rd. (#3) 1x1

J. If scale remains balanced, proceed to K. If one side is heavier, proceed to L.

K. THE MARBLE, HELD IN HAND, IS THE HEAVIEST MARBLE. done

L. THE MARBLE ON THE SIDE THAT LOWERED IS THE HEAVIEST MARBLE. done



I think I got that right.

Short answer, no. How do you the marble you're wieghing is a normal wieghted marble or a differently wieghted marble? The left side could go down and the right side could go up and you have two possibilities. The right side marble is the odd marble and is lighten than the rest or the left side marble is odd marble and heavier than the rest.

There's a better solution. Good try.

Actually, I think he's right. He'll know if it's heavier or lighter than "normal" based on the second weighing. If the original heavier group ends up being equal when split in half then you know the offending marble is lighter than normal.



Nevermind, I see this was already covered.

 

[pearlJammer]

Originally posted by: atrowe

you are a prisoner in a foreign land. your fate will be determined by a little game. there are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)?

this is probably not right but oh well...

put 1 white marble in one jar and the remaining 49 in the other along with the 50 black?

jar #1: 1 white
jar #2: 49 white, 50 black

you have a 50% chance outright before you even pull a marble... if you are given the "wrong" jar to pick from, you'll still have a 49.5% chance...

 

[BooneRebel]

you are a prisoner in a foreign land. your fate will be determined by a little game. there are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)?

I'm no probability whiz, but here's how I see it:

You need to pull a white marble. Put 1 white marble in one of the jars, placing the remaining 49 white and 50 black marbles in the other jar.

Your chances are 50/50 of receiving the jar with the one white marble. (you live)
If you receive the other jar, you still have a 50% (well, 49 out of 99) of receiving a white marble (you live)

So, if you do it this way, you're looking at 50%+24.74747474...% or roughly a 74.5% chance of selecting a white marble. Still, not a game that I would want to bet my life on.

DOH! Beat to the punch by the pearljammer...

 

[SagaLore]

Originally posted by: atrowe

Originally posted by: Lounatik
This is a brainteaser for all the old farts out there. Lets see if we have any braincells left!

On the Beatles "White Album" (Technically called "The Beatles") there are 29 songs. Supposing these songs are on, what us oldtimers call "vinyl", you would have two albums/four sides. Okay,on side one you have as follows:

1. Back in the USSR
2. Dear Prudence
3. Glass onion
4. Ob la di ob la da
5. Wild honey pie
6. Continuing story of bungalow Bill
7. While my guitar gently weeps
8. Happiness is a warm gun

On side two:

9. Martha my dear
10. I'm so tired
11. Blackbird
12. Piggies
13. Rocky raccoon
14. Don't pass me by
15. Why don't we do it in the road

on side three:

16. I will
17. Julia
18. Yer blues
19. Mother nature's son
20. Everybody's got something to hide except me and my monkey
21. Sexy Sadie
22. Helter skelter
23. Long long long

and finally side four:

24. Revolution
25. Honey pie
26. Savoy truffle
27. Cry baby cry
28. Revolution 9
29. Goodnight

Now the riddle is this: The total album time is 93 minutes. Your job is to tell me how many grooves are on each side of each record. Side one has x grooves,side two has x etc.

There is exactly one long spiraled groove on each side of each record. Next one:

you are a prisoner in a foreign land. your fate will be determined by a little game. there are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)?

I would put all but one white marble in with the black marbles. Which means one jar has 1 white marble and the other jar has 49 white marbles and 50 black marbles. So then if they give me the nearly empty jar, I live. If they give me the full jar, I have a 49 out of 99 chance of living.

The maximum probability is going to be something like 3/4 chance of living.

 

[SagaLore]

Originally posted by: atrowe
Nobody's interested? I'll post another:


you have two ropes, each of which takes one hour to burn completely. both of these ropes are nonhomogeneous in thickness, meaning that some parts of the ropes are chunkier than other parts of the rope. using just these nonhomogeneous ropes and a lighter, accurately time 45 minutes.

Note: Some clarification on what is meant by nonhomogeneous. For instance, maybe a particular section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. then it would take 10 minutes to burn off the remaning 7/8, since we know that the whole rope takes an hour to burn off. that's just an example; we don't know any such ratios beforehand. The point is, if you look at one of your ropes and cut it into pieces, you have no clue how long any individual piece will take to burn off.

Fold one of them in half, so each end is touching each other, and the entire rope is parallel. Light it, this will give you 30 minutes. Take the other rope, fold it, then fold it again. Creases/ends are touching and rope is parallel. Burn one end. This gives you 15 minutes. Which makes 45 minutes.

 

[BooneRebel]

Originally posted by: SagaLore

Originally posted by: atrowe
Nobody's interested? I'll post another:


you have two ropes, each of which takes one hour to burn completely. both of these ropes are nonhomogeneous in thickness, meaning that some parts of the ropes are chunkier than other parts of the rope. using just these nonhomogeneous ropes and a lighter, accurately time 45 minutes.

Note: Some clarification on what is meant by nonhomogeneous. For instance, maybe a particular section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. then it would take 10 minutes to burn off the remaning 7/8, since we know that the whole rope takes an hour to burn off. that's just an example; we don't know any such ratios beforehand. The point is, if you look at one of your ropes and cut it into pieces, you have no clue how long any individual piece will take to burn off.

Fold one of them in half, so each end is touching each other, and the entire rope is parallel. Light it, this will give you 30 minutes. Take the other rope, fold it, then fold it again. Creases/ends are touching and rope is parallel. Burn one end. This gives you 15 minutes. Which makes 45 minutes.

I was thinking along the same lines but I don't think this would work. Take the 50minute superchunky piece of rope as an example. If you fold the rope in half and burn it, one "half" will quickly burn away but the second half would continue to burn for 50+ minutes until the 'chunk' was gone. If you had some sort of timing mechanism (not a clock, but some sort of measured counter) you could take the average between the time to burn each half. But that's overcomplicating the question I think.