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Riddle and Puzzle Thread - Page 6

post #91 of 180
Quote:
Originally Posted by Ernest Jones View Post

My Precious?

 

Somehow I knew you would get it Ernie.  I also knew that you (being on the east coast) would see it before most.

post #92 of 180

Another oldie, but a goodie......

 

You're traveling through the woods and come to a bridge.....complete with the obligatory troll.  The troll tells you that not only do you have to answer his riddle to cross the bridge, if you fail, he's going to eat you.  The good news is that if you answer correctly, not only may you cross his bridge, you'll also leave with wealth beyond your wildest dreams.

 

He points to 5 large chests, each filled with gold bars.  All the bars are visually identical, but the troll tells you that 4 of the chests contain fake bars and only one contains real gold.  The ONLY difference between the bars is that while each fake bar weighs exactly 10 lbs, each real gold bar weighs exactly 10 lbs, 1 oz.  Your challenge is to choose the chest with the real bars.  Do so and you'll be allowed to take it with you as you continue your journey.  Fail, and you get eaten.

 

The individual bars are far too heavy to feel the weight difference, but the troll offers you a scale, accurate to the ounce, to help you choose.  Unfortunately, it's an old "penny" type scale that only gives you a single weight printed on a slip of paper, and the troll informs you that you're only allowed to use the scale one time. 

 

With only one use of the scale, how do you determine which of the 5 chests contains the real gold?

post #93 of 180
Thread Starter 
Put one bar from chest 1, two bars from chest 2, three bars from chest 3, four from #4, and five from #5 on the scale.

 

Then if the thing weighs 15 pounds 3 ounces the real gold is in chest #3. Stupid troll. He could have put one or two real gold bars randomly in each of the chests and you'd really have no way of knowing. :)

post #94 of 180
Quote:
Originally Posted by iacas View Post
 
Put one bar from chest 1, two bars from chest 2, three bars from chest 3, four from #4, and five from #5 on the scale.

 

Then if the thing weighs 15 pounds 3 ounces the real gold is in chest #3. Stupid troll. He could have put one or two real gold bars randomly in each of the chests and you'd really have no way of knowing. :)

 

Trolls have never been known for being that smart!

post #95 of 180

You guys are missing a puzzle:  Here's a couple oldies, depending on your math skills:

 

 

 

 

Harder - A cylindrical hole, 6 inches long, has been drilled straight through the centre of a solid sphere. What is the volume remaining from the original sphere?  (visualize a bead shape, or a wedding band shape - BTW - there is an equation for the volume of a dome cap shape.  Of course there are equations for a sphere and a cylinder.)

 

Like a dork, I solved this one explicitly the first time out - but there's an easier way to attack it.

 

 

 

Easy - What volume of dirt is in a hole of the following dimensions:  1 foot by 3 feet by 2 feet?

post #96 of 180
Quote:
Originally Posted by rehmwa View Post
 

You guys are missing a puzzle:  Here's a couple oldies, depending on your math skills:

 

 

 

 

Harder - A cylindrical hole, 6 inches long, has been drilled straight through the centre of a solid sphere. What is the volume remaining from the original sphere?  (visualize a bead shape, or a wedding band shape - BTW - there is an equation for the volume of a dome cap shape.  Of course there are equations for a sphere and a cylinder.)

 

Like a dork, I solved this one explicitly the first time out - but there's an easier way to attack it.

 

 

 

Easy - What volume of much dirt is in a hole of the following dimensions:  1 foot by 3 feet by 2 feet?

Skipping the first one.

 

Easy - well if it's a hole there shouldn't be any dirt in it.

post #97 of 180
Quote:
Originally Posted by rehmwa View Post
 

You guys are missing a puzzle:  Here's a couple oldies, depending on your math skills:

 

 

 

 

Harder - A cylindrical hole, 6 inches long, has been drilled straight through the centre of a solid sphere. What is the volume remaining from the original sphere?  (visualize a bead shape, or a wedding band shape - BTW - there is an equation for the volume of a dome cap shape.  Of course there are equations for a sphere and a cylinder.)

 

Like a dork, I solved this one explicitly the first time out - but there's an easier way to attack it.

 

 

 

Easy - What volume of dirt is in a hole of the following dimensions:  1 foot by 3 feet by 2 feet?

Harder:  Is this a trick?  I feel like it must be because you didn't give us any info on the diameter of the cylinder.

 

Easy:  Ernie already got it ... and it reminds me of a favorite that my dad told me as a kid:  If a plane crashes directly on the California/Nevada border, where do you bury the survivors?

post #98 of 180
Quote:
Originally Posted by Golfingdad View Post
 

I feel like it must be because you didn't give us any info on the diameter of the cylinder.

 

Seems like it doesn't it? 

 

(no hints just yet - I had to solve it the hard way to realize there is an intuitive hint that makes sense after the fact - but there are smarter people here. let's see if they come up with it without help)

post #99 of 180
Quote:
Originally Posted by Golfingdad View Post

Harder:  Is this a trick?  I feel like it must be because you didn't give us any info on the diameter of the cylinder.

Easy:  Ernie already got it ... and it reminds me of a favorite that my dad told me as a kid:  If a plane crashes directly on the California/Nevada border, where do you bury the survivors?

Even my then 12 year old daughter got that one. You don't bury survivors....
post #100 of 180
Quote:
Originally Posted by rehmwa View Post
 

You guys are missing a puzzle:  Here's a couple oldies, depending on your math skills:

 

 

 

 

Harder - A cylindrical hole, 6 inches long, has been drilled straight through the centre of a solid sphere. What is the volume remaining from the original sphere?  (visualize a bead shape, or a wedding band shape - BTW - there is an equation for the volume of a dome cap shape.  Of course there are equations for a sphere and a cylinder.)

 

Like a dork, I solved this one explicitly the first time out - but there's an easier way to attack it.

 

 

 

Easy - What volume of dirt is in a hole of the following dimensions:  1 foot by 3 feet by 2 feet?

 

Brute force for the two cases for where rc < rs:

All units are in inches,

rs= radius of sphere,

rc= radius of cylinder,

h= rs-sqrt(rs^2-rc^2),

 

and 2*rs-2*h > 6 inches

Volume remaining is: 4 / 3 * pi * rs^3 - pi * rc^2 * 6 - pi * h / 6 * ( 3 * rc^2 + h^2 )

 

and 2*rs-2*h <= 6 inches

Volume remaining is: 4 / 3 * pi * rs^3 - pi * rc^2 * (2 * rs - 2 * h) - 2 * pi * h / 6 * ( 3 * rc^2 + h^2 )

 

If rc > rs then there is no volume remaining in the sphere.

 

Edit: Let me see if I can simplify it. . .


Edited by Lihu - 12/16/13 at 12:17pm
post #101 of 180
Thread Starter 
Quote:
Originally Posted by rehmwa View Post
 

Harder - A cylindrical hole, 6 inches long, has been drilled straight through the centre of a solid sphere. What is the volume remaining from the original sphere?  (visualize a bead shape, or a wedding band shape - BTW - there is an equation for the volume of a dome cap shape.  Of course there are equations for a sphere and a cylinder.)

 

That's not really a riddle, though. It's more like a math problem.

post #102 of 180
Quote:
Originally Posted by Lihu View Post
 

 

Brute force for the two cases for where rc < rs:

All units are in inches,

rs= radius of sphere,

rc= radius of cylinder,

h= rs-sqrt(rs^2-rc^2),

 

and 2*rs-2*h > 6 inches

Volume remaining is: 4 / 3 * pi * rs^3 - pi * rc^2 * 6 - pi * h / 6 * ( 3 * rc^2 + h^2 )

 

and 2*rs-2*h <= 6 inches

Volume remaining is: 4 / 3 * pi * rs^3 - pi * rc^2 * (2 * rs - 2 * h) - pi * h / 6 * ( 3 * rc^2 + h^2 )

 

If rc > rs then there is no volume remaining in the sphere.

 

Edit: Let me see if I can simplify it. . .

Yeah, this is supposed to be a "riddle" thread ... not a math/physics/science thread. :beer:  I leave those hats at the office. ...

 

... Of course, I'm usually browsing TST at the office, soooo I guess I am excuse-less either way. :-P

post #103 of 180
Quote:
Originally Posted by Golfingdad View Post
 
Quote:
Originally Posted by Lihu View Post
 

 

Brute force for the two cases for where rc < rs:

All units are in inches,

rs= radius of sphere,

rc= radius of cylinder,

h= rs-sqrt(rs^2-rc^2),

 

and 2*rs-2*h > 6 inches

Volume remaining is: 4 / 3 * pi * rs^3 - pi * rc^2 * 6 - pi * h / 6 * ( 3 * rc^2 + h^2 )

 

and 2*rs-2*h <= 6 inches

Volume remaining is: 4 / 3 * pi * rs^3 - pi * rc^2 * (2 * rs - 2 * h) - pi * h / 6 * ( 3 * rc^2 + h^2 )

 

If rc > rs then there is no volume remaining in the sphere.

 

Edit: Let me see if I can simplify it. . .

Yeah, this is supposed to be a "riddle" thread ... not a math/physics/science thread. :beer:  I leave those hats at the office. ...

 

... Of course, I'm usually browsing TST at the office, soooo I guess I am excuse-less either way. :-P


I think I made a mistake for the second and third cases. You need to multiply the cap volume by 2 for the second case, and the third case needs to be >= and not >.

Oh, well. I used Solidworks to confirm a couple cases to a few digits.

post #104 of 180
Quote:
Originally Posted by Lihu View Post
 


I think I made a mistake for the second and third cases. You need to multiply the cap volume by 2 for the second case, and the third case needs to be >= and not >.

Oh, well. I used Solidworks to confirm a couple cases to a few digits.

Well this is fun...

post #105 of 180
Quote:
Originally Posted by Ernest Jones View Post
 
Quote:
Originally Posted by Lihu View Post
 


I think I made a mistake for the second and third cases. You need to multiply the cap volume by 2 for the second case, and the third case needs to be >= and not >.

Oh, well. I used Solidworks to confirm a couple cases to a few digits.

Well this is fun...


Yeah, okay. I need to get back to work. Simplification is turning out to be not so simple. I'll leave it for somebody who has more time. . .

post #106 of 180
Quote:
Originally Posted by 14ledo81 View Post

OK. I've got another one. Should be short/easy for the right person. A fairly famous riddle.

"What have I got in my pocket?"

I have to leave for work, so I won't be able to answer for a bit. I think someone will get it though.

 

 

Quote:
Originally Posted by Ernest Jones View Post

My Precious?

I have never heard this one before AND ... I don't get it.  Can one of you guys explain this one?

post #107 of 180
Quote:
Originally Posted by Golfingdad View Post
 

 

 

I have never heard this one before AND ... I don't get it.  Can one of you guys explain this one?

No, and hopefully no one else explains it either. You just need to get out more!

post #108 of 180
Quote:
Originally Posted by Golfingdad View Post
 
Quote:
Originally Posted by 14ledo81 View Post

OK. I've got another one. Should be short/easy for the right person. A fairly famous riddle.

"What have I got in my pocket?"

I have to leave for work, so I won't be able to answer for a bit. I think someone will get it though.

 

 

Quote:
Originally Posted by Ernest Jones View Post

My Precious?

I have never heard this one before AND ... I don't get it.  Can one of you guys explain this one?

 

It's in the "Hobbit"


Gollum and Bilbo are in the Goblin caves and Bilbo and Gollum play a game so Bilbo can escape. They pitch riddles back and forth until Bilbo decides to use this one and says "What's in my pocket?", when earlier he found Gollum's ring (His precious which is the one ring, of course).

 

Bilbo uses it to escape later on, because it turns the wearer invisible.

 

 

EDIT: Sorry Ernest. . .

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