time matters in this challenge
-
hey cp'ians. how u all doing.. well i got another math challenge ..and this was easy enough to be solved pretty quickly...but sometimes, depending on how u think, it can take a long time ...its actually tricky...lets see how many of you get it and how fast...btw, our friend Paul Ingles was the one i challenged first, and then we decided to have it on CP as well...lol ok here: A cylinder 108cm high has a circumference of 24cm. A string makes exactly 6 complete turns round the cyclinder while its two ends touch the cylinder's top and bottom. How long is the string in cm? oOoo...LOL
-
180 cm "I will find a new sig someday."
-
( (24*6)^2 + 108^2)^(1/2) = 180 "I will find a new sig someday."
-
( (24*6)^2 + 108^2)^(1/2) = 180 "I will find a new sig someday."
:omg: Clever!
I'd wear a miniskirt and pimp myself for an extra ten grand a year. - David Wulff
-
( (24*6)^2 + 108^2)^(1/2) = 180 "I will find a new sig someday."
-
Imagine a really wide but short right-angled triangle cut out of a piece of paper. It's 24x6 across its base and 108 tall. Then wrap it around the cylinder. The hypotenuse describes the path that the string will take as it wraps around the cylinder. From there, it just Pythagoras's Theorem...
I'd wear a miniskirt and pimp myself for an extra ten grand a year. - David Wulff
-
hey cp'ians. how u all doing.. well i got another math challenge ..and this was easy enough to be solved pretty quickly...but sometimes, depending on how u think, it can take a long time ...its actually tricky...lets see how many of you get it and how fast...btw, our friend Paul Ingles was the one i challenged first, and then we decided to have it on CP as well...lol ok here: A cylinder 108cm high has a circumference of 24cm. A string makes exactly 6 complete turns round the cyclinder while its two ends touch the cylinder's top and bottom. How long is the string in cm? oOoo...LOL
I assumed one end of string at the top and one at the bottom. you wrap around the top 6 times and drop the string to the bottom. (24 * 6)+ 108 = 252 BW "I'm coming with you! I got you fired, it's the least I can do. Well, the least I could do is absolutely nothing, but I'll go you one better and come along!" - Homer J. Simpson
-
hey cp'ians. how u all doing.. well i got another math challenge ..and this was easy enough to be solved pretty quickly...but sometimes, depending on how u think, it can take a long time ...its actually tricky...lets see how many of you get it and how fast...btw, our friend Paul Ingles was the one i challenged first, and then we decided to have it on CP as well...lol ok here: A cylinder 108cm high has a circumference of 24cm. A string makes exactly 6 complete turns round the cyclinder while its two ends touch the cylinder's top and bottom. How long is the string in cm? oOoo...LOL
I'm going against the group and saying 144 cm (6*24) ;) --Mike-- "Adventure. Excitement. A Jedi craves not these things." -- Silent Bob 1ClickPicGrabber - Grab & organize pictures from your favorite web pages, with 1 click! My really out-of-date homepage Sonork-100.19012 Acid_Helm
-
hey cp'ians. how u all doing.. well i got another math challenge ..and this was easy enough to be solved pretty quickly...but sometimes, depending on how u think, it can take a long time ...its actually tricky...lets see how many of you get it and how fast...btw, our friend Paul Ingles was the one i challenged first, and then we decided to have it on CP as well...lol ok here: A cylinder 108cm high has a circumference of 24cm. A string makes exactly 6 complete turns round the cyclinder while its two ends touch the cylinder's top and bottom. How long is the string in cm? oOoo...LOL
ﻡﺟﻧ_Najm wrote: How long is the string in cm? Ok, I may be way off here, but we know that it is the total height of 108 with one end at top and one end at bottom, so it is safe to say they meet at the middle. We also have to assume that for that string which is now actually seen as two pieces (the upper portion and the lower portion that meet in the middle both have to go around 6 times, thus we get this: 108 + (2*(24*6)) = 396
Nick Parker
Not everything that can be counted counts, and not everything that counts can be counted. - Albert Einstein
-
ﻡﺟﻧ_Najm wrote: How long is the string in cm? Ok, I may be way off here, but we know that it is the total height of 108 with one end at top and one end at bottom, so it is safe to say they meet at the middle. We also have to assume that for that string which is now actually seen as two pieces (the upper portion and the lower portion that meet in the middle both have to go around 6 times, thus we get this: 108 + (2*(24*6)) = 396
Nick Parker
Not everything that can be counted counts, and not everything that counts can be counted. - Albert Einstein
Not exactly. This problem can be solved easier when it's 2 dimensional. Assume that the cilinder is made out of very thin layers of plastic. Unwind the cilinder so you get a rectangle of 180cm height and 6x24cm=144cm width. The string now starts at the top-left corner and goes to the bottom-right corner of the rectangle. Since, according to Pythagoras x^2 = y^2 + y^2, the length of the string is equal to squareroot(180^2+144^2) which is squareroot(53136) = 230.51247 cm. Easy, isn' it ? Goh, I'm really missing the mathematics courses at University. Enjoy life, this is not a rehearsal !!!
-
( (24*6)^2 + 108^2)^(1/2) = 180 "I will find a new sig someday."
Pythagoras is dead. Long live the king! -- This space for rent.
-
Imagine a really wide but short right-angled triangle cut out of a piece of paper. It's 24x6 across its base and 108 tall. Then wrap it around the cylinder. The hypotenuse describes the path that the string will take as it wraps around the cylinder. From there, it just Pythagoras's Theorem...
I'd wear a miniskirt and pimp myself for an extra ten grand a year. - David Wulff
Taka Muraoka wrote: From there, it just Pythagoras's Theorem Right:) "I will find a new sig someday."
-
Pythagoras is dead. Long live the king! -- This space for rent.
Jörgen Sigvardsson wrote: Long live the king Elvis or Carl? (If my memory is correct.) "I will find a new sig someday."
-
Not exactly. This problem can be solved easier when it's 2 dimensional. Assume that the cilinder is made out of very thin layers of plastic. Unwind the cilinder so you get a rectangle of 180cm height and 6x24cm=144cm width. The string now starts at the top-left corner and goes to the bottom-right corner of the rectangle. Since, according to Pythagoras x^2 = y^2 + y^2, the length of the string is equal to squareroot(180^2+144^2) which is squareroot(53136) = 230.51247 cm. Easy, isn' it ? Goh, I'm really missing the mathematics courses at University. Enjoy life, this is not a rehearsal !!!
sqrt((24*6)^2+108^2) = 180 Patje wrote: Goh, I'm really missing the mathematics courses at University ... and the use of the University calculator? :-D
The opinions expressed in this communication do not necessarily represent those of the author (especially if you find them impolite, discourteous or inflammatory).
-
sqrt((24*6)^2+108^2) = 180 Patje wrote: Goh, I'm really missing the mathematics courses at University ... and the use of the University calculator? :-D
The opinions expressed in this communication do not necessarily represent those of the author (especially if you find them impolite, discourteous or inflammatory).
Yep, you're right. I typed 180 instead of 108. What's the use of a good calculator if you can't type. Shame on me. Seems my math is better than my typing. The ironical part of the story is that the result is 180, just the thing I mistyped. Moral of the story: just mistype the input to get the output. :cool::cool: Enjoy life, this is not a rehearsal !!!