Math Credit Challenge - Page 2

dAnceguy117 · 11-8-2007, 09:56 PM

Quote:

Originally Posted by MagicCarpetRide

Actually they could be typed just as fast....

I type letters faster than I type numbers haha

ledwix · 11-8-2007, 10:12 PM

A runner sprints around a circular track of radius 100.0 meters at a constant speed of 7.00 meters per second. The runner's friend is standing at a distance of 200.0 meters from the center of the track. How fast is the distance between the friends changing when the distance between them is 200.0 meters?

silvercomet1525 · 11-8-2007, 11:56 PM

Quote:

Originally Posted by Reach

D:

It should proceed, after 2491...3599, 4757, 5767...etc.

You missed a value >_>...!

Okay... maybe we're not thinking of the same thing? I kind of made this up, but it does have mathematical logic behind it.

MrRubix · 11-8-2007, 11:59 PM

Reach: 56980, found manually. Working on the formula to make sure it accounts for every case.
EDIT: Let X denote the sum and let sub-n denote the first n elements to sum: Xn = [sum from i=1 to i=n(n+1)/2] i^2

So if you wish to find X(3), that means you sum from i=1 to 3*(3+1)/2 or 6, so you sum the first 6 squares: 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2, or 91, which is indeed equal to the first 3 terms summed. Verify with X(10): sum from i=1 to i=10(10+1)/2 or 55. The sum of the first 55 squares is equal to 56980.

Credits!

Tass: LOL if I solved p=np I'd want a lot more than FFR credits.

**MagicCarpetRide** · 11-9-2007, 09:01 AM

Quote:

Originally Posted by ledwix

A runner sprints around a circular track of radius 100.0 meters at a constant speed of 7.00 meters per second. The runner's friend is standing at a distance of 200.0 meters from the center of the track. How fast is the distance between the friends changing when the distance between them is 200.0 meters?

There is no such thing as a circular track.
I win.

**Reach** · 11-9-2007, 01:41 PM

Quote:

Originally Posted by silvercomet1525

Okay... maybe we're not thinking of the same thing? I kind of made this up, but it does have mathematical logic behind it.

No, you made a mistake. To be more specific, IIRC you skipped the prime 61 after 59 and went straight to 67.

And well done Rubix. Yea, the problem is designed such that it can be solved without any expertise in mathematics actually.

This one is tricky.

Assume that gravity on or near the airless surface of a certain planet called Reach is such that any free-falling object will reach a velocity of 10 meters per second after one second, 20 meters per second after two seconds, 30 meters per second after three seconds, and so forth. Suppose a frictionless pulley (i.e., a wheel with a grooved rim that, once in motion, does not slow down due to friction), having negligible mass, is attached to a sturdy beam at the top of a vertical, airless mine shaft, and an inelastic rope 1000 meters in length but of negligible weight is threaded through the pulley with identical buckets of negligible weight dangling at each end. In one bucket are fifty stones weighing 1 kilogram apiece and an alien named Rubix weighing 50 kilograms, while in the other bucket is a single stone weighing 50 kilograms and an alien named Keeley weighing 50 kilograms. Suppose Rubix commences dropping stones out of its bucket at a rate of one per second. The instant the first stone is dropped Keeley commences climbing the rope directly above its own bucket at a rate of one meter of rope per second. If Rubix starts at rest at a position 20 meters below Keeley, how many stones will Rubix have dropped the instant the two buckets are precisely side by side?

edit: I gave you 1000 credits Rubix.

MrRubix · 11-9-2007, 01:57 PM

I see the pattern too in silvercomet's sequence and I agree -- it starts to not work.

I know how to do the pulley problem too I think -- I love these kind of physics problems. Will solve it when I get back (picking up some drinks)

silvercomet1525 · 11-9-2007, 04:19 PM

Okay, I guess it was a bit obvious to you huh? Yeah, you got the pattern, I just made a mistake.

**albaneenesk8r** · 11-9-2007, 09:25 PM

mr rubix + FFR = ______

**Reach** · 11-9-2007, 09:56 PM

Quote:

Originally Posted by MrRubix

I see the pattern too in silvercomet's sequence and I agree -- it starts to not work.

I know how to do the pulley problem too I think -- I love these kind of physics problems. Will solve it when I get back (picking up some drinks)

I will be impressed if you can! It's quite the mental gym. Though there are harder problems surely (I'm sure I could make that same problem drastically harder, but then I wouldn't know how to solve it myself XD), I've never met anyone that could actually solve it...at least without a second try.

**MixMasterLar** · 11-9-2007, 10:31 PM

if you can go metric = english then Im out 100 credz rofl....but I'll try

Ok, how much dirt (measured in km)is in a hole if the hole is:

13 feet deep
8 feet wide
9 feet long.

**Reach** · 11-9-2007, 10:34 PM

Quote:

Originally Posted by MixMasterLar

if you can go metric = english then Im out 100 credz rofl....but I'll try

Ok, how much dirt (measured in km)is in a hole if the hole is:

13 feet deep
8 feet wide
9 feet long.

omg none...!

Problems · 11-9-2007, 10:40 PM

This is really a very simple problem, but many people find it very counter-intuitive.

You are driving by car to a particular destination, and our only assumption is that you are free to drive at any speed you choose - no traffic jams or anything like that. For the first half of the journey (and by half we mean half the overall distance between the starting point and your finishing point) you drive at 20 miles per hour. You then realise that this is all taking much too long, and that you are going to be late. You therefore decide that you will increase your speed so that your overall average speed for the whole journey will be 40 miles per hour. How fast do you have to drive for the remaining part of your journey in order for your average speed for the whole journey to be 40 miles per hour?

Also, I know this is a pretty popular problem so don't google it. It'll spoil the fun. And this problem is on about an eigth grade level if you read it correctly. If you can't get this, which I very highly doubt, i'll give a hint.

MrRubix · 11-9-2007, 11:01 PM

Problems: Impossible to do unless you somehow possess a teleporter.

Credits please

MrRubix · 11-9-2007, 11:04 PM

Reach: There are tons of things going on assuming I am doing it right. I am just hoping I am not making an incorrect assumption somewhere.

**Reach** · 11-9-2007, 11:05 PM

LOL.

v2= infinity (where acceleration is infinite, starting from exactly the halfway point).

>__>

edit: gl rubix

MrRubix · 11-9-2007, 11:07 PM

Problems · 11-9-2007, 11:08 PM

Haha, pretty much. That one was incredibly easy. I'll give you a harder one later. Credits sent.

**Tasselfoot** · 11-9-2007, 11:08 PM

lucky for us, i have a teleporter!

edit: also will give the physics problem to my buddy, who is much better at solving those kinds of problems than i.

Simonsmp · 11-9-2007, 11:21 PM

why isnt it 60 mph?

also, p=np
n=1

11-8-2007, 10:12 PM	#22
ledwix Giant Pi Operator Join Date: Mar 2006 Location: California Age: 35 Posts: 2,878	Re: Math Credit Challenge A runner sprints around a circular track of radius 100.0 meters at a constant speed of 7.00 meters per second. The runner's friend is standing at a distance of 200.0 meters from the center of the track. How fast is the distance between the friends changing when the distance between them is 200.0 meters?

11-8-2007, 11:59 PM	#24
MrRubix FFR Player Join Date: Dec 1969 Location: New York City, New York Posts: 8,340	Re: Math Credit Challenge Reach: 56980, found manually. Working on the formula to make sure it accounts for every case. EDIT: Let X denote the sum and let sub-n denote the first n elements to sum: Xn = [sum from i=1 to i=n(n+1)/2] i^2 So if you wish to find X(3), that means you sum from i=1 to 3(3+1)/2 or 6, so you sum the first 6 squares: 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2, or 91, which is indeed equal to the first 3 terms summed. Verify with X(10): sum from i=1 to i=10(10+1)/2 or 55. The sum of the first 55 squares is equal to 56980. Credits! Tass: LOL if I solved p=np I'd want a lot more than FFR credits. Last edited by MrRubix; 11-9-2007 at 02:06 AM..*

11-9-2007, 01:57 PM	#27
MrRubix FFR Player Join Date: Dec 1969 Location: New York City, New York Posts: 8,340	Re: Math Credit Challenge I see the pattern too in silvercomet's sequence and I agree -- it starts to not work. I know how to do the pulley problem too I think -- I love these kind of physics problems. Will solve it when I get back (picking up some drinks) Last edited by MrRubix; 11-9-2007 at 02:03 PM..

11-9-2007, 10:31 PM	#31
MixMasterLar Beach Bum Extraordinaire Join Date: Aug 2006 Location: The Emerald Coast Posts: 5,224	Re: Math Credit Challenge if you can go metric = english then Im out 100 credz rofl....but I'll try Ok, how much dirt (measured in km)is in a hole if the hole is: 13 feet deep 8 feet wide 9 feet long. __________________ Facebook / Youtube / Twitter .

11-9-2007, 10:40 PM	#33
Problems FFR Player Join Date: Apr 2005 Posts: 513	Re: Math Credit Challenge This is really a very simple problem, but many people find it very counter-intuitive. You are driving by car to a particular destination, and our only assumption is that you are free to drive at any speed you choose - no traffic jams or anything like that. For the first half of the journey (and by half we mean half the overall distance between the starting point and your finishing point) you drive at 20 miles per hour. You then realise that this is all taking much too long, and that you are going to be late. You therefore decide that you will increase your speed so that your overall average speed for the whole journey will be 40 miles per hour. How fast do you have to drive for the remaining part of your journey in order for your average speed for the whole journey to be 40 miles per hour? Also, I know this is a pretty popular problem so don't google it. It'll spoil the fun. And this problem is on about an eigth grade level if you read it correctly. If you can't get this, which I very highly doubt, i'll give a hint.

11-9-2007, 11:01 PM	#34
MrRubix FFR Player Join Date: Dec 1969 Location: New York City, New York Posts: 8,340	Re: Math Credit Challenge Problems: Impossible to do unless you somehow possess a teleporter. Credits please

11-9-2007, 11:04 PM	#35
MrRubix FFR Player Join Date: Dec 1969 Location: New York City, New York Posts: 8,340	Re: Math Credit Challenge Reach: There are tons of things going on assuming I am doing it right. I am just hoping I am not making an incorrect assumption somewhere.

11-9-2007, 11:05 PM	#36
Reach FFR Simfile Author Join Date: Jun 2003 Location: Canada Age: 39 Posts: 7,471	Re: Math Credit Challenge LOL. v2= infinity (where acceleration is infinite, starting from exactly the halfway point). >__> edit: gl rubix __________________

11-9-2007, 11:07 PM	#37
MrRubix FFR Player Join Date: Dec 1969 Location: New York City, New York Posts: 8,340	Re: Math Credit Challenge rofl

11-9-2007, 11:08 PM	#38
Problems FFR Player Join Date: Apr 2005 Posts: 513	Re: Math Credit Challenge Haha, pretty much. That one was incredibly easy. I'll give you a harder one later. Credits sent.

11-9-2007, 11:08 PM	#39
Tasselfoot Retired BOSS Join Date: Jul 2003 Location: Widget Heaven Age: 42 Posts: 25,185	Re: Math Credit Challenge lucky for us, i have a teleporter! edit: also will give the physics problem to my buddy, who is much better at solving those kinds of problems than i. __________________ RIP

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)