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08-03-2005, 02:58 PM
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#16 | | Member
Join Date: Jun 2005 Location: NY
Posts: 488
| If it takes 4 men, 8 hours to dig 1 hole - THEN::
That means it takes 1 man 2 hours to dig 1/16 of a hole.
This means it takes 2 men 2 hours to dig 1/8 of a hole.
This means it takes 2 men 16 hours to dig 1 complete hole.
Then, it's obvious that it takes 2 men 8 hours to dig 1/2 a hole. | 
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08-03-2005, 03:01 PM
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#17 | | Member
Join Date: Mar 2005
Posts: 691
| hmm... i dont understand your reasoning eternity... could you perhaps make it a bit clearer. | |
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08-03-2005, 03:01 PM
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#18 | | Junior Member
Join Date: Jun 2005 Location: Earth
Posts: 96
| I'm sorry, but this was a smartass question. There's really no such thing as a half of a hole. A hole is a hole.  | |
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08-03-2005, 03:07 PM
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#19 | | Member
Join Date: Jun 2005 Location: NY
Posts: 488
| Yeah but a hole can still be quantified.
You can have a hole the size of a lake or a hole the size of a small house.
In any case, I'm sure my prior answer is correct.
Don't have time to explain right now but I'll try and get to it later. | |
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08-03-2005, 03:09 PM
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#20 | | Member
Join Date: May 2005
Posts: 932
| here was my reasoning..
If it takes 4 men, 8 hours to dig 1 hole - THEN::
That means it takes 1 man 2 hours to dig a hole.
This means it takes two men 1 hour to dig a hole.
This means it takes two men half hour to dig half a hole.
im not good at math so i'd appreciate it if someone could tell me why my logic is flawed. | |
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08-03-2005, 03:15 PM
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#21 | | Senior Member
Join Date: Aug 2004 Location: Student @ JHU (Baltimore, MD) / Houston, TX
Posts: 2,144
| Lol.. obviously, the question was posted in jest, but jai6638, here's why your method doesnt work:
If it takes 4 men 8 hours to do a job, then the total work done is 32 man-hours and it would take 1 man 32 hours to do the work. In this case, we want to do "half" a job, so the total time is 16 man-hours. If two men were working on the job, then both men would work for 8 hours (for a total of 16 man-hours) to do half the job. | |
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08-03-2005, 03:16 PM
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#22 | | Junior Member
Join Date: Jul 2005
Posts: 237
| Think about it. If it takes 1 man 2 hours to dig a hole, how can it take 4 men MORE than 2 hours to dig THE SAME hole? Won't 4 men do the job FASTER than 1 man? (Remember, we're talking about 1 hole here...) | |
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08-03-2005, 03:22 PM
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#23 | | Member
Join Date: Mar 2005
Posts: 535
| Guys - this could be a real SAT question:
Think about it - it's an inverse proportion problem...
It takes 4 men 8 hours to dig 1 hole.
How long does it take 2 men to dig the same hole?
4*8=2x
X = 16 hours
If we want to complete only half of the hole, divide the total time for the entire hole by half; it takes 8 hours to complete half of the hole.  | |
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08-03-2005, 03:50 PM
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#24 | | Junior Member
Join Date: Jun 2005 Location: Earth
Posts: 96
| I agree that holes can be quantified, but if you really read into the question that deeply, then you'd agree that I said a "half of a hole" which implies a half of any random hole, not necessarily the hole previously stated. And I could've sworn your answer used to say 16/5 of an hour...
It's good to see that everyone is flexing his math muscles, anyway.
Oh, and I'm working on the Barron's Math IIC book right now, and the questions on probability are getting my man-panties in a bind. They seem completely random. How many of these questions appear on the test? | |
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08-03-2005, 05:15 PM
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#25 | | Member
Join Date: Aug 2004 Location: Chicago
Posts: 570
| It takes (4)(8) = 32 man_hours to dig 1 hole
In general, (#men)(#hours) should = 32, to dig 1 hole.
So we need 32/2 = 16 man_hours for half a hole.
With 2 men, we need (2)(#hours) = 16
i.e. #hours = 16/2 = 8 for 2 men to dig half a hole.
(P.S. Oh, oh- I missed pg. 2, and TanMan's virtually identical post above. Never mind...) | |
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08-03-2005, 06:40 PM
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#26 | | Member
Join Date: Jun 2005 Location: NY
Posts: 488
| Optimizerdad and Programmer1234 have just confirmed that my prior posted answer was correct.
I rest my case.
Thank you gentleman.
Now, another question.......make it hard this time......... | |
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08-03-2005, 06:56 PM
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#27 | | Member
Join Date: Jun 2005 Location: NY
Posts: 488
| One of Euler's famous mathamatical relationships is below:
e^(i*pi) + 1 = 0
***where e = 2.714...
***where "^" means "to the power of"
***where i = imaginary number = sqrt(-1)
***where pi = 3.14159...
Prove Euler's formula and show that it really equals 0.
Whoever can prove this in less than 1 day without having a course on complex numbers is a genius. Furthermore, note that the above formula contains all the famous and remarkable numbers in exactly 1 equation. It contains, "e", "i", "pi", "1", and "0" all in one single equation. That's pretty awesome. I know that this formula is correct but I don't know the proof for this damn thing.
Last edited by eternity_hope2005; 08-03-2005 at 07:11 PM.
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08-03-2005, 07:09 PM
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#28 | | Junior Member
Join Date: Jun 2005 Location: Earth
Posts: 96
| I read the proof at Wikipedia, but I'm not going to say anything since I cheated. It seems kind of random though... You'd really need to be familiar with Euler's concepts to get the correct answer.
Last edited by StepIntoThisReality; 08-03-2005 at 07:09 PM.
Reason: Clarification.
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08-03-2005, 09:42 PM
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#29 | | Member
Join Date: Jun 2005 Location: NY
Posts: 488
| The last question was way too hard and advanced for this forum. It's just beyond your level (at least until you take Calculus II in college).
Don't worry about the question. If you're still interested in the answer, someone answered it on another forum. The URL is below: http://talk.collegeconfidential.com/...61#post1103861
In anycase, just move on. If anybody else is interested in presenting another question, feel free...
Last edited by eternity_hope2005; 08-03-2005 at 09:50 PM.
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08-03-2005, 10:00 PM
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#30 | | Senior Member
Join Date: Aug 2004 Location: Student @ JHU (Baltimore, MD) / Houston, TX
Posts: 2,144
| All I can say is Yuck! - bring back bad memories of complex numbers in Calc III  | |
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