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By Apocalypse_Now (Apocalypse_Now) on Saturday, July 10, 2004 - 01:36 pm: Edit |

Working alone, Holmes puts two coats of paint on a wall, one before lunch and one after lunch. One day Holmes began at the usual time, and two hours before lunch he was joined by Watson, who paints at the rate of 600 square meters per day, and who left when they finished the first coat. Holmes promptly began the second coat, and then be broke for lunch at the usual time. One hour before his normal quitting time, Holmes had painted a second coat everywhere except where Watson had painted that morning. If each man works at a constant rate, what was the area of the wall in square meters?

Sorry it's so wordy.

By Tongos (Tongos) on Saturday, July 10, 2004 - 05:28 pm: Edit |

heres one that might end up, but if it doesnt its still cool. x^x=5?

By Crypto86 (Crypto86) on Saturday, July 10, 2004 - 07:11 pm: Edit |

For the x^x = 5 problem, I represented that using logs:

log x (5) = X

Now since we don't know what the base is, I'll use the change-of-base formula to change base "x" into base 10:

log a (X) --> log b (x) = log b (X) / log b (a)

log x (5) --> log 10 (5) = log 10 (5) / log 10 (x)

By properties of logs, log x (A) / log x (B) = log x (A-B), so log 10 (5) / log 10 (x) = log 10 (5/x)

So now the expression of x^x = 5 can be written as log 10 (5/x) = x, which can be written as (5/x) = 10^x.

I can't get past here! :D

By Tongos (Tongos) on Saturday, July 10, 2004 - 07:23 pm: Edit |

well, if you want a full on solution, visit my solution at the cafe, its under the thread, "nice approximations of x^x=5". there was this huge conspiracy over the problem a week ago. you can look under calculating pi or "i was intrigued" to see the progress that we made on the solution (might as well been nothing!). but, yeah, it was a big deal, mainly because nobody knew how to solve it algebraically, and note that the solution is not algebraic either.

By Devious (Devious) on Saturday, July 10, 2004 - 07:34 pm: Edit |

What's the final answer to the original question? I wanna check to see if I got it right.

By Tongos (Tongos) on Saturday, July 10, 2004 - 07:39 pm: Edit |

2.129..... AND some imaginary set AND no negative solutions, AND no other real solution!!!!!!!!!!!!!!!!!!!!!!!!!!!!

By Apocalypse_Now (Apocalypse_Now) on Saturday, July 10, 2004 - 07:39 pm: Edit |

Devious, it's 900 m^2 i believe

By Tongos (Tongos) on Saturday, July 10, 2004 - 07:43 pm: Edit |

what does that mean?

By Tongos (Tongos) on Saturday, July 10, 2004 - 07:44 pm: Edit |

oh, sorry, your talking about the the first question question

By Devious (Devious) on Saturday, July 10, 2004 - 07:49 pm: Edit |

900? I see where I messed up. Anyway, here's what I've gotten so far:

It takes Holmes N hours to cover the wall with one coat of paint, thus it takes him 2N to cover the wall with two coats. Let's call the total area of the wall A m^2. We gather from this that Holmes paints at A/N m^2/h. Watson starts at (N-2) hours, and paints at 25 m^2/h.

They both paint for a period of time (V hours), during which Watson paints X m^2 and Holmes paints Y m^2, covering the wall with one coat of paint. Watson leaves at (N-2+V) hours, and Holmes starts painting again. Holmes covers 2Y m^2 during (2N-1) hours.

Something note-worthy is that they both cover A m^2 in (N-V) hours.

Here are the equations I've got:

[1] A = Y + X

[2] A/N = 2Y/(2N-1)

[3] 25 = X/(N-2+V)

[4] A/N = Y/(N-V)

[5] A = 25N - 50N^2 (got this one from manipulating the info from the last line I typed out)

Solving [2] and [4] simultaneously gives V=0.5 hours. Using that value and [3], we get X=25N-37.5.

Now all I need is another equation in Y and I'd be able to solve this, but I can't seem to get one. Oh well, here's the final equation I managed to get:

50N^2 + Y - 37.5 = 0

I hope some of this is right...

By Apocalypse_Now (Apocalypse_Now) on Saturday, July 10, 2004 - 07:58 pm: Edit |

Devious, that might work out if you slug your way through it. However, there is a much simpler, more elegant solution.

By Devious (Devious) on Saturday, July 10, 2004 - 08:11 pm: Edit |

Probably, heh. Care to share it?

By Feuler (Feuler) on Sunday, July 11, 2004 - 12:21 am: Edit |

Is that the exact wording of the question? It seems a little ambiguous. What I've got:

Holmes spent exactly one hour less painting than he does normally, and painted the normal amount minus twice the area Watson painted. Thus Watson painted the area of wall that normally would take Holmes 30 minutes to do.

Now, when you say Watson paints 600 m^2/day, do you mean he painted 600 m^2 on that day, or he would paint 600 m^2 if he worked as long a day as Holmes normally works, or what?

By Devious (Devious) on Sunday, July 11, 2004 - 01:14 am: Edit |

Feuler, it means that Watson can paint 600 m^2 if he works an entire day.

By Feuler (Feuler) on Sunday, July 11, 2004 - 05:16 am: Edit |

The problem is ambiguous as written.

Let t be the amount of time both Holmes and Watson work, x be the area of the wall in m^2, and h and w be the amount of wall, in m^2, Holmes and Watson can paint in one hour. As I demonstrated above, Watson does in t hours what Holmes can do in 1/2 hours, wt=h/2, so h/w=2t. However, h/w is also x/300, so x/300=2t, and x=600t.

Now, suppose d is the number of hours in a "work day" which is so conveniently left ambiguous in the problem statement. Since Holmes can paint the wall in d/2 hours, hd/2=x. Also, 2h+t(h+w)=x, and from above, wt=h/2, so x=h(5/2+t), so hd/2=h(5/2+t), and d=5+2t, and t=(d-5)/2, so x=600t=300d-1500.

If you make the assumption that a work day is 8 hours, you get 900 m^2, but this is not given. Any value of d that gives a positive t will lead to perfectly valid values for all other variables and a different value for the area of the wall.

By Apocalypse_Now (Apocalypse_Now) on Sunday, July 11, 2004 - 10:20 am: Edit |

Feuler, you are making the problem unnecessarily complicated. It really does not require anything more than gradeschool arithmetic, so you don't need all those equations and algebra tricks.

I can assure you that the problem is NOT ambiguous as written. You did not include all of the given information in your equations, and that is why it SEEMS ambiguous. Specifically, the part about Watson starting two hours before lunch. (If i had not given you this information then the problem would, in fact, be ambiguous)

Your first idea (about Holmes painting Watson's section in half hour) is correct and is a key step in solving it. So take another stab at it if you like; I'll post the solution later today or sometime tommorrow.

By Tongos (Tongos) on Sunday, July 11, 2004 - 02:19 pm: Edit |

why would you spend so much time on just an easy problem? i really don't understand.

By Feuler (Feuler) on Sunday, July 11, 2004 - 03:24 pm: Edit |

Oh, I see. I read it as Watson joining Holmes after two hours, rather than two hours before lunch, so I not only did not consider given information, I assumed information that was not given (that happened to be true anyway). The question does indeed take about 5 seconds as written.

Boy, do I feel stupid.

By Loop123 (Loop123) on Monday, July 12, 2004 - 03:20 pm: Edit |

Could someone explain the concise solution to the first question?

I, like Feuler, found the question ambiguous. However, I'm sure that I've gone wrong due to popular opinion on this thread. I too realized that Holmes could complete Watson's work in 1/2 hour but, I'm unable to proceed. Thanks guys.

By Feuler (Feuler) on Monday, July 12, 2004 - 04:55 pm: Edit |

Since Watson did the equivalent of Holmes working for 30 minutes, they finished 30 minutes before lunch, thus Watson worked for 90 minutes and therefore works at 1/3 the speed of Holmes, so Holmes can paint 600*3=1800 m^2 in a day, making each coat 900 m^2.

By Loop123 (Loop123) on Monday, July 12, 2004 - 05:06 pm: Edit |

Wow I feel like an idiot. That's just one of those problems that does that to you! Still, it seems as though it should be specified that the 600m^2 that Watson can do in one "day" is indeed in the same timeframe as Holmes.

By Apocalypse_Now (Apocalypse_Now) on Monday, July 12, 2004 - 09:54 pm: Edit |

It's interesting that you mention that loop; This problem was actually disqualified later on because contestants had similar complaints about the assumptions one needed to make, due to the funky wording. Nonetheless, i thought it was a pretty good problem.

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