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 September 15th, 2019, 09:04 AM #1 Newbie   Joined: Jan 2017 From: London Posts: 18 Thanks: 0 Algebraic Puzzle If x men working x hours a day for x days produce x articles (not necessarily a whole number of articles) find out how many articles are produced by y men working y hours a day for y days Can anyone help in how to approach this? September 15th, 2019, 09:30 AM #2 Senior Member   Joined: Aug 2012 Posts: 2,412 Thanks: 754 y, by symmetry. There's no difference between x and y in this question. Thanks from topsquark and mathsonlooker September 15th, 2019, 11:11 AM #3 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 Doesn't match the physics of the problem, though. $(x~men)(x~hr/dy)(x~dy)=x^3~man-hours$. Let $a = f(mh)$ be the number of articles produced. Theoretically, it should be proportional to man-hours. $a = r \cdot mh$ $x = r \cdot x^3$ suggests $r = x^{-2}$. Thus, $\displaystyle f(y^3) = \frac{y^3}{x^2}$. Example: x = 2, y = 5. 2 men*2 hours a day*2 days = 8 man-hours. 2 articles in this time means r = 0.25 articles/man-hour. (5 men)*(5 hr/dy)*(5 dy) = 125 man-hours (0.25 art/mh)*(125 mh) = 31.25 articles tl;dr: The problem doesn't make sense if postulated "for all x," implying x is a particular value that happens to make the statement true. Thanks from mathsonlooker September 15th, 2019, 11:43 AM   #4
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I've attached the scanned question q.8
Attached Images 20190915_201426.jpg (90.4 KB, 15 views) September 15th, 2019, 12:34 PM #5 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 I'm sticking with my answer. It fits the question and also the theme of polynomial expressions in the neighbouring questions. Also, pet peeve of mine, but, "scanned," and, "photographed," are not the same things. Scanned images tend not to have as much distortion, out-of-plane focus issues, and shadows of your cell phone on top of them. (Educating the next generation before they come to uni and have us ask for scanned documents in their projects and lab reports.) September 15th, 2019, 02:30 PM   #6
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Quote:
 Originally Posted by DarnItJimImAnEngineer Doesn't match the physics of the problem, though. $(x~men)(x~hr/dy)(x~dy)=x^3~man-hours$. Let $a = f(mh)$ be the number of articles produced. Theoretically, it should be proportional to man-hours. $a = r \cdot mh$ $x = r \cdot x^3$ suggests $r = x^{-2}$. Thus, $\displaystyle f(y^3) = \frac{y^3}{x^2}$. Example: x = 2, y = 5. 2 men*2 hours a day*2 days = 8 man-hours. 2 articles in this time means r = 0.25 articles/man-hour. (5 men)*(5 hr/dy)*(5 dy) = 125 man-hours (0.25 art/mh)*(125 mh) = 31.25 articles tl;dr: The problem doesn't make sense if postulated "for all x," implying x is a particular value that happens to make the statement true.
I was actually thinking along these lines too but then realised that it cannot work for any value of x. The relationship between x and y is r (0.25) really but this could be different for different x's September 15th, 2019, 03:19 PM #7 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timey-wimey stuff. My problem is that the problem starts with "x workers" producing "x articles" and that x is "not necessarily a whole number of articles" meaning that the number of workers may also not be a whole number! I'm thinking this problem is either seriously badly written or that there a massive typo. -Dan Thanks from skeeter and mathsonlooker September 16th, 2019, 05:11 AM #8 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 You've never heard of a family with 2.3 children? They grew up and started working. Thanks from topsquark September 16th, 2019, 01:14 PM   #9
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What's the latest consensus? I admit I could never get through a grade school math curriculum these days.

If we assume that they mean that this is true for one particular value of x, does the question make more sense? I couldn't quite follow the earlier exposition. I'd be interested in seeing a clear explanation of this problem. I agree that because they printed this in a text, they must mean for you to do some kind of calculation.

There is an old story.

Quote:
 Schoolmaster: Suppose x is the number of sheep in this problem Pupil: But, Sir, suppose x is not the number of sheep (I asked professor Wittgenstein if this is not a profound philosophical joke, and he said it was.)
https://philosophy.stackexchange.com...ein-joke/53439 September 17th, 2019, 12:33 AM   #10
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Quote:
 Originally Posted by topsquark My problem is that the problem starts with "x workers" producing "x articles" and that x is "not necessarily a whole number of articles" meaning that the number of workers may also not be a whole number! I'm thinking this problem is either seriously badly written or that there a massive typo. -Dan
Actually, the number of workers can be non-integer, even complex number, depending on the universe.   Last edited by skipjack; September 17th, 2019 at 08:05 AM. Tags algebraic, puzzle Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Kavesat Math 1 February 12th, 2017 06:25 AM burgess Algebra 1 June 30th, 2014 12:38 AM Chikis Elementary Math 8 March 2nd, 2014 10:07 AM empiricus Art 11 March 6th, 2011 07:42 AM

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