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Women, sacks and cats algebra

  1. Mar 24, 2016 #1
    1. The problem statement, all variables and given/known data
    In West Cornwall, I met w women each carrying s sacks, each containing c cats. However, a third of the women then lost half their sacks, and the remaining women lost half their cats from a third of their sacks. How many cats remained, it terms of w, s and c?

    2. The attempt at a solution

    w.s.c - (1/3w.1/2s) - (2/3w.1/2c.1/3s)

    . = multiplication?
  2. jcsd
  3. Mar 24, 2016 #2


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    Almost. Consider the case w=3, s=2, c=100, and ignore the cat losses for now (so just 1/3 of the women loses 1/2 their sacks). Can w.s.c - (1/3w.1/2s) be true?

    * is the usual (ascii) multiplication sign.
  4. Mar 24, 2016 #3
    Hallsoflvy why is (w/3)(s/2) sacks so wsc/6 cats. I do not understand this part...

    (w/3)(s/2) = 2w3s/6
  5. Mar 24, 2016 #4

    Ray Vickson

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    The correct result is as HallsofIvy says:
    [tex] \frac{w}{3} \frac{s}{2} = \frac{ws}{6} [/tex]
    You multiply the numerators together to get the new numerator, and multiply the denominators together to get the new denominator. Why would you think otherwise?
    Last edited: Mar 24, 2016
  6. Mar 24, 2016 #5
    (w/3)(s/2) = (ws/6)


    (2/3w)(s/3)(c/2) = wsc/9
  7. Mar 24, 2016 #6
    How did Hallsoflvy get to 11 wsc/6

    I get (w/3)(s/2)(c/1) = wsc/6
    (2/3w)(s/3)(c/2) = wsc/9

    wsc/6 + wsc/9 = 5wsc/18
    Last edited: Mar 24, 2016
  8. Mar 25, 2016 #7


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    Subtract that from the total, and you get the same result as me.
  9. Mar 25, 2016 #8
    swc - 5wsc/18 = 13wsc/18 (is this correct?)
  10. Mar 25, 2016 #9


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    You need to get into the habit of testing answers with some real numbers. This sort of problem should be easy to check. Try, for example:

    w = 3, s = 6, c = 10

    It's even a good idea to work out the problem with these actual numbers first, to establish the pattern of the solution. Then, extend your specific solution using 3, 6, 10 to a general one using w, s, c.
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