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    I'm just working through the June 2007 OCR D1 paper
    http://www.ocr.org.uk/download/pp_07..._jun_l_gce.pdf

    but I can't work out why for question 2iv) the minimum feasible value of f is 20!

    Please help!
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    (Original post by xsweetcherrypiex)
    I'm just working through the June 2007 OCR D1 paper
    http://www.ocr.org.uk/download/pp_07..._jun_l_gce.pdf

    but I can't work out why for question 2iv) the minimum feasible value of f is 20!

    Please help!
    Although 20 is correct, I can't see how you can just "write down" as the question asks, without doing some calculating.

    From the question, 4f is >= 4d (since f>=d), which is in turn is >= g, which is >= 40.

    if f took it's minimum value you'd have 4f = 4d = g, and since you know f+g+d = 120 it follows by substituting into that second equation that f=20.

    It's possible there is another way of seeing it that I've overlooked.
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    (Original post by ghostwalker)
    Although 20 is correct, I can't see how you can just "write down" as the question asks, without doing some calculating.

    From the question, 4f is >= 4d (since f>=d), which is in turn is >= g, which is >= 40.

    if f took it's minimum value you'd have 4f = 4d = g, and since you know f+g+d = 120 it follows by substituting into that second equation that f=20.

    It's possible there is another way of seeing it that I've overlooked.
    Oh wow. I would never have thought to calculate the value of f like that! That's really clever!!! Thank you so much - you're really good at maths - you helped me with that bearings question as well!!

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    (Original post by xsweetcherrypiex)
    Oh wow. I would never have thought to calculate the value of f like that!
    It's not obvious that that's what they're looking for, IMHO.

    PS: You have to rep 40 other people, before you can rep the same person again, so don't worry about it.
 
 
 
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