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    'A gardener has been asked to lay a patio 3m longer than its width. Each slab is a square of side 0.5m and costs £4. The gardener has been asked to spend more than £220 but less than £448. Find the two possible amounts he could spend.'

    What I've done: Each slab costs £4. Divided each amount by 4 to give 55 slabs and 112 slabs. Each slab is 0.25 m sqd. So 55 slabs is 13.75 m sqd and 112 slabs is 28 m sqd. x(x+3) = x sqd + 3x.
    x sqd + 3x > 13.75
    x sqd + 3x < 28
    But they don't factorise, and formula gives long decimals, so don't think this is the route to solving...
    Would really appreciate help here.
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    (Original post by Oldie)
    'A gardener has been asked to lay a patio 3m longer than its width. Each slab is a square of side 0.5m and costs £4. The gardener has been asked to spend more than £220 but less than £448. Find the two possible amounts he could spend.'

    What I've done: Each slab costs £4. Divided each amount by 4 to give 55 slabs and 112 slabs. Each slab is 0.25 m sqd. So 55 slabs is 13.75 m sqd and 112 slabs is 28 m sqd. x(x+3) = x sqd + 3x.
    x sqd + 3x > 13.75
    x sqd + 3x < 28
    But they don't factorise, and formula gives long decimals, so don't think this is the route to solving...
    Would really appreciate help here.
    Try doing it in terms of the number of slabs along the width instead of the width. If n is the number of slabs, then n(n+6) are used and the cost is 4n(n+6). You'll find the numbers drop out much nicer this way.
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    (Original post by Gregorius)
    Try doing it in terms of the number of slabs along the width instead of the width. If n is the number of slabs, then n(n+6) are used and the cost is 4n(n+6). You'll find the numbers drop out much nicer this way.
    Thanks Gregorius. In fact, I'd tried that as well, and did find that they factorised nicely.
    x sqd + 6x > 55
    x sqd + 6x < 112
    I get x values of -11 and 5, and -14 and 8.
    I rejected the negative values as that would give a negative number of slabs.
    So we're left with x= 5 slabs, or 8 slabs.
    But if I'm right up to that point, not sure where to go next.
    If x = 5 slabs, then if area = x sqd + 6x, then one area would be 55 slabs, and 55x4 = £220, which is going round in circles.
    I've spent hours on this, approaching it in different ways, and know there's something daft I'm doing/not doing - what is it?
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    (Original post by Oldie)
    Thanks Gregorius. In fact, I'd tried that as well, and did find that they factorised nicely.
    x sqd + 6x > 55
    x sqd + 6x < 112
    I get x values of -11 and 5, and -14 and 8.
    I rejected the negative values as that would give a negative number of slabs.
    OK so far...

    So we're left with x= 5 slabs, or 8 slabs.
    [/QUOTE]

    Whoa! we are told that the gardener must spend more than £200 and less than £448. So we have

    n^2 + 6n - 55 > 0

    If we solve n^2 + 6n - -55 == 0 then we get n = 5 (discarding the negative root). Therefore n must be larger than 5 (look at the graph of the quadratic, if this is not obvious). Similarly, from the other equation, n must be less than 8.
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    (Original post by Gregorius)
    OK so far...

    So we're left with x= 5 slabs, or 8 slabs.
    Whoa! we are told that the gardener must spend more than £200 and less than £448. So we have

    n^2 + 6n - 55 > 0

    If we solve n^2 + 6n - -55 == 0 then we get n = 5 (discarding the negative root). Therefore n must be larger than 5 (look at the graph of the quadratic, if this is not obvious). Similarly, from the other equation, n must be less than 8.[/QUOTE]

    Thank you for being patient with me, Gregorius, but yes I'd even got to that point as well (in one of the many different approaches I've used for this!). I'd just used 'equals' here as an intermediate step - as so used to using it when factorising - and realise this is inequalities. (I do use graphs for looking at where y is greater than/less than,etc).
    I can see that n must be larger than 5 and less than 8. BUT...how do I then get to the two values of n that will give me the two different expenditures the question asks for? The answers give two exact amounts rather than inequalities - £288 and £364.
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    (Original post by Oldie)
    I can see that n must be larger than 5 and less than 8. BUT...how do I then get to the two values of n that will give me the two different expenditures the question asks for? The answers give two exact amounts rather than inequalities - £288 and £364.
    n is an integer larger than 5 and less than 8 - therefore n = 6 or n = 7. Substitute back into 4n(n+6) and you get £288 & £364.
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    (Original post by Gregorius)
    n is an integer larger than 5 and less than 8 - therefore n = 6 or n = 7. Substitute back into 4n(n+6) and you get £288 & £364.
    Of course!! How daft of me. So sorry. I've done that so much in 'easier' questions - can't think how I didn't see that that's what I had to do next.

    Thank you so so much Gregorious.
 
 
 
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