hope you can help with this question, its a gcse paper (16 year olds) and I think i may have either made a mistake, been doing it the wrong way, or simply found the answer but me being paranoid thinks the working is too long and answer not simple enough(adsbygoogle = window.adsbygoogle || []).push({});

thnx for the help

1. The problem statement, all variables and given/known data

The length of a side of the square base of a pyramid isxmetres.

The vertical height isymetres.

Both measurements are correct to the nearest metre.

question) Find an expression for the difference between the upper bound and the lower bound of the volume of the pyramid.

Give your answer in its simplest form.

2. Relevant equations

volume of pyramid = (1/3)base x height

3. The attempt at a solution

my notation: (lb) means lower bound of, so (lb)y would mean the lower bound of y, and so on.

p.s. I know some (well alot) are redundant but I stuck them in to make the workings easier to read. Also, i should really learn Latex but I havnt got round to yet so please try to read my non-latex stuck lol.

(lb)x = x - (1/2) x (1/10)x

(lb)x = x - (1/20)x

(lb)x = (19/20)x

therefore: (ub)x = (21/20)x

(lb)y = (19/20)y

(ub)y = (21/20)y

(lb)base = (lb)x^2

(lb)base = (19x/20)^2

therefore: (ub)base = (21x/20)^2

(lb)volume = (1/3) x (lb)b x (lb)h

v = (1/3) x (19x/20)^2 x (19y/20)

(ub)volume = (1/3) x (21x/20)^2 x (21y/20)

difference = (ub)vol - (lb)vol

d = (1/3) x (21x/20)^2 x (21y/20) - (1/3) x (19x/20)^2 x (19y/20)

d = (21y x (21x/20)^2)/60 - (19y x (19x/20)^2) /60

d = (21y x (21x/20)^2 - 19y x (19x/20)^2) /60

is this the simplest form? seem rather complex to me.

I tried factorising the brackets and doing all sorts but I couldn't get it any cleaner than this.

thnx for your help

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# Quite a long 'bounds' question

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