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    (Original post by Nator)
    I got A.
    spot on

    could you please explain it to me?

    are we trying to make H the subject of the formulae, but how do we do it in terms of R?
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    Your first stage is correct, here is how I done it:

    2Pir^2 + 2pirh = pir^2h
    Now you have common terms, so they can be cancelled out to get:
    2r + 2h = rh
    Then it's just standard factorising:
    2r = rh - 2h
    2r = h(r - 2)
    h = 2r / r-2

    Enjoy
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    (Original post by Nator)
    Your first stage is correct, here is how I done it:

    2Pir^2 + 2pirh = pir^2h
    Now you have common terms, so they can be cancelled out to get:
    2r + 2h = rh
    Then it's just standard factorising:
    2r = rh - 2h
    2r = h(r - 2)
    h = 2r / r-2

    Enjoy
    Amazingly brillaint. how did you spot that?
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    (Original post by section 2 prep)
    spot on

    could you please explain it to me?

    are we trying to make H the subject of the formulae, but how do we do it in terms of R?
    Explained under your response, basically if you know what you're doing you can do such questions in under 30 secs which keeps you good for time
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    (Original post by section 2 prep)
    Amazingly brillaint. how did you spot that?
    Because you want to get rid of as much as possible to get the answer easily, also you can judge by the simplicity of the options, and you got constants in there like pi which was an obvious hint to me to cancel it out, alongside r.
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    (Original post by Nator)
    Because you want to get rid of as much as possible to get the answer easily, also you can judge by the simplicity of the options, and you got constants in there like pi which was an obvious hint to me to cancel it out, alongside r.
    http://www.admissionstests.cambridge...ion_2_2010.pdf

    Try question 1
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    (Original post by Nator)
    Your first stage is correct, here is how I done it:

    2Pir^2 + 2pirh = pir^2h
    Now you have common terms, so they can be cancelled out to get:
    2r + 2h = rh
    Then it's just standard factorising:
    2r = rh - 2h
    2r = h(r - 2)
    h = 2r / r-2

    Enjoy
    Damn, internet cut out and I got beaten to it
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    E?
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    (Original post by Nator)
    Because you want to get rid of as much as possible to get the answer easily, also you can judge by the simplicity of the options, and you got constants in there like pi which was an obvious hint to me to cancel it out, alongside r.
    http://www.admissionstests.cambridge...ion_2_2010.pdf

    i normally am ok with chemistry/numbers but question 6 has really confused me
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    (Original post by Normandy114)
    E?
    (Original post by Normandy114)
    Damn, internet cut out and I got beaten to it
    nator is by far better than me, currently.

    you are obviously better than me

    i have never learnt q1 at gcse, can you tell me where to learn the knowledge?^_^

    again, spot on it's E

    try question 6 please do explain!
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    (Original post by section 2 prep)
    http://www.admissionstests.cambridge...ion_2_2010.pdf

    i normally am ok with chemistry/numbers but question 6 has really confused me
    Got D.
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    By the balanced equation on 2, you can see there is a 1:2 ratio between C and CO, so the mass of CO = 2 x (12 + 16) = 56g.
    Now you use this mass value on equation 3, so convert it to moles by doing mass/Mr, and you get 56/28 = 2 moles, and since there ratio is 3:3, i.e. the same between CO and CO2, you stick to this value of 2 moles, and multiply it by the Mr of CO2, therefore the mass of CO2 = 2 x (32 + 12) = 88g
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    (Original post by section 2 prep)
    nator is by far better than me, currently.

    you are obviously better than me

    i have never learnt q1 at gcse, can you tell me where to learn the knowledge?^_^

    again, spot on it's E

    try question 6 please do explain!
    For one I knew that hypothalamus deals with heat (only know this due to AS psychology), that hairs relax, and that capillaries do not move.
    For 6 I get D?
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    (Original post by section 2 prep)
    there's the official one, and the isc medical one.

    not sure about spec, i need to find like a site that goes through everything in section 2 we must know...

    i also need to brush my maths skills..
    Could I get links to those?

    (Original post by section 2 prep)
    i've read you've bought the genomics book. Are you planning on quoting these books in your ps?
    Probably not.


    (Original post by Nator)
    There is the big Kaplan one which goes over the old spec, however no books which go over the new current spec, just past papers and the official BMAT book should be enough prep.
    Could you give me an online link to it? So that I can order it.
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    (Original post by JChoudhry)
    Could I get links to those?



    Probably not.




    Could you give me an online link to it? So that I can order it.
    Sure, here's a link:http://www.amazon.co.uk/exec/obidos/...SIN=190581206X
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    (Original post by Normandy114)
    For one I knew that hypothalamus deals with heat (only know this due to AS psychology), that hairs relax, and that capillaries do not move.
    For 6 I get D?
    so is it an unfair question?:O

    i'll digg it somewhere!

    could you please explain D to me. I.E the logic behind it
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    (Original post by section 2 prep)
    so is it an unfair question?:O

    i'll digg it somewhere!

    could you please explain D to me. I.E the logic behind it
    I explained it above
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    (Original post by JChoudhry)
    Could you give me an online link to it? So that I can order it.
    You know the issue with the prices of these right? :rolleyes:

    http://cgi.ebay.co.uk/Kaplan-BMAT-Bi...#ht_500wt_1156

    Most people say it's not worth it though.
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    (Original post by Nator)
    By the balanced equation on 2, you can see there is a 1:2 ratio between C and CO, so the mass of CO = 2 x (12 + 16) = 56g.
    Now you use this mass value on equation 3, so convert it to moles by doing mass/Mr, and you get 56/28 = 2 moles, and since there ratio is 3:3, i.e. the same between CO and CO2, you stick to this value of 2 moles, and multiply it by the Mr of CO2, therefore the mass of CO2 = 2 x (32 + 12) = 88g
    absolutley astounding.

    how long did it take to work that out - be honest?

    ALL of the CO, is used in step 3. so 56Grams of it

    Very good logic tbh, very good

    so tell me, if it was a 3:2 ratio...how would you go about that?

    56grams = 28 2

    3:2 ratio

    (12+32) x 2/1.5 = mass of Co2?

    see, when i read it, i completely got blitzed. I did not use logic - but even so, it is very good that you saw that...good find imo
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    (Original post by Nator)
    You know the issue with the prices of these right? :rolleyes:

    http://cgi.ebay.co.uk/Kaplan-BMAT-Bi...#ht_500wt_1156

    Most people say it's not worth it though.
    http://www.admissionstests.cambridge...ion_2_2010.pdf

    What about number 12? ^_^
 
 
 
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