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    So, I used the Search function and found some helpful threads about rough UMS - raw mark conversion (some users were kind enough to post spreadsheets for estimate calculations etc), but I was wondering, how is the 'cap' mark decided, where you get full UMS?

    I read it was double the mark between grade boundaries; but wouldn't that mean in some cases you would need full raw marks to get full UMS?

    For example, if the max raw mark is 90, and the boundaries are

    A = 72
    B = 63
    C = 54 etc, the difference between each grade is 9, so (9 x 2) + 72 = 90?


    The reason I ask is because I'm resitting an A2 English Language module (ENB6, Language Development) which is out of 70 raw marks, and worth 120 UMS. I need 101 UMS points to get an A overall...would do that equate to in raw marks? 60/70? :confused:
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    You double the difference between the A and B boundary to get the cap-mark.

    And yes, it's not uncommon for full UMS to only be given for full raw marks; that occurs in many relatively easy papers like the chemistry one I just sat.
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    (Original post by DJkG.1)
    You double the difference between the A and B boundary to get the cap-mark.

    And yes, it's not uncommon for full UMS to only be given for full raw marks; that occurs in many relatively easy papers like the chemistry one I just sat.

    Right, I see, thankyou
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    But if the cap-mark is higher than the normal full marks for the paper, then normal full marks becomes 100% UMS, which effectively changes the rate of exchange at the A boundary.
 
 
 

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