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    what a piece of cake. i reckon i got an A or a B,alot easier then non calc. How the hell can you not know how to do simultanous equations lol

    for the sim question for those who got stuck.. You have to equal the Y+ to the Y- ie 3y and -3y as an example so you do that by multiplying one line of the equation so the Y becomes a non value..

    Then your left with X's and numbers and its simply a case of drag and drop, simple, now i can enjoy summer :-) i dont believe i failed and hopefully will redeem a B.


    For the hexagon that was tricky, i knew each triangle should be 60 degrees. no matter what, you should work it out vertically first and use the inside 2cm as a radius. Pythagurus and pie round it, i can't remember the figure i got but basically you got 6 triangles.

    Anyway, the common mistake people would of made was to use 1/3 and not 1/2. when you pi the radius squared you divide the total by 6 and you basically got one area of one triangle. The shape is NOT a cone. You add the total triangles together to calculate total area :-)
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    paper 6? That was paper 19 for us-so that's the discrpancy for the sim.eq. i hope u lot realise tht the question said 'by eliminating y' luckily i did tht anyway but a friend bought it to my attention. Oh well still gt the answer right! :-)
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    What surds one!!!!!!!! lool
    Omg that was so hard I did NOT have time every question took ages
    I'm sooo scared non calculator was eay easier!
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    can anyone guess that wot marks do we need to get A*
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    hey im jst chekin if my accnt works lol
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    If last year was 65% for A* and this year was easier id say around 75% but then agen i culd b wrong. UMS is confusing
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    yay it dus! now on to maths....yer jst did the edexcel higher....was nt that bad....pretty easy. i thought it wud b rly hard cus the other paper wasn't tht bad either lol. so far i no i got 2 questions wrong (but i got the right working)

    hmm chem 2moro...and dt

    o n ye i didn't have a surds/complete the sq question either!
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    i reckon abt 90% for an a* hehe (nt rele)
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    lol i bloody hope not :p:
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    for the lorry one i got 19 because wasnt it how many could he get on safely? so i did lowest bound for lorry capacity and highest for gold bars, or is it 23? lol
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    (Original post by magnum)
    i did intermediate...

    the question that i couldnt get my head around was the one about...

    sale price. 20% off.
    the sale price was £220

    what was the normal price?

    i thought its 20% of 220 then add it back on... but it went wrong
    Wasn't it 275 pounds?

    It was 20 per cent off, of which this person bought the item hence, in fact he bought the item for 80 per cent fo the original price.

    So 220/80 x 100 should overall come to 275 pounds.
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    I though the paper was quite good. A lot easier than previous years' papers.
    For the lorry one I got 17:

    1500kg mak to 2sf
    60kg bars to 1sf

    therefore Lower bound mak weight = 1150kg
    and Upper bound bar weight = 65kg

    so max amount = 1150/65
    = 17.69 so max 17bars can be carried.

    This should be right - but now I am worrying if the question said that the weights were correct to 1sf/2sf or 1dp/2dp....
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    can anyone remember the simultaineous equation???
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    (Original post by sanderslr)
    I though the paper was quite good. A lot easier than previous years' papers.
    For the lorry one I got 17:

    1500kg mak to 2sf
    60kg bars to 1sf

    therefore Lower bound mak weight = 1150kg
    and Upper bound bar weight = 65kg

    so max amount = 1150/65
    = 17.69 so max 17bars can be carried.

    This should be right - but now I am worrying if the question said that the weights were correct to 1sf/2sf or 1dp/2dp....
    you want the greatest number of bars so you divide the upper bound by the lower bound so its 1250/55 which gives you 22.727272... so 22 bars can be carried.
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    what did every1 get for 19c, the last one on the hexagon pryamid question???
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    (Original post by .x.g.x.)
    you want the greatest number of bars so you divide the upper bound by the lower bound so its 1250/55 which gives you 22.727272... so 22 bars can be carried.
    yes xgx is rite
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    (Original post by .x.g.x.)
    you want the greatest number of bars so you divide the upper bound by the lower bound so its 1250/55 which gives you 22.727272... so 22 bars can be carried.
    you don't want the greatest number of bars. the question said the greatest number of bars which can be carried SAFELY so you had to do lower bound divided by upper bound which gives 17. you have to assume the lorry can carry the lowest weight and the bars are the highest weight to be positive that it is safe.

    but there was a lot of confusion over that question; it was really badly worded.
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    also for our paper - paper 6 non modular edexcel people - the one WITHOUT surds or complete the square

    Did our simultaneous equation say eliminate y?
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    (Original post by cinderella1288)
    you don't want the greatest number of bars. the question said the greatest number of bars which can be carried SAFELY so you had to do lower bound divided by upper bound which gives 17. you have to assume the lorry can carry the lowest weight and the bars are the highest weight to be positive that it is safe.

    but there was a lot of confusion over that question; it was really badly worded.
    no you do the greatest amount it could carry (upper bound), over the lightest wieght of the bar (lower bound)

    upper/lower, = 22bars when rounded, some people rounded to 23 which although theorectically is right, its wrong because 23 bars would make it too heavy.
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    Yes, I got the correct answer on the lorry as 17... Safely meant without risks. You need to presume the lorry is a minimum weight, and all the bars are maximum weight... equalling 17, as someone has done above.
 
 
 
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