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    (Original post by DingDong!)
    For 5 part 3, should be a less than sign, not less than or equal to surely? The question how much shorter should it be for it to be the SHORTEST as in the only short one, that's how I interpreted it.
    I interpreted as needing less that equals to because it asks for it to be a shortest path not the shortest path. They done a dirty one like this in a past paper which I got wrong, so I'm hoping for better fortune this time round.
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    (Original post by metaltron)
    I interpreted as needing less that equals to because it asks for it to be a shortest path not the shortest path. They done a dirty one like this in a past paper which I got wrong, so I'm hoping for better fortune this time round.
    It says for "it to become part of a shortest path" , the word "it" referring to CE. :confused:
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    (Original post by DingDong!)
    It says for "it to become part of a shortest path" , the word "it" referring to CE. :confused:
    I still interpret that as needing a less than or equals sign. I find this really ambiguous but certainly to me it seems like a less than or equals sign is required because 'a' shortest path suggest that there can be two shortest paths.
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    (Original post by DingDong!)
    It says for "it to become part of a shortest path" , the word "it" referring to CE. :confused:
    Also refer to question 5ii) on the June 11 paper.
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    (Original post by alow)
    Just put it in my pocket and walked out
    Well done, sir...well done :yep:
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    (Original post by metaltron)
    work so far, will be continued if paper re-uploaded:

    unofficial mark scheme

    1i)  24 \ 57 \ 9 \ 31 \ 16 \ 4

     24 \ 9 \ 57 \ 31 \ 16 \ 4

     24 \ 9 \ 31 \ 57 \ 15 \ 4

     24 \ 9 \ 31 \ 15 \ 57 \ 4

     24 \ 9 \ 31 \ 15 \ 4 \ 57

    circles where necessary to indicate comparisons. (2 marks)

    ii)  pass \ 2 \ : \ 9 \ 24 \ 15 \ 4 \ 31 \ 57

     pass \ 3 \ : \ 9 \ 15 \ 4 \ 24 \ 31 \ 57

    (2 marks)

    iii) pass 1 - 4 swaps, pass 2 - 3 swaps, pass 3 - 2 swaps , pass 4 - 1 swap, pass 5 -1 swap (2 marks)

    2i)a) square plus a loop at one vertex for example (1 mark)

    b) max vertex order for simple graph with 4 vertices is 3, but is eulerian so order has to be 2 (as 0 means one vertex is not connected). Hence total order = 4x2 = 8. 8/2 = 4, so 4 arcs, not 5 arcs. (2 marks)

    ii)a) total order = 2 x no. Of arcs = 2 x 10 = 20 (1 mark)

    b) max - 6, min - 2, and correct graph drawn (3 marks)

    c) two vertices of order 4, six of order 2. (1 mark)

    3i) correct trace through the algorithm, yielding outputs 6 and 90. (6 marks)

    ii) after you fill in the space in the answer booklet, you will get f d and e the same as in part i). From this it follows that g = 18 /6 = 3 and m = 3 x 30 = 90. The same as in part i). (2 marks)

    iii) f = 4 and m = 24. (2 marks)

    4i)  a - b -c-d-e-f-g
    weight = 51 km
    82km - 51km = 31km

    so:  a-b-c-d-e-f-g-f-b-a

    with weight 82km. (4 marks)

    ii) 51 - 5 - 7 + 8 = 47km.
    Go back through d so:

     a-b-c-e-f-g-d-a (2 marks)

    iii) nn starts  a - b-c-d-e-f-g

    stalls as can't return to a.

     b - a - c - d - e -f-g-b

    weight = 76km.

     b - a - c - d - e - g - f - b

    weight = 69km. (6 marks)

    5i)  a - b - f - g

    weight = 31km. (5 marks)

    ii) 25 hours (2 marks).

    Iii)  11 + 15 + ce \leq 31

     ce \leq 5

    8 - 5 = 3km (2 marks)

    iv) odd nodes a,b,c and f.
    All repeated arc lengths are 32km.
    32 + 224 - 11 - 8 = 237km. (6 marks)

    6i) attached using wolfram (4 marks)

    ii)  (6,3) ,  \ (4,0), \  (3,0) \ and \  (4,4.5)

    at (6,3) p = 54
    at (4,0) p = 20
    at (0,3) p = 24
    at (4,4.5) p = 56.

    So at optimal point x = 4, y = 4.5, p = 56. (4 marks)

    iii) when x = 0, max y = 3, p = 24.
    When x = 1, max y = 3, p = 29
    when x = 2, max y = 3, p =34
    when x = 3, max y = 4, p = 47
    when x = 4, max y = 4, p = 52
    when x = 5, max y = 3, p = 49
    when x = 6, max y = 3, p = 54.

    So optimal point is when x = 6, y = 3 and p = 54. (4 marks)

    iv) simplex is designed for inequalities with a less than or equals sign. (1 mark)

    v) correct simplex tableau (2 marks)

    vi) first pivot 8 in y-column.
    Show how rows calculated.
    P = 24, x = 0, y = 3.
    Pivot is 4.5 in x-column.
    Verify that optimum point is when x =4, y = 4.5 and p = 56.

    thank you!! Made me feel a lot better :d
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    (Original post by Zakee)
    Same, I am really annoyed. Honestly! I mean, I missed out the second iteration on the last question, I missed out 3 marks for one of the explain questions, I missed out drawing a cycle as I thought I'd come back to it, but I never did. Honestly, I thought I could get back to them. D1 should be 2 hours long. not one hour and a half. It's absurd because it's not truly testing mathematical ability by trying to squeeze in so much. I'm sure I could've answered it if I was actually able to pace myself without being frenzied from trying to stay within the damn time limits. I mean why do they have to make questions so long-winded.

    I really needed a good mark on this, and I think I've blown it majorly.
    Yes, I know what you mean! To be honest the paper was fairly simple, nothing that was beyond our ability, but the timing caused the problem, not the actual maths. I mean I went at quite a fast pace, but still left out nearly the whole first page. I promised myself I'd come back to it, but just did not have the time to in the end. I'm sure you will still get a good grade! The only reason I'm not too worried about it is because the grade boundaries are always ridiculously low so hopefully our marks gets pushed up
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    thought the final pass of bubble sort should have 0 swap?
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    (Original post by missypink<3)
    Simplex should have given you (4,4.5) which corresponded to P=56. This was the same as in 6(i) and (ii).
    Yeah, I must've made a mistake somewhere. Was running out of time, so I had to rush through question 6 :l Hopefully I get some method marks for evaluating the correct coordinates.

    (Original post by DingDong!)
    For 5 part 3, should be a less than sign, not less than or equal to surely? The question how much shorter should it be for it to be the SHORTEST as in the only short one, that's how I interpreted it.
    I interpreted it the same way lol. But I think I remember doing a past paper where there was a similar question. I think for it to be the shortest route, it has to match the value of the current shortest route. Not quite sure.
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    Hi everyone. I also found the timing impossible for that exam. Q3 where you had to carry out the algorithm took forever. And how did everyone do Q4i/ii where you had to get 81 and 82? I knew you had to take away 51, but then what?...
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    (Original post by jenq)
    thought the final pass of bubble sort should have 0 swap?
    Pass 4: 9 4 15 24 ... 1 swap

    Pass 5: 4 9 15 24 ... 1 swap
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    For all the questions 4 and 5 I didn't put km i just wrote the numbers. Will i lose marks for that?
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    (Original post by Naik100)
    Hi everyone. I also found the timing impossible for that exam. Q3 where you had to carry out the algorithm took forever. And how did everyone do Q4i/ii where you had to get 81 and 82? I knew you had to take away 51, but then what?...
    The way I did it was to find a route to get back to A from G with a distance of 31km (82-51) for part 4(i) which happened to be GFBA and got the cycle: ABCDEFGFBA = 82km.

    Similarly in part (ii), you do the same thing. But at first, you need essentially delete the arcs from D in your minimum spanning tree and and connect it again using CE since it wanted you to find the minimum spanning tree for ABCEFG. Once you did that, you get the weight to be 47km for the tree. Again, 81-47 = 34km. Therefore, find a route getting back to A from the end node of your tree that also visits the farmhouse at D since it wanted you to visit all 7 nodes exactly once before you return to A. I ended up with the route ABCEFGDA = 81km.

    Although I got there in the end, who knows if I actually got the method correct lol. Judging by my mistakes in the paper, don't take this as correct!
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    (Original post by JimmyA*)
    For all the questions 4 and 5 I didn't put km i just wrote the numbers. Will i lose marks for that?
    I reckon you should be ok. I remember mark schemes in previous papers saying "condone X" without units.
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    (Original post by metaltron)
    Pass 4: 9 4 15 24 ... 1 swap

    Pass 5: 4 9 15 24 ... 1 swap
    i must have misread the question and thought it meant the fifth pass was the final pass

    will you get follow through marks for 6ii and 6iii if you draw the graph for 6i wrong? if not, i would have lost 12 marks on that question.... great
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    (Original post by jenq)
    i must have misread the question and thought it meant the fifth pass was the final pass

    will you get follow through marks for 6ii and 6iii if you draw the graph for 6i wrong? if not, i would have lost 12 marks on that question.... great
    Pretty sure you will get at least a few follow through marks, but you will lose all of the accuracy marks.
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    48/72 for 80 UMS I think
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    (Original post by missypink<3)
    48/72 for 80 UMS I think
    Why do you say that? Is that a maths teacher who's given that estimate to you? If you're right, which I hope you are, I should scrape an A, but I somehow doubt boundaries will be that generous! x
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    I think it wud be about 54/55 for an A
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    (Original post by Arsenal121)
    I think it wud be about 54/55 for an A
    It will be lower than that as that was January's and that was an easier paper and had resitter's and further maths people taking it. This has alot of AS students taking it and was a lot harder paper due to how long it was meaning lot's of people missed out questions completely so it will probably be around 48-50.
 
 
 
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