The Student Room Group

D1 Jan 2012 Edexcel Post-Exam Discussion - Solutions and paper in first post

Scroll to see replies

it should be 155.
they dont change the question in between
Original post by .Username.
Thats where you went wrong mate, the shortest remaining was 14 and not 18, thats why it is only a 1 increase



Original post by .Username.
Look at the route inspection diagram you went for the DE straight which was 18, but there was a shorter way of 14 by doing DC(6) then CE (8)


ahh dangaroonies, that makes sense, thanks guys! Looks like my A grade depends on boundaries and how generous the markers are feeling, I somehow mucked up prims/kruskals and got 149, so hoping I read my calculator wrong but my working was correct! And well, B) for postmans was only 2 marks so hopefully my explanation of repeating the canal + the shortest remaining odd vertices will save me and get 1 mark! Then I got 115 for Djikstras somehow for A) so not sure how many I'll lose out of the 7.
I may be clutching at straws but we'll see. And then you have the definitions... My regurgitations will either make perfect sense or get 0. I think if we assume 90% is an A, that's 68 which means we can drop 7 marks and get an A... Not confident.
Reply 82
Well, that's stupid and misleading. I know the questions have normally asked you to go back to the starting point, but this one didn't . I even started doing that as well. Well, wonderful. There goes my 90UMS I need.
Reply 83
Original post by Prolite
I got 115 tooo??

dang man

I am defo getting U in this paper


i know is annoying just hoping for low grade boundaires like in the low 60 regions
Reply 84
For the postie question I wrote two answers stating it would increase if Brett were to start and finish at the same node, but would decrease if he were to start and finish at two different nodes. i then did the calculations to prove this I think, stating that you would get 26 + 129 if start and finish at the same one, and something in the mid 130s if he were to start and finish at two different nodes.

A bipartite graph consists of two sets of nodes. Nodes in one set may only connect to nodes in the other set.

A matching is the pairing of some elements in one set to the pairing of some elements in the other set.

My dummy descriptions were along the lines of yours, and I said something about two edges cannot come from one event and then finish at the same event as the algorithm wouldn't work or something.

Would the answers above get me the marks? Thanks.
Well there's got to be hard and misleading questions otherwise everyone will get 85/90+
Reply 86
what do you think the grade boundaries are?
Reply 87
Original post by Mike-nificent
Does anyone know if I'll lose marks by stating rejections in Prims (I got the actual answer but I stated nodes that I rejected) :s-smilie:


I don't think you will, i did the same thing in a sort of better safe than sorry way, as long as you have the correct answer i don't think it should matter.
ohhhh come on sia. that is not a good way of giving hope
Reply 89
Lovely stuff, should be 75/75. Wrote about starting and finishing at seperate nodes but then crossed it out and put it incrases by 1 mile :smile: rest was pretty easy, nice paper
Im not sure about the postie because you are kind of covering two answers, so im not sure how theyll mark that, (its like circling 3 out of 4 answers when they ask you to circle 2), the definitions seem alright and dummy is fine i think
Reply 91
I don't know what was wrong with me but I just could not draw the simplest inequality.. :mad: It kept going off with a very steep gradient, so I knew it was wrong but I tried plugging in values so many times and it wouldn't change. I must have been making the same mistake over and over. Quite embarassing really. :frown:

But yeah, how many marks would I lose if I managed to draw one inequality right with the correct part shaded, but not the other? This means that my feasible region was wrong, and the profit I got was also wrong. However, I did manage to draw the profit line correctly.

I'm glad that semi-eularian thing is wrong. I did what your markscheme did. :colondollar:

Also, I didn't say that it had increased by 1 mile exactly. I said that one was greater than the other, so it has been increased. That's fine right?
(edited 12 years ago)
Reply 92
I can only get one mark for that question now, for identifying the odd nodes. I'm very annoyed. I don't want to have to retake this paper in the summer because of one question that, I thought, was intentionally changed a little to make you think. It asked us to use the route inspection problem, yeah that normally leads us to a Eularian graph, but it didn't tell us we had to finish at the same place, and my answer of 139 technically crosses all canals and is smaller than yours.

So, there's seven marks gone. What a stupid and unfair question.
(edited 12 years ago)
Reply 93
Original post by pepeeglesfield
I got the exact same...there were 2 possible paths you could repeat with a length of 10


I completelybagree.....either BD could have been repeated, starting at F and finishing at E, or vice versa
OR DF could have been repeated, starting at B and finishing at E, or vice versa

In either case, a length of 10 is repeated, yielding a minimum length of 139.

Then in the second part, nodes F and B now became even, leaving D and E as the only odd nodes, thus making the graph semi-eulerian. This means no repeats are now necessary as you can start at D and end at E or vice versa. However, an arc of length 12 has been added, meaning the route is now 141 miles, i.e. Longer than before.

I see no problem with my logic here :smile:
Does anyone know, say I get a B in D1 but 90%+ in C1, C2 and S1 and then get 90%+ in C3 and C4, is that enough for my A* maths or do you HAVE to get A's in everything, and so would my D1 stop me even if my others are averaging 90%?
(edited 12 years ago)
Original post by conzag123
I completelybagree.....either BD could have been repeated, starting at F and finishing at E, or vice versa
OR DF could have been repeated, starting at B and finishing at E, or vice versa

In either case, a length of 10 is repeated, yielding a minimum length of 139.

Then in the second part, nodes F and B now became even, leaving D and E as the only odd nodes, thus making the graph semi-eulerian. This means no repeats are now necessary as you can start at D and end at E or vice versa. However, an arc of length 12 has been added, meaning the route is now 141 miles, i.e. Longer than before.

I see no problem with my logic here :smile:


But your starting at finishing at two different nodes, can you show where on the question this is indicated.
noooooo. my guess is
100 ums 73
90 ums 65
80 ums 62
Reply 97
Original post by .Username.
But your starting at finishing at two different nodes, can you show where on the question this is indicated.


Can you show where on the question it doesn't indicate this?
although username is actually the master of ums
That seems very low especially for d1 :wink:

Quick Reply

Latest