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AQA Mathematics MD01 Decision 1 – Friday 24th June [Exam Discussion Thread]

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Original post by Olsmarto
On the kruskals question I said it had to be k> or equal to 10, as you could choose either 10 therefore it can still be true if it was equal to 10, or am I wrong?


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I wrote this!!


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Original post by Olsmarto
On the kruskals question I said it had to be k> or equal to 10, as you could choose either 10 therefore it can still be true if it was equal to 10, or am I wrong?


Posted from TSR Mobile


Glad I wasn't the only one that did this lol, I also did k ≥ 10
Original post by tajtsracc
I did this for the semi-eulerian tree


I'm sure it said 5 edges?
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Original post by -jordan-
If there are 2 vertices with order five (I think it was) then there are two vertices that connect to every other vertex in the network, therefore it's impossible that there's one with an order of one because of the two vertices that connects to all the others. It's a hard one to phrase!


I thought it was as well! I put something about the number of odd vertices being even, not sure if it's correct or not.
Reply 224
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Original post by fpmaniac
I put profit as 520. used 2x=y and 2x+y=720? or something like that....


Yeah £520 is correct
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Original post by sufiyan1999
I'm sure it said 5 edges?


It was 5 vertices
Reply 226
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Original post by Olsmarto
On the kruskals question I said it had to be k> or equal to 10, as you could choose either 10 therefore it can still be true if it was equal to 10, or am I wrong?


Original post by beth000
I wrote this!!


Original post by tajtsracc
Glad I wasn't the only one that did this lol, I also did k ≥ 10


I think the answer is x>10x > 10 because if x=10x = 10 was chosen, there would be no guarantee that CD would be included in the tree as CD has a weight of 10 as well.
(edited 7 years ago)
Original post by sam_97
I think the answer is x>10x > 10 because if x=10x = 10 was chosen, there would be no guarantee that CD would be included in the tree as CD has a weight of 10 as well.


I think they'll accept both. I can just see it saying "Condone x>10" at the side of the mark scheme. Just depends how many people disagreed I guess.
(edited 7 years ago)
Original post by sam_97
I think the answer is x>10x > 10 because if x=10x = 10 was chosen, there would be no guarantee that CD would be included in the tree as CD has a weight of 10 as well.


I'm fairly sure it didn't specify that CD had to be chosen, just that it was in that case. Therefore x could equal 10 and CD was still chosen.
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Original post by -jordan-
I think they'll accept both. I can just see it saying "Condone x>10" at the side of the mark scheme. Just depends how many people disagreed I guess.


Yeah I agree, they will probably accept either. They don't seem to be too harsh when it comes to inequalities in decision.

Original post by Lumberjack1
I'm fairly sure it didn't specify that CD had to be chosen, just that it was in that case. Therefore x could equal 10 and CD was still chosen.


As -jordan- said above, they'll probably accept both.
Original post by sam_97
The maximum profit for the last question was definitely £520, just checked on Wolfram Alpha. I somehow ended up with £515 in the exam so I'm not too sure where I went wrong there!


Could you send me the equations/the working for the last question please? Need to understand where I went wrong to see how many marks I lost, thanks.
Original post by hi-zen-berg
any *UMS ideas for 65 raw?


~84
(edited 7 years ago)
Original post by fpmaniac
d1 img.png The vertex question part ii) was something along these lines where the graph was semi eulerian and had 5 vertices and was a tree


I just did five vertices in a 'W' pattern
\/\/
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Original post by Chickenslayer69
Could you send me the equations/the working for the last question please? Need to understand where I went wrong to see how many marks I lost, thanks.


The equations were:

x≥5 x \geq 5

y≥5y \geq 5

2x+3y≤722x + 3y \leq 72

x+y≤32x + y \leq 32

x≥2yx \geq 2y

We had to maximise profit, which was given by P=15x+20yP = 15x + 20y.
(edited 7 years ago)
With the Chinese postman problem, i believe you were forced to end your journey at the vertex D. Therefore when choosing the lowest route to repeat, you needed not to have chosen one containing the vertex D. Because the lowest repeating route contained D, you needed to chose the next highest one.
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Original post by billdjango99
With the Chinese postman problem, i believe you were forced to end your journey at the vertex D. Therefore when choosing the lowest route to repeat, you needed not to have chosen one containing the vertex D. Because the lowest repeating route contained D, you needed to chose the next highest one.


That's correct. The vertices B and D needed to have an odd degree, so the shortest route between the other two odd vertices (12) had to be added to the total weight of the edges in the graph. The start vertex was B.
(edited 7 years ago)
64/63 for A?
Original post by fpmaniac
d1 img.png The vertex question part ii) was something along these lines where the graph was semi eulerian and had 5 vertices and was a tree


I don't think that's a correct tree. A tree is a graph without cyclic routes, and this graph is cyclic
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Original post by tajtsracc
It was x≥2yx \geq 2y not 2x≥y2x \geq y

"Twice as many as x as y". So if y was 50 then x would be 100.


Yeah it was, my mistake. See edit
Is there a D2 thread anywhere?

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