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

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Well the variation between £500/£505 can be explained by a sight variation in lines because both my values were values which i rounded down. This was necessary as rounding up from e.g. 10.5 is not right as it would be outside the feasible region.

However the general consensus seems to be £520 so I'm not very confident. If it's £505 i should get the marks though due to reading errors and vise versa for you, but i don't it will be either :frown:

**** it what can you do right, i mean ppl in my sch got x=18 y=8 whicb is close to 20 and 10 but yh i musy have ****ed up somewhere

Reply 201

sam_97

Original post by Hjyu1

Why is kruskals wrong there is not other information to say k between any other boundary

Original post by sufiyan1999

Apparently according to people on this its 10 <K<14??
Not sure. But i got the 10

I got x > 10 as well. I'm not sure where the 14 is coming from. If x was greater than 14, there would be no effect on the edges chosen and CD would still be part of the tree.

Reply 202

sufiyan1999

Original post by sam_97

I got x > 10 as well. I'm not sure where the 14 is coming from. If x was greater than 14, there would be no effect on the edges chosen and CD would still be part of the tree.

That's what I got have no clue why 14 is put in.

Reply 203

Hjyu1

Original post by sufiyan1999

That's what I got have no clue why 14 is put in.

There's no information to indicate X is less than 14 as if either the 14 edge are X edge are chosen it forms a cycle so yeah I think X>10 is right

Reply 204

-jordan-

Original post by sufiyan1999

That's what I got have no clue why 14 is put in.

Agreed, x>10

Reply 205

sam_97

Original post by yelash

Well the variation between £500/£505 can be explained by a sight variation in lines because both my values were values which i rounded down. This was necessary as rounding up from e.g. 10.5 is not right as it would be outside the feasible region.However the general consensus seems to be £520 so I'm not very confident. If it's £505 i should get the marks though due to reading errors and vise versa for you, but i don't it will be either

Original post by Zyzz_aesthetic

**** it what can you do right, i mean ppl in my sch got x=18 y=8 whicb is close to 20 and 10 but yh i musy have ****ed up somewhere

The answer to this question was definitely £520 (x=24, y=8), I checked on Wolfram Alpha.

I ended up with x=21 and y=10 in the exam, which gave a profit of £515. I'm not sure where I went wrong to be honest!

Reply 206

-jordan-

Original post by sam_97

21 and 10 comes from the wrong region, above x=2y instead of below. Sure it wasn't this? How many marks was it for the maximising profit anyway?

Reply 207

sam_97

Original post by Hjyu1

There's no information to indicate X is less than 14 as if either the 14 edge are X edge are chosen it forms a cycle so yeah I think X>10 is right

Those were exactly my thoughts during the exam.

(edited 7 years ago)

Reply 208

sam_97

Original post by -jordan-

21 and 10 comes from the wrong region, above x=2y instead of below. Sure it wasn't this? How many marks was it for the maximising profit anyway?

I'm not too sure what I was thinking to be honest. I think it was 2 marks for finding the values of x and y, and 1 mark for calculating the maximum profit.

Reply 209

-jordan-

Original post by sam_97

I'm not too sure what I was thinking to be honest. I think it was 2 marks for finding the values of x and y, and 1 mark for calculating the maximum profit.

It's usually only one mark for correct indication of the feasible region, the rest are for the lines. So if it was that then you haven't dropped too many.

Reply 210

sam_97

Original post by -jordan-

It's usually only one mark for correct indication of the feasible region, the rest are for the lines. So if it was that then you haven't dropped too many.

I'm sure my feasible region etc was correct, I think I just read off the values wrong. What did you put for the last question on graph theory by the way?

Reply 211

-jordan-

Original post by sam_97

I'm sure my feasible region etc was correct, I think I just read off the values wrong. What did you put for the last question on graph theory by the way?

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!

Reply 212

fpmaniac

Original post by sam_97

I'm sure my feasible region etc was correct, I think I just read off the values wrong. What did you put for the last question on graph theory by the way?

I put profit as 520. used 2x=y and 2x+y=720? or something like that....

Reply 213

fpmaniac

That was an easy paper tbh. I hate tracing algorithms so much and it didnt come up as I mess up on that everytime!!!! So yay for me

Reply 214

1/2(p^2-p)

Original post by fpmaniac

That was an easy paper tbh. I hate tracing algorithms so much and it didnt come up as I mess up on that everytime!!!! So yay for me

sames i thought that the paper was fairly easy and i was made up when there was no algorithms

Reply 215

fpmaniac

The vertex question part ii) was something along these lines where the graph was semi eulerian and had 5 vertices and was a tree

Reply 216

Lumberjack1

Original post by Hjyu1

There's no information to indicate X is less than 14 as if either the 14 edge are X edge are chosen it forms a cycle so yeah I think X>10 is right

I put x is more than or equal to 10 as I thought that even if x was 10 CD could still be chosen in Kruskals. Anyone agree?

Reply 217

sufiyan1999

Original post by Lumberjack1

I put x is more than or equal to 10 as I thought that even if x was 10 CD could still be chosen in Kruskals. Anyone agree?

Disagree because if k was 10 you could opt for it which would mean you cant choose the edge with 14 as a cycle would form

Reply 218

Hjyu1

Original post by fpmaniac

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 think it was simple so can't have a repeated edge just draw 5 vertices then a straight line through them

Reply 219

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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