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Edexcel D1 question help, please!

Can someone please help me on question 7, thanks, I worked out the intersection of 3x=2y and 5x + 4y = 80, I got what the mark scheme has but their x is my y and my x is their y so I don't know what I'm doing wrong?, I figured that if you workout the intersection of y > 6 and 5x + 4y = 80 then you get the mark schemes answer but,isn't the y> 6 a inequality and not a line? therefore it doesn't make sense to find an intersection. PLEASE HELP, I'M WORRIED ABOUT THE EXAM, thanks!! =D
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-ms/6689_01_msc_20100618.pdf

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-qp/6689_01_que_20100527.pdf
Reply 1
Original post by Skilled
Can someone please help me on question 7, thanks, I worked out the intersection of 3x=2y and 5x + 4y = 80, I got what the mark scheme has but their x is my y and my x is their y so I don't know what I'm doing wrong?, I figured that if you workout the intersection of y > 6 and 5x + 4y = 80 then you get the mark schemes answer but,isn't the y> 6 a inequality and not a line? therefore it doesn't make sense to find an intersection. PLEASE HELP, I'M WORRIED ABOUT THE EXAM, thanks!! =D
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-ms/6689_01_msc_20100618.pdf

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-qp/6689_01_que_20100527.pdf


Hi, y>6 is already on the graph & shaded in & the optimal point is the intersection point of 3x=2y & 5x+4y=80. :smile:
Reply 2
y isn't included so you have to use integer testing to find the correct point which works.
Reply 3
Original post by Fortitude
Hi, y>6 is already on the graph & shaded in & the optimal point is the intersection point of 3x=2y & 5x+4y=80. :smile:


I worked out the intersection of 3x=2y and 5x + 4y=80 but I got (7,11) but the mark scheme says (11,7)
Reply 4
Original post by Dilzo999
y isn't included so you have to use integer testing to find the correct point which works.


sorry but i dont understand what you mean
Reply 5
Original post by Skilled
I worked out the intersection of 3x=2y and 5x + 4y=80 but I got (7,11) but the mark scheme says (11,7)


Hi, You were right, I just realised that there's a vertex at y=6 so that would be the optimal point hence it'd be the intersection of y=6 & 5x+4y=80, sorry for mistake :colondollar: which gives you (11.2, 6) (I used simultaneous equations to calculate this or you could read from the graph but mine's not that clear) Now as you said, you can't include y=6 so the next integer would be 7 hence then answer (11,7) & you could also test out other integers which are close to it as well to find the optimal solution but be careful to see if they're feasible or not. :biggrin:
Reply 6
Original post by Fortitude
Hi, You were right, I just realised that there's a vertex at y=6 so that would be the optimal point hence it'd be the intersection of y=6 & 5x+4y=80, sorry for mistake :colondollar: which gives you (11.2, 6) (I used simultaneous equations to calculate this or you could read from the graph but mine's not that clear) Now as you said, you can't include y=6 so the next integer would be 7 hence then answer (11,7) & you could also test out other integers which are close to it as well to find the optimal solution but be careful to see if they're feasible or not. :biggrin:


Thanks, I get it now! =D , I have not seen a tricky question like this before
Reply 7
Original post by Skilled
Thanks, I get it now! =D , I have not seen a tricky question like this before


You're welcome :biggrin:
Reply 8
Original post by Fortitude
You're welcome :biggrin:


Please help

When it says I can start and end at any point in a Chinese postman question, how do you workout those points? and if it gives you a start and tells you to end at any point then what do you do? what if it gives you an end and tell you to choose a start?

Thanks in advance
Reply 9
Original post by Skilled
Please help

When it says I can start and end at any point in a Chinese postman question, how do you workout those points? and if it gives you a start and tells you to end at any point then what do you do? what if it gives you an end and tell you to choose a start?

Thanks in advance


I'll try my best to explain but if you have a specific Q please post it as it'd probably make more sense. Anyway for eg you've found that you need to repeat the arcs A,C,D,G:

AC + DG = 45 + 20 =65
AD + CG = 22 + 46 = 68
AG + CD = 35 + 29 = 64

If you're told to start at G & you need to find where to finish & which arcs to repeat pick the arcs which give you the shortest path, here it will be AD (don't include DG becuase G is your starting vertex so no need to repeat) & therefore you'll finish at C. :smile:
Reply 10
Original post by Fortitude
I'll try my best to explain but if you have a specific Q please post it as it'd probably make more sense. Anyway for eg you've found that you need to repeat the arcs A,C,D,G:

AC + DG = 45 + 20 =65
AD + CG = 22 + 46 = 68
AG + CD = 35 + 29 = 64

If you're told to start at G & you need to find where to finish & which arcs to repeat pick the arcs which give you the shortest path, here it will be AD (don't include DG becuase G is your starting vertex so no need to repeat) & therefore you'll finish at C. :smile:


yup, i understand, one last doubt, what if i had to choose a start and end point for this example
Original post by Skilled
yup, i understand, one last doubt, what if i had to choose a start and end point for this example


Then you just look at which path is the shortest, here it's DG (20) therefore you'd repeat DG & start & finish at A & C
Reply 12
Original post by Fortitude
Then you just look at which path is the shortest, here it's DG (20) therefore you'd repeat DG & start & finish at A & C

thanks!, I can't give you rep because it doesn't let me =(
Original post by Skilled
thanks!, I can't give you rep because it doesn't let me =(


You're welcome :biggrin:
Reply 14
Here are some notes I created! Feel free to download them :smile:

D1.pptx
Reply 15
Original post by Fortitude
You're welcome :biggrin:


When it gives you an equation eg. 6x + 9y = P max, then does the graph cross at (0,6) and (9,0) ? is it always the coordinates of each other?
Original post by Skilled
When it gives you an equation eg. 6x + 9y = P max, then does the graph cross at (0,6) and (9,0) ? is it always the coordinates of each other?


Is this an objective line?
Reply 17
Original post by Fortitude
Is this an objective line?


yup
Original post by Skilled
When it gives you an equation eg. 6x + 9y = P max, then does the graph cross at (0,6) and (9,0) ? is it always the coordinates of each other?


What I've been taught is to choose a no that is a multiple of 6 & 9 for eg 360 & put in into the equation so 6x + 9y = 360 & use this to work out coordinates so here it could be (0, 40) & (60, 0). It doesn't matter which no you use as long as it's a multiple & it obvs depends on scale of the axis because it'd be no use using 360 if the scale on the axis was from 2550 to 8000:tongue:
Reply 19
Original post by Fortitude
What I've been taught is to choose a no that is a multiple of 6 & 9 for eg 360 & put in into the equation so 6x + 9y = 360 & use this to work out coordinates so here it could be (0, 40) & (60, 0). It doesn't matter which no you use as long as it's a multiple & it obvs depends on scale of the axis because it'd be no use using 360 if the scale on the axis was from 2550 to 8000:tongue:


Thanks a lot

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