ATAR Notes: Forum
Archived Discussion => Mathematics and Science => 2012 => End-of-year exams => Exam Discussion => Victoria => Further Mathematics => Topic started by: Will T on November 02, 2012, 02:46:22 pm
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These are the answers I believe to be correct, but please feel free to debate :)
Also, if you're arguing against an answer, please provide working out and logical reasoning to support your choice. Don't just say "It was A!".
Networks - Question 5: According to Graph Theory, an Eulerian cycle will always exist in an undirected graph where all the vertices have even degree (I was unaware of this). Furthermore (no pun intended), an Eulerian cycle/path can still exist even if the undirected graph is non-planar (as is the case with complete graphs with n number of vertices where n is > 4). This means Question 5 is actually incredibly simple. If you don't however, believe this to be true, allow all the vertices of the pentagon, heptagon and nonagon graphs to be labelled alphabetically in a clockwise direction (i.e. A, B, C, D......).
They would all have at least the following Eulerian cycles (many will exist)
For the complete graph with n vertices where n = 5, an Eulerian cycle will be: A, B, C, D, E, A, C, E, B, D, A.
For the complete graph with n vertices where n = 7, an Eulerian cycle will be: A, B, C, D, E, F, G, A, C, E, G, B, D, F, A, D, G, C, F, B, E, A.
For the complete graph with n vertices where n = 9, an Eulerian cycle will be: A, B, C, D, E, F, G, H, I, A, C, E, G, I, B, D, F, H, A, D, G, A, E, H, B, E, I, C, F, I, D, H, C, G, B, F, A.
Core:
1. E.
2. B.
3. E.
4. D.
5. D.
6. B.
7. D.
8. A.
9. B.
10. D.
11. C.
12. A.
13. E.
Module 1 - Number patterns:
1. D.
2. E.
3. A.
4. D.
5. C.
6. E.
7. D.
8. B.
9. A.
Module 2 - Geometry and trigonometry (courtesy of StumbleBum):
1. C.
2. D.
3. B.
4. B.
5. D.
6. D.
7. C.
8. C.
9. D.
Module 3 - Graphs & relations:
1. B.
2. A.
3. D.
4. A.
5. C.
6. D.
7. A.
8. A.
9. D.
Module 4 - Business mathematics (courtesy of StumbleBum):
1. C.
2. C.
3. E.
4. D.
5. A.
6. C.
7. D.
8. D.
9. A.
Module 5 - Networks & decision mathematics:
1. E.
2. A.
3. C.
4. D.
5. D. (See above for details).
6. A.
7. C.
8. C.
9. D. The explanation for this can be found in this thread: Can someone explain networks question 9?.
Module 6 - Matrices:
1. D.
2. B.
3. A.
4. E.
5. D.
6. C.
7. B.
8. B.
9. B.
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for netwroks 5 is D and 7 is D, and yes 9 is D in networks , core looks fine
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pretty sure q7 in networks is C (line 2 = 240 + 110)
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pretty sure q7 in networks is C (line 2 = 240 + 110)
its D, line 2 gives u 240 + 110 + 50 which is 400
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pretty sure q7 in networks is C (line 2 = 240 + 110)
its D, line 2 gives u 240 + 110 + 50 which is 400
50 shouldn't be counted, that arrow is moving towards the source, not the sink.
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pretty sure q7 in networks is C (line 2 = 240 + 110)
its D, line 2 gives u 240 + 110 + 50 which is 400
50 shouldn't be counted, that arrow is moving towards the source, not the sink.
ooops.. omg
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Not too sure but... Is Graphs + Rel Q.7 E?
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Shouldn't Question 7, networks be:
240 + 110 = 350 on Line 2
Answer C.
190 is flowing back in, 260 is flowing back in, 50 is flowing back in?
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Mine for Geometry and trigonometry:
1.C
2.D
3.B
4.B
5.D
6.D
7.C
8.C
9.D
Mine for Business mathematics:
1.C
2.C
3.E
4.D
5.A
6.C
7.D
8.D
9.A
Did you want to add these to the original post?
ALSO, i had all the same matrices. So they should be right.
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Not too sure but... Is Graphs + Rel Q.7 E?
Don't think so, the gradient of the transformed graph and the original graph should be the same (5/2), which gives either answer A or B, and is A since the transformed graph is y vs (1/x)
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Not too sure but... Is Graphs + Rel Q.7 E?
I'm fairly sure it is A. How did you get E?
The gradient of the line is 5/2. So k in y = k/x should equal 5/2.
So shouldn't it be y = 5/2x?
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Shouldn't Question 7, networks be:
240 + 110 = 350 on Line 2
Answer C.
190 is flowing back in, 260 is flowing back in, 50 is flowing back in?
I think so Yendall, read the above posts ^_^
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Not too sure but... Is Graphs + Rel Q.7 E?
I'm fairly sure it is A. How did you get E?
The gradient of the line is 5/2. So k in y = k/x should equal 5/2.
So shouldn't it be y = 5/2x?
Hmm... That's interesting. But the point (2,5)doesn't lie on the line y=5/2x, does it??
Poop. I was really hoping for 40/40. :(
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OH! I think I may have substituted 2 for x, not 2 for 1/x :(
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Not too sure but... Is Graphs + Rel Q.7 E?
I'm fairly sure it is A. How did you get E?
The gradient of the line is 5/2. So k in y = k/x should equal 5/2.
So shouldn't it be y = 5/2x?
Hmm... That's interesting. But the point (2,5)doesn't lie on the line y=5/2x, does it??
Poop. I was really hoping for 40/40. :(
It doesn't but the x-value 2 is the value after the transformations have taken place.
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Not too sure but... Is Graphs + Rel Q.7 E?
I'm fairly sure it is A. How did you get E?
The gradient of the line is 5/2. So k in y = k/x should equal 5/2.
So shouldn't it be y = 5/2x?
AHHHHHH!! I SEE!!! That is such an amateur assumption! Damn!! :( 39/40 :(
Hmm... That's interesting. But the point (2,5)doesn't lie on the line y=5/2x, does it??
Poop. I was really hoping for 40/40. :(
It doesn't but the x-value 2 is the value after the transformations have taken place.
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ALSO, i had all the same matrices. So they should be right.
I can confirm this as well, those answers are correct :)
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Isnt Q5 Networks D?
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I also have networks Q5 D and Q9 is C, anyone else get C for Q9?
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Isnt Q5 Networks D?
yes it should be
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Isnt Q5 Networks D?
yes it should be
How is it D? minimum completion time is 18, so you only need to reduce 2 hours, thus 2 activities?
wrong question hahaahah
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I OBJECT:
Question 8 in Networks:
If you forward and backward scan this network, the minimum completion time is 18 | 18. It says the duration of each activity can be reduced by one hour.
To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is:
If the completion time is 18, you need to reduce it by 2 hours. If you reduce 2 activities by one hour each (as that's the limit) shouldn't the answer be B?
This is given that I've forward scanned and backward scanned correctly, which I believe I have. If I haven't, please subject me to the highest form of shame.
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I have completed solutions for Number patterns, I found it incredibly easy and am regretting not doing it on the exam.
We now have rough solutions for all the Modules and Core, so hopefully if we can have some debate we will have a correct set of solutions between us. :)
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I think there are 2 critical paths so if you reduce F and H then you gotta reduce E as well
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I OBJECT:
Question 8 in Networks:
If you forward and backward scan this network, the minimum completion time is 18 | 18. It says the duration of each activity can be reduced by one hour.
To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is:
If the completion time is 18, you need to reduce it by 2 hours. If you reduce 2 activities by one hour each (as that's the limit) shouldn't the answer be B?
This is given that I've forward scanned and backward scanned correctly, which I believe I have. If I haven't, please subject me to the highest form of shame.
Sorry, but it should be C.
There are two critical paths : A-F-H and B-C-F-H (18 hours each), as well as the path A-E-G (17 hours).
To reduce the critical paths to 16, H must be reduced, and reducing A and B then brings all the paths over 16 hours down to 16 hours. :) (hard to explain sorry!)
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Can anyone explain how to do Q11 in core?
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I also have networks Q5 D and Q9 is C, anyone else get C for Q9?
i also got c !
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I OBJECT:
Question 8 in Networks:
If you forward and backward scan this network, the minimum completion time is 18 | 18. It says the duration of each activity can be reduced by one hour.
To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is:
If the completion time is 18, you need to reduce it by 2 hours. If you reduce 2 activities by one hour each (as that's the limit) shouldn't the answer be B?
This is given that I've forward scanned and backward scanned correctly, which I believe I have. If I haven't, please subject me to the highest form of shame.
Sorry, but it should be C.
There are two critical paths : A-F-H and B-C-F-H (18 hours each), as well as the path A-E-G (17 hours).
To reduce the critical paths to 16, H must be reduced, and reducing A and B then brings all the paths over 16 hours down to 16 hours. :) (hard to explain sorry!)
dang haha
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Can anyone explain how to do Q11 in core?
Add all the long-term average rainfalls together (=216.6), then divide by 4. This gives the seasonal average. Then, dividing the average for spring (61.3) by the seasonal average gives the seasonal index, roughly equal to 1.13.
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For question 5 of networks.
For a complete graph to have an Euler Circuit the vertices must all be of even degree when transformed to planar. [I wasn't meant to say that, Planar is irrelevant]
Network 1 has 4 vertices of odd degree
Network 2 has 5 vertices of even degree
Network 3 has 7 vertices of even degree
Network 4 has 9 vertices of even degree
Therefore, 3 of the 4 graphs can have an Euler Circuit.
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Is question 8 in graphs and relations not B? Because it says at least 5 loaves of white bread are made for each loaf of brown bread...so wouldn't that mean that one white = five or more brown?
aka w =< 5b?
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can someone clarify graphs and relations q7? I got E
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How can you transform a complete graph into a planar graph? I believe it is impossible when the number of vertices is greater than 4. I can see how you can transform the complete graph with No. of vertices = 4. But the Pentagon, Heptagon and Nonagon shapes are not able to be transformed into a planar graph?
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can someone clarify graphs and relations q7? I got E
Me too, but turns out it's wrong, the k value is 5/2
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How can you transform a complete graph into a planar graph? I believe it is impossible when the number of vertices is greater than 4. I can see how you can transform the complete graph with No. of vertices = 4. But the Pentagon, Heptagon and Nonagon shapes are not able to be transformed into a planar graph?
Euler circuits can exist on non-planar graphs. I don't know why I said planar transformation, must've been thinking of something else. The fact that three of the graphs have non-odd vertices shows that an Eulerian circuit is possible.
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Euler circuits exist when all vertices have an even degree.
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Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.
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Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.
The source is down the bottom, not to the left. 120 is flowing out of the cut.
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Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.
The cut is suppose to be drawn in a way that if you remove those lines, you can't get from one end to the next. If you cut out the lines in line 1, you could still get from the town to the freeway.
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Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.
You made the same mistake as me! I feel slightly better now knowing I wasn't the only one.
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Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.
The cut is suppose to be drawn in a way that if you remove those lines, you can't get from one end to the next. If you cut out the lines in line 1, you could still get from the town to the freeway.
Oh crap you're right! Damnnn :-[
What's the general consensus on Networks Q9? I've seen some places say D and some E. I have a feeling it's E though.
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Guys i'm pretty sure networks question 7 is A.
Line 1 was 250+60=310. 300 and 120 don't count towards the cut.
The cut is suppose to be drawn in a way that if you remove those lines, you can't get from one end to the next. If you cut out the lines in line 1, you could still get from the town to the freeway.
Oh crap you're right! Damnnn :-[
What's the general consensus on Networks Q9? I've seen some places say D and some E. I have a feeling it's E though.
I didn't even know what to do with that question haha I said 17 :p
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I drew a really detailed diagram and found that D worked, so it can't be E. I'll redraw it and upload it soonish.
Edit: Attached an image of Q9 Networks
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The last question in business maths is C, not A
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nooo whaat D:
for graphs and relations - question 8
isnt it A??? i have a formula that i got from my teacher that says
"for each x there is at least 5y" and then its 2y<(including arrow)x
:(
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I OBJECT:
Question 8 in Networks:
If you forward and backward scan this network, the minimum completion time is 18 | 18. It says the duration of each activity can be reduced by one hour.
To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is:
If the completion time is 18, you need to reduce it by 2 hours. If you reduce 2 activities by one hour each (as that's the limit) shouldn't the answer be B?
This is given that I've forward scanned and backward scanned correctly, which I believe I have. If I haven't, please subject me to the highest form of shame.
Sorry, but it should be C.
There are two critical paths : A-F-H and B-C-F-H (18 hours each), as well as the path A-E-G (17 hours).
To reduce the critical paths to 16, H must be reduced, and reducing A and B then brings all the paths over 16 hours down to 16 hours. :) (hard to explain sorry!)
I may be wrong but just trying to understand... Why can't F and H each be crashed by one hour? That would mean your shortest path is 7+4+5 = 16, by only crashing two (answer B)?
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The last question in business maths is C, not A
I got the same thing. A doesn't show that the loan was fully repayed (as implied in the question), while C does.
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The last question in business maths is C, not A
No, its A. If you put all the information into the calculator, then you end up with a final value of about $2. As this is not 0, it can't be C.
the question states that "he planed to repay the loan fully", however in reality he doesn't fully pay it off.
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nooo whaat D:
for graphs and relations - question 8
isnt it A??? i have a formula that i got from my teacher that says
"for each x there is at least 5y" and then its 2y<(including arrow)x
:(
Yeh I'm probably wrong but it's the only question I got wrong that I didn't understand
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Question 5 networks, you can test it on the second diagram (the pentagon sorta diagram) and an Euler circuit exists. So either way, the answer can't be 0.
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nooo whaat D:
for graphs and relations - question 8
isnt it A??? i have a formula that i got from my teacher that says
"for each x there is at least 5y" and then its 2y<(including arrow)x
:(
Yeh I'm probably wrong but it's the only question I got wrong that I didn't understand
ah im so silly!
i meant isnt q8 suppose to be B??
i got B too D: i hope thats right.... sighs
i made so many silly mistakes no more 100% T^T
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The last question in business maths is C, not A
No, its A. If you put all the information into the calculator, then you end up with a final value of about $2. As this is not 0, it can't be C.
the question states that "he planed to repay the loan fully", however in reality he doesn't fully pay it off.
um, where'd you get the final value of $2 from?
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I OBJECT:
Question 8 in Networks:
If you forward and backward scan this network, the minimum completion time is 18 | 18. It says the duration of each activity can be reduced by one hour.
To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is:
If the completion time is 18, you need to reduce it by 2 hours. If you reduce 2 activities by one hour each (as that's the limit) shouldn't the answer be B?
This is given that I've forward scanned and backward scanned correctly, which I believe I have. If I haven't, please subject me to the highest form of shame.
Sorry, but it should be C.
There are two critical paths : A-F-H and B-C-F-H (18 hours each), as well as the path A-E-G (17 hours).
To reduce the critical paths to 16, H must be reduced, and reducing A and B then brings all the paths over 16 hours down to 16 hours. :) (hard to explain sorry!)
I may be wrong but just trying to understand... Why can't F and H each be crashed by one hour? That would mean your shortest path is 7+4+5 = 16, by only crashing two (answer B)?
You're right, that would bring both critical paths to 16, but a new critical path is then formed (A-E-G), which is 17. You must then crash one of those events to make 16. :)
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I OBJECT:
Question 8 in Networks:
If you forward and backward scan this network, the minimum completion time is 18 | 18. It says the duration of each activity can be reduced by one hour.
To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is:
If the completion time is 18, you need to reduce it by 2 hours. If you reduce 2 activities by one hour each (as that's the limit) shouldn't the answer be B?
This is given that I've forward scanned and backward scanned correctly, which I believe I have. If I haven't, please subject me to the highest form of shame.
Sorry, but it should be C.
There are two critical paths : A-F-H and B-C-F-H (18 hours each), as well as the path A-E-G (17 hours).
To reduce the critical paths to 16, H must be reduced, and reducing A and B then brings all the paths over 16 hours down to 16 hours. :) (hard to explain sorry!)
I may be wrong but just trying to understand... Why can't F and H each be crashed by one hour? That would mean your shortest path is 7+4+5 = 16, by only crashing two (answer B)?
The critical path would have then been BCEG I think which has a completion time of 17.
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I OBJECT:
Question 8 in Networks:
If you forward and backward scan this network, the minimum completion time is 18 | 18. It says the duration of each activity can be reduced by one hour.
To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is:
If the completion time is 18, you need to reduce it by 2 hours. If you reduce 2 activities by one hour each (as that's the limit) shouldn't the answer be B?
This is given that I've forward scanned and backward scanned correctly, which I believe I have. If I haven't, please subject me to the highest form of shame.
Sorry, but it should be C.
There are two critical paths : A-F-H and B-C-F-H (18 hours each), as well as the path A-E-G (17 hours).
To reduce the critical paths to 16, H must be reduced, and reducing A and B then brings all the paths over 16 hours down to 16 hours. :) (hard to explain sorry!)
I may be wrong but just trying to understand... Why can't F and H each be crashed by one hour? That would mean your shortest path is 7+4+5 = 16, by only crashing two (answer B)?
Exactly what I thought. As long as they are on either critical path it shouldn't matter should it?
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nooo whaat D:
for graphs and relations - question 8
isnt it A??? i have a formula that i got from my teacher that says
"for each x there is at least 5y" and then its 2y<(including arrow)x
:(
Yeh I'm probably wrong but it's the only question I got wrong that I didn't understand
ah im so silly!
i meant isnt q8 suppose to be B??
i got B too D: i hope thats right.... sighs
i made so many silly mistakes no more 100% T^T
Yeah I was thinking about that question for like 5 minutes....somebody clarify!
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The last question in business maths is C, not A
No, its A. If you put all the information into the calculator, then you end up with a final value of about $2. As this is not 0, it can't be C.
the question states that "he planed to repay the loan fully", however in reality he doesn't fully pay it off.
um, where'd you get the final value of $2 from?
Put all the information into the calculator...
EDIT: Oh yeh, you can just make up information from what has been given, and then you change it to suit the new problem were he forgets to pay for one month and you end up with something that isnt 0
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Question 5 networks, you can test it on the second diagram (the pentagon sorta diagram) and an Euler circuit exists. So either way, the answer can't be 0.
I see now. I have found a multitude of Eulerian circuits for the pentagon-shaped complete graph. I will try a similar process for the heptagon and nonagon. This means the range of possible answers is now 1-3 so alternatives B, C or D could be correct at the moment.
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nooo whaat D:
for graphs and relations - question 8
isnt it A??? i have a formula that i got from my teacher that says
"for each x there is at least 5y" and then its 2y<(including arrow)x
:(
Yeh I'm probably wrong but it's the only question I got wrong that I didn't understand
ah im so silly!
i meant isnt q8 suppose to be B??
i got B too D: i hope thats right.... sighs
i made so many silly mistakes no more 100% T^T
Yeah I was thinking about that question for like 5 minutes....somebody clarify!
I thought it was A???
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I OBJECT:
Question 8 in Networks:
If you forward and backward scan this network, the minimum completion time is 18 | 18. It says the duration of each activity can be reduced by one hour.
To complete this project in 16 hours, the minimum number of activities that must be reduced by one hour each is:
If the completion time is 18, you need to reduce it by 2 hours. If you reduce 2 activities by one hour each (as that's the limit) shouldn't the answer be B?
This is given that I've forward scanned and backward scanned correctly, which I believe I have. If I haven't, please subject me to the highest form of shame.
Sorry, but it should be C.
There are two critical paths : A-F-H and B-C-F-H (18 hours each), as well as the path A-E-G (17 hours).
To reduce the critical paths to 16, H must be reduced, and reducing A and B then brings all the paths over 16 hours down to 16 hours. :) (hard to explain sorry!)
I may be wrong but just trying to understand... Why can't F and H each be crashed by one hour? That would mean your shortest path is 7+4+5 = 16, by only crashing two (answer B)?
You're right, that would bring both critical paths to 16, but a new critical path is then formed (A-E-G), which is 17. You must then crash one of those events to make 16. :)
Thank you! That's very helpful :)
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I'm not debating the answer and I've skimmed read so I don't think anyone else has asked the question...
Question 5, Core, why answer D? I had histogram. I feel good, I ran out of time and randomly shaded 3 answers, one of them was correct, so lost 2 on time, and this Core question I also lost. So looks like 37/40. Bloody hope the cut off is 36.
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Question 5 networks, you can test it on the second diagram (the pentagon sorta diagram) and an Euler circuit exists. So either way, the answer can't be 0.
I see now. I have found a multitude of Eulerian circuits for the pentagon-shaped complete graph. I will try a similar process for the heptagon and nonagon. This means the range of possible answers is now 1-3 so alternatives B, C or D could be correct at the moment.
Its definitely 3.
On a complete graph with n vertices, the degree of each vertex is (n-1) (since it connects to all other vertices)
The graphs have 4,5,7,9 vertices respectively.
for n-1 to be even, n has to be odd.
For a Eulerian circuit to exist, degree of every vertex has to be even.
So Eulerian circuits exist in K5,7,9 so the answer is 3
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I'm not debating the answer and I've skimmed read so I don't think anyone else has asked the question...
Question 5, Core, why answer D? I had histogram. I feel good, I ran out of time and randomly shaded 3 answers, one of them was correct, so lost 2 on time, and this Core question I also lost. So looks like 37/40. Bloody hope the cut off is 36.
Yeah same here. I went with C - Histogram, but it makes sense that it's time series plot, because it's obviously talking about the time of the temperature.
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I'm not debating the answer and I've skimmed read so I don't think anyone else has asked the question...
Question 5, Core, why answer D? I had histogram. I feel good, I ran out of time and randomly shaded 3 answers, one of them was correct, so lost 2 on time, and this Core question I also lost. So looks like 37/40. Bloody hope the cut off is 36.
I had Histogram as well. Apparently time series is more suitable, but I think both would work fine.
I saw it as two sets of numeric data, and the list of possibilities was:
- Histograms
- Frequency polygons
- Cumulative frequency polygons
- Boxplots
- Stemplots
- Dotplots
I didn't think Time Series Plots would even come into the equation when summarising two sets of Numerical Data.
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I'm not debating the answer and I've skimmed read so I don't think anyone else has asked the question...
Question 5, Core, why answer D? I had histogram. I feel good, I ran out of time and randomly shaded 3 answers, one of them was correct, so lost 2 on time, and this Core question I also lost. So looks like 37/40. Bloody hope the cut off is 36.
I had Histogram as well. Apparently time series is more suitable, but I think both would work fine.
I saw it as two sets of numeric data, and the list of possibilities was:
- Histograms
- Frequency polygons
- Cumulative frequency polygons
- Boxplots
- Stemplots
- Dotplots
I didn't think Time Series Plots would even come into the equation when summarising two sets of Numerical Data.
Histograms count frequency/percentage frequency though. You don't count a discrete amount of degrees, its a continuous variable.
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The last question in business maths is C, not A
No, its A. If you put all the information into the calculator, then you end up with a final value of about $2. As this is not 0, it can't be C.
the question states that "he planed to repay the loan fully", however in reality he doesn't fully pay it off.
But as the question doesn't state the original balance of the loan (as far as I can see), isn't it correct to assume that he does repay it in full? I'm quite possibly missing something here, feel free to give me the details if I'm wrong.
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The last question in business maths is C, not A
No, its A. If you put all the information into the calculator, then you end up with a final value of about $2. As this is not 0, it can't be C.
the question states that "he planed to repay the loan fully", however in reality he doesn't fully pay it off.
But as the question doesn't state the original balance of the loan (as far as I can see), isn't it correct to assume that he does repay it in full? I'm quite possibly missing something here, feel free to give me the details if I'm wrong.
Nope, pick a random number and percentage and chuck it on TVM and fiddle with it and you won't get a fully repayed loan. It won't be 0 because 260 equal payments got it to zero, correct? That was consistent, equal payments. When you skip a payment, his principal increases slightly, because nothing has been paid back but the interest rate has increased it a little bit. Thus, even though he double the payment to 520, it does not take into account that slight increase when the fourth payment was missed. Does that make sense?
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The last question in business maths is C, not A
No, its A. If you put all the information into the calculator, then you end up with a final value of about $2. As this is not 0, it can't be C.
the question states that "he planed to repay the loan fully", however in reality he doesn't fully pay it off.
But as the question doesn't state the original balance of the loan (as far as I can see), isn't it correct to assume that he does repay it in full? I'm quite possibly missing something here, feel free to give me the details if I'm wrong.
Nope, pick a random number and percentage and chuck it on TVM and fiddle with it and you won't get a fully repayed loan. It won't be 0 because 260 equal payments got it to zero, correct? That was consistent, equal payments. When you skip a payment, his principal increases slightly, because nothing has been paid back but the interest rate has increased it a little bit. Thus, even though he double the payment to 520, it does not take into account that slight increase when the fourth payment was missed. Does that make sense?
Unfortunately it does, hahahaha. Thanks for the info :)
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You're welcome :)
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hey how is question 2 of the core, B? isnt it E
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Question 5 networks, you can test it on the second diagram (the pentagon sorta diagram) and an Euler circuit exists. So either way, the answer can't be 0.
I see now. I have found a multitude of Eulerian circuits for the pentagon-shaped complete graph. I will try a similar process for the heptagon and nonagon. This means the range of possible answers is now 1-3 so alternatives B, C or D could be correct at the moment.
Its definitely 3.
On a complete graph with n vertices, the degree of each vertex is (n-1) (since it connects to all other vertices)
The graphs have 4,5,7,9 vertices respectively.
for n-1 to be even, n has to be odd.
For a Eulerian circuit to exist, degree of every vertex has to be even.
So Eulerian circuits exist in K5,7,9 so the answer is 3
I am in agreement with you now, I just had to check some stuff, thanks.
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hey how is question 2 of the core, B? isnt it E
No, its a bar chart, not a percentage frequency graph. So its not 35%, but actually 35/214 which is 16%
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hey how is question 2 of the core, B? isnt it E
No, its a bar chart, not a percentage frequency graph. So its not 35%, but actually 35/214 which is 16%
god damnit
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I thought histograms were only for one variable? and the other axis being frequency of that variable?
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Bro thats right. If you're talking about the temperature one it should be a Time-Series plot.
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39/40, got q2 on networks wrong, dont ask how.. haha :)
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Trig question 3 anyone??
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Trig question 3 anyone??
The shape is constructed from a rectangle with net area 1000cm squared. So the surface area around the cylinder, which is given by 2 x pi x r x h will equal 1000, therefore r is approximately 8.0 cm.
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explain further?
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Well if the answers in the OP are correct, I got 39/40!
Got question 7 in Networks wrong ::)
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shouldn't 4 for networks be E as there are four areas in total but the area outside of the network is also considered a region?
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shouldn't 4 for networks be E as there are four areas in total but the area outside of the network is also considered a region?
I drew a connected planar graph from the adjacency matrix. It had 3 faces and then the external outside face made it 4 faces in total. Euler's formula also verifies this.
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shouldn't 4 for networks be E as there are four areas in total but the area outside of the network is also considered a region?
I drew a connected planar graph from the adjacency matrix. It had 3 faces and then the external outside face made it 4 faces in total. Euler's formula also verifies this.
noticed i drew an extra edge. my bad haha
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Not too sure but... Is Graphs + Rel Q.7 E?
I'm fairly sure it is A. How did you get E?
The gradient of the line is 5/2. So k in y = k/x should equal 5/2.
So shouldn't it be y = 5/2x?
It was a poorly asked question
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Not too sure but... Is Graphs + Rel Q.7 E?
I'm fairly sure it is A. How did you get E?
The gradient of the line is 5/2. So k in y = k/x should equal 5/2.
So shouldn't it be y = 5/2x?
It was a poorly asked question
How is it poorly asked, it seems quite clear to me.
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Can someone explain q12 on core.
I got E.
this is because well autum was 48.9 for 2011. I assumed since that's WELL below its average it won't follow the same SI which was 1.01.
Wouldn't u then plug in 48.9 for the autumn value in the table get a new SI and go from there.
Or did I over think it?
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Can someone explain q12 on core.
I got E.
this is because well autum was 48.9 for 2011. I assumed since that's WELL below its average it won't follow the same SI which was 1.01.
Wouldn't u then plug in 48.9 for the autumn value in the table get a new SI and go from there.
Or did I over think it?
Yeh you overthought it...
you just divide 48.9 by 1.01, the seasonal index for autumn, and you get 48.4 (A).
The question didn't state anywhere to use this figure to calculate new seasonal indices, and in doing so you would have defeated the whole purpose of actually getting a deseasonalised value because you would have ended up with a value equal to the yearly average.
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Shouldn't question 9 for business be C because A implies that his balance went up in month 4 however he didn't not pay any money so the balance would stay the same as shown in C.
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Shouldn't question 9 for business be C because A implies that his balance went up in month 4 however he didn't not pay any money so the balance would stay the same as shown in C.
He didn't pay any money, however that makes the balance go up because interest was still charged on that month.