My Solutions (Core; Graphs; Networks; Matrices)

[Will T]

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.

[Tonychet2]

for netwroks 5 is D and 7 is D, and yes 9 is D in networks , core looks fine

[oppalovesme]

pretty sure q7 in networks is C (line 2 = 240 + 110)

[Tonychet2]

pretty sure q7 in networks is C (line 2 = 240 + 110)

its D, line 2 gives u 240 + 110 + 50 which is 400

[oppalovesme]

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.

[Tonychet2]

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

[Varunchka]

Not too sure but... Is Graphs + Rel Q.7 E?

[Yendall]

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?

[StumbleBum]

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.

[oppalovesme]

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)

[Will T]

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?

[oppalovesme]

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 ^_^

[Varunchka]

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.

[Varunchka]

OH! I think I may have substituted 2 for x, not 2 for 1/x

[julie9300]

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.