VCE General & Further Maths Question Thread!

[j0yce]

I'll try explain this as best as i can, in the first transition matrix, you see that 10% of B changes to C before C withdraws, That's clear right? You can see from the matrix.

Next, in the second matrix, which depicts what would happen after Mr Choi withdraws, 10% would split equally, 5% will go to A (10% to 15%) and 5% to B (80% to 85%), you can see how the 10% would split equally between A and B, right? since 5% of a total 10% goes to B, 5/10 * 100 = 50%

Oh I get it! Thank you so much! I was struggling with this question for so long

[TheMereCat]

Oh I get it! Thank you so much! I was struggling with this question for so long

You're very welcome

[n.a]

Question 3 from VCAA 2010 Exam 1, please!

[bedigursimran]

You can't substitute the coordinates to find k here, because it's plotted as H against 1/d^2, if you had the original graph, which would be H against D, you would be able to use it to find the coefficient. Since you have the transformed graph, you'll need to find the K using the gradient of the straight line.

Thanks dude! That helped me so much

[MightyBeh]

Question 3 from VCAA 2010 Exam 1, please!

Covered this one too, link.

[bedigursimran]

Oh I get it! Thank you so much! I was struggling with this question for so long

When my teacher told me, I was surprised how easy it was too haha.

[bedigursimran]

I've explained this one before! I think Latex is being weird though so if the worded explanation isn't enough I'll re-do the working out.

Although I'd also like to add since doing that the first time I've run into this rule which I think was mentioned in the itute and VCAA solutions:
(x/a) + (y/b) = 1,
where a and b are the x and y intercepts, respectively. (pretty sure it's relevant )

Definitely above 90%. No more than a couple of marks each exam, it gets pretty competitive above 40

EDIT; Fixed the link

The images aren't loading for me

[MightyBeh]

The images aren't loading for me

Here's copied but without the latex, hopefully that makes it readable

Spoiler

This one was actually kind of tough - I checked out the examiner's report and the itute solutions and neither of them really gave a clear, step by step explanation on how to do the question - although itute was a bit better than the VCAA.

Because Z is maximised at the points M and N, we can say that Z(m) and Z(n) are equal. For that to make much sense though, we need to know the coordinates of M and N. They're pretty clear on the graph as long as you work out the the interval is 5 (and maybe have a ruler to double check your points line up) but if you really wanted to you could set up some simultaneous equations to find them, too.

In any case, the coordinates of M and N are (20,50) and (45,25) respectively. We can set up two equations/functions for Z with this information:
Z = 20a + 50b (M)
Z = 45a + 25b (N)

Because we know Z has to be equal for both M and N to find a and b, we can quickly throw out some of the options that would give us something ridiculous - this means we don't have to worry about either of the options with negative values (A, E) and can probably get away without checking that D fits our rules because the numbers are pretty different to A, B and C. (Note: With multiple choice questions, you can usually eliminate two options after reading the question or minimal working out: in this case it was D and E)

So that leaves us with option B and option C to check, like so:
B. Z = 45*15 + 25*15 =1050
B. Z = 20*15 + 50*15=1050
We can already see that these are equal (and therefore, B is the answer), but let's check C to be sure we haven't messed up somewhere:
C. Z = 45*15 + 25*25 =1300
C.  Z = 20*15 + 50*25 =1550

So B is our answer - of course some of the constraints I set were maybe a little too general (option A could've probably been a reasonable choice) but in the end, there's only five equations to check so realistically you probably wouldn't need to set any at all.

[Zealous]

Just want to say, awesome work to all of you who are actively helping out in this thread! Especially MightyBeh and TheMereCat - you seem to be on top of everything.  I'm sure your contributions are helping many more than just the people who ask the question.

[bedigursimran]

Here's copied but without the latex, hopefully that makes it readable

Spoiler

This one was actually kind of tough - I checked out the examiner's report and the itute solutions and neither of them really gave a clear, step by step explanation on how to do the question - although itute was a bit better than the VCAA.

Because Z is maximised at the points M and N, we can say that Z(m) and Z(n) are equal. For that to make much sense though, we need to know the coordinates of M and N. They're pretty clear on the graph as long as you work out the the interval is 5 (and maybe have a ruler to double check your points line up) but if you really wanted to you could set up some simultaneous equations to find them, too.

In any case, the coordinates of M and N are (20,50) and (45,25) respectively. We can set up two equations/functions for Z with this information:
Z = 20a + 50b (M)
Z = 45a + 25b (N)

Because we know Z has to be equal for both M and N to find a and b, we can quickly throw out some of the options that would give us something ridiculous - this means we don't have to worry about either of the options with negative values (A, E) and can probably get away without checking that D fits our rules because the numbers are pretty different to A, B and C. (Note: With multiple choice questions, you can usually eliminate two options after reading the question or minimal working out: in this case it was D and E)

So that leaves us with option B and option C to check, like so:
B. Z = 45*15 + 25*15 =1050
B. Z = 20*15 + 50*15=1050
We can already see that these are equal (and therefore, B is the answer), but let's check C to be sure we haven't messed up somewhere:
C. Z = 45*15 + 25*25 =1300
C.  Z = 20*15 + 50*25 =1550

So B is our answer - of course some of the constraints I set were maybe a little too general (option A could've probably been a reasonable choice) but in the end, there's only five equations to check so realistically you probably wouldn't need to set any at all.

That's very helpful. Thanks.

[bedigursimran]

Can you guys help me with question 7 Matrices from Further Exam 1 2013?

[AngelWings]

Can you guys help me with question 7 Matrices from Further Exam 1 2013?

First you need to consider how you would get the mean in a set of data. (Think back to Core Module.)

Spoiler

sum of data divided by number of data pieces

Then, you need to see how you would combine the matrices you are given to get the set of data. Since the matrices S and R are filled with 1s, you know that multiplication would just spit out the same number... and we know that with matrix multiplication you must add these together.

In terms of the choices:

Spoiler

Automatically ditch E. Option E would make no sense in terms of the matrix dimensions. 15 x 1, 3 x 15, 1 x 3 - you can't realistically multiply these together.
Same with C: 3 x 15, 1 x 3, 15 x 1
So you're left with A, B and D.
A wouldn't make sense because we want it per class, not per student. (Did you read "class" in the question?)
B wouldn't make sense because it would give a matrix with dimensions: 3 x 15, 15 x 3 = a 3 x 3 matrix. Going back to the question, we want the mean of each class, so we want the sum of the fifteen numbers per line ie. 1 number. Thus, this is incorrect also, as it receives 3 numbers, but we want 1.

Thus we receive the last option D.

Checking:
To find the mean, we know that there are 15 pieces of data per row and using the mean formula, you should receive the fact that we need to divide the sum by 15. Thus 1/15 should exist outside the matrix in our answer. (D has this, so it's a good sign. E, too, but...) Anyway, you can apply the mean formula on Class A on your calculator and you'd find that D will give you what you want rather than E.

EDIT: Please see the attached file. I realise screwed up the dimensions and did it correctly there.

If that doesn't quite make sense, please PM me and I'll try to explain again.

[epichedgehog]

Hi, could someone please explain VCAA 2013 Exam 2 Graphs and Relations question 4? specifically, 4ei and 4eii? I've read the examiner's report but I don't really understand it still.
Thanks!

[MightyBeh]

Hi, could someone please explain VCAA 2013 Exam 2 Graphs and Relations question 4? specifically, 4ei and 4eii? I've read the examiner's report but I don't really understand it still.
Thanks!

For 4ei, you take your answers from the last part (4d/4c) and apply the rules they give you in the question. ($60 per powered, $30 per unpowered)
The cost function that comes out of that is:
cost = 60x + 30y

Then we just sub in our two minimums from 4d to find the lower cost;
(2,9) cost = 60(2) + 30(9) = $390
(3,8) cost = 60(3) + 30(8) = $420

So clearly (2,9) is the real minimum of the function, with the lowest cost being $390

For 4eii,
I'm not 100% sure on the logic, but here's itute's solutions which are a bit clearer than the VCAA's.

[TheMereCat]

Hi, could someone please explain VCAA 2013 Exam 2 Graphs and Relations question 4? specifically, 4ei and 4eii? I've read the examiner's report but I don't really understand it still.
Thanks!

For this question, you'd obviously have to refer to the solution region in the graph, you can see that there are some integer points that could be solutions in your graph right?

Since in this context we're talking about an integer number of things (power and unpowered camps) we really can't have the maximum at the corner vertex, since we're more than likely to get a decimal amount of camp sites, which is impossible.

By looking at the solution region, you can see the smallest integer number of powered and unpowered camp sites will occur at either 2,9 or 3,8. Anything beyond below that is a decimal or not in the solution region so it has to be one of those two points. using the objective function 60x+30y you find that the point 2,9 gives the minimum cost which is $390.

For question 4 eii you have to understand that since all girls have to use one campsite and all boys have to use one campsites,  it means that there are 24 boys at powered campsite and 24 girls and unpowered camp sites. powered campsites carry 6 people, 24/6 = 4, so X must equal to 4 powered campsites required is = 4. Unpowered campsites carry 4 people, 24/4 = 6.  unpowered campsites required would be 6.

So look at the solution region, you see that 4,6 isn't in that region and it has to match constraints y => 2x, to match this constraint y= 2* 4 and so y must = 8. so the points we have now are 4,8

Now just find the cost using the objective function 60x+30y, and you get 480$.



It's a bit long but i hope you understand..


EDIT: So MightyBeh beat me to it