VCE Methods Question Thread!

[keltingmeith]

Am I correct in thinking that checkpoints is basically a collection of past vcaa exams, so doing it before starting vcaa practice exams would "spoil" them?

Not quite. There's so many questions, it's impossible to have them memorised by the time you're doing practice exams. Particularly if you focus on checkpoints now, and try to not to touch them about a month (even just a week or two) before you start doing practice exams.

[snowisawesome]

Thanks guys for the help

Can you rote learn methods as a subject and get 35 raw without too much understanding?

Sine, what's your opinion on this?

A metal worker is required to cut a circular cylinder from a solid sphere of radius 5 cm. A cross-section of the sphere and the cylinder is shown in the diagram? (don't have diagram sorry)
a. Express r in terms of h, where r cm is the radius of the cylinder and h cm is the height of the cylinder. Hence show that the volume, V cm^3 of the cylinder is given by V = (1/4)(pi)(h)(100-h^2)
I attempted the question, could anyone please tell me where i've gone wrong
V = (1/4)(pi)(h/2)((100-(h^2/2))

An open tank is to be made from a sheet of metal 84cm by 40cm by cutting congruent squares of side length x cm from each of the corners.
State the maximal domain for V when it is considered as a function of x.
Find the values of x for which the capacity of the tank is 10 litres
Find the maximum capacity of the tank in cubic centimetres
Could someone please explain how to do these?
Thanks

[jazzycab]

A metal worker is required to cut a circular cylinder from a solid sphere of radius 5 cm. A cross-section of the sphere and the cylinder is shown in the diagram? (don't have diagram sorry)
a. Express r in terms of h, where r cm is the radius of the cylinder and h cm is the height of the cylinder. Hence show that the volume, V cm^3 of the cylinder is given by V = (1/4)(pi)(h)(100-h^2)
I attempted the question, could anyone please tell me where i've gone wrong
V = (1/4)(pi)(h/2)((100-(h^2/2))

For this one, I imagine you've made a mistake in the way that you've defined r in terms of h, or your algebraic manipulation of the expression, but it's far too difficult to tell without both the diagram and your working out.

I imagine the diagram is such that four points of the cylinder (that make up a cyclic quadrilateral) touch the surface of the sphere. In this case, the radius of the sphere can be drawn from the centre of the sphere to one of these points, which can, in turn, be connected to the centre of the base of the cylinder (the cylinder radius, r), then to the centre of the sphere (half of h) forming a right-angled triangle.
Using Pythagoras, you would get:

Working through the algebra and arithmetic from here should give the equation stated in the answers.

An open tank is to be made from a sheet of metal 84cm by 40cm by cutting congruent squares of side length x cm from each of the corners.
State the maximal domain for V when it is considered as a function of x.
Find the values of x for which the capacity of the tank is 10 litres
Find the maximum capacity of the tank in cubic centimetres
Could someone please explain how to do these?
Thanks

For this one here, I can only assume the 'tank' is a rectangular prism (given that the metal sheet is rectangular and squares are cut from the corners).
We are cutting x from 2 places on each side of the rectangle. This means each base dimension is reduced by 2x (and the tank height would be the cut length, x)
That is:

The maximal domain can be found by considering the physical restrictions on the dimensions (i.e. each dimension must be positive)

The expression:

is the actual volume (or capacity) of the tank (however, it is in cubic centimetres). So to answer the second part, you need to convert (1 cubic millimetre = 1 millilitre).

To find the maximum capacity, Find where the derivative of the volume function is equal to zero.

[snowisawesome]

V = (1/4)(pi)(h/2)((100-(h^2/2))
This was my answer which was wrong. I didn't show any working out for it

Also, for the question where it asked for the maximal domain for V, the answer said (0,20)
but if 84-2x>0
then -2x> -84
then x<42
so then wouldn't the domain be (0,42)

How similar are the exercise questions and chapter review questions in mathsquest, since i have access to worked solutions for the exercise questions, but not for the chapter review questions?

[jazzycab]

V = (1/4)(pi)(h/2)((100-(h^2/2))
This was my answer which was wrong. I didn't show any working out for it

How did you get that expression? I guess, given that you have everything in terms of h, the question is how did you get r in terms of h? That is probably where you've made the error.

Also, for the question where it asked for the maximal domain for V, the answer said (0,20)
but if 84-2x>0
then -2x> -84
then x<42
so then wouldn't the domain be (0,42)

Although this is true, you need to consider ALL dimensions to identify the overall domain.
We also have that 40-2x>0 and x>0
This gives x<20 intersecting with x>0 and x<42
The net result is (0,20) for your domain.

How similar are the exercise questions and chapter review questions in mathsquest, since i have access to worked solutions for the exercise questions, but not for the chapter review questions?

I'm not super familiar with MathsQuest, but typically in most textbooks the review questions are somewhat comparable with the exercise questions, but with far less repetition and often more emphasis on extended response

[HighSchoolerRS]

Hi guys, I was hoping somebody could help me do this probability question. I was able to get the right answer for part c but I cannot seem to get the other parts.
Thank you!

[TheAspiringDoc]

Hi guys, I was hoping somebody could help me do this probability question. I was able to get the right answer for part c but I cannot seem to get the other parts.
Thank you!

For a and b, odd/even is just determined by the final digit, so it's just 50% for each, I think.
Since you can do c, just use the conditional probability formula to solve d
Edit: I was wrong. Good pickup!

[HighSchoolerRS]

For a and b, odd/even is just determined by the final digit, so it's just 50% for each, I think.
Since you can do c, just use the conditional probability formula to solve d

I initially thought that it would be 50% too but then I realised that the first digit cannot be 0 and the question just got confusing after that. The answer for a is supposed to be 25/49 and 24/29 for b, according to the textbook but I don't understand why.

[RuiAce]

I initially thought that it would be 50% too but then I realised that the first digit cannot be 0 and the question just got confusing after that. The answer for a is supposed to be 25/49 and 24/29 for b, according to the textbook but I don't understand why.

(6 for the first digit to cater for no zero, and then the rest is straightforward.)


It is certainly possible to consider the number of even 4-digit numbers that can be formed first. To do this, however, you must address the first digit problem as you said. Therefore, you would have to break the cases up - 0 being the last digit, and 0 not being the last digit.

[Lear]

(I don’t think this is against forum rules but if it is mods feel free to remove this)
Does anyone have vcaa methods exams with solutions from 1990s-2005?

[Sine]

(I don’t think this is against forum rules but if it is mods feel free to remove this)
Does anyone have vcaa methods exams with solutions from 1990s-2005?

From the vcaa page

Each year from 1997 to 2009 schools have been sent a CD ROM containing that year’s VCE exams and GAT. Please ask the librarian or the VCE Coordinator at your school about access to past exam papers on this CD ROM. Exams prior to 1997 are not available.

so ask your school.

I've got the mid to late 90's exams too but only in hard copy :/ 

[Lear]

Thank you Sine, I will ask my library.

[jazzycab]

Thank you Sine, I will ask my library.

I have 2002 exams onwards (pdf). If you have any luck with earlier ones, can you let me know please? My library doesn't have them and doesn't seem to know why the disclaimer is on the VCAA website. Mind you, the library staffing and general staffing has changed significantly in the last decade or two.

[snowisawesome]

This is a worked example in the mathsquest 12 methods textbook, for exercise 1.4
worked example 12.b.
2sin^2(θ) + 3sin(θ) - 2=0
Let A = sin(θ)
2A^2 + 3A - 2 = 0
(2A-1)(A+2) = 0
A = 1/2, A = -2,
so
sin(θ) = 1/2 or sin(θ) = -2

for sin(θ) = 1/2
base = pi/6
θ = pi/6, 5pi/6

sin(θ) = -2
there is no solution as -1 ≤ sin(θ)  ≤ 1
can someone please explain why -1 ≤ sin(θ)  ≤ 1
thanks

[smamsmo22]

Hi, I just had a quick question/inquiry about parameters (in the subject of linear simultaneous equations). We went over them very briefly in class and I understand how to use the calculator to formulate an answer etc but I was wondering if we needed to be able to formulate this answer by hand? Edrolo's explanation only covered calculator use but I wasn't sure if we'd ever have to do it by hand, as I came across a question asking for simultaneous equation solutions which included parameters and naturally started doing it by hand but had some difficulties with working it out clearly.
If anyone knows if working them out by hand is necessary, and if so, any basic tips of what to look for (I understand a lot of it does just involve rearranging) to start off doing so, that would be super helpful! Thanks