VCE Methods Question Thread!

[DBA-144]

Also, sometimes schools don't go exactly by the study design (I had hypergeometric probability in one of my SACs this year :O ) so it can often be good to ask the 'could this be relevant?' questions to your teacher if it's to prepare for SACs

Just something to note- hypergeometric distribution is still in study design, I think, as it can come as part of discrete probability distributions.

[AlphaZero]

Just something to note- hypergeometric distribution is still in study design, I think, as it can come as part of discrete probability distributions.

Actually, knowing the distribution itself is not in the study design. VCAA expect that you go through all the possible ways each event can occur when sampling without replacement for small sample sizes. (This is why they choose sample sizes no bigger than 4). However, using the hypergeometric distribution is permitted (which is why it's in the appendices). It was also part of the study design before 2006, but no longer is.

It definitely doesn't hurt to learn it though! It's rather simple and is often much faster calculations wise

[peachxmh]

Hi guys, I'm aware that this question has popped up on the forums previously as someone else asked before but I'm asking again because I didn't really understand the explanation they gave:

Find the maximal domain of . The answer is x < -2 or x ≥ 1.

So far, based on my knowledge I made the denominator > 0 (i.e. x + 2 > 0), which gets me x > -2. However, what I don't get is how the answer can then be x<-2? Also, to get x ≥ 1 would I have to equate the numerator to ≥ 0 and solve for x? Thanks :3

[Sine]

Hi guys, I'm aware that this question has popped up on the forums previously as someone else asked before but I'm asking again because I didn't really understand the explanation they gave:

Find the maximal domain of . The answer is x < -2 or x ≥ 1.

So far, based on my knowledge I made the denominator > 0 (i.e. x + 2 > 0), which gets me x > -2. However, what I don't get is how the answer can then be x<-2? Also, to get x ≥ 1 would I have to equate the numerator to ≥ 0 and solve for x? Thanks :3

The denominator of this expression can actually be less than zero, however in this scenario the numerator would also need to be negative (less than zero) in order for the negative signs to cancel out and give you a positive number within the square root.

hope this helps

[VanillaRice]

Hi guys, I'm aware that this question has popped up on the forums previously as someone else asked before but I'm asking again because I didn't really understand the explanation they gave:

Find the maximal domain of . The answer is x < -2 or x ≥ 1.

So far, based on my knowledge I made the denominator > 0 (i.e. x + 2 > 0), which gets me x > -2. However, what I don't get is how the answer can then be x<-2? Also, to get x ≥ 1 would I have to equate the numerator to ≥ 0 and solve for x? Thanks :3

EDIT: Sine has provided an answer above which is more specific to the method you described

I'm not sure of the methods described in the other posts, but this is how I would approach this question (I apologise in advance if this is the same method):
Perform long division on the fraction underneath the square root:

Then, sketch the graph of \(y = 1 - \frac{3}{x+2}\), which is a hyperbola. Since your original expression contained a square root, we want all values of x such that y is non-negative (i.e. touching, or above the x-axis). You can determine these x-values visually from your graph.

Hope that helps - post if you're still stuck

[aspiringantelope]

Two tanks contain equal amounts of water. They are connected by a pipe and 3000 litres of water is pumped from one tank to the other. One tank then contains 6 times as much water as the other. How many litres of water did each tank contain originally?

Can I please get a full detailed solution

I know
two pipes (t and r)
t = r (Litres) therefore the ratio is 1:1
t - 3000 = r + 3000
becomes 1:6
Can someone please help me now?

[MB_]

One tank then contains 6 times as much water as the other.

From this you know r + 3000 is 6 times t - 3000 so adjust your equation to 6(t-3000)=r+3000. You should be able to solve from there.

[aspiringantelope]

From this you know r + 3000 is 6 times t - 3000 so adjust your equation to 6(t-3000)=r+3000. You should be able to solve from there.

Ok thanks! Got it!

[peachxmh]

I have another question lol - any help would be appreciated! I don't really get the notation of the question, hence I have no idea what I need to do to solve it :/

If A = (-6,1) and B = (6, -5), find the coordinates of P where P ∈ AB and AP : PB = 2:1

Edit: Following on from my previous question, how do you find the magnitude of ∠BAC, correct to one decimal place, if C = (4, 11)?

[Eric11267]

I have another question lol - any help would be appreciated! I don't really get the notation of the question, hence I have no idea what I need to do to solve it :/

If A = (-6,1) and B = (6, -5), find the coordinates of P where P ∈ AB and AP : PB = 2:1

P is a point along the line that joins the points A and B. The ratio of the lengths of AP to PB is 2:1

[AlphaZero]

I have another question lol - any help would be appreciated! I don't really get the notation of the question, hence I have no idea what I need to do to solve it :/

If A = (-6,1) and B = (6, -5), find the coordinates of P where P ∈ AB and AP : PB = 2:1

Hey there, this visual should help. Think about it as: the point \(P\) is two-thirds of the way along the line segment \(AB\) from the point \(A\).



Edit: as for your 2nd question, use this visual. You can form two right-angled triangles like so. Add the blue and red angles to get the required result

[aspiringantelope]

A shopkeeper buys a crate of eggs at $1.50 per dozen. He buys another crate, containing 3 dozen more than the first crate, at $2.00 per dozen. He sells them all for $2.50 a dozen and makes $15 profit. How many dozens were there in each of the crates?

Thanks

[peachxmh]

Sorry for asking so many questions (trying to finish off my Methods holiday homework!) but there's another question I'm confused with:

Consider the system of equations and state the values of 'a' for which there is a unique solution:
1. x + 2y - z = 2
2. 2x + 5y - (a +2)z = 3
3. -x + (a - 5)y + z = 1

Context - There was a question before this one asking you to find x, y and z in terms of 'a' which I did (and checked with the solutions so they're correct): , and (btw this is unrelated to the q but how do you put all the fractions on the same line using LaTeX instead of them being on different lines?).

The solutions are all real numbers except a=3 and a=0. To my understanding, you equate the gradients of the lines to each other and also the y-intercept, and then exclude the resulting values to remove the cases where the lines are parallel and when they are the same. How do you do this with three simultaneous equations?

So far, I've figured out the solution a=3 (but I'm not sure I used the right method) - I equated the gradient of the third equation to the gradient of the first equation, and got a=3, but tried using this same method for the other lines, however I didn't get a=0 and got different answers.

Thank you so much in advance and sorry this is so long!

[aspiringantelope]

Also:
A boy is 24 years younger than his father. In two years time the sum of their ages will be 40. Find the present ages of father and son.

I can do this with guess and check but I can't do it through algebra -_-
I do
b = f - 24
b + 2 = f - 22
b + 2 + f - 22 = 40
b + f -20 = 40
b + f = 60
f - 24 + f = 60
2f - 24 = 60
2f = 84
f = 42
The answer is supposed to be 30 for father and 6 for son
f = b + 24
f + 2 = b + 26
f + 2 + b + 26 = 40
f + b + 28 = 40
f + b = 12
b + 24 + b = 12
2b + 24 = 12
2b = -12
b = -6
Negative?

[PhoenixxFire]

@aspiring hopefully this helps get you started on both of those questions. It’s been a long time since I’ve done this sort of maths though so someone say something if I messed it up haha

You may have to click on the image to unstretch it

Spoiler

Edit: lol I did screw it up, one sec I’ll fix it
Edit2: never mind, MB beat me to it. Just add 4 to that side of the equation.