VCE General & Further Maths Question Thread!

[Jackie Chan]

Hi everyone,

If a question comes up in the finance section of the exam and asks for us to round to the nearest whole number (for payments) must we always round up? For example let's say you get 12.3 payments on the finance solver, do you have to go up to 13 if the question asks to the nearest whole number?
Thanks

I would say yes because only 12 payments will not pay off the whole loan; you would still have a bit left to owe back (that 0.3).

[zofromuxo]

Hi everyone,

If a question comes up in the finance section of the exam and asks for us to round to the nearest whole number (for payments) must we always round up? For example let's say you get 12.3 payments on the finance solver, do you have to go up to 13 if the question asks to the nearest whole number?
Thanks

Always round up

[Peter Griffin]

Need help for Graphs and the types of data which are used for each!
Using mainly the Australian Bureau of Statistics and the Cambridge textbook I have made this list of notes; however, when faced with multiple options in exam 1, I sometimes make a mistake as the data can possibly fall within more than one graph and its quite frustrating! So I am looking for someone with the perfect notes who can help me with this issue.

Two-Way Frequency table: categorical data

Bar Chart: Categorical data

Stacked or Segmented Bar Chart: Comparing two or more categorical variables
Frequency Table: Categorical or discrete numerical data

Dot Plots: Categorical and discrete numerical data

Stem and Leaf Plot: Discrete and continuous numerical data

Back to Back Stem and Leaf Plot: Categorical (i.e. Sex) Discrete and continuous numerical data

Histograms: Ungrouped Discrete and continuous numerical data

Box Plot: Categorical and discrete numerical data

Parallel Box Plot: Numerical data sets

Scatter Plot: Numerical data

Repost need help..

[Jackie Chan]

Repost need help..

I relied on the Cambridge Textbook's notes on this topic for these types of questions and sometimes it just doesn't work out as more than one graph is appropriate according to their notes. I think when it comes to this, you just have to use common sense and actually picture whether a type of graph would suit the data.

[Peter Griffin]

I relied on the Cambridge Textbook's notes on this topic for these types of questions and sometimes it just doesn't work out as more than one graph is appropriate according to their notes. I think when it comes to this, you just have to use common sense and actually picture whether a type of graph would suit the data.

Alright, cheers! Its the only type of question that annoys me because it isn't only one correct answer.. I hope the exam makes it easy in choosing which!

[zxcvbnm18]

Hi is there a quicker way to do multiple choice question 8 on the 2016 sample further maths exam 1 ?

[SMOKIE]

hey boys can anyone help me with some questions
Q5 and Q8 for graphs and relations
Q4 for matrices
Q24 from finance

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2017/nht/2017FM1-nht-w.pdf

[BlinkieBill]

Hey y'all,
just wanted to ask what are some of the more difficult concepts vcaa tests us on that I should revise?
Im doing networks and matrices, but it would be nice if someone could also tell me what some of the harder concepts in the financial module are?
thanksssssssss.

[BlinkieBill]

also, just wanna confirm that we dont need to know the sum of terms equation that was preivously required in number patterns?

and ... how can i use normCdf to check my normal distribution answers on my CAS??

[Opengangs]

hey boys can anyone help me with some questions
Q5 and Q8 for graphs and relations
Q4 for matrices
Q24 from finance

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2017/nht/2017FM1-nht-w.pdf

Q5: Clearly the  inequalities compose of: x <= 7, 2 <= y <= 6.
To find the equations of the lines, we find the slope by using the rise over run formula.
One of the lines has a gradient of 3, meaning the equation of the line is: y = 3(x - 3), and so y = 3x - 9
Now, we know that the inequality of that line is: y <= 3x - 9, since the shaded area is always sitting below the line.

Rearranging, we get: 3x - y => 9 (D)

Q8: We define the rate of change to be the slope at any given point, so we first find the rate for the first 15 minutes; this tells us about the rate at which BOTH hoses are leaving: 400/15 = 80/3 (this rate of change is const).
Now, we need to find the time it takes to fill 600L of water, and since we have the rate, we just need to find the time it takes.

80/3 = 600/t
t = 1800/80 = 45/2 = 22.5
We are given that it takes 9 minutes less to fill the full 600L than a, so let 22.5 = a - 9
a = 31.5

Now, we just need to find the slope between 15 and 31.5 minutes.
Thus, rate of change = (600 - 400)/(31.5 - 15) = 200/16.5 = 12.121212... which is approximately 12 minutes (A)

MODULE 1 - MATRICES
Q4: (A) clearly works: m_{1 1} = 2(1) + 1 = 3
(B): By considering i as 1, we need to verify that each element follows the rule.
m_{1 1} = 3 works
m_{1 2} = 2(1) + 2 = 4 works
m_{1 3} = 2(1) + 3 = 5 works
Thus, (B) also works.

It follows that (C) and (E) satisfies the rule, meaning that the matrix (D) doesn't.

SECTION A - RECURSION AND FINANCIAL MODELLING
Q24:
What we want to find is the principal amount - that is, P₀.
Consider the compound interest formula: P(1 + r/100)^n, being compounded annually.
To reduce it down to monthly installments, we need to convert the rate and the value of n so that we are compounding them over monthly installments.
P(1 + r/1200)^(n/12)
After six months, we get: P(1 + r/1200)^6 = 4418.80 (equation 1)
After two years, we get: P(1 + r/1200)^24 = 4862.80 (equation 2)

Substitute equation 1 into equation 2: 4418.80(1 + r/1200)^18 = 4862.80
~1.1 = (1 + r/1200)^18
Solving for r should give you 6.400089204... rounded to 6.4.

Now, substitute that into either equation 1 or equation 2.
P(1 + r/1200)^6 = 4418.80
Solving for P should give you $4, 279.9998904... ~~ $4, 280 (C)

[jemma.clayton]

Need some help with a matrices question from th 2011 VCAA exam 1.

Matrix A is a 3x3 matrix. Seven of the elements in matrix A are zero. Matrix B contains six elements, none of which are zero. Assuming the matrix product AB is defined, the minimum number of zero elements in the product matrix AB is:
A. 0
B. 1
C. 2
D. 4
E. 6

[Maya24]

I need some help with this question. It is from the 2011 exam.

[mtDNA]

I need some help with this question. It is from the 2011 exam.

Do x/seasonal index, so here do x/0.8. This will give you 1.25x. So to correct you increase by 25%

[hmdeadas]

Do you suggest i put my practice sacs / exmas i have done for maths in my bound reference book?

[zofromuxo]

Do you suggest i put my practice sacs / exmas i have done for maths in my bound reference book?

I would put only the questions you struggle with in your bound reference and not all the practice sacs/exams in it.