VCE General & Further Maths Question Thread!

[Patches]

Depends on the question. Some will say x can be reduced by 3 hours, some others will say y can be reduced to 3 hours. It'll be pretty clear which applies, I would think.

[jimmy22]

Hi,
What is the main thing to look for when trying to see if a time series has cyclic trend?
thanks

[Huntyy]

Range divided by 6. You see, this method could only be applied if your data distribution was approximately symmetric with no extreme values (or outliers).

Just did this question myself and got it wrong. So the range is 16 from the histogram, if you divide that by 6 you get a standard deviation of 2.33 which would be rounded to option b (incorrect answer) rather than the correct answer, c. Also why do you divide the range by six, is it something to do with the number of columns? Which is 14 in total?

Please help, thanks

[Patches]

Dividing the range by four or six to find the standard deviation only works if it's completely symmetrical - definitely would not do it in the exam. Just put the values into your calculator and calculate it there.

[tae]

For business maths.. struggling with reducing balance q's. From VCAA 2008.

Question 8
A loan of $300 000 is taken out to finance a new business venture.
The loan is to be repaid fully over twenty years with quarterly payments of $6727.80.
Interest is calculated quarterly on the reducing balance.
The annual interest rate for this loan is closest to:
A. 4.1%
B. 6.5%
C. 7.3%
D. 19.5%
E. 26.7%

Answer is B :\

[plato]

For business maths.. struggling with reducing balance q's. From VCAA 2008.

Question 8
A loan of $300 000 is taken out to finance a new business venture.
The loan is to be repaid fully over twenty years with quarterly payments of $6727.80.
Interest is calculated quarterly on the reducing balance.
The annual interest rate for this loan is closest to:
A. 4.1%
B. 6.5%
C. 7.3%
D. 19.5%
E. 26.7%

Use your TVM with:

N = 80 (20 lots o 4 quarters)
I% = ?? (the interest rate you want to find)
PV = 300000 (positive as money comes to you)
PMT = -6727.80 (negative cos you have to pay it out)
FV = 0 (nothing left to pay after 20 years)
P/Y = 4 (number of payments per year)
C/Y = 4 (number of interest calculations per year)

[Zealous]

Got a quick question =)

VCAA FM2011 Exam 1
Geometry and Trigonometry Q9
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011furmath1-w.pdf Page 18

How can we be sure that triangles ABC and ABD are the similar triangles? or a better question: is there some way we can completely disprove the triangles ABC and BCD are similar (given we are given one angle and one side length)?
It does make sense that ABC and ABD would be similar, but when you look at BCD, with a bit of flipping etc. you can get a triangle that looks very similar to ABD, which sort of led me on the wrong track.

[serendipity21]

Can someone please help me with this question? Crashing questions are my weakness

[plato]

Got a quick question =)

VCAA FM2011 Exam 1
Geometry and Trigonometry Q9
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011furmath1-w.pdf Page 18

How can we be sure that triangles ABC and ABD are the similar triangles? or a better question: is there some way we can completely disprove the triangles ABC and BCD are similar (given we are given one angle and one side length)?
It does make sense that ABC and ABD would be similar, but when you look at BCD, with a bit of flipping etc. you can get a triangle that looks very similar to ABD, which sort of led me on the wrong track.

Angle ABD = angle ACB = theta
The angle at A is common to triangles ABC and ABD
ie angle ABD = angle ABC
Now that the two triangles ABC and ABD have two angles in common, it must mean that the third angle is also the same for each triangle.
ie angle ADB = angle ABC

With three angles in common, these triangles must then be similar.

[Zealous]

Angle ABD = angle ACB = theta
The angle at A is common to triangles ABC and ABD
ie angle ABD = angle ABC
Now that the two triangles ABC and ABD have two angles in common, it must mean that the third angle is also the same for each triangle.
ie angle ADB = angle ABC

With three angles in common, these triangles must then be similar.

Awesome! thanks plato.

[plato]

Can someone please help me with this question? Crashing questions are my weakness

There are five paths.
BGIJ - 17
BDFJ - 18
BDCEHJ - 19
AFJ - 14
ACEHJ - 15

You could assume you cannot reduce J to zero but, if you could, it means every one of these paths is reduced by one and the critical path would be 18 long.

You need to reduce an activity on the critical path, BDCEHJ.
You can only reduce one of C, E or H by one hour, otherwise you make BDFJ become a critical path of 18 which means only a one hour reduction.

D cannot be reduced below zero and so you must reduce B which will also reduce paths BGIJ, BDFJ and BDCEHJ. But you cannot reduce these paths to be less than 15 which, if you did, is where ACEHJ would become the single, new critical path (and also activity B would disappear with zero time).

So – reducing B (and the project) by four hours will make two critical paths, ACEHJ and BDCEHJ, both 15 hours long. This is a four hour reduction and the answer is D.

[tcstudent]

Hey everybody, could i please get help with this exam

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011furmath2-w.pdf

Q2) Graphs and relations about Mike's Hike, i dont understand the whole question

Thank you.

[serendipity21]

There are five paths.
BGIJ - 17
BDFJ - 18
BDCEHJ - 19
AFJ - 14
ACEHJ - 15

You could assume you cannot reduce J to zero but, if you could, it means every one of these paths is reduced by one and the critical path would be 18 long.

You need to reduce an activity on the critical path, BDCEHJ.
You can only reduce one of C, E or H by one hour, otherwise you make BDFJ become a critical path of 18 which means only a one hour reduction.

D cannot be reduced below zero and so you must reduce B which will also reduce paths BGIJ, BDFJ and BDCEHJ. But you cannot reduce these paths to be less than 15 which, if you did, is where ACEHJ would become the single, new critical path (and also activity B would disappear with zero time).

So – reducing B (and the project) by four hours will make two critical paths, ACEHJ and BDCEHJ, both 15 hours long. This is a four hour reduction and the answer is D.

Plato you are amazing! Explained it better than my teacher!

Thank you so much!

[serendipity21]

Hey everybody, could i please get help with this exam

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011furmath2-w.pdf

Q2) Graphs and relations about Mike's Hike, i dont understand the whole question

Thank you.

Hey everybody, could i please get help with this exam

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011furmath2-w.pdf

Q2) Graphs and relations about Mike's Hike, i dont understand the whole question

Thank you.

Hi tcstudent!
The graph is showing the distance Mike travels. During the hours where the graph is in a slope he's travelling at a constant speed, and where the graph is just a horizontal line he is stationary (as his distance is not changing.

a) Average speed is given by the total distance he travels/total time taken.
He travels 16km in 7 hours so, 16/7=2.3 km/h

b) The equation with the variables they want you to find is the equation of the second slope (between 3 to 7 hours).
You can use simultaneous equations to solve this.
The variables from the graph are t (for the x axis) and d for the y axis, (as the equation is in the form y=mx+c). So pick two points on the line, say (5,13) and (3,10) and sub them in to the equation. (Pick clear points so you don't have to estimate).
The two equations I got from these points are:
13=a5+b  and 10=a3+b. Plug them in to your CAS and you get a=3/2 and b=11/2 which in decimals are 1.5 and 5.5.

c) I made a mistake with this one first time I did it.
Sketching the line for the distance traveled by Katie first is a good thing to do. This also helps with the next question!
The line intersects Michael's line at t=2. Therefore they meet at 2 hours.
However just to be super careful incase it's not like 2.1 or something, double check in the exam if you have time!
Use the equation for the first slope in Michael travels (d=5t, get the five from the rise/run of the line) and make it equal the the equation for Katies distance travelled.
Plug in 5t=-3t+16 into the cas to get t=2 hrs.

d) As you can see from the lines on the graph, the distance between Michael and Katie becomes 3km somewhere between 1-2 hours until they finally meet at 2.
So since the distance is the y value here, makes Katies distance-Michael's=3 to find t when their distance apart is 3
-3t+16-5t=3
solve on CAS, t=1.625
After two when they're moving away from each other, you can see just from the graph that at t=3, Katie is at 7km and Michael is at 10km from his destination, difference of 3km.
So they are only 3km near each other from t=1.625 and t=3
3-1.625=1.38 km

Hope that helps!

[Colokid]

help with this question please