VCE General & Further Maths Question Thread!

[LiquidPaperz]

thanks guys. also had this question bth q3 and q4, idk if im doing something stupid but i dont know how the answers are C and D respectively

[BLACKCATT]

Pretty sure that is a further question, and it's not in current study design anymore (from like year 2000? lol).

[LiquidPaperz]

yeah i just found that out, thanks

btw can someone help me out with these sort of questions, i know how to find the equations, but if it asks us when t=4, for example, why cant we allocate 2 or 1.9999 to the first equation and then the remaining, either 2 or 2.01 etc to the second one, why do we sub the 4 into the second one? because there is different speeds (in this case) for the two equations, it was in the graphs and relations module if you wanted to know http://www.vcaa.vic.edu.au/Documents/exams/mathematics/03fmexam2.pdf

thanks

[Moderator Action] Split from Methods board and moved to Further. LiquidPaperz please keep your Further questions over here!

[Billion]

I believe this question was asked by someone else, but they never received an answer.
Thanks.

[AngelWings]

@ The person who asked the question which Billion has posted:

I believe this question was asked by someone else, but they never received an answer.
Thanks.

The y-axis just seems to be mislabeled (perhaps due to the fickle nature of computers). Count the boxes (1 box = 1 delivery) and use that to find the median. Tedious, but that's about all I can suggest.

If you're still unsure...

Spoiler

The total adds up to 49 deliveries.
Median should be th term. (Please excuse that type of notation.) Therefore, 25^th value.
The 25th value is in the 3^rd category or the 10-15 one. This eliminates option E.
Note that this is a histogram, where the categories are continuous...
which means that we use the mid-point of this category. i.e. midpoint of 10 and 15 = 12.5
I believe the answer you are looking for is C.

Please correct me if I am wrong, as I can't recall everything at the moment and have left my reference book elsewhere.

[Billion]

@ The person who asked the question which Billion has posted:The y-axis just seems to be mislabeled (perhaps due to the fickle nature of computers). Count the boxes (1 box = 1 delivery) and use that to find the median. Tedious, but that's about all I can suggest.

If you're still unsure...

Spoiler

The total adds up to 49 deliveries.
Median should be th term. (Please excuse that type of notation.) Therefore, 25^th value.
The 25th value is in the 3^rd category or the 10-15 one. This eliminates option E.
Note that this is a histogram, where the categories are continuous...
which means that we use the mid-point of this category. i.e. midpoint of 10 and 15 = 12.5
I believe the answer you are looking for is C.

Please correct me if I am wrong, as I can't recall everything at the moment and have left my reference book elsewhere.

Hey, thanks for taking the time to answer the question.
However I too, got C. But the solutions say B.
(I did this practice exam today, and I remember the question was asked before, but can't find out if someone figured it out)

The solutions are attached, it talks about stuff from units 1 and 2 (I didn't do those units). However it switches mid-sentence and says the mean is $11, would that be the problem?
Thanks.

[myanacondadont]

I'm fairly sure this question is on a page previous somewhere haha, someone answered it iirc? Anyhow I remember how to do yet I remember people condemning it for such specificity.

I believe the answer is B, if not please ignore this.

Total deliveries = 49 therefore median = 25th value.

If we count from the left hand side of the graph, the 25th value occurs in the 10-15 region. This is where people were saying it shouldn't require this type of working out;
 
So counting from the left hand side, we get to 23 before it enters the 10-15 region. This means there is 2 points of data left over before we find the 'true' median.

Therefore since there is 10 values in the 10-15 region we can (much to my disgust) say

which = 1. So we can conclude it is 1 dollar into the 10-15 region. Hence 11$.

edit: just restating that I seriously doubt vcaa would do a question like that - seems way too uhh, confusing?

[LiquidPaperz]

how do you know AM = AN do the 45 degree angles always mean this?

[Zealous]

how do you know AM = AN do the 45 degree angles always mean this?

Well yes - provided there is also a right angle. The triangle has a right angle and a 45 degree angle which means it is a right angled isosceles triangle - two sides of equal length and one hypotenuse. It might be worthwhile looking up the conditions for special types of triangles, it can be super useful in exams.

[LiquidPaperz]

is there anything specific youd have a link to Zealous?

Thanks

[Snorlax]

Both these questions are from the VCAA 2013 E1 (Module 3)
Q7) I would like to know a method for questions like these.


Q8) Don't understand it at all...

[myanacondadont]

Q7:  So we have to find the constant. We know it's already H versus 1/d^2.

If we consider that then we have to find the relationship between , where k is the unknown.

Soooo, they give us the point (0.44, 25.7). Remember if they give you information like this - you're usually going to have to use it.
In this case we can consider that 25.7 is unaltered (as there is no transformation on the y axis)
But with 0.44 it is ALREADY altered ()

Therefore to find the relationship it becomes

Subbing in values :                  

Now we have to put it back into the relationship. HOWEVER be careful as we have to put it into
Therefore answer is A. Not sure if I explained this adequately.

[Billion]

Hi, I got the correct answer, however the assessor's report says there was a "relatively routine computation"
I calculated the correct answer by extending the triangle. Is there another, more efficient method?
Thanks.

[LiquidPaperz]

with the corner principle for graphs and relations, if they are all integers - say for a given context - is their ANY possible way to maximise the  objective function from points within. eg. if corners are 1,8    6,2    and    1,2     , is there any way it can be 4,3 for example?

[LiquidPaperz]

Hi, I got the correct answer, however the assessor's report says there was a "relatively routine computation"
I calculated the correct answer by extending the triangle. Is there another, more efficient method?
Thanks.

Hey man, this is how i do it, i dont think there is another simple way

you like the colours haha?