VCE Further Maths Exam 1 Discussion and Solutions

[KidClutch]

Are significant figures different to decimal places? eg. would the number 1.2345 to three significant figures be 1.235 or 1.23

Yes. They are different. 3 sig figs is 1.23

[hmdeadas]

The mean score last year was 50.9/80. You got 54/80. Also, this year's mean is almost certainly much lower than last year's, which means you will get in the low to mid 30s if you do equally well in exam 2. But for now, do as many previous exam 2s as possible in preparation for monday! Its still possible for you to get high 30s if you do well enough in it

Thank You! Makes me feel a bit better haha. Only aiming for a 32/33 so hopefully it works out :-)

[hmdeadas]

Can someone explain question 5 on NETWROKS (maths exam), I got 4 but can't get question5. ALSO question6 netwroks please, i looked at the solutions but it still doesn't make sense, how do i handle these types of questions
Thanks

[Educator]

Hi all, new user.
regarding core question 11,
Could someone explain to me why a seasonal index can never be negative?
For instance

Q1: 20
Q2: -50
Q3: -75
Q4: 10

What are the seasonal indices.

TIA

[addictwithatextbook]

Can someone explain question 5 on NETWROKS (maths exam), I got 4 but can't get question5. ALSO question6 netwroks please, i looked at the solutions but it still doesn't make sense, how do i handle these types of questions
Thanks

Questions 5 essentially asks the number of activities on the critical path(s). If you complete forward and backward scanning on the activity network for the question, you will find that activities C-F-H-M have the same EST and LST, and thus are on the only critical path. As any activity on the critical path cannot be delayed without increasing the project's minimum time, and that there are four activities on the critical path, the answer is 4 (C).

Question 6 made me muck up a few times. A good way to approach this question is to label the degree of each vertex, and it will result in the top, centre, and bottom two vertices being odd, making a total of four odd vertices. The question asks how many ways an edge can be removed to make an Eulerian trail. An Eulerian trail can only occur if there are two odd vertices. Removing any edge connecting two odd vertices will decrease their vertex to an even. (For example, removing the edge between the top vertex and centre vertex will decrease their vertex to 2 and 4 respectively). This will result in only two odd vertices left, allowing for an Eulerian trail to occur. In the graph, there are five edges connecting only the odd degree vertices, therefore there are 5 ways (E) to remove an edge for an Eulerian trail to be possible.

[dec.hargreaves]

Questions 5 essentially asks the number of activities on the critical path(s). If you complete forward and backward scanning on the activity network for the question, you will find that activities C-F-H-M have the same EST and LST, and thus are on the only critical path. As any activity on the critical path cannot be delayed without increasing the project's minimum time, and that there are four activities on the critical path, the answer is 4 (C).

Question 6 made me muck up a few times. A good way to approach this question is to label the degree of each vertex, and it will result in the top, centre, and bottom two vertices being odd, making a total of four odd vertices. The question asks how many ways an edge can be removed to make an Eulerian trail. An Eulerian trail can only occur if there are two odd vertices. Removing any edge connecting two odd vertices will decrease their vertex to an even. (For example, removing the edge between the top vertex and centre vertex will decrease their vertex to 2 and 4 respectively). This will result in only two odd vertices left, allowing for an Eulerian trail to occur. In the graph, there are five edges connecting only the odd degree vertices, therefore there are 5 ways (E) to remove an edge for an Eulerian trail to be possible.

i agree. question 6 was annoying at first but eventually got it

[darcypeaceee]

Could someone please explain the working out for question 8 matrices?
Thank you in advance

[Paolo Grande]

Does anybody have worked solutions that they can post?

[Educator]

Hi,
Regarding my earlier query, are we now all agreed that seasonal indices can be negative and therefore D cannot be a correct answer to Core question 11?

[dec.hargreaves]

Hi,
Regarding my earlier query, are we now all agreed that seasonal indices can be negative and therefore D cannot be a correct answer to Core question 11?

This questions seems to have been argued a bit. Which answer would you have chosen? I chose D because I think the rest can definitely be negative

[Educator]

The question does not have a correct answer because all 5 options can definitely be negative. I will contact the VCAA to alert them to the fact. Everyone should get full marks for this question.

[dec.hargreaves]

How can a seasonal index be negative? This is new to me

[Educator]

Try calculating the seasonal indices for the following raw data:

Q1: 10
Q2: 30
Q3: -5
Q4: 20

[dec.hargreaves]

Okay. So this is how you would do it (correct me if I'm wrong):
Seasonal Index = value for season/seasonal average
10+30+(-5)+20 = 55
55/4 = 13.75 (seasonal average)
10/13,75 = 0.7272727... SI-1
30/13.75 = 2.1818181... SI-2
-5/13.75 = -0.3636363... SI-3
20/13.75 = 1.4545454... SI-4
SI-1 + SI-2 + SI-3 + SI-4 = 4.000000001
seasonal indices can be negative.
thanks for opening this up to me. I thought that because they usually have to do with sales made in a season (year, quarter, month) they had to be positive because you can't have negative amount of sales

[Educator]

Exactly, thanks for doing the numbers.
That's why I opened it up for discussion.