1. ## Re: International Baccalaureate Marathon 2016

Im not much of a Physics person but just found this definition of magnetic dipole: "A magnetic dipole is the limit of either a closed loop of electric current or a pair of poles as the dimensions of the source are reduced to zero while keeping the magnetic moment constant."

Im going to say the magnetic dipole is not 2 since the average of those values is 2.000 0012

2. ## Re: International Baccalaureate Marathon 2016

Originally Posted by davidgoes4wce
Im not much of a Physics person but just found this definition of magnetic dipole: "A magnetic dipole is the limit of either a closed loop of electric current or a pair of poles as the dimensions of the source are reduced to zero while keeping the magnetic moment constant."

Im going to say the magnetic dipole is not 2 since the average of those values is 2.000 0012
That's still pretty close to 2, isn't it?

3. ## Re: International Baccalaureate Marathon 2016

Not sure if there are any IB markers in here but I came across a question like this tonight, which is equivalent to a Band 5 HSC question:

'A circular disc is cut into twelve sectors whose angles are in an arithmetic sequence. The angle of the largest sector is twice the angle of the smallest sector. Find the size of the angle of the smallest sector.'

The answer was 20 degrees, I used the answer in radians which was $\frac{\pi}{9}$

Would I get penalized if I left my answer in radians , even though it didn't specifically mention either?

4. ## Re: International Baccalaureate Marathon

***Renaming this thread since it is 2017, and stickying it for easy access***

5. ## Re: International Baccalaureate Marathon

Just confirming this was an error from TZ1 Mathematics Standard SL 2015 paper.

I learnt something new today , TZ1 is the Americas time zone.

6. ## Re: International Baccalaureate Marathon

^^ Because part (a) is wrong, part (b) which follows is also wrong.

7. ## Re: International Baccalaureate Marathon

Back to back mistakes on this paper. Part (i) is wrong .

If I was an IB marker, I would have penalized 1 mark in Q3 and 1 mark in Q4. Working out step by step was correct.

8. ## Re: International Baccalaureate Marathon

I came across this question tonight and have seen two answers:

9. ## Re: International Baccalaureate Marathon

Here are the 2 different solutions:

Which one is the correct one?

10. ## Re: International Baccalaureate Marathon

Also if anyone has an IB past exam papers or tests from their schools, it would be great if you could share it whether it be a school in NSW, Australia wide or internationally.

Cheers

11. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
Here are the 2 different solutions:

Which one is the correct one?
$\noindent If the cost per guest is G(n), then the total cost for n guests is nG(n). So the answer that did 45\times G(45) is correct.$

12. ## Re: International Baccalaureate Marathon

There are 15 schools in NSW that offer the IB. I would be keen to know which Options, as part of the Maths HL. Would be keen to know which Options each of the 15 schools offer?

13. ## Re: International Baccalaureate Marathon

Came across this question tonight in the Maths textbook Eg 11.7 , part(b):

Can anyone give some confirmation as to whether there is a mistake in the solutions for Q 10 (a)?

As you can see the questions are the exact same as the Worked Example part (b) but one answer has the $\pm \sqrt symbol$

14. ## Re: International Baccalaureate Marathon

I guess my thinking is why wouldn't you have a $\pm$ sign for the worked example as well?

15. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
I guess my thinking is why wouldn't you have a $\pm$ sign for the worked example as well?
$\noindent We don't need then \pm sign because we are only asked to find \emph{a} desirable vector (so only need one).$

16. ## Re: International Baccalaureate Marathon

Came across this question this morning,

Maths SL Cambridge EX 6E, Q 2 (ii), I think there is a typo in the question or something.

The answer for (ii) is n=7

$u_n=u_1 \ r^{n-1}$

$In this question , u_{1}=20, r=\frac{5}{2}, u_n=4882.8125$

I tried entering in 8125=Un as a whole number as well and it gave me a decimal number.

17. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
Came across this question this morning,

Maths SL Cambridge EX 6E, Q 2 (ii), I think there is a typo in the question or something.

The answer for (ii) is n=7

$u_n=u_1 \ r^{n-1}$

$In this question , u_{1}=20, r=\frac{5}{2}, u_n=4882.8125$

I tried entering in 8125=Un as a whole number as well and it gave me a decimal number.
The last term is 4882.8125 as you have established already, then you got confused and thought that the last term is 8125 because you mis-read the comma (,) as a decimal point (.)

18. ## Re: International Baccalaureate Marathon

Well I tried subbing in Un=4882.8125, then tried Un=8215 both didn't get me the answer which was n=7/

This was my working out:

$u_n=u_{1} r^{n-1}$

$4882.8125=20 \ (\frac{5}{2})^{n-1}$

$244.140625 =(\frac{5}{2})^{n-1}$

$Taking logs of both sides log \ 244.140625 =(n-1) \ log (\frac{5}{2})$

$\frac{log \ 244.140625}{log \frac{5}{2}}+1=n$

$\therefore n=7$

OK, I eventually got the answer, after going through it again.

19. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
Well I tried subbing in Un=4882.8125, then tried Un=8215 both didn't get me the answer which was n=7/

This was my working out:

$u_n=u_{1} r^{n-1}$

$4882.8125=20 \ (\frac{5}{2})^{n-1}$

$244.140625 =(\frac{5}{2})^{n-1}$

$Taking logs of both sides log \ 244.140625 =(n-1) \ log (\frac{5}{2})$

$\frac{log \ 244.140625}{log \frac{5}{2}}+1=n$

$\therefore n=7$

OK, I eventually got the answer, after going through it again.
I agree that u_n = 4882.8125 yields n = 7, but I'm still confused as to why you tried u_n = 8125.

20. ## Re: International Baccalaureate Marathon

Originally Posted by He-Mann
I agree that u_n = 4882.8125 yields n = 7, but I'm still confused as to why you tried u_n = 8125.
I thought it was a typo in the book.

21. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
I thought it was a typo in the book.
No typo, ahah.

22. ## Re: International Baccalaureate Marathon

I came across this question today Question 3(i):

Here was my attempted working out:
$\overrightarrow{AB}= \begin{pmatrix} 1\\-2\\-2 \end{pmatrix} \ \overrightarrow{AC}= \begin{pmatrix} 4\\0\\-1 \end{pmatrix} \ \overrightarrow{BC}= \begin{pmatrix} 3\\2\\1 \end{pmatrix}$

$cos \ \theta =\frac{\overrightarrow{AB}.{\overrightarrow{AC}}}{ | \overrightarrow{AB} | | \overrightarrow{AC}|}=\frac{6}{3 \sqrt{17}}$

$\theta=61^\circ$

$cos \ \theta =\frac{\overrightarrow{AB}.{\overrightarrow{BC}}}{ | \overrightarrow{AB} | | \overrightarrow{BC}|}=\frac{-3}{3\sqrt{14}}$

$\theta=105.5^\circ$

$cos \ \theta =\frac{\overrightarrow{AC}.{\overrightarrow{BC}}}{ | \overrightarrow{AC} | | \overrightarrow{BC}|}=\frac{11}{\sqrt{238}}$

$\theta=44.5^\circ$

I tried to visualize it in MATLAB when drawing it and I can see that the angles are all acute angles. Just a bit confused why my calculations work out to be 105.5 degrees? (I understand that the 3rd angle should be 74.5 degrees as the answer)

23. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
I came across this question today Question 3(i):

Here was my attempted working out:
$\overrightarrow{AB}= \begin{pmatrix} 1\\-2\\-2 \end{pmatrix} \ \overrightarrow{AC}= \begin{pmatrix} 4\\0\\-1 \end{pmatrix} \ \overrightarrow{BC}= \begin{pmatrix} 3\\2\\1 \end{pmatrix}$

$cos \ \theta =\frac{\overrightarrow{AB}.{\overrightarrow{AC}}}{ | \overrightarrow{AB} | | \overrightarrow{AC}|}=\frac{6}{3 \sqrt{17}}$

$\theta=61^\circ$

$cos \ \theta =\frac{\overrightarrow{AB}.{\overrightarrow{BC}}}{ | \overrightarrow{AB} | | \overrightarrow{BC}|}=\frac{-3}{3\sqrt{14}}$

$\theta=105.5^\circ$

$cos \ \theta =\frac{\overrightarrow{AC}.{\overrightarrow{BC}}}{ | \overrightarrow{AC} | | \overrightarrow{BC}|}=\frac{11}{\sqrt{238}}$

$\theta=44.5^\circ$

I tried to visualize it in MATLAB when drawing it and I can see that the angles are all acute angles. Just a bit confused why my calculations work out to be 105.5 degrees? (I understand that the 3rd angle should be 74.5 degrees as the answer)

$\noindent The relevant triangle's angle is the angle between \overrightarrow{BA} and \overrightarrow{BC} but you did the angle between \overrightarrow{AB} and \overrightarrow{BC} instead. Since you negated the first vector (and left the second one unchanged), the angle you got is the supplement of the desired value.$

24. ## Re: International Baccalaureate Marathon

$\overrightarrow{BA}= \begin{pmatrix} -1\\2\\2 \end{pmatrix}$

$cos \ \theta= \frac{\overrightarrow{BA} . \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|}=\frac{1}{\sqrt{14}}$

$\theta =74.5 ^\circ$

That's the standard 1st year University level Linear Algebra question. I wouldn't have got that right if that had come up in an exam or test paper if they required working out. I learnt something new today.

If I could maybe break it into simpler words, I would have looked at the directions of the vectors and then factored into account the angles of the two vectors which they lie between.

25. ## Re: International Baccalaureate Marathon

^^ Building on from what I have said, if the vectors are meeting in opposite directions, I need to reverse one of them so that either they meet together or leaving the same point at the one time.

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