IB Maths Marathon (2 Viewers)

davidgoes4wce · Feb 3, 2017

Re: International Baccalaureate Marathon

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

davidgoes4wce · Feb 3, 2017

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.

davidgoes4wce · Feb 4, 2017

Re: International Baccalaureate Marathon

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

davidgoes4wce · Feb 4, 2017

Re: International Baccalaureate Marathon

Here are the 2 different solutions:

Which one is the correct one?

davidgoes4wce · Feb 9, 2017

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

InteGrand · Feb 9, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
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.$

davidgoes4wce · Feb 12, 2017

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?

davidgoes4wce · Feb 13, 2017

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 $$

davidgoes4wce · Feb 13, 2017

Re: International Baccalaureate Marathon

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

InteGrand · Feb 13, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
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).$

davidgoes4wce · Feb 13, 2017

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.

He-Mann · Feb 13, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
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 (.)

davidgoes4wce · Feb 13, 2017

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.

He-Mann · Feb 13, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
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.

davidgoes4wce · Feb 13, 2017

Re: International Baccalaureate Marathon

He-Mann said:
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.

He-Mann · Feb 13, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
I thought it was a typo in the book.

No typo, ahah.

davidgoes4wce · Feb 14, 2017

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)

InteGrand · Feb 14, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
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.$

davidgoes4wce · Feb 14, 2017

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.

davidgoes4wce · Feb 14, 2017

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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