1. ## Re: International Baccalaureate Marathon

Was going through worked example from Cambridge tonight and there was a mistake in the working out for part (c), vector BC should be -2-k. Small mistake but its important that I point it out to all IBers in 2017:

2. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
Was going through worked example from Cambridge tonight and there was a mistake in the working out for part (c), vector BC should be -2-k. Small mistake but its important that I point it out to all IBers in 2017:

Wow, how much maths do you do a day?

3. ## Re: International Baccalaureate Marathon

Originally Posted by boredofstudiesuser1
Wow, how much maths do you do a day?
Not that much. I just went to the gym for about an 1.5 hour, also did a bit of driving and I went out to the library for a bit of fresh air.

4. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
Was going through worked example from Cambridge tonight and there was a mistake in the working out for part (c), vector BC should be -2-k. Small mistake but its important that I point it out to all IBers in 2017:

What do you mean? The vectors b and c have no k, so vector BC would have no k.

5. ## Re: International Baccalaureate Marathon

Originally Posted by InteGrand
What do you mean? The vectors b and c have no k, so vector BC would have no k.
I meant vector AB, should have been -2-k.

6. ## Re: International Baccalaureate Marathon

$\noindent Oh yeah, they accidentally did -2-2k instead.$

7. ## Re: International Baccalaureate Marathon

$Points A,B and C have position vectors A \begin{pmatrix} 1\\-19\\5 \end{pmatrix}, B \ \begin{pmatrix} 3.2\\3.6\\2 \end{pmatrix} , C \ \begin{pmatrix} -6\\-15\\7 \end{pmatrix}$

$The respective magnitudes: |\overrightarrow{AB}|=\sqrt{524.6} , |\overrightarrow{AC}|=\sqrt{69} , |\overrightarrow{BC}|=\sqrt{455.6} , |\overrightarrow{CB}|=\sqrt{455.6}$

$Find the Area of the triangle ABC$

$I preceded to find vectors \overrightarrow{AB}, \overrightarrow{AC} and \overrightarrow{BC}$

$This is a diagram I drew and then I preceded to work out the individual angles of each of the vertices.$

$\overrightarrow{AB}=\begin{pmatrix} 2.2\\22.6\\-3 \end{pmatrix} \ , \ \overrightarrow{AC}= \begin{pmatrix} -7\\4\\2 \end{pmatrix} \ , \overrightarrow{BC}= \begin{pmatrix} -9.2\\-18.6\\5 \end{pmatrix} \ , \overrightarrow{CB}= \begin{pmatrix} 9.2\\18.6\\-5 \end{pmatrix}$

$cos \ \theta=\frac{\overrightarrow{AB} \ . \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|}=\frac{69}{\sqrt{524.6} \times \sqrt{69}} \implies \theta=68.7^\circ$

$cos \ \beta=\frac{\overrightarrow{BC} \ . \overrightarrow{AC}}{|\overrightarrow{BC}| |\overrightarrow{AC}|}=0 \implies \beta=90^\circ$

$cos \ \alpha=\frac{\overrightarrow{AB} \ . \overrightarrow{CB}}{|\overrightarrow{AB}| |\overrightarrow{CB}|} \implies cos \ \alpha=\frac{455.6}{\sqrt{524.6} \times {\sqrt{455.6}}}=21.3^\circ$

$As a test I then worked out the individual areas of the triangles, use either of the 3 methods and I didn't get the same values. Well one of them was close to the actual answer of 88.7$

$A=\frac{1}{2} |\overrightarrow{BC}| |\overrightarrow{AC}| \ sin \beta=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{69} \ sin \ 90^\circ=88.7 \ units^{2}$

$A=\frac{1}{2} |\overrightarrow{AC}| |\overrightarrow{AB}| \sin \theta=\frac{1}{2} \times \ 13 \ \sqrt{524.6} \ sin \ 68.7^\circ=88.6 \ units ^{2}$

$A=\frac{1}{2} |\overrightarrow{CB}| |\overrightarrow{AB}| \ sin \alpha=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{524.6} \times \ sin \ 21.3^\circ =88.8 units^{2}$

$Now I spent roughly a couple of hours doing this question tonight, to calculate the area of a triangle in vector form (I'm actually not sure how to word it properly), but what I take from doing a couple of questions today , is you either have to multiply 2 vectors coming in or 2 vectors coming out together to get the required area. So what I mean by this is, at point A in my diagram, both of the vectors are going out. At point C, both of the vectors are coming in. At point B, I had to reverse one of the vectors so that it showed them both coming in.$

8. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
$Points A,B and C have position vectors A \begin{pmatrix} 1\\-19\\5 \end{pmatrix}, B \ \begin{pmatrix} 3.2\\3.6\\2 \end{pmatrix} , C \ \begin{pmatrix} -6\\-15\\7 \end{pmatrix}$

$The respective magnitudes: |\overrightarrow{AB}|=\sqrt{524.6} , |\overrightarrow{AC}|=\sqrt{69} , |\overrightarrow{BC}|=\sqrt{455.6} , |\overrightarrow{CB}|=\sqrt{455.6}$

$Find the Area of the triangle ABC$

$I preceded to find vectors \overrightarrow{AB}, \overrightarrow{AC} and \overrightarrow{BC}$

$This is a diagram I drew and then I preceded to work out the individual angles of each of the vertices.$

$\overrightarrow{AB}=\begin{pmatrix} 2.2\\22.6\\-3 \end{pmatrix} \ , \ \overrightarrow{AC}= \begin{pmatrix} -7\\4\\2 \end{pmatrix} \ , \overrightarrow{BC}= \begin{pmatrix} -9.2\\-18.6\\5 \end{pmatrix} \ , \overrightarrow{CB}= \begin{pmatrix} 9.2\\18.6\\-5 \end{pmatrix}$

$cos \ \theta=\frac{\overrightarrow{AB} \ . \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|}=\frac{69}{\sqrt{524.6} \times \sqrt{69}} \implies \theta=68.7^\circ$

$cos \ \beta=\frac{\overrightarrow{BC} \ . \overrightarrow{AC}}{|\overrightarrow{BC}| |\overrightarrow{AC}|}=0 \implies \beta=90^\circ$

$cos \ \alpha=\frac{\overrightarrow{AB} \ . \overrightarrow{CB}}{|\overrightarrow{AB}| |\overrightarrow{CB}|} \implies cos \ \alpha=\frac{455.6}{\sqrt{524.6} \times {\sqrt{455.6}}}=21.3^\circ$

$As a test I then worked out the individual areas of the triangles, use either of the 3 methods and I didn't get the same values. Well one of them was close to the actual answer of 88.7$

$A=\frac{1}{2} |\overrightarrow{BC}| |\overrightarrow{AC}| \ sin \beta=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{69} \ sin \ 90^\circ=88.7 \ units^{2}$

$A=\frac{1}{2} |\overrightarrow{AC}| |\overrightarrow{AB}| \sin \theta=\frac{1}{2} \times \ 13 \ \sqrt{524.6} \ sin \ 68.7^\circ=88.6 \ units ^{2}$

$A=\frac{1}{2} |\overrightarrow{CB}| |\overrightarrow{AB}| \ sin \alpha=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{524.6} \times \ sin \ 21.3^\circ =88.8 units^{2}$

$Now I spent roughly a couple of hours doing this question tonight, to calculate the area of a triangle in vector form (I'm actually not sure how to word it properly), but what I take from doing a couple of questions today , is you either have to multiply 2 vectors coming in or 2 vectors coming out together to get the required area. So what I mean by this is, at point A in my diagram, both of the vectors are going out. At point C, both of the vectors are coming in. At point B, I had to reverse one of the vectors so that it showed them both coming in.$
The area values you got are all very close. Haven't checked your calculations but if they're all right, the small discrepancies would just be due to the fact that you used rounded values for the angle.

$\noindent To calculate \sin \theta exactly as a surd here, you can use \sin \theta =\sqrt{1 -\cos^{2}\theta} and use the value you found for \cos \theta.$

9. ## Re: International Baccalaureate Marathon

Not sure what the best way to study for sequences and series, whether it be HSC or IB but should we tell the students to memorise the formulae (even though the formuale sheet is provided)?

10. ## Re: International Baccalaureate Marathon

$Anyway I came across this question tonight and it has me a bit stumped. The first term is 1, sum of the first 6 terms is 1.24992. Find the value of the common ratio if:$

$S_n=\frac{u_1(r^n-1)}{r-1}$

$1.24992=\frac{1(r^6-1)}{r-1}$

$I tried using my graphics calculator and it gave me an r value=1$

11. ## Re: International Baccalaureate Marathon

OK problem solved if you guys are familiar with the Casio fx-9860G AU, you should leave the fraction to the right and not cross multiply in Solver.

$Don't enter it in the calculator as$

$1.24992(r-1)=1(r^6-1) it won't solve it for an unknown reason.$

12. ## Re: International Baccalaureate Marathon

This is a Band 7 IB question (equivalent to a HSC Band 6 aiming for 95%+ on paper)

With regards to (b) , would it be OK to not write the answer involving sequences? i.e doing it manually in order to get the marks

$The following was the solution:$

$t_n=2+2 \sum_{k=1}^n h_k$

$t_8=2+16(1-0.8^8)=15.3 \ metres$

My way I did it was just using the 'common sense method':

$Total distance ball has traveled before 9th bounce : =2+2(1.6)+2(1.28)+2(1.024)+2(0.892)+2(0.655536)+2 (0.524288)+2(0.4194304)+2(0.33554432)=15.46 \ metres$

Notice both answers do give slightly different decimal values .

13. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
This is a Band 7 IB question (equivalent to a HSC Band 6 aiming for 95%+ on paper)

With regards to (b) , would it be OK to not write the answer involving sequences? i.e doing it manually in order to get the marks

$The following was the solution:$

$t_n=2+2 \sum_{k=1}^n h_k$

$t_8=-2+16(1-0.8^8)=15.3 \ metres$

My way I did it was just using the 'common sense method':

$Total distance ball has traveled before 9th bounce : =2+2(1.6)+2(1.28)+2(1.024)+2(0.892)+2(0.655536)+2 (0.524288)+2(0.4194304)+2(0.33554432)=15.46 \ metres$

Notice both answers do give slightly different decimal values .
0.892 and 0.655536 are wrong. By doing it manually like this you are greatly increasing the number of points at which you can mess up the calculation. And of course this won't be feasible if 9 is replaced by 1000.

The correct answer is the expression t8 (although the 2 is +2, not -2), which in decimal form is exactly 15.31564544.

Ps. alarm bells should be ringing in your head at "slightly different decimal values". The only step in which any rounding is involved is at the very end, so these answers should coincide exactly prior to rounding, which they clearly don't.

14. ## Re: International Baccalaureate Marathon

OK yep I see it

should be 0.8192 metres and 0.65536 metres

15. ## Re: International Baccalaureate Marathon

Originally Posted by seanieg89
0.892 and 0.655536 are wrong. By doing it manually like this you are greatly increasing the number of points at which you can mess up the calculation. And of course this won't be feasible if 9 is replaced by 1000.

The correct answer is the expression t8 (although the 2 is +2, not -2), which in decimal form is exactly 15.31564544.

Ps. alarm bells should be ringing in your head at "slightly different decimal values". The only step in which any rounding is involved is at the very end, so these answers should coincide exactly prior to rounding, which they clearly don't.
I'm aware that there was a slight error with my working out but in terms of 'method' of getting to the answer, there are 2 methods in order to get to the right answer.

Would my method be suitable in order to answer a question like this, whether it be HSC and/or IB?

16. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
I'm aware that there was a slight error with my working out but in terms of 'method' of getting to the answer, there are 2 methods in order to get to the right answer.

Would my method be suitable in order to answer a question like this, whether it be HSC and/or IB?
Depends what you mean by suitable.

a) Will the answer be the same?
Yes, obviously. Summing a geometric series manually will give you the same result as applying the formula for the sum of a geometric series.

b) Will markers treat the solution as being equally valid?
I don't know the marking criteria, but I can't imagine any reason why not.

c) Do I recommend manually summing geometric/arithmetic series?
Definitely not, unless the series is only a couple of terms long. As I said in my previous post, you are greatly increasing the number of places at which you can make mistakes. The method is also slower (assuming proficiency with both), and not feasible for series with many terms. You need to know how to sum general arithmetic and geometric series for the course anyway, so why not use this knowledge?

17. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
$Points A,B and C have position vectors A \begin{pmatrix} 1\\-19\\5 \end{pmatrix}, B \ \begin{pmatrix} 3.2\\3.6\\2 \end{pmatrix} , C \ \begin{pmatrix} -6\\-15\\7 \end{pmatrix}$

$The respective magnitudes: |\overrightarrow{AB}|=\sqrt{524.6} , |\overrightarrow{AC}|=\sqrt{69} , |\overrightarrow{BC}|=\sqrt{455.6} , |\overrightarrow{CB}|=\sqrt{455.6}$

$Find the Area of the triangle ABC$

$I preceded to find vectors \overrightarrow{AB}, \overrightarrow{AC} and \overrightarrow{BC}$

$This is a diagram I drew and then I preceded to work out the individual angles of each of the vertices.$

$\overrightarrow{AB}=\begin{pmatrix} 2.2\\22.6\\-3 \end{pmatrix} \ , \ \overrightarrow{AC}= \begin{pmatrix} -7\\4\\2 \end{pmatrix} \ , \overrightarrow{BC}= \begin{pmatrix} -9.2\\-18.6\\5 \end{pmatrix} \ , \overrightarrow{CB}= \begin{pmatrix} 9.2\\18.6\\-5 \end{pmatrix}$

$cos \ \theta=\frac{\overrightarrow{AB} \ . \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|}=\frac{69}{\sqrt{524.6} \times \sqrt{69}} \implies \theta=68.7^\circ$

$cos \ \beta=\frac{\overrightarrow{BC} \ . \overrightarrow{AC}}{|\overrightarrow{BC}| |\overrightarrow{AC}|}=0 \implies \beta=90^\circ$

$cos \ \alpha=\frac{\overrightarrow{AB} \ . \overrightarrow{CB}}{|\overrightarrow{AB}| |\overrightarrow{CB}|} \implies cos \ \alpha=\frac{455.6}{\sqrt{524.6} \times {\sqrt{455.6}}}=21.3^\circ$

$As a test I then worked out the individual areas of the triangles, use either of the 3 methods and I didn't get the same values. Well one of them was close to the actual answer of 88.7$

$A=\frac{1}{2} |\overrightarrow{BC}| |\overrightarrow{AC}| \ sin \beta=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{69} \ sin \ 90^\circ=88.7 \ units^{2}$

$A=\frac{1}{2} |\overrightarrow{AC}| |\overrightarrow{AB}| \sin \theta=\frac{1}{2} \times \ 13 \ \sqrt{524.6} \ sin \ 68.7^\circ=88.6 \ units ^{2}$

$A=\frac{1}{2} |\overrightarrow{CB}| |\overrightarrow{AB}| \ sin \alpha=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{524.6} \times \ sin \ 21.3^\circ =88.8 units^{2}$

$Now I spent roughly a couple of hours doing this question tonight, to calculate the area of a triangle in vector form (I'm actually not sure how to word it properly), but what I take from doing a couple of questions today , is you either have to multiply 2 vectors coming in or 2 vectors coming out together to get the required area. So what I mean by this is, at point A in my diagram, both of the vectors are going out. At point C, both of the vectors are coming in. At point B, I had to reverse one of the vectors so that it showed them both coming in.$
$Not sure if you know about the \textit{vector cross product}, but if you do one of the geometric interpretations for the magnitude of the cross product between two vectors corresponds to the area of a triangle.$

$If two sides of a triangle are formed by the vectors \overrightarrow{u} and \overrightarrow{v} the area \mathcal{A} of the triangle is given by$

$\mathcal{A} = \frac{1}{2} \|\overrightarrow{u} \times \overrightarrow{v}\|$

$So if the two sides of a triangle are formed by the vectors \overrightarrow{AB} and \overrightarrow{AC} the area \mathcal{A} of the triangle with vertices A, B, and C will be given by$

$\mathcal{A} = \frac{1}{2} \|\overrightarrow{AB} \times \overrightarrow{AC}\|.$

$Now, for your example, we have \overrightarrow{AB} = \langle 3.2 - 1,3.6 + 19, 2-5 \rangle = \langle 2.2,22.6,-37 \rangle and \overrightarrow{AC} = \langle -6 - 1, -15 + 19, 7 - 5 \rangle = \langle -7, 4, 2 \rangle. Note here for brevity I am using triangluar brackets to denote a vector.$

$Finding the cross product between these two vectors.$

\begin{align*}\overrightarrow{AB} \times \overrightarrow{AC} &=\begin{array}{|ccc|}\hat{e}_x & \hat{e}_y & \hat{e}_z\\2.2 & 22.6 & -3\\-7 & 4 & 2\end{array}\\&= \hat{e}_x (22.6 \cdot 2 + 3 \cdot 4) - \hat{e}_y (2.2 \cdot 2 - (-3) \cdot (-7)) + \hat{e}_z (2.2 \cdot 4 + 22.6 \cdot 7)\\&= \langle 57.2, -16.6,167 \rangle \end{align*}

$Now finding the magnitude of this vector we have$

$\|\overrightarrow{AB} \times \overrightarrow{AC}\| = \sqrt{57.2^2 + (-16.6)^2 + 167^2} = \sqrt{31436.4}.$

$So for the area we have$

$\mathcal{A} = \frac{1}{2} \sqrt{31436.4} \doteq 88.65 units^2$

18. ## Re: International Baccalaureate Marathon

$Not sure if you know about the \textit{vector cross product}, but if you do one of the geometric interpretations for the magnitude of the cross product between two vectors corresponds to the area of a triangle.$

$If two sides of a triangle are formed by the vectors \overrightarrow{u} and \overrightarrow{v} the area \mathcal{A} of the triangle is given by$

$\mathcal{A} = \frac{1}{2} \|\overrightarrow{u} \times \overrightarrow{v}\|$

$So if the two sides of a triangle are formed by the vectors \overrightarrow{AB} and \overrightarrow{AC} the area \mathcal{A} of the triangle with vertices A, B, and C will be given by$

$\mathcal{A} = \frac{1}{2} \|\overrightarrow{AB} \times \overrightarrow{AC}\|.$

$Now, for your example, we have \overrightarrow{AB} = \langle 3.2 - 1,3.6 + 19, 2-5 \rangle = \langle 2.2,22.6,-37 \rangle and \overrightarrow{AC} = \langle -6 - 1, -15 + 19, 7 - 5 \rangle = \langle -7, 4, 2 \rangle. Note here for brevity I am using triangluar brackets to denote a vector.$

$Finding the cross product between these two vectors.$

\begin{align*}\overrightarrow{AB} \times \overrightarrow{AC} &=\begin{array}{|ccc|}\hat{e}_x & \hat{e}_y & \hat{e}_z\\2.2 & 22.6 & -3\\-7 & 4 & 2\end{array}\\&= \hat{e}_x (22.6 \cdot 2 + 3 \cdot 4) - \hat{e}_y (2.2 \cdot 2 - (-3) \cdot (-7)) + \hat{e}_z (2.2 \cdot 4 + 22.6 \cdot 7)\\&= \langle 57.2, -16.6,167 \rangle \end{align*}

$Now finding the magnitude of this vector we have$

$\|\overrightarrow{AB} \times \overrightarrow{AC}\| = \sqrt{57.2^2 + (-16.6)^2 + 167^2} = \sqrt{31436.4}.$

$So for the area we have$

$\mathcal{A} = \frac{1}{2} \sqrt{31436.4} \doteq 88.65 units^2$

I did some vector cross product before , a large chapter was devoted to that in Calculus Stewart book. I remember when I did 2nd year Engineering, I spent a lot of time doing those questions. It's a very efficient way of calculating an area.

19. ## Re: International Baccalaureate Marathon

$Cambridge question$

$Q1. Kenny is offered a choice of two investment plans, each requiring an initial investment of \10,000.$

$Plan A offers a fixed return of \800 per year.$

$Plan B offers a return of 5 \% each year, reinvested in the plan$

$Over what period of time is plan A better than plan B ?$

$I graphed the two functions respectively , A_n=10000+800n , B_n=10000(1.05)^n$

The answer in the book states that ' ..for the first 19 years $A_n > B_n$

Want to check if I could say something along the lines of '18 years and 11 months $A_n > B_n$ or if we could leave the answer as a decimal (with 1 d.p), 18.4 years?

$P.S I do know that 0.838 of a year is roughly 10.056 months$

20. ## Re: International Baccalaureate Maths Marathon

If you were pedantic, then expressing your answer in years only is more appropriate since our question was with respect to years. i.e. maintain consistency of units

21. ## Re: International Baccalaureate Maths Marathon

I wasn't sure if I got the wording of this question right:

My initial thinking was to set the first derivative equal to 0 and then from that determine the intervals where it was decreasing. Reflecting back on it I see that they have set the 2nd derivative to less than zero, then determined the x-values of the decreasing gradient. (I could understand had they worded it 'concave down') . What are your guys thoughts on this question?

Q10. This was the solution:

$I did something along the lines of :$

$\frac{dy}{dx}=7-2x-3x^2$

$determine where \frac{dy}{dx} < 0$

22. ## Re: International Baccalaureate Maths Marathon

Another point to make out to IBers for 2017, there was a mistake in the Oxford SL book by Laurie Buchanan, Jim Fensom, Ed Kemp, Paul La Rondie and Jill Stevens.

The image is a bit small but the question was

$log_{5} (x-2)=3$

in the following line , they then wrote:

$log_{5} (2x-1)=3$

23. ## Re: International Baccalaureate Maths Marathon

Originally Posted by davidgoes4wce
I wasn't sure if I got the wording of this question right:

My initial thinking was to set the first derivative equal to 0 and then from that determine the intervals where it was decreasing. Reflecting back on it I see that they have set the 2nd derivative to less than zero, then determined the x-values of the decreasing gradient. (I could understand had they worded it 'concave down') . What are your guys thoughts on this question?

Q10. This was the solution:

$I did something along the lines of :$

$\frac{dy}{dx}=7-2x-3x^2$

$determine where \frac{dy}{dx} < 0$
the question asked you to find where the function's gradient is decreasing, not where the function itself is decreasing.

24. ## Re: International Baccalaureate Maths Marathon

Just my opinion but comparing the IB Cambridge v HSC Cambridge (Yr 11/12 ) Extension books, the IB Cambridge is a far superior book. The book which is co-authored by 4 Cambridge University graduates, is written in a reader friendly way. I get comments from students that read the HSC Cambridge book and say it is too 'theoretical' and 'hard to understand' and I totally understand where they are coming from. You would think that be the opposite, with Cambridge Uni being a higher rank and the overall general consensus is a book written by Cambridge grads would be of higher mathematical literature.

I decided to do a bit of a comparison of the books:

WORKED EXAMPLES-
IB Cambridge- every step is explained in detail. With their graphs, they utilize different colours to break down key components. They keep things short and brief with their explanations.
HSC Cambridge- every example is black and yellow. Their explanations can be like essays sometimes.

EXPLANATION OF CONCEPTS:
IB Cambridge-Despite being 670 pages worth of content their explanations are brief and directly to the point. It's the kind of book where if you read a particular exercise it won't take you more longer than 5-15 minutes to understand what is going on, by just reading theory alone. They also include different colouring to emphasize their key points as well as bolded font to highlight some important formula. They give things like 'Exam Hints' and when linking it up to some history they keep it simple. They also provide an End of Chapter Review which goes through Key formula.
HSC Cambridge- It's a 637 page book but sometimes I feel its 6370 pages. Some exercises can go up to as much as 8 key points per exercise. I read chapter 8 of the Year 12 Cambridge and it seemed to me more of an English essay, personally feel they need to be more direct to the point and feel the need to make it more 'fun ' for the students to read. No end of chapter review summary is provided.

EXERCISES:
IB Cambridge- drill questions provide practice of new methods, they colour-code their questions to a certain IB Grade (i.e Band 4,5,6,7), questions predominantly exam-style
HSC Cambridge-breakdown is Basic, Development and Extension. Development questions are most similar to Exam style questions. Some of the exercises which have graphs, they use are hard-to-read background for the graphs.

AUTHORS:
IB Cambridge- Fannon, Kadelburg, Wooley, Stephen Ward are all Cambridge University graduates and teach both the IB and A Level Mathematics in the UK.
HSC Cambridge- Pender studied at Sydney University, Macquarie University & Bonn University. David Sadler - studied at Sydney Uni and UNSW. Julia Shea studied at University of Tasmania. David Ward studied at UNSW.

No offence to the universities in New South Wales or Australian universities but the Cambridge University maths school is better than any of the Australian University mathematics schools and the rankings prove this.

25. ## Re: International Baccalaureate Maths Marathon

It goes without saying that Cambridge University has been the pre-eminent university in mathematics in the UK (since the days of Isaac Newton) just as Princeton is, in the U.S..

Page 5 of 6 First ... 3456 Last

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•