Maths Challenge

Hi, do feel free to try out our Maths Challenge (Secondary 4 / age 16 difficulty):

Source: Anderson E Maths Prelim 2011

If you have solved the problem, please email your solution to mathtuition88@gmail.com .

(Include your name and school if you wish to be listed in the hall of fame below.)

Students who answer correctly (with workings) will be listed in the hall of fame. 🙂

Hall of Fame (Correct Solutions):

1) Ex Moe Sec Sch Maths teacher Mr Paul Siew

2) Queenstown Secondary School, Maths teacher Mr Desmond Tay

3) Tay Yong Qiang (Waiting to enter University)

The ideal Singapore JC subject combination for applying to Medicine

Why Additional Maths (A Maths) is important for entering Medicine:

Pathway: A Maths (O Level) –> H2 Maths (A Level) –> NUS Medicine

Source: http://sgforums.com/forums/2297/topics/439605

Quote: While NUS and NTU Medicine does not (officially) require H2 Maths (ie. ‘A’ level Maths), some other (overseas) Medical schools might. And not having H2 Maths might (unofficially) disadvantage your chances, even for NUS and NTU.

Therefore (assuming you intend to fight all the way for your ambition), your safest bet would be to (fight for the opportunity) to take both H2 Bio and H2 Math. The ideal Singapore JC subject combination for applying to Medicine (in any University) is :

H2 Chemistry, H2 Biology, H2 Mathematics

Source: http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=40&t=12228

Quote: pre-requisites for nus medicine will be H2 Chem and H2 bio or physics.

as for what’s best,
H2 math is almost a must since without it you’ll be ruling out a lot of ‘back-up courses’

Ad: Maths Group Tuition 2014

Source: http://www.youtube.com/watch?v=BVsIAa2XNKc

Source: http://en.wikipedia.org/wiki/M%C3%B6bius_strip

The Mobius Strip is a really interesting mathematical surface with just one side. It is easy to make, and cutting it produces many surprising effects! 🙂

The Möbius strip or Möbius band (UK /ˈ m ɜr b i ə s / or US /ˈ m oʊ b i ə s /; German: [ˈmøːbi̯ʊs]), also Mobius or Moebius, is a surface with only one side and only one boundary component. The Möbius strip has the mathematical property of being non-orientable. It can be realized as a ruled surface. It was discovered independently by the German mathematicians August Ferdinand Möbius and Johann Benedict Listing in 1858.^[1]^[2]^[3]

A model can easily be created by taking a paper strip and giving it a half-twist, and then joining the ends of the strip together to form a loop. In Euclidean space there are two types of Möbius strips depending on the direction of the half-twist: clockwise and counterclockwise. That is to say, it is a chiral object with “handedness” (right-handed or left-handed).

The Möbius band (equally known as the Möbius strip) is not a surface of only one geometry (i.e., of only one exact size and shape), such as the half-twisted paper strip depicted in the illustration to the right. Rather, mathematicians refer to the (closed) Möbius band as any surface that is homeomorphic to this strip. Its boundary is a simple closed curve, i.e., homeomorphic to a circle. This allows for a very wide variety of geometric versions of the Möbius band as surfaces each having a definite size and shape. For example, any closed rectangle with length L and width W can be glued to itself (by identifying one edge with the opposite edge after a reversal of orientation) to make a Möbius band. Some of these can be smoothly modeled in 3-dimensional space, and others cannot (see section Fattest rectangular Möbius strip in 3-space below). Yet another example is the complete open Möbius band (see section Open Möbius band below). Topologically, this is slightly different from the more usual — closed — Möbius band, in that any open Möbius band has no boundary.

It is straightforward to find algebraic equations the solutions of which have the topology of a Möbius strip, but in general these equations do not describe the same geometric shape that one gets from the twisted paper model described above. In particular, the twisted paper model is a developable surface (it has zero Gaussian curvature). A system of differential-algebraic equations that describes models of this type was published in 2007 together with its numerical solution.^[4]

The Euler characteristic of the Möbius strip is zero.

Stanford University Research: The most important aspect of a student’s ideal relationship with mathematics

Source: Taken from Research by Stanford, Education: EDUC115N How to Learn Math

This word cloud was generated on August 9th based on 850 responses to the prompt “Please submit a word that, in your opinion, describes the most important aspect of a student’s ideal relationship with mathematics.”

H2 Maths 2012 A Level Solution Paper 2 Q6; H2 Maths Group Tuition

6(i)

$H_0: \mu=14.0 cm$

$H_1: \mu\neq 14.0 cm$

(ii)

$\bar{x}\sim N(14,\frac{3.8^2}{20})$

For the null hypothesis not to be rejected,

$Z_{2.5\%}<\frac{\bar{x}-14}{3.8/\sqrt{20}}<Z_{97.5\%}$

$-1.95996<\frac{\bar{x}-14}{3.8/\sqrt{20}}<1.95996$ (use GC invNorm function!)

$12.3<\bar{x}<15.7$ (3 s.f.)

(iii) Since $\bar{x}=15.8$ is out of the set $12.3<\bar{x}<15.7$ , the null hypothesis would be rejected. There is sufficient evidence that the squirrels on the island do not have the same mean tail length as the species known to her.

(technique: put in words what $H_1$ says!)

H2 Maths A Level 2012 Solution, Paper 2 Q5; H2 Maths Tuition

5(i)(a)

$P(\text{patient has the disease and test positive})=0.001(0.995)=9.95\times 10^{-4}$

$P(\text{patient does not have the disease and he tests positive})=(1-0.001)(1-0.995)=4.995\times 10^{-3}$

$P(\text{result of the test is positive})=9.95\times 10^{-4}+4.995\times 10^{-3}=5.99\times 10^{-3}$

(b)

Let A=patient has disease

Let B=result of test is positive

$\displaystyle\begin{array}{rcl}P(A|B)&=&\frac{P(A\cap B)}{P(B)}\\ &=&\frac{(0.001)(0.995)}{5.99\times 10^{-3}}\\ &=&0.166 \end{array}$

Note that the probability is surprisingly quite low! (This is called the False positive paradox, a statistical result where false positive tests are more probable than true positive tests, occurring when the overall population has a low incidence of a condition and the incidence rate is lower than the false positive rate. See http://en.wikipedia.org/wiki/False_positive_paradox)

(ii)

$\displaystyle P(A|B)=\frac{(0.001)p}{(0.001)p+(1-0.001)(1-p)}=0.75$

By GC, $p=0.999666$ (6 d.p.)

List of JCs in Singapore; H2 Maths Tuition

Source: http://en.wikipedia.org/wiki/List_of_schools_in_Singapore#Junior_Colleges_.28JC.29

Junior Colleges (JC)

These offer two-year courses leading to the GCE A-level examination.

Code	Zone	College Name			Established	Address	Type	Special Programmes
Code	Zone	English	Chinese	Abb.	Established	Address	Type	Special Programmes
0705	North	Anderson Junior College	安德逊初级学院	AJC	1984	4500 Ang Mo Kio Avenue 6	Government
7001	West	Anglo-Chinese School (Independent) IB World School	英华中学 (自主)	ACS(I)-IBDP	2004 (IBDP)	121 Dover Road	Independent	IP, MEP
0803	West	Anglo-Chinese Junior College	英华初级学院	ACJC	1977	25 Dover Close East	Government-Aided	MEP, DEP(TSD), LEP (EL)
0802	South	Catholic Junior College	公教初级学院	CJC	1975	129 Whitley Road	Government-Aided	LEP (EL)
3101	East	Dunman High School	德明政府中学	DHS	2005 – IP	10 Tanjong Rhu Road	Autonomous	IP, MEP, BSP, LEP (CL), AEP
0806	Central	Hwa Chong Institution	华侨中学	HCI	1974	661 Bukit Timah Road	Independent	IP, HP, LEP (CL), AEP, BSP
0713	North	Innova Junior College	星烁初级学院	IJC	2005	21 Champions Way	Government	LEP (ML)
0703	West	Jurong Junior College	裕廊初级学院	JJC	1981	800 Corporation Road	Government	LEP (CL)
0712	East	Meridian Junior College	美廉初级学院	MJC	2003	21 Pasir Ris Street 71	Government
0908	West	Millennia Institute	励仁高级中学	MI	2004	60 Bukit Batok West Avenue 8	Government	DTP
0805	North	Nanyang Junior College	南洋初级学院	NYJC	1978	128 Serangoon Avenue 3	Government-Aided	LEP (CL), AEP
0712	Central	National Junior College	国家初级学院	NJC	1969	37 Hillcrest Road	Government	IP, HP, AEP, MEP, STaR
7801	West	NUS High School of Mathematics and Science	新加坡国立大学附属数理中学	NUSHS	2005	20 Clementi Ave 1	Independent	IP, DIP
0711	West	Pioneer Junior College	先驱初级学院	PJC	1999	21 Teck Whye Walk	Government
0704	South	Raffles Institution	莱佛士初级学院	RI	1826	10 Bishan Street 21	Independent	IP, HP, LEP (JL), LEP (EL), MEP, TSD
3103	West	River Valley High School	立化中学	RVHS	1956 2006 – IP	6 Boon Lay Avenue	Autonomous	IP, BSP
0710	North	Serangoon Junior College	实龙岗初级学院	SRJC	1988	1033 Upper Serangoon Road	Government
0804	South	Saint Andrew’s Junior College	圣安德烈初级学院	SAJC	1978	55 Potong Pasir Avenue 1	Government-Aided
0709	East	Tampines Junior College	淡滨尼初级学院	TPJC	1986	2 Tampines Avenue 9	Government	LEP (ML), TSD
0702	East	Temasek Junior College	淡马锡初级学院	TJC	1977	22 Bedok South Road	Government	IP, HP, LEP (CL), MEP
0706	East	Victoria Junior College	维多利亚初级学院	VJC	1984	20 Marine Vista	Government	IP, HP, TSD, NAV
0708	North	Yishun Junior College	义顺初级学院	YJC	1986	3 Yishun Ring Road	Government

Centralised Institutes (CI)

The only centralised institute is Millennia Institute (MI), which offers a three-year course leading to the GCE A-level examination in arts, science, and commerce.^[3]

How to use tables in CASIO FX-9860G Slim (H2 Maths Tuition)

Tables in CASIO FX-9860G Slim

The most popular Graphical Calculator for H2 Maths is currently the TI-84 PLUS series, but some students do use Casio Graphical Calculators.

The manual for CASIO FX-9860G Slim is can be found here:

http://edu.casio.com/products/fx9860g2/data/fx-9860GII_Soft_E.pdf

The information about Tables and how to generate a table is on page 121.

Generating tables is useful to solve some questions in sequences and series, and also probability. It makes guess and check questions much faster to solve.

h2 tuition gc — Picture of the CASIO FX-9860G Slim Calculator

MI H2 Maths Prelim Solutions 2010 Paper 2 Q7 (P&C)

MI H2 Maths Prelim 2010 Paper 2 Q7 (P&C)

Question:
7) Two families are invited to a party. The first family consists of a man and both his parents while the second family consists of a woman and both her parents. The two families sit at a round table with two other men and two other women.
Find the number of possible arrangements if
(i) there is no restriction, [1]
(ii) the men and women are seated alternately, [2]
(iii) members of the same family are seated together and the two other women must be seated separately, [3]
(iv) members of the same family are seated together and the seats are numbered. [2]

Solution:

(i) (10-1)!=9!=362880

(ii) First fix the men’s sitting arrangement: (5-1)!

Then the remaining five women’s total number of arrangements are: 5!

Total=4! x 5!=2880

(iii) Fix the 2 families (as a group) and the 2 men: (4-1)! x 3! x 3!

(3! to permute each family)

By drawing a diagram, the two women have 4 slots to choose from, where order matters: $^4 P_2$

Total = $(4-1)! \times 3! \times 3! \times ^4 P_2 = 2592$

(iv)

We first find the required number of ways by treating the seats as unnumbered: $(6-1)!\times 3!\times 3! =4320$

Since the seats are numbered, there are 10 choices for the point of reference, thus no. of ways = $4320 \times 10 =43200$

H2 JC Maths Tuition Foot of Perpendicular 2007 Paper 1 Q8

One of my students asked me how to solve 2007 Paper 1 Q8 (iii) using Foot of Perpendicular method.

The answer given in the TYS uses a sine method, which is actually shorter in this case, since we have found the angle in part (ii).

Nevertheless, here is how we solve the question using Foot of Perpendicular method.

(Due to copyright issues, I cannot post the whole question here, so please refer to your Ten Year Series.)

Firstly, let F be the foot of the perpendicular.

Then, $\vec{AF}=k\begin{pmatrix}3\\-1\\2\end{pmatrix}$ ——– Eqn (1)

$\vec{OF}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17$ ——– Eqn (2)

From Eqn (1), $\vec{OF}-\vec{OA}=\begin{pmatrix}3k\\-k\\2k\end{pmatrix}$

$\begin{array}{rcl}\vec{OF}&=&\vec{OA}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\ &=&\begin{pmatrix}1\\2\\4\end{pmatrix}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\ &=&\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\end{array}$

Substituting into Eqn (2),

$\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17$

$14k+9=17$

$k=4/7$

Substituting back into Eqn (1),

$\displaystyle\vec{AF}=\frac{4}{7}\begin{pmatrix}3\\-1\\2\end{pmatrix}$

$\displaystyle|\vec{AF}|=\frac{4}{7}\sqrt{14}$

JC Junior College H2 Maths Tuition

If you or a friend are looking for Maths tuition: O level, A level H2 JC (Junior College) Maths Tuition, IB, IP, Olympiad, GEP and any other form of mathematics you can think of.

Experienced, qualified (Raffles GEP, NUS Maths 1st Class Honours, NUS Deans List) and most importantly patient even with the most mathematically challenged.

So if you are in need of the solution to your mathematical woes, drop me a message!

Tutor: Mr Wu

Email: mathtuition88@gmail.com

Website: Singapore Maths Tuition | Patient and Dedicated Maths Tutor in Singapore