## 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)

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

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

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

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

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 /ˈmɜrbiəs/ or US /ˈmbiəs/; 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.

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.

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}} $-1.95996<\frac{\bar{x}-14}{3.8/\sqrt{20}}<1.95996$ (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

### Junior Colleges (JC)

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

Code Zone College Name Established Address Type Special Programmes
English Chinese Abb.
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.

## 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.

## 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, 
(ii) the men and women are seated alternately, 
(iii) members of the same family are seated together and the two other women must be seated separately, 
(iv) members of the same family are seated together and the seats are numbered. 

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 tuitionO 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