Right brain training

This post is a review on Right brain training, and also a list of resources that one can research on regarding to the popular method of Right brain training.

(Source: https://theconversation.com/mondays-medical-myth-you-can-selectively-train-your-left-or-right-brain-4704)

When it comes to New Year’s resolutions, getting your body in shape often tops the list. But what about your brain?

Top Seller on Amazon.com on **Right brain training**

If your left or right brain is feeling a little flabby, there’s a wide range of books, teaching programs, and even a Nintendo DS game, purporting to train your left and/or right brain. Indeed, if you Google “right brain training”, you’ll score 53,900,000 hits.

These products are based on the belief that the left and right hemispheres are polar opposites. The left brain is often characterised as your intelligent side: rational, logical and analytic. In contrast the right brain is stereotyped as the “touchy-feely” hemisphere, viewed as artistic, creative, and emotive.

Such left and right brain stereotypes have led theorists to suggest that people can be classified according to their “hemisphericity”. If you’re a logical, rational scientist, for instance, you’re left-brained. But creative types, from artists to writers, are right-brained.

Based on my teaching experience, I do find that left-handers (right-brained) students tend to be very creative and usually excel at arts and humanities. However, their math skills can be good too, especially with practice. This shows that the human brain is like a muscle, it gets better with practice and use.

Did you know our Prime Minister Lee Hsien Loong is left-handed too? Barack Obama is also left-handed. Prime Minister Lee Hsien Loong is very good at math, so this should dispel any myths that left-handed students are not good at math.

(Source: http://psychology.about.com/od/cognitivepsychology/a/left-brain-right-brain.htm)

Left Brain vs Right Brain

Understanding the Myth of Left Brain and Right Brain Dominance

The Right Brain

According to the left-brain, right-brain dominance theory, the right side of the brain is best at expressive and creative tasks. Some of the abilities that are popularly associated with the right side of the brain include:

Recognizing faces
Expressing emotions
Music
Reading emotions
Color
Images
Intuition
Creativity

The Left Brain

The left-side of the brain is considered to be adept at tasks that involve logic, language and analytical thinking. The left-brain is often described as being better at:

Language
Logic
Critical thinking
Numbers
Reasoning

Also, check out the above Youtube video to check if you are a right-brained or left-brained person!

The Right Brain vs Left Brain test … do you see the dancer turning clockwise or anti-clockwise?

If clockwise, then you use more of the right side of the brain and vice versa.

Most of us would see the dancer turning anti-clockwise though you can try to focus and change the direction; see if you can do it.

LEFT BRAIN FUNCTIONS
uses logic
detail oriented
facts rule
words and language
present and past
math and science
can comprehend
knowing
acknowledges
order/pattern perception
knows object name
reality based
forms strategies
practical
safe

RIGHT BRAIN FUNCTIONS
uses feeling
“big picture” oriented
imagination rules
symbols and images
present and future
philosophy & religion
can “get it” (i.e. meaning)
believes
appreciates
spatial perception
knows object function
fantasy based
presents possibilities
impetuous
risk taking

Right Brain Training Test

Take the test to see if you are right-brained or left-brained!

Any comments or websites about Right brain training to share? Leave your comments below!

Right Brain Training Video

Watch this free brain training video and follow the instructions to increase your brain power. This is an online “game” that really works to improve brain function. You can actually feel it work!

Brain Training can increase your brain power just like weight training can increase your strength. Use this exercise to work out your brain. Bookmark this video and come back and practice with variations on the basics as discussed in the video.

Right Brain Training Books

A Whole New Mind: Why Right-Brainers Will Rule the Future

Right-Brained Children in a Left-Brained World: Unlocking the Potential of Your ADD Child

Drawing on the Right Side of the Brain Workbook: The Definitive, Updated 2nd Edition

Bishan-Ang Mo Kio area to get new JC in 2017

Source: http://news.asiaone.com/news/edvantage/bishan-ang-mo-kio-area-get-new-jc-2017

Bishan-Ang Mo Kio area to get new JC in 2017

The site for the new JC at the junction of Sin Ming Avenue and Marymount Road.

Lee Jian Xuan

The Straits Times

Saturday, Jan 04, 2014

SINGAPORE – A new junior college that will open in 2017 for students from three Integrated Programme (IP) schools will likely be built on the site of the Asian Golf Academy near Bishan.

A statement on the Ministry of Education (MOE) website says the new campus will be at the junction of Sin Ming Avenue and Marymount Road, where the driving range is located.

The area is also zoned for an educational institution, according to the Urban Redevelopment Authority’s Draft Master Plan 2013.

Singapore’s 20th school to offer a JC programme will take in IP students from Catholic High School, CHIJ St Nicholas Girls’ School and the Singapore Chinese Girls’ School. It will also admit more than 100 students from other secondary schools who have completed their O levels.

It will be the newest JC since Innova JC in Woodlands was completed in 2005.

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)

Maths Skills to be a Good Lawyer

Doctor and Lawyer are the top two favourite careers in Singapore. On the surface, Lawyers seem not to need much maths, but recent research shows that Mathematics skills and thinking may be crucial to becoming a better Lawyer.

Source: http://news.illinois.edu/news/13/0403numeracy_ArdenRowell.html

There is a “highly significant relationship” between law students’ math skills and the substance of their legal analysis, according to research from Arden Rowell, a professor of law and the Richard W. and Marie L. Corman Scholar at Illinois.

CHAMPAIGN, Ill. — The stereotype of lawyers being bad with numbers may persist, but new research by two University of Illinois legal scholars suggests that law students are surprisingly good at math, although those with low levels of numeracy analyze some legal questions differently.

According to research from Arden Rowell and Jessica Bregant, there is a “highly significant relationship” between law students’ math skills and the substance of their legal analysis, suggesting that legal analysis – and by extension, legal advice – may vary with a lawyer’s native math skills.

“What the research shows is that math matters to lawyers more – and for different reasons – than people have realized,” said Rowell, a professor of law and the Richard W. and Marie L. Corman Scholar at Illinois. “People are only now starting to pay attention to the fact that lawyers and judges who are bad at math can make mistakes that ruin people’s lives. That implicates numeracy as a neglected but potentially critical aspect of legal education, because it’s not something that law schools have traditionally focused on when selecting students.”

Undergraduate Study in Mathematics (NUS)

Maths Group Tuition to start in 2014!

If you are interested in Mathematics, do consider to study Mathematics at NUS!

Source: http://ww1.math.nus.edu.sg/undergrad.aspx

Quote:

Undergraduate Study in Mathematics (NUS)

Overview

The Department of Mathematics at NUS is the largest department in the Faculty of Science. We offer a wide range of modules catered to specialists contemplating careers in mathematical science research as well as to those interested in applications of advanced mathematics to science, technology and commerce. The curriculum strives to maintain a balance between mathematical rigour and applications to other disciplines.

We offer a variety of major and minor programmes, covering different areas of mathematical sciences, for students pursuing full-time undergraduate studies. Those keen in multidisciplinary studies would also find learning opportunities in special combinations such as double degree, double major and interdisciplinary programmes.

Honours graduates may further their studies with the Graduate Programme in Mathematics by Research leading to M.Sc. or Ph.D. degree, or with the M.Sc. Programme in Mathematics by Course Work.

3D Trigonometry Maths Tuition

Solution:

(a) Draw a line to form a small right-angled triangle next to the angle $18^\circ$

Then, you will see that

$\angle ACD=90^\circ-18^\circ=72^\circ$ (vert opp. angles)

$\angle BAC=180^\circ-72^\circ=108^\circ$ (supplementary angles in trapezium)

By sine rule,

$\displaystyle \frac{\sin\angle ABC}{30}=\frac{\sin 108^\circ}{40.9}$

$\sin\angle ABC=0.697596$

$\angle ABC=44.23^\circ$

$\angle ACB=180^\circ-44.23^\circ-108^\circ=27.77^\circ=27.8^\circ$ (1 d.p.)

(b) By Sine Rule,

$\displaystyle\frac{AB}{\sin\angle ACB}=\frac{30}{\sin 44.23^\circ}$

$AB=\frac{30}{\sin 44.23^\circ}\times\sin 27.77^\circ=20.04=20.0 m$ (shown)

(c)

$\angle BCD=\angle ACD-\angle ACB=72^\circ-27.77^\circ=44.23^\circ$

By Cosine Rule,

$BD^2=40.9^2+50^2-2(40.9)(50)\cos 44.23^\circ=1242.139$

$BD=35.24=35.2 m$

(d)

$\displaystyle\frac{\sin\angle BDC}{40.9}=\frac{\sin 44.23^\circ}{35.24}$

$\sin\angle BDC=0.80957$

$\angle BDC=54.05^\circ$

angle of depression = $90^\circ-54.05^\circ=35.95^\circ=36.0^\circ$ (1 d.p.)

(e)

Let X be the point where the man is at the shortest distance from D. Draw a right-angle triangle XDC.

$\displaystyle\cos 72^\circ=\frac{XC}{50}$

$XC=50\cos 72^\circ=15.5 m$

Tips on attempting Geometrical Proof questions (E Maths Tuition)

Tips on attempting Geometrical Proof questions (O Levels E Maths/A Maths)

1) Draw extended lines and additional lines. (using pencil)

Drawing extended lines, especially parallel lines, will enable you to see alternate angles much easier (look for the “Z” shape). Also, some of the more challenging questions can only be solved if you draw an extra line.

2) Use pencil to draw lines, not pen

Many students draw lines with pen on the diagram. If there is any error, it will be hard to remove it.

3) Rotate the page.

Sometimes, rotating the page around will give you a fresh impression of the question. This may help you “see” the way to answer the question.

4) Do not assume angles are right angles, or lines are straight, or lines are parallel unless the question says so, or you have proved it.

For a rigorous proof, we are not allowed to assume anything unless the question explicitly says so. Often, exam setters may set a trap regarding this, making the angle look like a right angle when it is not.

5) Look at the marks of the question

If it is a 1 mark question, look for a short way to solve the problem. If the method is too long, you may be on the wrong track.

6) Be familiar with the basic theorems

The basic theorems are your tools to solve the question! Being familiar with them will help you a lot in solving the problems.

Hope it helps! And all the best for your journey in learning Geometry! Hope you have fun.

“There is no royal road to Geometry.” – Euclid

Animation of a geometrical proof of Phytagoras... — Animation of a geometrical proof of Pythagoras theorem (Photo credit: Wikipedia)

Parallelogram Maths Tuition: Solution

Solution:

(a) We have $\angle APQ=\angle ARQ$ (opp. angles of parallelogram)

$AP=RQ$ (opp. sides of parallelogram)

$AR=PQ$ (opp. sides of parallelogram)

Thus, $\triangle APQ\equiv\triangle QRA$ (SAS)

Similarly, $\triangle ABC\equiv\triangle CDA$ (SAS)

$\triangle CHQ\equiv\triangle QKC$ (SAS)

Thus, $\begin{array}{rcl}\text{area of BPHC}&=&\triangle APQ-\triangle ABC-\triangle CHQ\\ &=&\triangle QRA-\triangle CDA-\triangle QKC\\ &=& \text{area of DCKR} \end{array}$