CoffeeCat:

The truth is that a tutor cannot make wonders because the tutor only sees the child for an extremely short time each week. As u r well aware of the \"practice makes perfect\" thingy in maths, perhaps try to explore options with the tutor like assigning him problems to keep the child working a few days every week?

And perhaps can try something like get the P5 child to \"supervise\" or help the P4 one and get the P4 one to \"supervise\" the P2? Sometimes teaching others will help to internalise and deepen your own understanding =).

Yes. Practice makes perfect. Unless your student sees the tutor for Maths only on Mon Wed Fri, it is very hard for tutors to monitor his progress closely. The limited time is used by the tutor to explain in details to the student, mostly on questions not solved correctly by student.

Some parents hire 1 tutor for Eng Maths Sci, just 1h 30m a week. 30minutes per subject is very rush. If student is scoring 90% and tutor given 30minutes to guide student on the missing 10%, it should be reasonable. But if student is scoring 50%, there's plenty for tutor to go through and 30minutes won't be sufficient.

Parents can offer to monitor child's progress. Check homework/task given by school and tutor. Help balance time spent on subjects between school and tuition. Do remember to include some playtime/rest for child. Just drop a note to tutor through SMS/Whatsapp/Email to request for higher or lower workload for child for the next lesson.

If I do this question, I would do it sushi’s method.


For teaching students, I would pick Ender’s method. They must fully understand the age difference is constant. P5 aren’t good with algebra yet.
Age question normally start with 3 labels. Past, Present and Future. Jane and Mom. Solution presented would be in table format.

Yet for fast learners, I could add in alternate way of solving with algebra so when they learn algebra in P6 they can relate to using it to solve questions.

hoddle:

Hi


Can anyone tell me how do I paste an image or else I have to type my length question. Questions that have diagram is impossible to pose the questions. Please advise

Try using a upload website such as imgur. U can google it.
Make sure the photo is less than 1024 pixels, u can check the resolution of the picture using paint in Windows.
When posting images, click the Img beside the URL button OR ,img, *image link* ,/img, where the commas are [ and ]

You're welcomed to contact me through skype if you require further assistance.
Skype ID: SwiftFormula

Ilye:

The figure consists of 3 squares. The total area of the figure is 296cm2. Assuming that all the length of the squares are in whole numbers, find

(a) The area of the smallest square
(b) The perimeter of the figure
http://i61.tinypic.com/2j2urex.png\">
please help ...:?:

I am going to illustrate using algebra.
(Let the length of the smallest square be a)
Length of square B = a+2
Length of square C = 2a+2

Area = (2a+2)^2 + (a+2)^2 + a^2 = 296 cm^2
(2a+2)(2a+2) + (a+2)(a+2) + axa = 296 cm^2

Can expand out and solve but the method is more of secondary school.
Since question states it is a whole number and 296 is relatively small.

Start with a = 4
(8+2)(8+2) + (4+2)(4+2) + 4x4 = 100 + 36 + 16 = 152 (Too small)

Try a = 6
14x14 + 8x8 + 6x6 = 196 + 64 + 36 = 296
Therefore a = 6 cm

a) Area of smallest square = 36cm^2
b) Perimeter = 14+14+14+6+6+2+8+8 = 72cm

That's many years ago when that question came out. If many students were able to solve then, I don't see the reasons many parents create a fuss over it and the news won't be reporting it. The reason is mainly because a HIGH % of the students have NEVER been exposed to such questions ever, neither did they expect something like that.


Right now Maths Olympiad classes are easily available in many schools compared to then. Students are exposed to uncommon questions because MoE changed the syllabus.

School teachers are being so restricted now, they can't cane/scold/punish students because it's their \"RiceBowl\". Last time school teachers were fearsome like tigers, now paper-tigers. They worry more about their career than student's education.

Educational standard since then had risen by quite abit, the gap between worst student and best student had been widened. As a result, parents become more KiaSu, they do not want their children to lose the race for knowledge. The end result is more tuition, more homework, more stress.

Right now, there's even 6 years IP offered after PSLE or choice to enter NUSH where education accelerates.

I'll give an example. It's comparable to the answer of 4947 back then.
http://swiftformula.blogspot.sg/p/question-9.html
How many % of the P6 students are able to solve it?
Even university graduates of my year have difficulty solving P6 MO questions.
Long gone are the days when 40/50cents could fill my tummy with a decent meal.

It is highly possible, a question similar to this is set by MoE.

Just take note that PSLE no longer test on simple word problem.
Methods and solution are not so straightforward anymore.

Unorthodox questions are getting more common nowadays as our academic standard gets higher.

Good school will go thru such questions.
Thats why papers SA1 SA2 of reputable schools are tougher than others.

That’s quite a distance away from me. Near driving centre.

Student can enter St Anthony primary then swiss cottage secondary.
Nearest to home = best unless student wants to go to top few secondary school.

(100% - something) refers to the empty space beside the bar, compared to 100%

So we find all the empty space of all the bars except last 1.
Take the last bar minus all the empty space = x% (Do note if x is negative, take x to be 0)

http://i.imgur.com/diQofmu.png\">
(*need to open image in new tab to view it fully)

If the question asks for 4 sports :

95% students play A
90% students play B
85% students play C
80% students play D

To get minimum x% students to play A&B&C :
we use (100%-95%) + (100%-90%) + (100-85%) = 30%
80% - 30% = 50%
x = 50%


So we use 100% minus each of the % (ABC) excluding the last % (D)
Then we add up the sum of 100% minus each (ABC).
Use the last % (D) to minus off the sum (which is 30% in this case)

Hi SAHMom, I try my best. My english isn’t very good.

I change to the question to :
95% students play A
90% students play B
85% students play C

To get minimum x% students to play A&B&C :
we use (100%-95%) + (100%-90%) = 15%
85%-15% = 70%
x = 70%