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    2. FrekiWang
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    Recent Best Controversial
    • RE: * Victoria JC (VJC)

      4 is the usual COP of VJC, which implies VJC will take in partial 4 pointers. Itmight be 80% or 20% of the 4 pointers who apply, no one knows and it varies over different years.

      posted in Tertiary Education - A-Levels
      F
      FrekiWang
    • RE: O Level Results 2013

      1. VJC’s COP(sci) has been 4, not all 4 pointers can get in.

      2. NJC’s COP(sci) has been 5, according to what I know, not all 5 pointers can get in. This year it may become 4 as the IP cohort size increases.
      3. YOU CAN APPEAL before 31 jan, please contact the individual JC for details. For NJC, the appealing exercise is conducted before the posting is out.

      posted in Tertiary Education - A-Levels
      F
      FrekiWang
    • RE: GCE O level result release on 10th January 2013

      kiasuursula:
      I am just wondering,

      if my HMT replaced my English in the L1,
      can i still minus the bonus 2 points when I apply for JC?
      yes, as long as you pass both eng and HMT.

      However, when HMT is used as L1, MT cannot be included in R5 at the same time.

      posted in Secondary Schools - Academic Support
      F
      FrekiWang
    • RE: Do I need to revise O-lvl stuff before JC starts?

      For math, better revise the following a math topic:

      partial fraction
      quadratic equation and inequality
      calculus(diff and int)

      posted in Tertiary Education - A-Levels
      F
      FrekiWang
    • RE: All About Math Olympiad Training & Questions

      MathIzzzFun:
      Absolutely Bo Chap:


      4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.

      (This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)

      Ans: \t13 A

      Visualize the figure as a hexagon with alternate 2cm and 1 cm sides... suppose the hexagon is ABCDEF
      AB=2cm
      BC=1cm
      CD=2cm
      DE=1cm
      EF=2cm
      FA=1cm

      Extend AB, DC to meet at X
      Extend BA, EF to meet at Y
      Extend FE CD to meet at Z

      Now XYZ is an equilateral triangle with sides 4 cm. If you divide each side into 4 x 1cm sections. Subdivide this triangle into 16 x 1 cm side equilateral triangle.

      The hexagon will consist of 13 such triangles.

      http://i50.tinypic.com/2a6ui2w.png\">



      cheers.

      First of all, 1cm x 1cm x 1cm is a wrong description of a triangle, it should be 1cm by 1cm by 1cm, but not 1cm times 1cm times 1cm. The same applies to a hexagon.

      Assuming the description is valid, a 2x2x2x1x1x1 hexagon is different from a 2x1x2x1x2x1 hexagon (shown in that diagram).

      The correct diagram is the one below:

      http://i45.tinypic.com/1zdtzdv.jpg\">

      posted in Mathematics
      F
      FrekiWang
    • RE: CCA Points.

      There are CCA points awarded from competitions that are not related to your CCA, subject to a cap of around 20 points or something like that.


      For competitions related to your CCA, it will be counted as the level of participation in your CCA.

      posted in Secondary Schools - Academic Support
      F
      FrekiWang
    • RE: Help! My DS is not motivated.

      since he said he will stop once school terms start, just take away his PS3 and Xbox once the school term starts(tmr).


      By the way, kids need at least one non-academic activity to release their stresses. You have to help him to find a replacement if you don’t want him to play PS3/Xbox (such as sports or tablegames or whatever) zero leisure time is neither practical nor possible at all.

      posted in Secondary Schools - Academic Support
      F
      FrekiWang
    • RE: COP 2012 - For Secondary Schools in 2013

      The increase in the COP of those ‘top’ schools is consequece of the dragon year.


      Based on the formula, 250 is the benchmark for the top 10% candidates. Last year, there were 45261 candidates, but this number has increased to 48333 this year, which means there are 300+ more candidates who have scored 250+ this year. As there is no significant increase in the vacancies in those ‘top’ schools, their COPs certainly go up.

      posted in Secondary Schools - Selection
      F
      FrekiWang
    • RE: Any overseas Sporean returned at JC level?

      kiwi68:
      I'm hoping to hear fm any Sporeans who've kids returning at JC level, how are your kids integrating and what went on prior to coming back?


      We're overseas now and might return end this year. ds1 is finishing his year 11/grade 10, and will be eligible for JC1 next year. Have read enough about SPERS, but would like to hear more about the integration and coping with the rigorous 2yrs in JC.

      Appreciating any sharing. TIA.
      Unlike primary/secondary schools, your ds1 is not guaranteed a position in a JC, unless he can pass the Jpact test offered by those JCs which still have vacancies. According to my knowledge, most of the 'better' JCs will not have vacancies to take in any more student after JAE intake.

      My suggestion will be to ask your ds1 to take 2012 O-level Examinations, since he is a Singaporean, he will be still considered under JAE admission. A L1R5 of 20 points or lesser is required to be admitted to a JC. If your ds1 cannot clear it, I do not think he would be able to cope even a JC takes him in.

      posted in Tertiary Education - A-Levels
      F
      FrekiWang
    • RE: O-Level Additional Math

      Xiao Hu:
      Hi FrekiWang,


      Need help on this question, I think it's 2010 O level Maths paper 1 question.
      Part2 looks innocuously easy, but it trips me up. Difficult to put my reasonings in writings.

      There are 2 parts to the question.
      #1. Factorise 90. This is easy, 90=2x3x3x5

      #2
      LCM of 6,15 and x is 90. What are possible values of x if x is odd?

      From #1, since 90=6x15, how can this be used to work out possible values of x?

      The answers given for x are 9 and 45.

      TIA,
      Xiao Hu.
      90=2*3^2*5
      6=2*3
      15=3*5
      Therefore, x=2^a * 3^2 * 5^b
      where a=0 or 1(reject as odd) and b=0 or 1
      when a=0,b=0: x=3^2=9
      when a=0,b=1: x=3^2*5^1=45

      posted in Secondary Schools - Academic Support
      F
      FrekiWang
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