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    O-Level Additional Math

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
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    • W Offline
      wkong
      last edited by

      FrekiWang:
      wkong:

      Hi I have two questions from my son:


      1. Find the smallest whole number that is not a factor of 1*2*3*4*5......*31

      2. Find the largest prime number that divides this number

      1*2*3....*59+1*2*3*4.....*60

      Thank you in advance.

      1. 37
      2. 61

      I have to say that your son's trainer (most probably maths olympiad) is a failure >_< as these are very simple questions at competition level.

      Thanks Freki for your help. My DS wasn't trained for maths olympiad. This is part of his sch maths homework given by his sch teacher.

      1 Reply Last reply Reply Quote 0
      • H Offline
        Herbie
        last edited by

        I hv 2Qn.


        1 factorise 3x^2+26x+51.
        HENCE OR OTHERWISE FIND TWO FACTORS OF 32651.

        2. If (a+b)^2=73 and ab=6.5, CALCULATE THE VALUE OF a^2+b^2

        Can help to show the working? Tq

        1 Reply Last reply Reply Quote 0
        • M Offline
          mum_sugoku
          last edited by

          Herbie:
          I hv 2Qn.


          1 factorise 3x^2+26x+51.
          HENCE OR OTHERWISE FIND TWO FACTORS OF 32651.

          2. If (a+b)^2=73 and ab=6.5, CALCULATE THE VALUE OF a^2+b^2

          Can help to show the working? Tq
          1. (3x^2+26x+51) = (3x+17)(x+3)

          32651=3*100^2 +26*100 +51 ==> x=100

          Hence factors are (3*100+17) ie 317, and (100+3) ie 103.

          2. (a+b)^2=a^2+2ab+b^2

          ==> 73 = a^2 + 2(6.5) +b^2
          hence a^2 + b^2 =73 - 2(6.5) = 60

          1 Reply Last reply Reply Quote 0
          • Xiao HuX Offline
            Xiao Hu
            last edited by

            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.

            1 Reply Last reply Reply Quote 0
            • F Offline
              FrekiWang
              last edited by

              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

              1 Reply Last reply Reply Quote 0
              • Xiao HuX Offline
                Xiao Hu
                last edited by

                FrekiWang:
                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

                Hi FrekiWang,
                Good one!! I could't have expessed the solution this way, I would have them in English.
                This is a bit tough for O level Maths, isn't it? It's only 1 mark. Crazy.

                Thanks very much, appreciate it.
                Xiao Hu.

                1 Reply Last reply Reply Quote 0
                • B Offline
                  Bestmark
                  last edited by

                  Hi


                  Any recommendation for maths tuition centre? which is better mavis, mind stretcher or smartlab or ms loi?

                  1 Reply Last reply Reply Quote 0
                  • J Offline
                    JadeDry
                    last edited by

                    Hello,


                    Can you please help me with the following question?

                    Make "t" the subject of the formula:

                    T= 2(pi) * (the square root of(( (t^2) + (k^2)) / 2gt))

                    I have checked the answer which is:
                    t = (+ or -) (1/ (2pi)) (The square root of((2gt(T2)) - ((4(pi squared))(k squared))

                    I do not understand how they have used the lower case "t" in the formula.

                    Is it correct?

                    Thanks in advance.

                    1 Reply Last reply Reply Quote 0
                    • B Offline
                      Belle2011
                      last edited by

                      Hello,

                      May I ask where can I get books/resources on math olympiad questions suitable for lower sec?
                      Thank-you.

                      cheers,
                      Belle.

                      1 Reply Last reply Reply Quote 0
                      • M Offline
                        mum_sugoku
                        last edited by

                        JadeDry:
                        Hello,


                        Can you please help me with the following question?

                        Make \"t\" the subject of the formula:

                        T= 2(pi) * (the square root of(( (t^2) + (k^2)) / 2gt))

                        I have checked the answer which is:
                        t = (+ or -) (1/ (2*pi)) *(The square root of((2gt(T*2)) - ((4*(pi squared))(k squared))

                        I do not understand how they have used the lower case \"t\" in the formula.

                        Is it correct?

                        Thanks in advance.
                        Ya by right \"t\" shouldn't be in there if it is the subject.. So I too think that answer is incorrect.

                        Anyway this is a seemingly tedious sum (I mean the working). You'll need to apply the standard formula x=[-b +- sq rt (b^2 -4ac) ]/2a to solve it.. Here's what I've got after working it out, not sure if it's correct though:

                        t ={ (T^2)g +- {sq rt[(T^4)(g^2) - 16(pi^4)(k^2)]} } / (4pi^2)

                        1 Reply Last reply Reply Quote 0

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