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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • N
      nounou
      last edited by

      http://i50.tinypic.com/54yk9l.jpg\"> Some please help! :?: Need to double confirm answer to this Q.

      Nanyang Prelim 2012 Q2.
      :thankyou: :thankyou: :thankyou:

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      • N
        nounou
        last edited by

        http://i46.tinypic.com/idhth5.jpg\">

        :?: Please help! :imdrowning:
        Nanyang Prelim 2012 Q18 :slapshead: 😢
        Answer given = 96
        :thankyou: :thankyou: :thankyou:

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        • T
          tianzhu
          last edited by

          Treelodge:
          http://i46.tinypic.com/idhth5.jpg\">

          :?: Please help! :imdrowning:
          Nanyang Prelim 2012 Q18 :slapshead: 😢
          Answer given = 96
          :thankyou: :thankyou: :thankyou:
          Hi

          I think you can use MD, but I’ll use letters of the alphabet so as to save some time.

          Number of representatives in B ------- 1 box
          Number of representatives in A ------- 1 box + 18
          Number of representatives in C ------- 1 box + 18 + 6

          School A
          Boys: Girls------- 1A:3A

          School B
          Boys: Girls ------- 1B:5B

          School C
          Boys: Girls ------- 2C:5C

          1A + 1B + 2C ------ 1p
          3A + 5B + 2C ------3p

          3A + 3B +6C ------- 3p

          3A + 3B +6C ------- 3A + 5B + 2C

          Hence, you’ll get 1C ------- 2B

          Now change (2C + 5C) to 14B and do a comparison.

          14B – 6B (1B+5B) ------- 24

          8B ------- 24

          B ------- 3

          C ------- 6

          Number of representatives in C ------- 7*6 ------ 42

          Number of representatives in A ------- 42 - 6 ------ 36

          Number of representatives in C ------- 18

          Total number ------- 96

          Best wishes

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          • T
            tianzhu
            last edited by

            Treelodge:
            Treelodge:

            Please help to solve this Q :yikes: :?:

            Q10 from Nanyang Prelim 2012. :gloomy:
            Answer given = 266 2/3 m
            :thankyou: :thankyou: :thankyou: [IMG]http://i49.tinypic.com/30nfdja.jpg\">

            http://i49.tinypic.com/30nfdja.jpg\">

            Hi

            I did a quick check and get an answer of 266 and 2/3 (266.67)

            I don't have time to prepare slide now, is the answer same at the ws?

            Best wishes

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            • T
              tianzhu
              last edited by

              Treelodge:
              http://i50.tinypic.com/54yk9l.jpg\"> Some please help! :?: Need to double confirm answer to this Q.

              Nanyang Prelim 2012 Q2.
              :thankyou: :thankyou: :thankyou:
              Hi

              You would like to double confirm he answer.

              My answer is 150.

              Please share the answer given in the WS ?

              Best wishes

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              • T
                tianzhu
                last edited by

                Treelodge:
                http://i48.tinypic.com/11t9wz6.jpg\"> Very pai seh! :imsorry: Another Q from Nanyang Prelim 2012. Q15. :gloomy: Someone please tell (lie to) me actual PSLE Maths is not that difficult. 😢

                How to do bi and bii? Answers given bi=21, 22, 51,52 bii=72
                :thankyou: :thankyou: :thankyou: [img]
                Hi

                Could we have a clearer image, please?

                Best wishes

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                • MathIzzzFunM
                  MathIzzzFun
                  last edited by

                  Treelodge:
                  http://i50.tinypic.com/54yk9l.jpg\"> Some please help! :?: Need to double confirm answer to this Q.

                  Nanyang Prelim 2012 Q2.
                  :thankyou: :thankyou: :thankyou:
                  area of shaded regions = 1/2 area of square = 1/2 x 10 x 10 = 50 cm²

                  ratio of shaded : unshaded --> 1: 3 = 50 : 150
                  Unshaded area --> 150 cm²

                  cheers.

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                  • MathIzzzFunM
                    MathIzzzFun
                    last edited by

                    Treelodge:
                    http://i46.tinypic.com/idhth5.jpg\">

                    :?: Please help! :imdrowning:
                    Nanyang Prelim 2012 Q18 :slapshead: 😢
                    Answer given = 96
                    :thankyou: :thankyou: :thankyou:
                    alternate approach using concept of multiples and a little Guess & Check..

                    School B, boys : girls --> 1 : 5 (total 6)
                    School A, boys : girls --> 1 : 3
                    School C, boys : girls --> 2 : 5
                    School A = School B + 18 = School B + 6 x 3
                    School C = School A + 6 = School B + 6 x 4
                    Total number of students in School B is a multiple of 6,
                    So, total number of students in School B is a multiple of 6 and total number of students in School C is also a multiple of 6

                    School C, boys : girls --> 2 : 5 --> 12 : 30
                    School A, boys : girls --> 1 : 4 --> 9 : 27 (6 less than C)
                    School B, boys : girls --> 1 : 5 --> 3 : 15 (18 less than A)
                    check --> boys: girls = (12+9+3) : (30+27+15) = 24:72= 1:3 √
                    So,
                    total number of students = 24 + 72 = 96

                    cheers.

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                    • K
                      ksme
                      last edited by

                      [quote="kuts"]HI Ksme,


                      A B C D A A B B C C D D A A A B B B C C C D D D …? 160th
                      What is the 160th letter

                      Thanks in advance!


                      Hi,
                      Please see the solution below. It’s not a very elegant way but it gives the answer in a reasonable time. By the way, where is this question from and how many marks is the question for?

                      Pattern :
                      ABCD ------------------- 4 (1 time each alphabet in a set of 4)
                      AABBCCDD---------------- 8 (2 times each alphabet in the set of 4)
                      AAABBBCCCDDD----------- 12 (3 times each alphabet in the set of 4)
                      AAAABBBBCCCCDDDD ------ 16 (4 times each alphabet in the set of 4)
                      etc ------ 20 (5 times each alphabet in the set of 4)
                      etc ------ 24 (6 times each alphabet in the set of 4)
                      etc ------ 28 (7 times each alphabet in the set of 4)
                      etc ------ 32 (8 times each alphabet in the set of 4)
                      etc ------ 36 (9 times each alphabet in the set of 4)

                      Adding the total number of alphabets for each set to come close to 160 letters = 4+8+12+16+20+24+28+32= 144.

                      The 160th letter is therefore belonging to the 9 times each alphabet set of 4.

                      160-144 = 16

                      Since A will be shown 9 times (16-9 = 7), therefore, the 160th letter will be B.

                      Hope the above helps.

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                      • MathIzzzFunM
                        MathIzzzFun
                        last edited by

                        Treelodge:
                        Treelodge:

                        Please help to solve this Q :yikes: :?:

                        Q10 from Nanyang Prelim 2012. :gloomy:
                        Answer given = 266 2/3 m
                        :thankyou: :thankyou: :thankyou: [IMG]http://i49.tinypic.com/30nfdja.jpg\">

                        http://i49.tinypic.com/30nfdja.jpg\">

                        Approach 1 --> using speed ratio = distance ratio

                        Tom's speed : John's speed = 160 : 200 --> 4 : 5
                        Divide the track into 18 units (=400m), so that each half is 9 units

                        At first meeting point, Tom would have run 4 units, John 5 units
                        From first meeting point to second meeting point, Tom would have run 8 units, John --> 10 units
                        From start to second meeting point, Tom ran --> 4 units+ 8 units = 12 units
                        Distance ran by Tom --> 12/18 x 400 m = 266 2/3 m

                        Approach 2 --> total distance /total speed concept

                        At first meeting point, total distance run by both = 400m/2 = 200m
                        From 1st meeting point to 2nd meeting point, total distance run by both = 400m
                        Total distance run by both from start to 2nd meeting point = 600 m
                        In 1 min, total distance run by both = (160m+200m)=360m
                        Time taken to cover 600m = 600/360= 5/3 min
                        In 5/3 min, distance run by Tom = 160 x 5/3 = 266 2/3 m

                        cheers.

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