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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • T Offline
      tianzhu
      last edited by

      Apphia:
      1) Mr Lee had 36 more guppies than goldfish. He sold 1/4 of the guppies and 1/5 of the goldfish. He found that he had sold 28 more guppies than goldfish. The guppies were sold $0.55 each and the goldfish at $0.85 each. What was the total amount he received?
      Hi

      It is suggested that MD be used to solve this question.

      Please follow these steps.

      Guppies ------ 20 units + 36
      Goldfish -------20 units

      Guppies sold ------- 5 units + 9
      Goldfish sold ------- 4 units

      He found that he had sold 28 more guppies than goldfish.
      Compare the MD

      1 unit ----- 19

      Guppies ------(5*19)+9 ----- 104
      Price of Guppies -------104*0.55 ------- 57.20

      Goldfish ------(4*19) ------- 76
      Price of Goldfish --------76*0.85 ------- 64.60

      (57.20+64.60) ------- 121.80

      Best wishes

      1 Reply Last reply Reply Quote 0
      • T Offline
        tianzhu
        last edited by

        Apphia:
        Hi TianZhu,


        Thank you for replying so quickly.

        Ever so grateful ! šŸ˜„
        Hi

        You're welcome.

        Best wishes

        1 Reply Last reply Reply Quote 0
        • A Offline
          Amberz
          last edited by

          iFruit:
          Amberz:

          The total of 4 numbers P, Q, R,and S was 291.

          When P was tripled, Q was halved, R was increased by 20 and S was decreased by 25, the four numbers became equal.
          Find the value of R. :?

          This is a backwards problem..

          Assume at the end, each number became 3u.

          Then at the beginning:

          P = 3u/3 = u
          Q = 3u x 2 = 6u
          R = 3u - 20
          S = 3u + 25

          u+6u+3u-20 + 3u + 25 = 291

          13u = 286--->u =22

          R = 3u - 20 = 46

          HTH.

          Got it thanks!

          1 Reply Last reply Reply Quote 0
          • T Offline
            tisha
            last edited by

            tianzhu:
            Puffer:

            Can anyone help?


            In planning for a party, Mrs Lim bought enough satay to allow her guests an average of 25 sticks each. If 2 more people joined the party, how many fewer sticks of satay will each of her guests be given?

            Thanks in advance.

            Hi

            Good Afternoon.

            I did a quick check; it looks like there are multiple answers.

            One is 10 and the other is 5.

            It is suggested that you use Tabulated list/Systematic listing to solve this question.

            Best wishes

            Can no other method that can be used?
            How to do tabulation, for this problem I'm not even sure where to start. :?

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            • A Offline
              atutor2001
              last edited by

              tisha:
              [quote]
              In planning for a party, Mrs Lim bought enough satay to allow her guests an average of 25 sticks each. If 2 more people joined the party, how many fewer sticks of satay will each of her guests be given?
              Can no other method that can be used?
              How to do tabulation, for this problem I'm not even sure where to start. :?[/quote]This is question is an example of the superiority of algebra over heuristic (but it is too complicated for Pr. However, I think GEP can still do it) Let me first present the algebraic method then the heuristic method.

              Algebraic method :
              Let the no. of guests be G
              Let the decrease in the average be d

              Before :
              No. of guests = G
              Average no. of satays = 25
              Total no. of satay = 25G

              After :
              No. of guests = (G+2)
              Average no. of satays = (25-d)
              Total no. of satay = (G+2)(25-d) = 25G + 50 - d(G+2)

              The Totals are the same, so we form the algebraic equation : 25G = 25G + 50 - d(G+2)
              By simplifying we will get : d(G+2) = 50

              The factors of 50 are : 1x50, 2x25, 5x10

              By matching d x (G+2) to the above factors:
              When d = 1; then (G+2) must be 50 so No. of guests, G = 48
              When d = 2; then (G+2) must be 25 so No. of guests, G = 23
              When d = 5; then (G+2) must be 10 so No. of guests, G = 8
              When d = 10; then (G+2) must be 5 so No. of guests, G = 3
              When d = 50; then (G+2) must be 1 so No. of guests, G = -1 (rejected)
              When d = 25; then new average becomes (25-55 = 0) (rejected)


              So there are 4 possible answers :
              New average = 24; Original No. of guest = 48
              New average = 23; Original No. of guest = 23
              New average = 20; Original No. of guest = 8
              New average = 15; Original No. of guest = 3


              Now come the interesting part of how to use heuristics to solve this question.
              It centers on the ability of the child to see that d(G+2) = 50

              The Pr kid must reason that :

              1. If the 2 new person is also getting 25 satays, then another 50 satays is needed.

              2. Since no new satays are going to be added, these 50 satays must be \"taken back\" by asking everybody to contribute equal amount (including the 2 new guests).

              3. This contribution will cause a drop in their original average of 25 i.e. this contribution is the actually the difference in the 2 averages.

              4. So the 50 satays = [Contribution x New Total no. of children]

              5. The child must then know the mathematical skill of breaking 50 into its factors and match them to [Contribution x New Total no. of guests] to get the different combinations of answers.

              My above reasonings actually comes from the algebraic solution. So algebra is a mathematical tool but models and heuristics are not. We can get the heuristic concept from algebra. Good luck!

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              • P Offline
                pensiveowl
                last edited by

                Could someone pls help us with this problem? It's taken from PROBLEM SOLVING MADE EASY - FRACTIONS (son and I couldn't make the given soln out)


                Clement spent 1/5 of his money on some pens, 1/4 of his remaining money on a book and 3/4 of the money he had left on three thumb drives. Each thumb drive cost 18 times as much as a pen he bought. How many pens did he buy?

                1 Reply Last reply Reply Quote 0
                • Q Offline
                  qic
                  last edited by

                  Pen - 1/5

                  Remain - 1-1/5=4/5
                  Book - 1/4x4/5 = 1/5
                  Had left - 1-1/5-1/5 = 3/5
                  3 thumb Drive - 3/4x3/5 = 9/20
                  1 thumb Drive = 3/20
                  Pen - 3/20/18=1/120
                  number of pens - 1/5 / 1/120 = 24

                  Pens: 24

                  1 Reply Last reply Reply Quote 0
                  • P Offline
                    pensiveowl
                    last edited by

                    qic:
                    Pen - 1/5

                    Remain - 1-1/5=4/5
                    Book - 1/4x4/5 = 1/5
                    Had left - 1-1/5-1/5 = 3/5
                    3 thumb Drive - 3/4x3/5 = 9/20
                    1 thumb Drive = 3/20
                    Pen - 3/20/18=1/120
                    number of pens - 1/5 / 1/120 = 24

                    Pens: 24
                    Thanks very much! - much simpler than the soln given in the workbook!

                    1 Reply Last reply Reply Quote 0
                    • A Offline
                      anneshirleygilbert
                      last edited by

                      Can somebody please help with this question?


                      Samad bouoght 4 similar pens and 6 similar files. Each pen cost $1.20 more than a file. If the total cost was $6.40 more than the total cost of the pens, how much did Samad spend in all?

                      1 Reply Last reply Reply Quote 0
                      • T Offline
                        tianzhu
                        last edited by

                        anneshirleygilbert:
                        Can somebody please help with this question?


                        Samad bouoght 4 similar pens and 6 similar files. Each pen cost $1.20 more than a file. If the total cost was $6.40 more than the total cost of the pens, how much did Samad spend in all?
                        Hi

                        A quick one, the answers are in odd denominations.

                        If the total cost was $6.40 more than the total cost of the pens ----- this means that 6 similar files cost $6.40

                        6F -------6.4
                        1F ----- 1.067

                        Total cost ----- 10F +4.8 -------15.47

                        If it's 8 files instead of 6 files, you'll get nice numbers.

                        Best wishes

                        1 Reply Last reply Reply Quote 0

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