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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • MathIzzzFunM Offline
      MathIzzzFun
      last edited by

      PiggyLalala:
      Need Help.

      Mary has some red pens and some blue pens in the pencil box. If she removes 1 red pen from the pencil box, then 3/8 of the remaining pens in the pencil box are red. If we put back the red pen and remove 3 blue pens instead, then 5/11 of the remaining pens in the pencil box are red. Find the number of red and blue pens.
      TIA

      Hi

      http://i55.tinypic.com/16a77te.jpg\">

      cheers.

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

        ksi:
        tianzhu:

        [quote=\"ksi\"]
        Hi tianzhu,
        Can you check if my solution above is correct based on the revised fraction by adanial? Thanks.

        Hi ksi

        The store sold 15 TV, 10 fans and 20 WM.

        But it had only 3 fans in the store.

        That’s why I think some numbers may not be in order in the question.

        Best wishes

        Yes you are right...the numbers are illogical. :salute:[/quote]Hi Tianzhu, Ksi,
        Ive checked the paper again... the numbers are correct. I'm puzzled how did the teacher solve this one.. lol
        thank for the help.

        1 Reply Last reply Reply Quote 0
        • corneyAmberC Offline
          corneyAmber
          last edited by

          adanial:

          Hi Tianzhu, Ksi,
          Ive checked the paper again... the numbers are correct. I'm puzzled how did the teacher solve this one.. lol
          thank for the help.
          Hi adaniel, do you know the answer for this question?

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

            adanial:

            Hi Tianzhu, Ksi,
            Ive checked the paper again... the numbers are correct. I'm puzzled how did the teacher solve this one.. lol
            thank for the help.
            Hi adanial

            Based on the original question, you can still arrive at a solution, however it defies logic.

            Consider a group of 3 TV, 2 fans and 4 WM.

            The cost per group is (3*1350) + (2*102) + (4*640) or 4050+204+2560.
            This adds up to $6814 per group.

            34070/6814 --- 5
            Hence, you’ve 5 groups.
            This means the shop sold 15 TV, 10 fans and 20 WM.

            3 units ----- 45
            4 units ----- 60

            The shop had 60 items in the beginning.
            3 parts or 36 are for TV, leaving a remainder of 24 for fans and WM.
            1/8 of this remainder are for fans, this means 3 fans and the remaining 21 are for WM.

            Here we’ve an awkward situation where the number of fans sold (10) is more than the number of fan in the store (3).

            If you change 3/5 to 1/3 and 12.5% to 25%, the question will be more logical.

            Best wishes

            1 Reply Last reply Reply Quote 0
            • C Offline
              Concerndad
              last edited by

              2/5 x 100% = 40% (% of TV)

              100% -40% = 60% (%of remainder)
              12.5% x 60% = 7.5% (% of fans)
              60% - 7.5% = 52.5% (% of washing machines wm)

              3 x 1350 = 4050 (3units of tv)
              2 x 102 = 204 (2units of fans)
              4 x 640 = 2560 (4units of wm)
              4050 +204 + 2560 = 6814

              34070/ 6814 = 5 (5sets)

              3units+2units+4units =9units
              9x5=45 (total no. of items sold off during sales)

              3units - 45
              1 units - 15
              4units - 60 (total items in the stores)

              52.5% of 60 = 31.5 ?? (Hmm…paiseh…is there something wrong in the question?)



              Question 2

              1st day - 20%
              2nd day - 60
              total for 1st and 2nd days - 20% + 60

              2units -> 20% + 60 (if can draw model to show will be clearer)
              1 unit -> 10% + 30
              7units -> 70% + 210

              therefore 30% -> 210
              10% -> 210/3 = 70
              100% -> 70 x 10 = 700

              the jigsaw puzzle consist of 700pcs.

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

                ksi:
                adanial:


                Hi Tianzhu, Ksi,
                Ive checked the paper again... the numbers are correct. I'm puzzled how did the teacher solve this one.. lol
                thank for the help.

                Hi adaniel, do you know the answer for this question?


                Hi Tianzhu, the answer is 21

                1 Reply Last reply Reply Quote 0
                • L Offline
                  linden2000
                  last edited by

                  Alternative solution to Q2


                  1st day
                  Completed - 2units
                  Uncompleted - 8units
                  Total 10 units

                  2nd day
                  Completed - 2 units
                  Uncompleted - 5 units
                  Total 7 units

                  Since the total number of pieces is unchanged, equate total units on both days to a common number (lowest common multiple).

                  So
                  1st day
                  Completed 2x7=14
                  uncompleted 8x7=56
                  total 70

                  2nd day
                  Completed 2x10=20
                  uncompleted 5x10=50
                  total 70

                  Change in number completed =20-14=6
                  so 6 units = 60
                  1 unit = 10
                  Total number of pieces = 70 units = 70x10= 700

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

                    Can some kind soul please help with the following?


                    http://img.photobucket.com/albums/v257/tbctbc/mathq.jpg\">
                    Sorry, I omitted the Y when drawing. It is above T.

                    THe above figure shows a rectangle PQRS. PYR is a straight line and XYZ is parallel to SRT. Given that PX=10cm, TR=9cm and ZR =4cm, find the shaded area of rectangle XYTS.

                    1 Reply Last reply Reply Quote 0
                    • C Offline
                      CJ-2011
                      last edited by

                      Any kind soul can explain the rationale behind the solution for question PSP102, pg 368 of "Problem Solving Processes in Maths (New Syllabus)" by Fabian Ng ?


                      The problem is listed below :-

                      At 0800, Noel left town A for town B while Jane left town B for town A. At 1600 their cars passed each other. 5 hour later Jane reached town A but Noel was 240km away from town B,
                      a) How long after they start before they passed each other
                      b)Find the distance between the two towns

                      Solution :
                      a) 1600 - 0800 = 0800
                      Each driver had driven for 8 hr when they passed each other.

                      b) 240 / (8-5) = 80
                      The total speed of the two drivers was 80 km/h

                      80 x 8 = 640
                      The two towns was 640 km apart
                      ----------------------------------------------------------------
                      I cannot comprehend and explain the part (b) on total speed. Is this some form of relative speed ? Is the concept of total speed taught in school ?

                      Thanks

                      1 Reply Last reply Reply Quote 0
                      • D Offline
                        dovetail
                        last edited by

                        blitz:
                        Can some kind soul please help with the following?


                        http://img.photobucket.com/albums/v257/tbctbc/mathq.jpg\">
                        Sorry, I omitted the Y when drawing. It is above T.

                        THe above figure shows a rectangle PQRS. PYR is a straight line and XYZ is parallel to SRT. Given that PX=10cm, TR=9cm and ZR =4cm, find the shaded area of rectangle XYTS.
                        first you need to identify that triangle PYX is similar to triangle YRT.
                        Reason: observe that the angles in the 2 triangles are all the same size.

                        Now, use ratio

                        PX/YT = XY/TR
                        10/4 = XY/9
                        XY = 90/4

                        Area of XYST = XY * 4
                        = 90/4 * 4
                        = 90

                        Ans. 90 cm square.

                        1 Reply Last reply Reply Quote 0

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