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    Q&A - P5 Math

    Scheduled Pinned Locked Moved Primary 5
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    • J Offline
      Jamesbond
      last edited by

      1)3/4 of chelsia’s money is the same as 2/3 of benson’s money. If chelsia had $18 less than benson, what is the total sum of money of chelsia and benson?


      2) There is a total of 620 apples and oranges at a stall. 3/8 of the oranges is equal to 2/5 of the apples. How many apples are there at the stall?

      3)Daniel bought 75 pencil and pens. 3/5 of the pencils is equal to 2/5 of the pens. How many pencils did daniel buy?

      Pl explain detail how to solve the prolems…Thanks in advance.

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

        Dear MathIzzzFun,


        Thank you very much for your help. My son and I can understand. Really appreciate it.

        1 Reply Last reply Reply Quote 0
        • MathIzzzFunM Offline
          MathIzzzFun
          last edited by

          Jamesbond:
          1)3/4 of chelsia's money is the same as 2/3 of benson's money. If chelsia had $18 less than benson, what is the total sum of money of chelsia and benson?


          2) There is a total of 620 apples and oranges at a stall. 3/8 of the oranges is equal to 2/5 of the apples. How many apples are there at the stall?

          3)Daniel bought 75 pencil and pens. 3/5 of the pencils is equal to 2/5 of the pens. How many pencils did daniel buy?

          Pl explain detail how to solve the prolems....Thanks in advance.
          Hi

          Q1. 3/4 of chelsia's money = 2/3 of benson's money
          --> 6/8 of Chelsia's money = 6/9 of Benson's money
          So,
          Chelsia's money --> 8 units
          Benson's money --> 9 units
          1 unit --> $ 18
          Total 17 units --> $ 18 x 17 = $ 306

          They had $306 altogether.

          You can use the same approach for Q2 and Q3.

          cheers.

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          • C Offline
            clblinym
            last edited by

            My son had some problem in solving the question below. Can anyone help? TIA http://i44.tinypic.com/33w43yr.png\">

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            • T Offline
              tianzhu
              last edited by

              Hi moderators.


              Please help to delete.

              Best wishes

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              • Xiao HuX Offline
                Xiao Hu
                last edited by

                tianzhu:
                clblinym:

                My son had some problem in solving the question below. Can anyone help? TIA http://i44.tinypic.com/33w43yr.png\">


                Hi

                Good Morning.

                Draw a double sided arrow to show 10cm on the top and left side of the square and you’ll be able to see that the length and breadth of the shaded area are 6cm and 4cm.

                Perimeter of rectangle = 52cm
                Breadth of rectangle = 10cm
                Hence, length of rectangle = (52 – 20)/2 = 16cm

                Area of shaded part = 6*4 = 24 sqcm

                Area of unshaded area = (10*10) + (16*10) – 24 = 236 sqcm

                Best wishes

                Hi Tianzhu,
                May I know how you get the Breadth of rectangle = 10cm? That seems to be key to answering the rest of the question.

                TIA,
                Xiaohu

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                • T Offline
                  tianzhu
                  last edited by

                  Hi moderators


                  Please help to delete.

                  Best wishes

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

                    tianzhu:
                    Xiao Hu:


                    Hi Tianzhu,
                    May I know how you get the Breadth of rectangle = 10cm? That seems to be key to answering the rest of the question.

                    TIA,
                    Xiaohu

                    Hi Xiaohu

                    Visualise this, move the rectangle upwards until its top edge touches the top edge of the square.

                    Best wishes

                    Hi Tianzhu,
                    Thx. But I was thinking why can't the breadth be > 10cm? What's stopping the breadth to be say 11cm or 12 cm etc?

                    TIA,
                    Xiaohu.

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                    • T Offline
                      tianzhu
                      last edited by

                      Hi moderators


                      Please help to delete.

                      Best wishes

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

                        tianzhu:
                        Xiao Hu:

                        Hi Tianzhu,

                        Thx. But I was thinking why can't the breadth be > 10cm? What's stopping the breadth to be say 11cm or 12 cm etc?

                        TIA,
                        Xiaohu.

                        Hi Xiaohu

                        The breadth of the shaded area is 4cm (10-6).

                        As the slide the rectangle upwards till its top edge touches the top of the square, it moves a distance of 6cm.

                        Therefore the breadth of rectangle is equal to 4+6 which is 10.

                        Alternatively, you may slide the square down till its lower side touches the lower side of the rectangle.

                        Best wishes

                        Hi Tianzhu,
                        If the rectangle's breadth is indeed 10, then yes, moving to the top or bottom will move 6cm up or 4cm down.

                        Please see the figure I draw, these are to the scale of mm. The square is same 10mmx10mm as the original question's. The rectangle is bigger at 20mmx 30mm. The perimeter of the rectangle is 100mm. So for this new diagram, the rectangle still moves a distance of 6mm if you slide the rectangle up but it does not make the breadth 10mm, right?

                        http://i44.tinypic.com/4uzbww.gif\">

                        Is there supposed to be another guess-&-check strategy in order to get the breadth of the rectangle, using the information of the given perimenter to be 52?


                        TIA,
                        Xiao Hu

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