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    Q&A - P4 Math

    Scheduled Pinned Locked Moved Primary 4
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    • T Offline
      tianzhu
      last edited by

      elisammom:
      Hi another question


      24 coins are used to form a square. The number of coins on each side of the square is the same. Find the number of conins on each side of the square.

      Ans: 7

      Pls help. Thanks
      Hi

      Perhaps, for questions of such genre, it’s useful to have a simple pictorial representation. It might help to organise your thoughts.

      The coins at the 4 corners of the square are counted twice.

      24 + 4 ----- 28

      28/4 ----- 7

      Please study this similar type of question that appeared in the P5 thread.

      Hope this helps.

      Best wishes

      http://farm6.staticflickr.com/5272/6897367566_8d8cd99d93_z.jpg\">

      1 Reply Last reply Reply Quote 0
      • R Offline
        ruohoo97
        last edited by

        elisammom:
        Pls help,


        For every 4 beads that A has, B has 2 more beads than her. They have 60 beads altogether. How many beads does B have?

        Thanks
        If A has 24, B has 36, how can that be B has 2 more beads than her? I can see that 😓

        1 Reply Last reply Reply Quote 0
        • R Offline
          ruohoo97
          last edited by

          elisammom:
          Hi another question


          24 coins are used to form a square. The number of coins on each side of the square is the same. Find the number of conins on each side of the square.

          Ans: 7

          Pls help. Thanks
          24 coins put at four coins at four corner of a square, left 20. 20 divide by 4 =5, so each side has 5 coins plus 2 coins at corner=7.

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          • MathIzzzFunM Offline
            MathIzzzFun
            last edited by

            elisammom:
            Hi another question


            24 coins are used to form a square. The number of coins on each side of the square is the same. Find the number of conins on each side of the square.

            Ans: 7

            Pls help. Thanks
            Hi

            hope this picture will help to tackle such questions more easily.

            http://i44.tinypic.com/ieiiqb.png\">

            cheers.

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

              ruohoo97:
              elisammom:

              Pls help,


              For every 4 beads that A has, B has 2 more beads than her. They have 60 beads altogether. How many beads does B have?

              Thanks

              If A has 24, B has 36, how can that be B has 2 more beads than her? I can see that 😓

              if ruohoo97, if you read it carefully again, For every 4 beads that A has, B has 2 more beads than her

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              • T Offline
                tianzhu
                last edited by

                elisammom:
                I used the guess and check method but the answer is not right.

                The correct answer is 36, just like chloecube done.

                I like to know why guess and check method is wrong in this question. Thanks
                Hi

                Yes, you may use GC.

                For every 4 beads that A has, B has 2 more beads than her; this means if A has 4 beads, B has 6 beads.

                Try it and you should arrive at 24 beads (A) and 36 beads(B).

                A common method used by students is “Grouping concept”.

                Another way, observe that for every incremental step, there is a difference of 10 beads (total number)

                4, 6 ----- 10
                8, 12 ------ 20

                We are looking for 60.

                60 is 6 times as many as 10

                Hence multiply initial numbers by 6.

                4*6 -----24
                6*6 ------ 36

                Best wishes

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                • R Offline
                  ruohoo97
                  last edited by

                  Yeh, I see. Thanks a lot!

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

                    Maths is interesting and fun if we can grab the concepts. its challenging and fun solving them

                    but i find it hard to put it across to my kids sometimes, they dun seem to understand or see it at my level

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                    • E Offline
                      elisammom
                      last edited by

                      tianzhu:
                      elisammom:

                      I used the guess and check method but the answer is not right.

                      The correct answer is 36, just like chloecube done.

                      I like to know why guess and check method is wrong in this question. Thanks

                      Hi

                      Yes, you may use GC.

                      For every 4 beads that A has, B has 2 more beads than her; this means if A has 4 beads, B has 6 beads.

                      Try it and you should arrive at 24 beads (A) and 36 beads(B).

                      A common method used by students is “Grouping concept”.

                      Another way, observe that for every incremental step, there is a difference of 10 beads (total number)

                      4, 6 ----- 10
                      8, 12 ------ 20

                      We are looking for 60.

                      60 is 6 times as many as 10

                      Hence multiply initial numbers by 6.

                      4*6 -----24
                      6*6 ------ 36

                      Best wishes


                      Thanks tianzhu
                      Ops I make a cal error;( yes now I see it;)

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

                        Hi all,


                        Please help!

                        Q. There are some chickens and horses at a farm. Altogether, there are 41 heads and 118 leags. How many chickens are there?

                        TIA

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