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    Q&A - P3 Math

    Scheduled Pinned Locked Moved Primary 3
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    • D Offline
      Daddy
      last edited by

      Thanks,


      But seem not correct. After they have given the marbles. All have equal amount of marbles. How??

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

        Sorry. I interpret the question wrongly.


        This shld be a "Work Backward" method. \t\t\t\t\t\t\t
        3 of them have equal share of 21, must be 7 marble each for Alfred, Ben and Chuck.\t\t\t\t\t\t\t
        \t\t\t\t\t\t
        1) Add back what they have at first:\t\t\t
        2) Then minus off what they receive from each other

        Alfred
        7 + 2(gives to Ben) - 1 (receives from Chuck) = 8

        Ben :
        7 + 3 (gives to Chuck) - 2 (receives from Alfred) = 8

        Chuck :
        7 + 1 (gives to Alfred) - 3 (receives from Ben) = 5
        \t\t\t\t

        \t
        Now check again,\t\t\t\t\t\t\t
        Alfred has 8 at first, gives away 2 to Ben and receives 1 from Chuck = 7
        Ben has 8 at first, gives away 3 to Chuck, and receives 2 from Alfred, = 7
        Chuck has 5 at first, receives 3 from Ben, and gives away 1 to Alfred = 7

        Hope is correct now.

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        • D Offline
          Dharma
          last edited by

          elkniwt:
          Hi,


          I was going thr' this paper with my P3 girl. Need help in this qn. Thanks

          Catholic High 2007 CA2 Q40.

          http://www.postimage.org/image.php?v=Ts2ds4FJ
          Hi elkniwt,

          You have 5 boxes, (top, bottom, left, right and middle)

          The unknown is the middle one in the 3rd diagram. For patterns you need to find the relationship of the numbers of 4 outer boxes and the number in the middle box. You need to manipulate the numbers using the 4 basic operations and see if you get the number in the middle box for the 1st and 2nd diagrams. If they are correct, then you have established the pattern and go get the middle number in the 3rd diagram.

          1st diagram
          Left box + Top box = 1/8 + ¾ = 7/8
          Right box + Bottom box = 3/8 + 1/8 = 4/8
          Middle box = 7/8 – 4/8 = 3/8

          2nd diagram
          Left box + Top box = 1/3 + 1/6 = 4/12 + 2/12 = 6/12
          Right box + Bottom box = 4/12 + 1/12 = 5/12
          Middle box = 6/12 – 5/12 = 1/12

          3rd diagram
          Left box + Top box = 1/3 + 1/3 = 3/9 + 3/9 = 6/9
          Right box + Bottom box = 3/9 + 1/9 = 4/9
          Middle box = 6/9 – 4/9 = 2/9

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          • D Offline
            Dharma
            last edited by

            Daddy:
            //Moderator's note: Topics merged.


            Hi,

            I have one difficult question to ask. Please help. Thank.

            \"Alfred, Ben and Chuck have some marbles. They have 21 marbles altogether. If Alfred gives Ben 2 marbles. Ben gives Chuck 3 marbles and Chuck gives Alfred 1 marbles, they will have equal numbers of marbles each. How many marbles does each of them have at first?\" :?

            Thanks
            daddy
            http://www.postimage.org/

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            • T Offline
              Tang
              last edited by

              "Alfred, Ben and Chuck have some marbles. They have 21 marbles altogether. If Alfred gives Ben 2 marbles. Ben gives Chuck 3 marbles and Chuck gives Alfred 1 marbles, they will have equal numbers of marbles each. How many marbles does each of them have at first?"


              21 / 3 = 7

              Alfred – 7 - 1 + 2 = 8 marbles

              Ben – 7 + 3 - 2 = 8 marbles

              Chuck – 7 + 1 - 3 = 5 marbles


              Suggested solution by Tang’s sister

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              • T Offline
                Tang
                last edited by

                "Gregory and Timothy has 192 stickers together. After Timothy gave 30 stickers to Gregory, Timothy had thrice as many sticker as Gregory. How many stickers did Gregory have at first?"


                4u – 192
                1u – 192 / 4 = 48

                G – 48 - 30 = 18 stickers

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                • D Offline
                  Daddy
                  last edited by

                  Thanks Devnahouse. I understood and going to teach my duaghter.

                  1 Reply Last reply Reply Quote 0
                  • Suz855S Offline
                    Suz855
                    last edited by

                    //Moderator's note: Topics merged


                    Hi,

                    Not sure this qn should even be classify under P3 maths, anyway looking for a simplify soln, do share if you have good suggestion. No algebra or simultaneous equation, thanks

                    Qn :
                    There were 84 apples and pears in a box at first. When 2/5 of the apples
                    and 2/3 of the pears were eaten, there were 40 fruits left.

                    a) How many apples were there at first
                    b) What fraction of the fruits left were pears?

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                    • T Offline
                      tutormum
                      last edited by

                      Just take the extra 2 balloons and add the 2 more she needs. That will be 4. If she gives 6 each at first, and 8 each later, that means she gives each student another 2 each. Since she have 4 balloons to give away, she has 2 students.

                      Check: If she gives 2 students 6 balloons, she’ll have given 12, 2 X 6 = 12, there is 2 extra, therefore she has 14 balloons at first.
                      If she gives 2 students 8 balloons, she’ll have to give 2 X 8 = 16. As she has only 14 balloons, she needs 2 more.
                      Let me give another example.
                      If Ken gives each child 10 sweets, he’ll have 3 left. If he gives each child 1 more, he needs another 5 sweets. How many children are there?
                      Just take 3 + 5 = 8. If he gives each 1 more, he’ll give away all 8 sweets, right? So there are 8 children.
                      Check: 8 X 10 = 80 + 3 = 83 sweets in total
                      If he gives 1 more each, he needs 5 more, that is he needs 83 + 5 = 88. Since 1 child will be given 11, which is originally 10 + 1 more. 88/11 = 8.
                      For this type of question, just add the extra and the required number and work from there.
                      Hope my explanation is clear.

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                      • T Offline
                        tutormum
                        last edited by

                        Ooops. Sorry, forgot to quote. My early reply is for the following question.


                        A teacher gave her students some balloons. If she gave 6 balloons to each student, she will have 2 balloons left. If she gave 8 balloons to each student, she will be short of 2 balloons. What is the smallest possible number of students she has?

                        Try to do this without using ALGEBRA!

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