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    Q&A - P3 Math

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

      tutormum:
      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!
      For 2 additional balloons given out to a student – Teacher will have 4 less balloons.
      (Difference between 2 balloons left and short of 2 balloons = 4 balloons)

      No. of students = 4/2 = 2

      (# If teacher has 4 less balloons after giving out 2 additional balloons to each student, then there must be 4/2 =2 students)

      1 Reply Last reply Reply Quote 0
      • K Offline
        kitty2
        last edited by

        Great to hear that your book will be out soon as it’s just on time for my P6 kid next year.I like your simple and elegant solution ,hopefully all your questions come with detailed answers.I can’t find an assessment book which has detail answers.


        All the best to you and thank you for your kindness.

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        • M Offline
          Muffins
          last edited by

          Dharma:
          tutormum:

          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!

          For 2 additional balloons given out to a student – Teacher will have 4 less balloons.
          (Difference between 2 balloons left and short of 2 balloons = 4 balloons)

          No. of students = 4/2 = 2

          (# If teacher has 4 less balloons after giving out 2 additional balloons to each student, then there must be 4/2 =2 students)

          Dharma, hey! 🙂

          I forgot that method...

          Short, simple, and easy to remember. You really are one of the best sum solvers on this site! :udaman:

          1 Reply Last reply Reply Quote 0
          • M Offline
            mamona
            last edited by

            Sakina has 15 coins. There are 20c coins and 50c coins. The total amount is $4.80. How many 20c coins are there?


            The above sum can be done using the guess method like

            10 20c coins = $2.00
            5 50c coins = $2.50

            then by increasing 20c coins number & decreasing the 50c coins number. But this method is very time consuming & a P3 student may not able to guess correctly which coin to increase & which one to decrease.

            So can any one let me know some method which can be used for these kind of sums.

            Thanks in advance.

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            • T Offline
              tutor_your_child
              last edited by

              mamona:
              Sakina has 15 coins. There are 20c coins and 50c coins. The total amount is $4.80. How many 20c coins are there?
              Perhaps you could explain the following method for P3 :
              The first step is to assume that all the coins are either 50c coins or 20c coins.
              Assuming all 15 coins are 20c coins, total value is only $3
              Find the difference compared to $4.80; i.e. $1.80.
              (easier to do show this in a table - see below)

              2nd step
              Since the total is less than $4.80, there should be more 50c coins and fewer 20c coins ( i.e. need more of those with higher value)
              Conversely if the total is more than $4.80, there should be fewer 50c coins and more 20c coins. (This scenario is not possible for we had assumed that all are 20c coins).

              3rd step
              The 3rd step cuts down the number of iterations required for the 'guess and check' method.

              For every 50c coin you increase, you have to decrease the 20c coin as well.
              As such, the value increased per coin is only 30c (50c-20c)
              As above, assuming all are 20c coins, total value is only $3, we are short of $1.80.
              $1.80 / 30c = 6
              i.e. increase 6 50c coins and decrease 6 20c coins
              As such, the number of 20c coins is 15-6=9

              4th step
              Complete in table to check
              http://www.postimage.org/

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              • M Offline
                mamona
                last edited by

                Many Many Thanks for your reply. Yes I understood the method. :lol:

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                • T Offline
                  tutor_your_child
                  last edited by

                  mamona:
                  Many Many Thanks for your reply. Yes I understood the method. :lol:

                  You are welcome!

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                  • O Offline
                    Ogima
                    last edited by

                    hello comrades,


                    need your help to solve the following P4 Maths without using ALGEBRA. :?:

                    detailed explanation much needed. thanks.

                    Q35) May had 60 more apples than June. After June gave 1/3 of her apples away, May had twice as many apples as June.

                    How many apples did they both have at first ?

                    1 Reply Last reply Reply Quote 0
                    • E Offline
                      elkniwt
                      last edited by

                      Ogima:
                      hello comrades,


                      need your help to solve the following P4 Maths without using ALGEBRA. :?:

                      detailed explanation much needed. thanks.

                      Q35) May had 60 more apples than June. After June gave 1/3 of her apples away, May had twice as many apples as June.

                      How many apples did they both have at first ?
                      It will be easier to see if you draw model. (I am not good at creating nice pictures, so let me try to explain the \"model\" I drew.)

                      May [----][----][----][----]
                      {indicate the last unit}<60>
                      June [----][----][////]

                      (the last unit for June is the given away part)
                      Each unit should be drawn same size.

                      From model, 1 unit = 60
                      May has 4 x 60 = 240
                      June has 3 x 60 = 180 initially

                      1 Reply Last reply Reply Quote 0
                      • J Offline
                        john.09525316
                        last edited by

                        Ogima:
                        hello comrades,


                        need your help to solve the following P4 Maths without using ALGEBRA. :?:

                        detailed explanation much needed. thanks.

                        Q35) May had 60 more apples than June. After June gave 1/3 of her apples away, May had twice as many apples as June.

                        How many apples did they both have at first ?

                        Let me try, please note it is not drawn in scale

                        a)assume June has 3 units, therefore May has 3 units +60

                        May has 3 unit + 60 more [ ][ ][ ][ 60 more ]
                        June has 3 unit [ ][ ][ ]


                        b) May gave away 1/3, that mean is 1 unit, hence the model became

                        May [ ][ ][ ][ 60 more ]
                        June [ ][ ]

                        May has twice as many apples as June, hence

                        [1 st Half ] [2nd half ]
                        ------------- -----------------
                        May [ ][ ] [ ][ 60 more]
                        June [ ][ ]


                        c) Let look at the half of it

                        May [ ][ 60 more ]
                        June [ ][ ]


                        Canceal 1 unit, the other unit is equal to 60, therefore

                        June has 60+60+60+60 = 240
                        May has 60+60+60 = 180

                        Total = 240+180 = 420 apples

                        They have total of 420 apples

                        I hope the above is able to help you.

                        Cheer!

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