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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • T Offline
      tianzhu
      last edited by

      LaiHeng:
      thanks .. is really clear explanation

      Hi

      You're welcome.

      Best wishes

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

        LaiHeng:
        2. A bag contains 200 blue balls and 450 yellow balls. A tin contains 850 blue and 340 yellow balls. After some blue and yellow balls are moved from the tin to the bag, the percentage of the blue balls in the bag becomes 50% and the blue balls in the tin becomes 70%.

        a)How many blue balls are moved from the tin to the bag?
        b)How many yellow balls are moved from the tin to the bag?
        Hi

        In answering this question, a student should note that the total number of blue and yellow balls remains the same in the before and after scenarios.

        Before
        Bag
        Blue ----- 200
        Yellow ----- 450

        Tin
        Blue ----- 850
        Yellow ----- 340

        Total number of blue balls ------ 200+850 ------1050
        Total number of yellow balls ------ 450+340 ------790

        After
        Bag
        Blue ----- 1 unit
        Yellow ----- 1 unit

        Tin
        Blue ----- 7 parts
        Yellow ----- 3 parts

        Here, we’ve two equations with two variables, and we cannot run away from simultaneous equations.

        Some students may find it difficult to grasp SE concepts; the challenge is to help the less mathematically inclined students deal with such handicaps.

        Needless to say, students who have learned SE from external courses or their parents would go ahead with SE.

        Students who are able to grasp SE concepts may use letters of the alphabets to solve the equations or work using units and parts to represent the quantities of the blue and yellow balls as shown above.

        Let’s begin.

        1 unit + 7 parts ------1050
        1 unit + 3 parts ------ 790
        4 parts ------ 260

        Tin
        Yellow ------ 3 part ----- 3*65 ------ 195
        Blue ------ 7 parts ---- 7*65 ----- 455

        a)How many blue balls are moved from the tin to the bag?
        850 – 455 ------ 395


        b)How many yellow balls are moved from the tin to the bag?
        340 – 195 ------- 145

        For those students who have difficulties understanding SE, you may solve it by visualising such equations in pictorial form.

        Please help to check the calculations as they had been done in a rush.

        Best wishes

        1 Reply Last reply Reply Quote 0
        • ozoraO Offline
          ozora
          last edited by

          tianzhu:
          Hi LaiHeng


          Good Morning.

          Here's the solution to Q1.

          Hope this helps.

          Best wishes

          http://farm6.static.flickr.com/5095/5490097769_cb20f73bac_b.jpg\">
          thanks
          the explanation for both questions are clear. I really got a lot to learn from you.

          1 Reply Last reply Reply Quote 0
          • ozoraO Offline
            ozora
            last edited by

            tianzhu:
            LaiHeng:

            2. A bag contains 200 blue balls and 450 yellow balls. A tin contains 850 blue and 340 yellow balls. After some blue and yellow balls are moved from the tin to the bag, the percentage of the blue balls in the bag becomes 50% and the blue balls in the tin becomes 70%.

            a)How many blue balls are moved from the tin to the bag?
            b)How many yellow balls are moved from the tin to the bag?

            Hi

            In answering this question, a student should note that the total number of blue and yellow balls remains the same in the before and after scenarios.

            Before
            Bag
            Blue ----- 200
            Yellow ----- 450

            Tin
            Blue ----- 850
            Yellow ----- 340

            Total number of blue balls ------ 200+850 ------1050
            Total number of yellow balls ------ 450+340 ------790

            After
            Bag
            Blue ----- 1 unit
            Yellow ----- 1 unit

            Tin
            Blue ----- 7 parts
            Yellow ----- 3 parts

            Here, we’ve two equations with two variables, and we cannot run away from simultaneous equations.

            Some students may find it difficult to grasp SE concepts; the challenge is to help the less mathematically inclined students deal with such handicaps.

            Needless to say, students who have learned SE from external courses or their parents would go ahead with SE.

            Students who are able to grasp SE concepts may use letters of the alphabets to solve the equations or work using units and parts to represent the quantities of the blue and yellow balls as shown above.

            Let’s begin.

            1 unit + 7 parts ------1050
            1 unit + 3 parts ------ 790
            4 parts ------ 260

            Tin
            Yellow ------ 3 part ----- 3*65 ------ 195
            Blue ------ 7 parts ---- 7*65 ----- 455

            a)How many blue balls are moved from the tin to the bag?
            850 – 455 ------ 395


            b)How many yellow balls are moved from the tin to the bag?
            340 – 195 ------- 145

            For those students who have difficulties understanding SE, you may solve it by visualising such equations in pictorial form.

            Please help to check the calculations as they had been done in a rush.

            Best wishes

            thanks for the prompt help.

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

              LaiHeng:
              thanks

              the explanation for both questions are clear. I really got a lot to learn from you.
              Hi

              You're welcome.

              Best wishes

              1 Reply Last reply Reply Quote 0
              • ozoraO Offline
                ozora
                last edited by

                Hi

                I managed to solve this question but I wonder is there a another method.
                The question is as follows: There are 80 apples and oranges in a basket. 2/5 of the oranges and 2/3 of the apples are eaten. In the end, they are 36 fruits left. How many oranges were there at first?

                I used SE concepts to solve it : 2/5o+ 2/3a=44----eq1
                3/50+ 1/3a=36-----eq2
                I managed to get 35 oranges.

                Is it possible to use model to get the answer? If so how? Thanks

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

                  LaiHeng:
                  Hi

                  I managed to solve this question but I wonder is there a another method.
                  The question is as follows: There are 80 apples and oranges in a basket. 2/5 of the oranges and 2/3 of the apples are eaten. In the end, they are 36 fruits left. How many oranges were there at first?

                  I used SE concepts to solve it : 2/5o+ 2/3a=44----eq1
                  3/50+ 1/3a=36-----eq2
                  I managed to get 35 oranges.

                  Is it possible to use model to get the answer? If so how? Thanks
                  Hi! Since model is algebra (including SE) in disguise, so we can still use the so-called model method to help the students visualise this problem instead of calling it SE. So instead of using \"O\" and \"A\", we can always use the pictorial form. Also, I think students are more familiar & comfortable manipulating whole numbers compared to fractions (SE or model).

                  Here's a quick diagram so to speak. We can always translate to \"models\" to aid visualisation. not sure if this will be helpful. cheers!

                  Original
                  oranges [ ] [ ] [ ] [ ] [ ]
                  apples { } { } { }

                  [ ] [ ] [ ] [ ] [ ] + { } { } { } = 80 fruits
                  or 5 [ ] + 3 { } = 80 fruits

                  after eating 2/5 oranges & 2/3 apples
                  oranges [ ] [ ] [ ]
                  apples { }

                  [ ] [ ] [ ] + { } \t\t = 36 fruits

                  [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] + { } { } { } \t= 36 x 3 = 108 fruits
                  |-------------9---------------| |----3-----|
                  or 9 [ ] + 3 { } = 108 fruits

                  subtracting original of 5 [ ] & 3 { }, we have:
                  therefore 4 [ ] = 108 - 80 = 28
                  1 [ ] = 7
                  therefore 5 [ ] = 5 x 7 = 35 oranges at first

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

                    LaiHeng:
                    Hi

                    I managed to solve this question but I wonder is there a another method.

                    The question is as follows: There are 80 apples and oranges in a basket. 2/5 of the oranges and 2/3 of the apples are eaten. In the end, they are 36 fruits left. How many oranges were there at first?

                    I used SE concepts to solve it : 2/5o+ 2/3a=44----eq1
                    3/50+ 1/3a=36-----eq2
                    I managed to get 35 oranges.

                    Is it possible to use model to get the answer? If so how? Thanks
                    Hi

                    Besides solving SE in algebraic manipulation, alphabet method or in pictorial form, you may also solve this question with Systematic Listing.

                    SL is less efficient but may offer a lifeline for less mathematically inclined students. It may take a bit more time but it’ll give one precious few marks as long as a student is able to have the patience to list down the equivalent fractions in a systematic manner.

                    But, use it as a last resort, .... imagine a striker or midfielder taking long shots at goals in the dying minutes of the game hoping for that elusive winning goal or equaliser.

                    Best wishes

                    1 Reply Last reply Reply Quote 0
                    • ozoraO Offline
                      ozora
                      last edited by

                      tianzhu:
                      LaiHeng:

                      Hi

                      I managed to solve this question but I wonder is there a another method.

                      The question is as follows: There are 80 apples and oranges in a basket. 2/5 of the oranges and 2/3 of the apples are eaten. In the end, they are 36 fruits left. How many oranges were there at first?

                      I used SE concepts to solve it : 2/5o+ 2/3a=44----eq1
                      3/5o+ 1/3a=36-----eq2
                      I managed to get 35 oranges.

                      Is it possible to use model to get the answer? If so how? Thanks

                      Hi

                      Besides solving SE in algebraic manipulation, alphabet method or in pictorial form, you may also solve this question with Systematic Listing.

                      SL is less efficient but may offer a lifeline for less mathematically inclined students. It may take a bit more time but it’ll give one precious few marks as long as a student is able to have the patience to list down the equivalent fractions in a systematic manner.

                      But, use it as a last resort, .... imagine a striker or midfielder taking long shots at goals in the dying minutes of the game hoping for that elusive winning goal or equaliser.

                      Best wishes

                      is alphabet method the same as my stated method?
                      thanks for answering Tianzhu.

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

                        LaiHeng:
                        is alphabet method the same as my stated method?
                        Hi LaiHeng

                        Good Morning.

                        No, I think your method is geared more towards manipulation of algebraic fractions.

                        Aiyo, memory jammed early in the morning, I cannot recall where this term ”alphabet method” surface from. Jialat, really a sign of old age.

                        Anyway , we are using the letters of the alphabet to represent units and parts because the fractions for apples and oranges are of different measures.

                        There are 80 apples and oranges in a basket. 2/5 of the oranges and 2/3 of the apples are eaten. In the end, they are 36 fruits left. How many oranges were there at first?

                        1O represents 1 unit of oranges and 1A represent 1 part of apples.

                        2O+2A ------ 44
                        3O+1A ------ 36

                        Representing the equations in this form avoids working with fractions in the calculations.

                        We have
                        1O+1A ------22
                        2O ------36-22-----14
                        1O ------7
                        5O ------ 35.

                        A student may translate these equations into pictorial method if he prefers.

                        Do not be too particular about the names of the terms, different books may call the same term different names. What’s more important is that a child is able to apply the concepts to solve questions.

                        Let the child take the method he/she is most comfortable with to the exam hall.

                        Best wishes

                        1 Reply Last reply Reply Quote 0

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