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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • D Offline
      Daddy
      last edited by

      :thankyou: Tianzhu..

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

        Daddy:
        Hi Tiazhu,

        Is my answer corrected?

        Jim and his sister shared some money. 1/3 of Jim's amount is 3/4 of his sister's share. If the difference in their share of money was $50, how much money did Jim have?

        Jim [][][][][][][][][]
        sister [][][][][][][][][][][][]
        |------> $50
        3u -> $50
        1u -> 50/3
        Jim -> 9u
        = 9 x 50/3
        = $150
        Hi

        1/3 of Jim’s share is equal to of 3/4 his sister.

        This means Jim has a total of 9 units and his sister has a total of 4 units

        Difference ----- 9 - 4 -------5

        5 units -------50
        1 unit ------10

        Jim ------9 units -------90

        Best wishes.

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

          Daddy:
          :thankyou: Tianzhu..

          Hi

          You're welcome.

          Wishing you the very best for PSLE 2011.

          Best wishes

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

            tianzhu:
            Daddy:

            Hi Tianzhu,


            I still not understand.
            Is it the chicken's egg did not do a compare, Stand alone model?
            Duck eggs compared with quail eggs, so their model units are the same.
            Am I corrected?

            Thanks

            In this question, there is a link or relationship between duck and quail eggs,. He sold 3 times as many quail eggs as duck eggs.
            So, if duck eggs are represented by 2 boxes, then quail eggs are represented by 6 boxes.

            The chicken eggs are of different sets of objects; hence the units are of different kind and measure.

            Best wishes

            To solve out the number of chickens originally, can we resort to algebra?

            So from the models we can solve for the Quail and Ducks, and simple deduction will give us the number of Chicken eggs remaining: 205 (all eggs) - 45 (duck) - 90 (Quail) = 70 (Chicken eggs remaining)

            So now we know 70 is 40% of all chicken eggs originally. Therefore, 70/0.4 = 175.

            Is this allowed in P5/P6?

            1 Reply Last reply Reply Quote 0
            • S Offline
              small
              last edited by

              Daddy:
              Hi Tiazhu,

              Is my answer corrected?

              Jim and his sister shared some money. 1/3 of Jim's amount is 3/4 of his sister's share. If the difference in their share of money was $50, how much money did Jim have?

              Jim [][][][][][][][][]
              sister [][][][][][][][][][][][]
              |------> $50
              3u -> $50
              1u -> 50/3
              Jim -> 9u
              = 9 x 50/3
              = $150
              Hi,
              The answer shouldbe Jimmy had $90

              Jimmy - 9 Units
              Sister - 4 Units

              (9-4) units = $50
              5 Units = $50
              1U = $10

              Jimmy = 9 x $10
              = $90

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

                meremortal:

                Is this allowed in P5/P6?
                Hi

                If you were talking about the use of algebra in PSLE, here’s the reply from MOE.

                http://www.moe.gov.sg/media/forum/2007/20070217.htm

                Best wishes.

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

                  Any solution to these 2 questions from Rosyths paper?

                  q1. Abdul , Bern and Chin were all standing in a straight line waiting for a race to start. Chin was 300m ahead of Bern and Bern was 100m ahead of Abdul. At 9 am, they started the race.Abdul overtook Bern in 5 mins and in another 5 mins, Abdul overtook Chin. If Bern’s speed is 150m/min, at what time did Bern overtake Chin?

                  q2. Alan had some sweets and chocolate. If he ate 1 sweet, the ratio of the number of sweets to the number of chocolate left in the bag would be 2:3.
                  If he ate 1 chocolate, the ratio of the number of sweets to the number of chocolate left in the bag would be 7:10. What was the ratio of the number of sweets to the number of chocolate he had in the bag ?

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

                    ozora:
                    Any solution to these 2 questions from Rosyths paper?

                    q1. Abdul , Bern and Chin were all standing in a straight line waiting for a race to start. Chin was 300m ahead of Bern and Bern was 100m ahead of Abdul. At 9 am, they started the race.Abdul overtook Bern in 5 mins and in another 5 mins, Abdul overtook Chin. If Bern's speed is 150m/min, at what time did Bern overtake Chin?

                    q2. Alan had some sweets and chocolate. If he ate 1 sweet, the ratio of the number of sweets to the number of chocolate left in the bag would be 2:3.
                    If he ate 1 chocolate, the ratio of the number of sweets to the number of chocolate left in the bag would be 7:10. What was the ratio of the number of sweets to the number of chocolate he had in the bag ?
                    Hi, i have the solution for q2.

                    It is in the link https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B81le9JxERBGNjlhYjdhZjctYjIwYi00MTIwLWI5YjctMmE2MzUzNTk2NzQ1&hl=en_GB

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

                      ozora:
                      Any solution to these 2 questions from Rosyths paper?

                      q1. Abdul , Bern and Chin were all standing in a straight line waiting for a race to start. Chin was 300m ahead of Bern and Bern was 100m ahead of Abdul. At 9 am, they started the race.Abdul overtook Bern in 5 mins and in another 5 mins, Abdul overtook Chin. If Bern's speed is 150m/min, at what time did Bern overtake Chin?
                      Hi

                      As in any question on “Speed”, it’s always useful to start with a simple diagram showing the relationship between the runners.Once done, things fall into places.

                      There are two ways to go about it.

                      Use the familiar Speed triangle learned in school.

                      Abdul and Bern
                      In 5 mins, Bern travelled (150*5) or 750m
                      In 5 mins, Abdul travelled (100+750) or 850m
                      Hence, Abdul’s speed is 850/5 or 170 m/min

                      Abdul and Chin
                      In 10 mins, Abdul travelled (170*10) or 1700m
                      Chin travelled (1700 - 400) or 1300m in 10 mins.
                      Chin’s speed is 130m/min

                      Bern and Chin
                      Difference between Bern and Chin’s speed ------- 20 m/min.
                      Chin was 300m ahead of Bern
                      300/20 ------ 15 mins

                      Bern overtook Chin at 9.15am

                      The second way, use “catch up”

                      Abdul and Bern
                      Head start ------ 100m
                      Time to catch up -------5 mins
                      Speed difference ------100/5 ------20

                      Abdul’s speed -------- 150+20 ------170m/min

                      Abdul and Chin
                      Head start ------ 400m
                      Time to catch up -------10 mins
                      Speed difference ------400/10 ------40

                      Chin’s speed -------- 170 - 40 ------130m/min

                      Bern and Chin
                      Head start ------ 300m
                      Speed difference ------ 20
                      Time to catch up ------- 300/20 ----- 15 mins

                      Bern overtook Chin at 9.15am

                      Best wishes

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

                        ozora:
                        Any solution to these 2 questions from Rosyths paper?

                        q1. Abdul , Bern and Chin were all standing in a straight line waiting for a race to start. Chin was 300m ahead of Bern and Bern was 100m ahead of Abdul. At 9 am, they started the race.Abdul overtook Bern in 5 mins and in another 5 mins, Abdul overtook Chin. If Bern's speed is 150m/min, at what time did Bern overtake Chin?

                        q2. Alan had some sweets and chocolate. If he ate 1 sweet, the ratio of the number of sweets to the number of chocolate left in the bag would be 2:3.
                        If he ate 1 chocolate, the ratio of the number of sweets to the number of chocolate left in the bag would be 7:10. What was the ratio of the number of sweets to the number of chocolate he had in the bag ?
                        Hi

                        Q2 is a variation of the \"total unchanged\" type of questions - just remove the 1 from the total and work the question with initial ratio of 2:3 and final ratio of 7:10. And then add \"1\" back to the total..

                        If 1 sweet is removed,
                        Sweets : Chocolate = 2:3 (initial)

                        If 1 chocolate eaten (assume that now the 1 chocolate is replaced by 1 sweet), Sweets : Chocolate = 7:10 (final)

                        Make the initial total and final total the same...

                        Initially, Sweets : Chocolate = 2:3 = 34u : 51u
                        In the end, Sweets : Chocolate : 7 : 10 = 35u : 50u
                        So, 1u =1
                        Number of sweets = 35
                        Number of chocolates = 51

                        Ratio of sweets : chocolates = 35 : 51

                        cheers.

                        1 Reply Last reply Reply Quote 0

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