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    Tutor MathsGuru: Ask me for your burning Maths questions!

    Scheduled Pinned Locked Moved Primary Schools - Academic Support
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    • K Offline
      KSP
      last edited by

      I need help on the following questions.


      1. Container A contains 250 red marbles and 200 blue marbles. Container B contains 600 red marbles and 150 blue marbles. How many red and blue marbles must be moved from Container A to Container B such that 25% of the marbles in Container A are red and 75% of the marbles in Container B are red? Ans: 350


      2. In a math competition, participants can obtain 4 possible award: Gold, Silver, Bronze and Participation. 3/7 of the participants obtained Gold awards, 1/4 of the them obtained Silver awards, and 1/6 of them obtained Bronze awards. Given that there were less than 100 participants taking the competition,
      a) How many participants obtained the awards for Participation? Ans: 13
      b) How many more participants obtained Gold awards then Bronze awards? Ans: 22

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

        starlight1968sg:
        Dharma:

        I’m no expert at models but I am able to explain to u what I have done.


        1.\tThe initial condition is ratio of red : yellow : blue = 4 : 2 : 1
        2.\tFinal condition is ratio of red : yellow : blue = 6 : 2 : 1
        3.\tNo red marbles are given out, so the no. of red marbles DO NOT change.
        4.\tNow u have initial ratio of 12: 6 : 3 and final ratio of 12 : 4 : 2
        5.\tFrom (4) u can see for every 1 unit reduction of blue marbles, there will be 2 units of reduction in yellow marbles, based on the ratios above.

        Thanks.
        Your solution is so short and clear. Earlier, I was struggling with the model diagram.
        Myself is lousy in model & ratio. So I can't expect my dd to do well too. :stupid:

        I personally never liked models … find it time consuming and u don’t get marks for it.

        Ratio ... your child needs to understand and master ...cause it is a very powerful tool to solve most PSLE problem solving qns.

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

          P5G:
          Pl help.


          Paper 1 Q13.
          http://www.orlesson.org/orp/09Ma/2009-Math-SA1-MGS.pdf


          Worksheet answer is (4) 135 [not likely].


          Please advise how to get (3) 125 as Benny's score?


          TIA.
          Hi P5G,

          Sorry for the late reply. Here's my answer.

          🙂
          MathsGuru

          http://www.postimage.org/image.php?v=aVjUVjJ

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

            I need help on the 2 questions below.


            1. The figure below is not drawn to scale. Given that the area of WXYZ is 40cm sq, find the area of the shaded part. Ans: 68cm sq

            http://www.postimage.org/image.php?v=aVjVKHJ



            2. The figure below is made up of two identical right-angled triangles, a small square and a big square. The parimeter of the shaded region is 50cm, and the total area of the two unshaded squares is 254.5cm sq. Find the total area of the two shaded right-angled triangles. Ans: 114.75cm sq

            http://www.postimage.org/image.php?v=gxNVMm9

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

              mmmouse:
              Anyone knows how to solve this Pr. 3 Maths question?


              5 erasers + 3 rulers = S$5.05
              3 erasers + 6 rulers = S$3.95
              2 erasers + 2 rulers = ???

              Thanks!
              Hi Mmmouse,

              Think there's probably a typo error in the question, cos I can't seem to get nice numbers for the answers and I think for P3, they haven't learnt rounding off yet.

              In any case, here's my solution. You can follow the steps. It's the same kind of steps to solve such questions. 🙂

              Cheers,
              MathsGuru

              http://www.postimage.org/image.php?v=aVjVXar

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

                KSP:
                I need help on the following questions.


                1. Container A contains 250 red marbles and 200 blue marbles. Container B contains 600 red marbles and 150 blue marbles. How many red and blue marbles must be moved from Container A to Container B such that 25% of the marbles in Container A are red and 75% of the marbles in Container B are red? Ans: 350


                2. In a math competition, participants can obtain 4 possible award: Gold, Silver, Bronze and Participation. 3/7 of the participants obtained Gold awards, 1/4 of the them obtained Silver awards, and 1/6 of them obtained Bronze awards. Given that there were less than 100 participants taking the competition,
                a) How many participants obtained the awards for Participation? Ans: 13
                b) How many more participants obtained Gold awards then Bronze awards? Ans: 22
                Q1
                Container A
                Red : 250 – 1u [25%]
                Blue : 200 – 1p [75%]

                (250 – 1u) x 3 = 200 – 1p
                3u – 1p = 550 …………..(1)

                Container B
                Red : 600 + 1u [75%]
                Blue : 150 + 1p [25%]

                (150 + 1p) x 3 = 600 + 1u
                3p – 1u = 150 …………..(2)

                8u = (1550 x 3) + 150 = 1800 ….solving (1) and (2)
                1u = 225 red marbles
                1p = 225 x 3 -550 = 125 blue marbles
                No. of red and blue marbles moved = 225 + 125 = 350


                Q2
                Gold : 36/84
                Silver : 21/84
                Bronze : 14/84

                a)\tNo. participants obtained Participation = 84 – 36 – 21 – 14 = 13
                b)\t36 – 14 = 22 participants received gold than bronze.

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

                  Dharma:


                  Q2
                  Gold : 36/84
                  Silver : 21/84
                  Bronze : 14/84

                  a)\tNo. participants obtained Participation = 84 – 36 – 21 – 14 = 13
                  b)\t36 – 14 = 22 participants received gold than bronze.
                  How do you get 84?

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

                    KSP:
                    I need help on the 2 questions below.


                    1. The figure below is not drawn to scale. Given that the area of WXYZ is 40cm sq, find the area of the shaded part. Ans: 68cm sq

                    http://www.postimage.org/image.php?v=aVjVKHJ
                    The shaded area consist of 2 triangles with height of 18cm and total base length of 12cm but the have an overlapping area of 40cm2 (i.e. Area of WXYZ)
                    Shaded area = (½ x 18 x 12)cm2 – Area of WXYZ = 108cm2 – 40cm2 = 68cm2

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

                      KSP:
                      I need help on the 2 questions below.



                      2. The figure below is made up of two identical right-angled triangles, a small square and a big square. The parimeter of the shaded region is 50cm, and the total area of the two unshaded squares is 254.5cm sq. Find the total area of the two shaded right-angled triangles. Ans: 114.75cm sq

                      http://www.postimage.org/image.php?v=gxNVMm9
                      Side of large square = 1u
                      Side of small square = 1p

                      1u + 1p = (50 – 6)/ 2 = 22

                      Slide up the triangle with the larger square up the slope and u get 2 squares and a rectangle.

                      Shaded area = [(1u + 1p) x (1u + 1p) – 254.5] / 2 = [(22 x22) – 254.5] /2 = 114.75

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                      • D Offline
                        Dharma
                        last edited by

                        KSP:
                        How do you get 84?

                        84 is the smallest number that is divisible by 4,6 and 7

                        1 Reply Last reply Reply Quote 0

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