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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • MathIzzzFunM Offline
      MathIzzzFun
      last edited by

      Hershey:
      MathIzzzFun:

      [quote=\"Hershey\"]Dear Parents, hope someone can help to solve this 2 questions from 2010 Catholic High Prelim paper.


      1) At a conference made up of speakers and participants, there were 20% more men than women. the ratio of male speakers to female speakers was 8:5. There was an equal number of male and female participants.

      a) Find the ratio of male speakers to male participants at the conference.

      b) Halfway, 40 male participants left the conference and another 60 male participants joined the conference. in the end, there was 3/4 as many male participants as female participants remaining behind. How many speakers were there at the conference.


      2. At 9am, John was cycling from Point A to Point B. At the same time, Mary was cycling from Point B to Point A using the same route as John. Cycling at a speed of 4km/h faster than Mary, John would pass Mary 300m away from the midpoint.

      a) what time did they pass each other?

      b) If John took another 3 mins to reach Point B, what time would Mary reach Point A?

      Really don't understand why kids are being drill on this kind of questions...even adults need to think....

      Thanks.

      Hi

      For Q1(b), the question should be \"Halfway, 40 male participants left the conference and another 60 female participants joined the conference.\"

      http://i52.tinypic.com/280ld0g.jpg\">

      For Q2, refer to http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=27&t=280&start=3470

      cheers.

      Thank you so much. Its my typo error on Q1b. :imsorry:[/quote]u r welcome 😄

      cheers.

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

        MathIzzzFun:
        stary:

        Hi all,


        Need help on average.

        The pupils in 2 classes, 1A and 1B did a Mathematics test. The average score of class 1A, which had 24 pupils, was 87. The average score of Class 1B was 80. If the average score of the two classes was 83, how many pupils were there in Class 1B?

        Ans key says the ans is 32, but I do not know how to get the explanation for the workings...

        Can someone explain to me the bold and underlined part?

        Total score of class 1A -> 87 x 24 = 2088
        83 x 24 = 1992
        2088 - 1992 = 96
        83 - 80= 3
        96 / 3 = 32


        Appreciate your assistance. 😄

        Hi

        http://i52.tinypic.com/2r2ryb4.jpg\">

        cheers.

        Thanks MathIzzzFun! :thankyou:

        1 Reply Last reply Reply Quote 0
        • P Offline
          pensiveowl
          last edited by

          Tang:
          tianzhu:

          [quote=\"trytry\"]A courier company charged $25 for every large parcel and $15 for every

          small parcel delivered safely. However, a penalty of $50 was charged for
          every damaged parcel regardless of size.
          This month, the company delivered 120 parcels of which 1/4 of them were
          small parcels. It collected a total of $2000 after paying a penalty for
          an equal numbers of large and small parcels.
          How many large pacels were delivered safely?
          (Nanyang 2010 SA1 P2 Q12)
          The answer provided was 83. But mine is 85. Which is correct?

          Hi

          83 is correct.

          Best wishes


          Hi,

          85 is the correct answer.


          85 x 25 = 2125

          25 x 15 = 375

          2125 + 375 = 2500

          5 + 5 = 10

          10 x 50 = 500

          2500 - 500 = $2000 (amount collected).



          Solution as follows:

          120/4 = 30 small parcels

          120 - 30 = 90 large parcels

          Assume all dilivered safely, amount collected

          90 x 25 + 30 x 15 = $2700



          2700 - 2000 = $700 (too much)


          Removing 1 large and 1 small parcels,

          amount collected less by 25 + 15 = 40

          penalty paid 50 + 50 = 100


          Total amount reduced by 40 + 100 = $140


          700/140 = 5, i.e. 5 large and 5 small parcels damaged.

          90 - 5 = 85 large parcels were delivered safely.[/quote]I had a little trouble understanding this.. until I wrote it out like this:
          The company delivered 120 x ¼ = 30 small parcels and 90 large parcels.
          It should have collected [(30 x 15) + (90 x 25)] = $2700
          It didn’t get paid for items damaged, moreover was charged $50 for @ damaged.
          For each pair of large and small item damaged:
          It didn’t collect 25 + 15 = 40
          On top of this, it was charged 50 + 50 = 100
          Thus, it lost $140 ($40 + $100) for each pair of large and small items damaged.
          700/140 = 5
          Thus, 5 large and 5 small items were damaged.
          90 – 5 = 85
          85 large parcels were delivered safely.

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

            Hi all,


            Have a question...

            Ali and Baba ran in a race. When Baba had completed the race, Ali had only run 5/8 of the distance. Baba's speed was 75 m/min faster than Ali's speed. Both of them did not change their speeds throughout the race.
            Waht was Ali's speed in m/min?

            Appreciate your help... :?

            1 Reply Last reply Reply Quote 0
            • W Offline
              wkong
              last edited by

              attachment deleted

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              • W Offline
                wkong
                last edited by

                attachment deleted

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                • W Offline
                  wkong
                  last edited by

                  attachment deleted

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                  • W Offline
                    wkong
                    last edited by

                    Need help for the following question:-


                    Qn4


                    http://i56.tinypic.com/xfdd9k.jpg\">

                    1 Reply Last reply Reply Quote 0
                    • W Offline
                      wkong
                      last edited by

                      Need help for the following question:-


                      Qn5

                      http://i55.tinypic.com/2nv9k7q.jpg\">

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

                        wkong:
                        Need help for the following question:-


                        Qn5

                        http://i55.tinypic.com/2nv9k7q.jpg\">
                        Area of unshaded -> 8.5 x 9.4 = 79.9 cm2
                        Area of triangle -> 17+15.4 = 261.8 cm2
                        Shaded -> 261.8 - 79.9 = 181.9 cm2

                        1 Reply Last reply Reply Quote 0

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