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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • K Offline
      kiasustar
      last edited by

      Hi anyone can help?


      The average height of a group of children was 139.4 cm. When Mr Tan measured and recorded the height of these children, he wrongly recorded one child's height as 192 cm when it shld have been 129 cm. As a result, Mr Tan recalculated the average height as 142.4 cm. How many children were there in the group?

      :rahrah: :thankyou:

      1 Reply Last reply Reply Quote 0
      • isetanI Offline
        isetan
        last edited by

        Hi, need help on this question:--



        http://i50.tinypic.com/1g65hi.jpg\">

        Two candles of different thickness were placed on a table. Candle B was placed on a stand of height 8 cm to reach the same height as Candle A. At 0630, Candle A was lit. One hour later, Candle B was lit. Both candles burnt down to the height marked 'X' at 1100. At 1600, Candle B burnt out while Candle A only burnt out at 2000. The rate of burning for each candle was constant throughtout.

        (a) Find the ratio of the time taken for Candle A to burn to \"X\" to the time taken for Candle B to burn to \"X\".

        (b) Find the original height of Candle A.

        TIA.

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

          kiasustar:
          Hi anyone can help?


          The average height of a group of children was 139.4 cm. When Mr Tan measured and recorded the height of these children, he wrongly recorded one child's height as 192 cm when it shld have been 129 cm. As a result, Mr Tan recalculated the average height as 142.4 cm. How many children were there in the group?

          :rahrah: :thankyou:
          Hi
          Total increase in overall height
          = 192-129 = 63
          = increase in average height x number of children
          increase in average height = 142.4-139.4=3
          Number of children =63/3 = 21

          cheers.

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

            isetan:
            Hi, need help on this question:--



            http://i50.tinypic.com/1g65hi.jpg\">

            Two candles of different thickness were placed on a table. Candle B was placed on a stand of height 8 cm to reach the same height as Candle A. At 0630, Candle A was lit. One hour later, Candle B was lit. Both candles burnt down to the height marked 'X' at 1100. At 1600, Candle B burnt out while Candle A only burnt out at 2000. The rate of burning for each candle was constant throughtout.

            (a) Find the ratio of the time taken for Candle A to burn to \"X\" to the time taken for Candle B to burn to \"X\".

            (b) Find the original height of Candle A.

            TIA.
            Hi

            I presume you are able to work out (a)

            Time taken for A to burn to X --> 4.5h
            Time taken for B to burn to X --> 3.5h
            Total time taken for A to burn out --> 13.5h
            Total time taken for B to burn out --> 8.5h
            4.5/13.5 A = 3.5/8.5 B
            A/B = 21/17
            A --> 21 units, B --> 17 units
            4 units --> 8 cm
            So, original height of A --> 21/4 x 8 = 42 cm

            cheers.

            1 Reply Last reply Reply Quote 0
            • C Offline
              cimman
              last edited by

              brastilava:
              Hi, please help in this question:


              Tin A, Tin B and Tin C are required to paint a complete wall. If only Tin A is used, we are short of 7 litres. If only Tin B is used, we are short of 5 litres. If only tin C is used, we are short of 3 litres. How many litres of paint is required to paint a complete wall?
              here's another solution using equations:

              let w = paint needed to paint one wall

              A + B + C = w
              A = w - 7 -- (short of 7 litres to paint the wall)
              B = w - 5
              C = w - 3

              (w - 7) + (w - 5) + (w - 3) = w
              3w - 15 = w
              2w = 15
              w = 7.5

              1 Reply Last reply Reply Quote 0
              • N Offline
                Neat
                last edited by

                tianzhu:
                Neat:

                Hi! Good evening,


                Please help with this question:

                Jason, Edward and Sam had a total of $837. Jason had the least amount of money. The ratio of Edward's money to Sam's money was 4:3 at first. Jason and Edward each spent 1/3 of their money. Given that the three boys had $648 left, how much did Jason have at first?

                Thanks

                Hi

                There are a few ways to arrive at the solution.

                1/3 of Jason and Edward’s money ------- 837 – 648 ------- 189

                Hence, their total money ------ 3*189 ------ 567

                Sam’s money@first ------- 837 – 567 ------- 270

                270/3 ------ 90

                Therefore, Edward’s money@first ------- 4*90 ------ 360

                837 – 360 – 270 ------ 207 (Jason’s money@first)

                Best wishes

                [quote]Thanks a lot, tianzhu.

                If unit method is to be used, which item is to be made constant?

                Regards[/quote]

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

                  Neat:

                  Thanks a lot, tianzhu.

                  If unit method is to be used, which item is to be made constant?

                  Regards
                  Hi

                  Good Morning.

                  I am trying to understand “which item is to be made constant”. Could you help to elaborate on it?

                  Alternatively, you may also present the solution in “Units and Parts”.

                  The relationship between Edward and Sam is given.

                  Edward ------ 4 units
                  Sam ------- 3 units

                  We differentiate between Edward/Sam and Jason by using parts for Jason.

                  1 unit + 1 part ------- 837 – 648 ------- 189

                  3 units + 3 parts ------ 3*189 ------ 567 (total amount)

                  Sam’s money@first ------- 837 – 567 ------- 270

                  3 units ------- 270
                  1 unit ------- 90

                  Therefore, Edward’s money@first ------- 4*90 ------ 360

                  837 – 360 – 270 ------ 207 (Jason’s money@first)

                  Best wishes

                  1 Reply Last reply Reply Quote 0
                  • N Offline
                    Neat
                    last edited by

                    tianzhu:
                    Neat:


                    Thanks a lot, tianzhu.

                    If unit method is to be used, which item is to be made constant?

                    Regards

                    Hi

                    Good Morning.

                    I am trying to understand “which item is to be made constant”. Could you help to elaborate on it?

                    Alternatively, you may also present the solution in “Units and Parts”.

                    The relationship between Edward and Sam is given.

                    Edward ------ 4 units
                    Sam ------- 3 units

                    We differentiate between Edward/Sam and Jason by using parts for Jason.

                    1 unit + 1 part ------- 837 – 648 ------- 189

                    3 units + 3 parts ------ 3*189 ------ 567 (total amount)

                    Sam’s money@first ------- 837 – 567 ------- 270

                    3 units ------- 270
                    1 unit ------- 90

                    Therefore, Edward’s money@first ------- 4*90 ------ 360

                    837 – 360 – 270 ------ 207 (Jason’s money@first)

                    Best wishes

                    [quote]Hi tianzhu!

                    Good morning.

                    Yes, this is the method that I am looking for.

                    As I understand, if unit transfer method is used, we must always equalized the units and parts.

                    I used the following to solve but was stucked. Now, I understand.


                    J E S

                    before ? 4u 3u
                    change -1p -1p -
                    after 2p ? 3u

                    Thanks a lot, tianzhu.
                    [/quote]

                    1 Reply Last reply Reply Quote 0
                    • isetanI Offline
                      isetan
                      last edited by

                      MathIzzzFun:
                      isetan:

                      Hi, need help on this question:--



                      http://i50.tinypic.com/1g65hi.jpg\">

                      Two candles of different thickness were placed on a table. Candle B was placed on a stand of height 8 cm to reach the same height as Candle A. At 0630, Candle A was lit. One hour later, Candle B was lit. Both candles burnt down to the height marked 'X' at 1100. At 1600, Candle B burnt out while Candle A only burnt out at 2000. The rate of burning for each candle was constant throughtout.

                      (a) Find the ratio of the time taken for Candle A to burn to \"X\" to the time taken for Candle B to burn to \"X\".


                      Thanks Tianzhu!
                      (b) Find the original height of Candle A.

                      TIA.

                      Hi

                      I presume you are able to work out (a)

                      Time taken for A to burn to X --> 4.5h
                      Time taken for B to burn to X --> 3.5h
                      Total time taken for A to burn out --> 13.5h
                      Total time taken for B to burn out --> 8.5h
                      4.5/13.5 A = 3.5/8.5 B
                      A/B = 21/17
                      A --> 21 units, B --> 17 units
                      4 units --> 8 cm
                      So, original height of A --> 21/4 x 8 = 42 cm

                      cheers.

                      1 Reply Last reply Reply Quote 0
                      • K Offline
                        ksoh3000
                        last edited by

                        Hi


                        Can anyone help me with this qn ?

                        -Grace cycled from home to the park at 7.05 a.m. at a constant speed of 300m/min. She passed Colin at 7.15 a.m. as he started cycling at a constant speed from his house to the park. At 7.30 am, Colin was 1.5km ahead of Grace. Colin reached the park at 7.40 am.
                        (a) Find Colin’s speed, giving yr answer in kilometres per hour .
                        (b)How far is Grace’s house from the park? Give yr answer in kilometres.

                        Tks

                        1 Reply Last reply Reply Quote 0

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