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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

      isetan:

      Tianzhu,

      Just clarify again....for the last sentence, why do you need to 'plus 4' (ie. 96+8+4).

      thanks.
      Hi

      He can buy 100 rulers.

      1 free ruler for every 12 rulers bought.

      100 = 96 + 4

      For 96 rulers, he gets 8 free

      So, (96 + 😎 + 4 ------- 108

      Best wishes

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

        Thanks! Understand now.

        1 Reply Last reply Reply Quote 0
        • 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

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