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    Q&A - P3 Math

    Scheduled Pinned Locked Moved Primary 3
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    • corneyAmberC Offline
      corneyAmber
      last edited by

      The cost of 7 erasers and 3 rulers is $5.05.

      The cost of 2 erasers and 6 rulers is $3.95
      Find the cost of 2 erasers and 2 rules.

      7 erasers + 3 rulers = 5.05 --------(1) x 2
      14 erasers + 6 rulers = 10.1 --------(2)
      2 erasers + 6 rulers = 3.95 ---------(3)

      (2)-(3),
      12 erasers = 6.15
      2 erasers = 1.025
      Then 6 rulers = 3.95-1.025 = 2.925
      2 rulers = 0.975

      So 2 erasers + 2 rulers = 1.025 + 0.975 = $2

      or (1) + (2)
      9 erasers + 9 rulers = $9
      2 erasers + 2 rulers = $2 (proportionately)

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

        ksi:
        The cost of 7 erasers and 3 rulers is $5.05.

        The cost of 2 erasers and 6 rulers is $3.95
        Find the cost of 2 erasers and 2 rules.

        7 erasers + 3 rulers = 5.05 --------(1) x 2
        14 erasers + 6 rulers = 10.1 --------(2)
        2 erasers + 6 rulers = 3.95 ---------(3)

        (2)-(3),
        12 erasers = 6.15
        2 erasers = 1.025
        Then 6 rulers = 3.95-1.025 = 2.925
        2 rulers = 0.975

        So 2 erasers + 2 rulers = 1.025 + 0.975 = $2

        or (1) + (2)
        9 erasers + 9 rulers = $9
        2 erasers + 2 rulers = $2 (proportionately)
        Thanks ksi. But I do not how to explain this simultaneous equation to a P3

        My initial solution is :
        7 Erasers + 3 Rulers = $5.05
        2 Erasers + 6 Rulers = $ 3.95

        Add them together we have
        9 Erasers + 9 Rulers = $5.05 + 3.95 = $9

        Each set of 1 Eraser 1 Ruler = $1
        Therefore 2 sets of Eraser & Ruler = $2.

        I'm thinking since the question ask for the price of 2 sets, the working out of each price is not necessary.

        Anyone to shed some lights? Thanks in advance.

        1 Reply Last reply Reply Quote 0
        • corneyAmberC Offline
          corneyAmber
          last edited by

          Actually the second method i gave is the same as yours and it is not necessary to find the cost for 1, you can directly give the answer for 2 of each item.


          Also in method 1, I was not thinking of simultaneous equation, I was using common sense but I guess it looked like simultaneous equation. :lol:

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          • D Offline
            Dharma
            last edited by

            Devnahouse:
            Hello,

            Can help me solve this one.

            The cost of 7 erasers and 3 rulers is $5.05.
            The cost of 2 erasers and 6 rulers is $3.95
            Find the cost of 2 erasers and 2 rules.

            Many thanks.
            http://www.postimage.org/image.php?v=TsKNGfA

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

              Dharma, Thanks for taking time to explain in model form. Thanks ksi for your further clarifications. Appreciated.

              1 Reply Last reply Reply Quote 0
              • B Offline
                buds
                last edited by

                Easiest method to solve this one?


                A snail climbed 1cm up a wall in Day 1,
                2cm in Day 2, 3cm in Day 3 and so on..

                Each day, it climbed 1cm more than it did
                the previous day. How many days did the
                snail take to climb 21cm?

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

                  buds:
                  Easiest method to solve this one?


                  A snail climbed 1cm up a wall in Day 1,
                  2cm in Day 2, 3cm in Day 3 and so on..

                  Each day, it climbed 1cm more than it did
                  the previous day. How many days did the
                  snail take to climb 21cm?
                  Hi Buds

                  Easiest way is by listing

                  Day 1 => 1
                  Day 2 => 1 + 2 = 3
                  Day 3 => 1 + 2 + 3 = 6
                  Day 4 => 1 + 2 + 3 + 4 = 10
                  Day 5 => 1 + 2 + 3 + 4 + 5 = 15
                  Day 6 => 1 + 2 + 3 + 4 + 5 + 6 = 21

                  The snail took 6 days to climb 21cm

                  But if we have a large number, eg number of days taken by the snail to climb 210cm; using listing method will be very time consuming.

                  We need to establish a pattern

                  This is a question a series of consecutive numbers starting from 1

                  1 + 2 + 3 + ….+ N = N x (N + 1)/2

                  So, to find no. of days for snail to climb 210cm

                  N x (N +1 ) / 2 = 210
                  N x ( N + 1 ) = 420

                  We know 20 x 21 = 420

                  N = 20

                  So, in 20 days the snail will be able to climb 210cm

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

                    Hi Dharma.


                    Sorry to be a buzzkill, but the actual question was 21 cm, not 210 cm…

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

                      Muffins:
                      Hi Dharma.


                      Sorry to be a buzzkill, but the actual question was 21 cm, not 210 cm....

                      Hi Muffins,

                      To climb 21cm

                      N x (N + 1) / 2 = 21

                      N x (N + 1) = 42
                      6 x 7 = 42
                      N = 6 days

                      By using 210cm I was trying to show that it will be easier to solve the problem by finding a pattern. The listing method will
                      be just too long.

                      1 Reply Last reply Reply Quote 0
                      • I Offline
                        imacsg
                        last edited by

                        Can you help to explain why is it wrong.


                        I am a 3 digit number.
                        My hundreds and tens digit are the same.
                        My ones digit is 4 more than my tens digit.
                        What number can I be?

                        My son writes 448, but the teacher marked it wrong and said the correct ans is 559. I thought the ans can be either 559, 448, 337, 226 or 115

                        Thanks

                        1 Reply Last reply Reply Quote 0

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