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    O-Level Additional Math

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
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    • C Offline
      chrisu
      last edited by

      Can ignore this, I already got the workings and answer. Thank you.

      chrisu:
      Please help A Sec.2 Maths question of Proportion.

      Q) The scale of map P is 1:p and the scale of map Q is 1:q. If the same area is represented as 32sq.cm. on map P and 50sq.cm. on map Q, calculate the ratio p:q.

      Kindly show workings as well.

      Thank you.

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

        Sec1 maths Question:

        Two race cars, Car X and Car Y, are at the starting point of a 2km track at the same time. Car X and Car Y make one lap every 60s and 80s respectively.
        How long, in minutes, will it take for the faster car to be 5 laps ahead of the slower car?

        Thanks all Gurus in advance.

        1 Reply Last reply Reply Quote 0
        • ChiefKiasuC Offline
          ChiefKiasu
          last edited by

          insanePaPa:
          Sec1 maths Question:

          Two race cars, Car X and Car Y, are at the starting point of a 2km track at the same time. Car X and Car Y make one lap every 60s and 80s respectively.
          How long, in minutes, will it take for the faster car to be 5 laps ahead of the slower car?

          Thanks all Gurus in advance.
          5 laps = 5 x 2km = 10km

          Car A's speed = 2/60 = 1/30 kps
          Car B's speed = 2/80 = 1/40 kps

          Let T be time taken for Car A to be 10km ahead of Car B. Hence
          (1/30)*T - (1/40)*T = 10
          => T(1/30 - 1/40) = 10
          => T(4-3/120) = 10
          => T(1/120) = 10
          => T = 1200s = 20mins

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

            ChiefKiasu:
            insanePaPa:

            Sec1 maths Question:

            Two race cars, Car X and Car Y, are at the starting point of a 2km track at the same time. Car X and Car Y make one lap every 60s and 80s respectively.
            How long, in minutes, will it take for the faster car to be 5 laps ahead of the slower car?

            Thanks all Gurus in advance.

            5 laps = 5 x 2km = 10km

            Car A's speed = 2/60 = 1/30 kps
            Car B's speed = 2/80 = 1/40 kps

            Let T be time taken for Car A to be 10km ahead of Car B. Hence
            (1/30)*T - (1/40)*T = 10
            => T(1/30 - 1/40) = 10
            => T(4-3/120) = 10
            => T(1/120) = 10
            => T = 1200s = 20mins

            Thanks a million chief.

            1 Reply Last reply Reply Quote 0
            • ChiefKiasuC Offline
              ChiefKiasu
              last edited by

              chrisu:
              Can ignore this, I already got the workings and answer. Thank you.

              chrisu:

              Please help A Sec.2 Maths question of Proportion.

              Q) The scale of map P is 1:p and the scale of map Q is 1:q. If the same area is represented as 32sq.cm. on map P and 50sq.cm. on map Q, calculate the ratio p:q.

              Kindly show workings as well.

              Thank you.

              1/p = sqrt(x/32) => p = sqrt(32/x)
              1/q = sqrt(x/50) => q = sqrt(50/x)

              => p/q = sqrt(32/50) = sqrt(16/25)
              => p:q = sqrt(16) : sqrt(25) = 4 : 5

              1 Reply Last reply Reply Quote 0
              • ChiefKiasuC Offline
                ChiefKiasu
                last edited by

                YumYum:
                Hi, need help with this sec1 Qn:


                The key to a code is given by three whole numbers. The key is transmitted from the headquarters by 3 agents, who were given the product of 2 of the 3 numbers. Suppose that the numbers transmitted are 432, 540 and 720. What is the key code?

                thanks.
                The hard part to the question is trying to understand what the heck it is saying.

                The code is A, B, C

                Agent X has A*B = 432
                Agent Y has B*C = 540
                Agent Z has A*C = 720

                X/Y => A/C = 432/540 => A = C*432/540
                Since A*C =720
                => C*C*432/540 = 720
                => C*C = 720 * 540/432 = 900
                => C = 30
                A = 720/C = 720/30 = 24
                B = 432/A = 432/24 = 18

                So the 3 numbers are 18, 24 and 30.

                1 Reply Last reply Reply Quote 0
                • ChiefKiasuC Offline
                  ChiefKiasu
                  last edited by

                  YumYum:
                  Hi, 1 more Qn:


                  Given that S =. 1
                  ___________________________________________________
                  1/21 + 1/22 + 1/23 + 1/24 + 1/25 + 1/26 +...1/30

                  Find the largest whole number smaller than S.

                  Thanks
                  Sorry... don't understand what you typed. Can you scan the question and post it as an image?

                  1 Reply Last reply Reply Quote 0
                  • Y Offline
                    YumYum
                    last edited by

                    Dear chief,


                    Thank you for your help.

                    For the 2nd Qn, never mind, it was covered in school.

                    Thanks 

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

                      Sec1 Qn:


                      A class has between 30 to 40 students. Each boy in the class brings 15 chocolates for a class party. The chocolates are shared equally among 20 girls and a teacher with no leftovers.
                      i) How many students are there in the class?
                      ii) How many chocolates does their teacher receive?
                      :thankyou:

                      1 Reply Last reply Reply Quote 0
                      • J Offline
                        Jtutor
                        last edited by

                        insanePaPa:
                        Sec1 Qn:


                        A class has between 30 to 40 students. Each boy in the class brings 15 chocolates for a class party. The chocolates are shared equally among 20 girls and a teacher with no leftovers.
                        i) How many students are there in the class?
                        ii) How many chocolates does their teacher receive?
                        :thankyou:
                        Hi insanePapa,

                        Since there are between 30-40 students in the class, there should be between 10-20 boys (after deducting 20 girls).

                        If each boy brings 15 chocolates for the party, and the total chocolates are shared among 20 girls + 1 teacher = 21 people, we must find the number of boys such that when it is multiplied by 15 chocolates, the total number of chocolates can be divided equally among 21 people.

                        15 = 3 x 5
                        21= 3 x 7
                        Hence, possible values for the number of boys will be 7, 14, 21, etc.
                        Since we know that the value should be between 10-20, the number of boys should be 14.

                        i) Number of students in the class = 20 girls + 14 boys = 34.

                        ii) Each of the 20 girls & 1 teacher will receive 15 x 14 / 21 = 10 chocolates. Hence, answer is 10.

                        Cheers!
                        Jtutor

                        1 Reply Last reply Reply Quote 0

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