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    O-Level Additional Math

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
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    • A Offline
      Alarmchain
      last edited by

      MathIzzzFun:
      WTK:

      Help please!


      Find the greatest number that will divide 171, 255 and 304 so as to leave the same remainder in each case.

      Kindly show the workings. Thank you.

      Let the greatest number be n, and the remainder be R

      171 = An + R
      255 = Bn + R
      304 = Cn + R
      where A,B,C are integers

      (B-A)n = 255 - 171 = 84 = 12 x 7
      (C-B)n = 304 - 255 = 49 = 7 x 7
      (C-A)n = 304 - 171 = 133 = 19 x 7

      n = 7

      The greatest number is 7

      cheers.

      Hi MathIzzzFun,

      From your results above, we get
      1) B - A = 12
      2) C - B = 7
      3) C - A = 19

      Can I trouble you to show me the steps, from these 3 sets of equations, how do we end up with A = 24, B = 36 and C = 43?

      Thank you.

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

        Alarmchain:
        MathIzzzFun:

        [quote=\"WTK\"]Help please!


        Find the greatest number that will divide 171, 255 and 304 so as to leave the same remainder in each case.

        Kindly show the workings. Thank you.

        Let the greatest number be n, and the remainder be R

        171 = An + R
        255 = Bn + R
        304 = Cn + R
        where A,B,C are integers

        (B-A)n = 255 - 171 = 84 = 12 x 7
        (C-B)n = 304 - 255 = 49 = 7 x 7
        (C-A)n = 304 - 171 = 133 = 19 x 7

        n = 7

        The greatest number is 7

        cheers.

        Hi MathIzzzFun,

        From your results above, we get
        1) B - A = 12
        2) C - B = 7
        3) C - A = 19

        Can I trouble you to show me the steps, from these 3 sets of equations, how do we end up with A = 24, B = 36 and C = 43?

        Thank you.[/quote]
        171/7= 24r3, A= 24

        do the same for 255 n 304

        cheers

        1 Reply Last reply Reply Quote 0
        • A Offline
          Alarmchain
          last edited by

          Yes, that was what I did.


          But I was wondering how the same values for A, B and C can be gotten from the 3 simultaneous equations.

          Thanks.

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

            Alarmchain:
            Yes, that was what I did.


            But I was wondering how the same values for A, B and C can be gotten from the 3 simultaneous equations.

            Thanks.
            255/7 = 36R3, B = 36
            304/7 = 43R3, C = 43

            or with A = 24 and
            1) B - A = 12
            2) C - B = 7

            you can also get B= 36, C = 43

            cheers.

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

              Hi! Can anyone recommend a good math practice book for Sec One Maths (IP school)?

              1 Reply Last reply Reply Quote 0
              • A Offline
                Alarmchain
                last edited by

                MathIzzzFun:
                Alarmchain:

                Yes, that was what I did.


                But I was wondering how the same values for A, B and C can be gotten from the 3 simultaneous equations.

                Thanks.

                255/7 = 36R3, B = 36
                304/7 = 43R3, C = 43

                or with A = 24 and
                1) B - A = 12
                2) C - B = 7

                you can also get B= 36, C = 43

                cheers.

                My apologies, MathIzzzFun. I may have caused some confusion with my unclear question.

                With 3 unknowns, we usually can find the unknowns using algebra, if we have 3 simultaneous equations.

                In this instance, the 3 equations were:
                1) B - A = 12
                2) C - B = 7
                3) C - A = 19

                Yes, if we know that A= 24, we can substitute it into equations 1 and 2 to find the values of B and C. However, we got A by doing 171 / 7 first, which is perfectly ok for this problem.

                But my question is, are we able to find A, B and C with just the above 3 simultaneous equations, without resorting to finding any one of them initially, by using 171 / 7 or 255 / 7 or 304 / 7.

                Very sorry, my question is not just about this problem per se, but to correct a possible misunderstanding on my part, that Y unknowns can be found if we have Y number of simultaneous equations.

                If it cannot be done, does it mean that when solving unknowns using simultaneous equations, it is not always true that all the Y unknowns can be found even if we have Y number of simultaneous equations given.

                Again, very sorry for any confusion caused.

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

                  Alarmchain:
                  MathIzzzFun:

                  [quote=\"Alarmchain\"]Yes, that was what I did.


                  But I was wondering how the same values for A, B and C can be gotten from the 3 simultaneous equations.

                  Thanks.

                  255/7 = 36R3, B = 36
                  304/7 = 43R3, C = 43

                  or with A = 24 and
                  1) B - A = 12
                  2) C - B = 7

                  you can also get B= 36, C = 43

                  cheers.

                  My apologies, MathIzzzFun. I may have caused some confusion with my unclear question.

                  With 3 unknowns, we usually can find the unknowns using algebra, if we have 3 simultaneous equations.

                  In this instance, the 3 equations were:
                  1) B - A = 12
                  2) C - B = 7
                  3) C - A = 19

                  Yes, if we know that A= 24, we can substitute it into equations 1 and 2 to find the values of B and C. However, we got A by doing 171 / 7 first, which is perfectly ok for this problem.

                  But my question is, are we able to find A, B and C with just the above 3 simultaneous equations, without resorting to finding any one of them initially, by using 171 / 7 or 255 / 7 or 304 / 7.

                  Very sorry, my question is not just about this problem per se, but to correct a possible misunderstanding on my part, that Y unknowns can be found if we have Y number of simultaneous equations.

                  If it cannot be done, does it mean that when solving unknowns using simultaneous equations, it is not always true that all the Y unknowns can be found even if we have Y number of simultaneous equations given.

                  Again, very sorry for any confusion caused.[/quote]In this case, although there are 3 equations :
                  1) B - A = 12
                  2) C - B = 7
                  3) C - A = 19

                  there are only 2 independent equations
                  eg with 1) and 2), we can get 3)
                  or with 2) and 3), we can get 1)
                  this is why the variables cannot be solved with the 3 equations.

                  In short, to solve for Y unknowns, you need Y independent equations.

                  cheers.

                  1 Reply Last reply Reply Quote 0
                  • A Offline
                    Alarmchain
                    last edited by

                    Thank you so much! It is very clear now.


                    BTW, is there a way for a child to see and immediately tell that equations given are dependent or independent, like some kind of rule of thumb thingy? Unlike this example where the dependence was quite easy to visualise, not sure if there are such "tricks" or "tips" to help a child for more complicated examples.

                    Thanks again!

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

                      Hi, please kindly help me on this question:-


                      Find the equation of the circle,C, which passes through the points (0,5) and
                      (4, -3) and has its centre lying on the line y=3x+2. Hence, determine by calculation, whether the point (6,1) lies inside or outside C.

                      Thank you very much.

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

                        S-H:
                        Hi, please kindly help me on this question:-


                        Find the equation of the circle,C, which passes through the points (0,5) and
                        (4, -3) and has its centre lying on the line y=3x+2. Hence, determine by calculation, whether the point (6,1) lies inside or outside C.

                        Thank you very much.
                        here are the pointers to the solution..

                        http://i46.tinypic.com/30navph.png\">

                        here are some websites that you can learn more about equations of circles (or you can google other sites)
                        - http://www.regentsprep.org/Regents/math/algtrig/ATC1/circlelesson.htm
                        - http://www.mathsisfun.com/algebra/circle-equations.html

                        cheers.

                        1 Reply Last reply Reply Quote 0

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