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    O-Level Additional Math

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
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    • K Offline
      kittybank
      last edited by

      Thank you very much, u r fast!

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

        Hi jie heng

        Can show how you derive at the above answers?

        Thanks!

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        • T Offline
          tutor4kids
          last edited by

          Hi,


          Could you help me solve part ii of the following question?

          A polynomial p(x) when divided by (x^2 - 1) leaves a quadratic remainder with no remainder. The constant term in p(x) is 10.
          Given that p(x) leaves a remainder of 630 and 112 when divided by (x - 4) and (x + 3) respectively, find
          i) an expression for p(x) in descending powers of x.
          ii) solutions when p(x) = x^2 - x + 10

          Ans: i) 3x^4 + x^3 - 13x^2 - x + 10
          ii) x = 0, -7/3, 2

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          • L Offline
            leeven
            last edited by

            1. A number is a called palindrome if it reads the same from left and from right. example 130031 is a palindrome. writing all palindroms we get 1,2,3,4,5,…9,11,22,…what is the 2007th palindrome. do you have formula to find the nth palindrome.


            please help

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            • F Offline
              FrekiWang
              last edited by

              tutor4kids:
              Hi,


              Could you help me solve part ii of the following question?

              A polynomial p(x) when divided by (x^2 - 1) leaves a quadratic remainder with no remainder. The constant term in p(x) is 10.
              Given that p(x) leaves a remainder of 630 and 112 when divided by (x - 4) and (x + 3) respectively, find
              i) an expression for p(x) in descending powers of x.
              ii) solutions when p(x) = x^2 - x + 10

              Ans: i) 3x^4 + x^3 - 13x^2 - x + 10
              ii) x = 0, -7/3, 2
              I assume 'quadratic remainder' is an error and it should be 'quadratic quotient'.
              (i)
              p(x) = Divisor x Quatient + Remainder = (x^2 -1)(ax^2+bx+c)+0 = ax^4 +bx^3 + cx^2 -ax^2 - bx - c = ax^4 + bx^3 + (c-a)x^2 - bx - c
              Since the constant term is 10, we have -c = 10, c = -10
              Thus, p(x)=ax^4 + bx^3 + (-10-a)x^2 - bx + 10
              According to the remainder thereom,
              p(4)=630 and p(-3)=112
              256a+64b+16(-10-a)-4b+10=630 and 81a-27b+9(-10-a)+3b+10=112
              240a+60b=780 and 72a-24b=192
              solve, we get a=3 and b=1
              therefore p(x)= 3x^4 + x^3 -13x^2 - x + 10
              (ii) p(x)=x^2 - x + 10
              3x^4 + x^3 - 13x^2 - x +10 = x^2 - x + 10
              3x^4 + x^3 - 14x^2 = 0
              x^2(3x^2 + x - 14) = 0
              x^2(3x + 7)(x - 2) = 0
              x^2 = 0 or 3x + 7 = 0 or x - 2 = 0
              x = 0 or x = -7/3 or x = 2

              PS: You may need more systematic help for this chapter, not just Q&A

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              • F Offline
                FrekiWang
                last edited by

                leeven:
                1. A number is a called palindrome if it reads the same from left and from right. example 130031 is a palindrome. writing all palindroms we get 1,2,3,4,5,....9,11,22,,....what is the 2007th palindrome. do you have formula to find the nth palindrome.


                please help
                First of all, I do not believe this is a secondary school question. The standard of this problem is for competition at upper secondary school level, e.g. SMO (Senior)

                So, here is my solution, assuming you have some basic knowledge about competition mathematics.

                There are 9 1-digit palindroms.
                There are 9 2-digit palindroms.
                For 3-digit palindroms. You can choose a number from 0 to 9 (10numbers here) and insert into a 2-digit palindrom to form a 3-digit palindrom. Therefore 9 2-digit palindroms can produce 9 x 10 = 90 3-digit palindroms.
                For 4-digit palindroms. You can choose a number from 00, 11 to 99(10numbers here) and insert into a 2-digit palindrom to form a 4-digit palindrom.
                Therefore 9 2-digit palindroms can produce 9 x 10 = 90 4-digt palindroms.

                Similarly, insert a number from 0 ... 9 into a 4-digit palindrom to form a 5-digit palindroms, there are 90 x 10 = 900 of them.
                Insert a number from 00 ... 99 into a 4-digit palindrom to form a 6-digit palindroms, there are 90 x 10 = 900 of them.

                Now we have 9 + 9 + 90 + 90 + 900 + 900 = 1998 palindroms in totoal, we need 9 more.
                Let's start counting the 7-digit palindroms from the least: (inserting 0..9 into 100001)
                1000001, 1001001, 1002001, 1003001, 1004001, 1005001, 1006001, 1007001, 1008001 - this is the 9th, so the answer is 1008001.

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                • H Offline
                  Herbie
                  last edited by

                  hi how to solve this qqn


                  a team of 8 men can build 3 swimming pools in 5 weeks. Find the no. Of men required to build 6 swimming pools in 4 weeks? Tq

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                  • T Offline
                    tianzhu
                    last edited by

                    Herbie:
                    hi how to solve this qqn


                    a team of 8 men can build 3 swimming pools in 5 weeks. Find the no. Of men required to build 6 swimming pools in 4 weeks? Tq
                    Hi

                    One way is to use compound units which in this question is \"man-weeks\"

                    3 swimming pool ------ 8*5 ------ 40 man-weeks
                    6 swimming pools ------ 2*40 ------- 80 man-weeks

                    80/4 ------ 20 men

                    Best wishes

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                    • H Offline
                      Herbie
                      last edited by

                      hi tianzhu, many thanks!


                      I hv another qn.

                      15 men working 8 hours a day can build a bridge in 30 days.
                      How long will 9 men working 8 hours a day take to buiild the same ridge assuming that the rate at which they work remains unchanged? My ans is 50 days.

                      B. How many hours a day must 25 men work to build the same bridhe in 16 days!

                      Tq

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

                        Herbie:

                        15 men working 8 hours a day can build a bridge in 30 days.
                        How long will 9 men working 8 hours a day take to buiild the same ridge assuming that the rate at which they work remains unchanged? My ans is 50 days.

                        B. How many hours a day must 25 men work to build the same bridhe in 16 days!
                        Hi

                        Your answer of 50 days for part A is right.

                        For part B,
                        1 bridge ------ 15*30*8 ------- 3600 man-hours

                        Number of men*number of days*number of hours ------- 3600 man-hours

                        25*16*h ------- 3600 where h refers to the number of hours the men must work.

                        h ------- 9 hours.

                        Best wishes

                        1 Reply Last reply Reply Quote 0

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