Page 12 - TOP ONE MATHS F3
P. 12

Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd





                                                                                   Mathematics  Form 3  Chapter 1  Indices       Mathematics  Form 3  Chapter 1  Indices
                                       1    –   2                                n       5   3
                                                                                                                                                                                         –1 2 2
                                                                                                                                                                                                  3 –3
                                       2
                          (g)  (12 × 27)  ÷ 8  3                           (b)  5  × 5 = 5  × 5                                      PT3 Standard Practice  1                   6.  A  (3a b )  × 7a b
                                            1     –   2                         5 n + 1  = 5 5 + 3                                                                                    = 9a b  × 7a b
                                                                                                                                                                                          –2 4
                                                                                                                                                                                                 3 –3
                                          3 2
                              = (2  × 3 × 3 )  ÷ (2 )  3                        n + 1 = 8                                                                                             = (9 × 7)a –2 + 3 4 – 3
                                                 3
                                  2
                                                                                                                                                                                                  b
                                   1     2   (1 + 3) ×  1                                                                   Section A
                                 2 ×   – 3 × –
                              = 2  2     3  × 3   2                                n = 7                                                                                              = 63ab

                                                                                                                                           1 4
                                                                                                                                        
                              = 2  × 3 2                                                                                      1.  m   =  m  5                                           0     3 2  3   6

                                 3
                                                                                                                                     4
                                                                                                                                 5
                              = 8 × 9                                      (c)  3 = 9 n                                                   1                                        B  7c  × 7  ÷ 49c  × 343c
                                                                                 2
                                                                                                                                                                                                   3
                                                                                                                                                                                         1
                              = 72                                              3  = 3 2n                                             = m  5  × 4                                     = 7c  × 1 ÷ 7 c  × 7c 2
                                                                                                                                                                                                  2 2
                                                                                                                                                                                         2
                                                                                 2
                                                                                                                                                                                                 3
                                                                                2 = 2n                                                = m  4 5                                        = 7 1 – 2 + 1 2 1  –   + 2
                                                                                                                                                                                              c
                                                                                                                                                                                                 2
                      12.  (a)  2x y  × 5xy 2                                   n = 1                                                                                                 = c
                                3 2
                              = 2 × 5x 3 + 1 2 + 2                                                                               Answer: C                                                 5     1
                                         y
                                                                                                                                                                                              1
                                   4 4
                              = 10x y                                      (d)  (b )  = b 9                                                                                       C   1 d  2 –  3  =  d  3
                                                                                  n 3
                                                                                                                                                                                      5
                                                                                                                                                                                              5
                                                                                                                                         
                                                                                                                                            2
                                                                                                                                                   –1 2
                                                                                                                                   –1 2
                                                                                  3n
                                                                                 b  = b 9                                     2.   7 k   ×  7k 1 3  =   7 k   × 7 k 3                         1 3
                                                                                                                                                            3 2
                                                                                                                                                                                                  d
                                                                                                                                                     3
                          (b)  81m n  ÷ 27m n                                    3n = 9                                               3 2        (7 ) k 3 2                                 =  
                                           4 2
                                  3 2
                                                                                                                                                                                              5
                                                                                                                                                   2 2
                                 4
                              = 3 m n  ÷ 3 m n                                     n = 3                                         (49k)            –1 + 3 – 3 2 +   –  3                         1
                                   3 2
                                          3
                                            4 2
                                                                                                                                                           3
                                        n
                              = 3 4 − 3 m 3 − 4 2 − 2                                                                                          = 7     k   2  2                             ≠   5d
                                                                                                                                                                                               3
                                3                                                                                                              = 7 k
                                                                                                                                                  –1 2
                              =                                            (e)  27(5 ) = 125(3 )                                                                                   D  e × e × e × e × e × e = e 6
                                                                                    n
                                                                                             n
                                m                                                                                                              =  k 2
                                                                                                                                                 1
                                                                                    n  =  125                                                  7                                Answer: C
                                                                                  5
                                                                                  3
                                                                                         27
                                         a
                                          4 –1
                          (c)  (3a b )  ×                                            3                                     Answer: A                                    Section B
                                 3 –1 2
                                                                                     n
                                                                                  5
                                                                                         5
                                         b
                                       b
                              = 9a 3 × 2 – 4 –1 × 2+1                             3   =   3                                   3.  3q  × 5q  = (3 × 5)q 4 + 5
                                                                                                                                   4
                                                                                                                                        5
                              = 9a b                                                n = 3                                                 = 15q 9                               1.    a  × a 3                 a 6
                                  2 –1
                                                                                                                                                                                       5
                                 a 2
                              = 9                                                                                                Answer: A
                                  b
                                                                                                                                                                                       (a )                    a 8
                                                                                                                                                                                        3 2
                                                                                                                                        –3k
                              (hk )  × h k                                                                                    4.   2 k + 1  × 2   = 32
                                 2 2
                                       3 –3
                          (d)                                                                                                       k + 1 –3k  5
                                  h k                                                                                              2      = 2                                                                   2
                                   4 –4
                                        k
                              = h 1 × 2 + 3 – 4 2 × 2 – 3 – (–4)                                                                    1 – 2k = 5                                         a –3                    a 3
                              = h 2 +3 – 4 4 – 3 + 4                                                                                   2k = –4
                                      k
                                                                                                                                                                                                               1
                              = hk 5                                                                                                    k = –2                                         3  2                  —
                                                                                                                                                                                        a
                                                                                                                                 Answer: D                                                                     a 3
                                         5
                                 2 –3 3
                                         2 4
                              (2r s )  × r s                                                                                                                                         3   4
                          (e)                                                                                                       y 6                                            3r s 
                                     1                                                                                        5.  A    = y 6 – 5                                     4               3
                                  8r s                                                                                              y 5                                         2.         × (16r s )
                                     2 2
                                                                                                                                                                                                 2 –6 2
                                                                                                                                       = y                                          5r  s
                                                                                                                                                                                      –3 7
                                2 3  2 × 3 +   –  1
                                          5
                                                                                                                                     7
                                                                                                                                                                                             1
                                                                                                                                         4
                              =    × r    2  2 s –3 × 3 + 4 – 2                                                                  B  y  × y  = y 7 + 4                                       —  × 4
                                2 3                                                                                                       = y 11                                    3 r 3  × 4 s   2   3      3      3
                                                                                                                                                                                      4 4
                                   5
                              = r 6 +   –  1 2 s –9 + 4 – 2                                                                      C  y  + y 8                                      =     5r  s     × 4 2 ×  2 × r 2 ×  2 × s –6 ×  2
                                                                                                                                                                                          –3 7
                                   2
                                                                                                                                     3
                                 8 –7
                              = r s                                                                                                                                                     3 2
                                                                                                                                 D  (y )  = 1                                     =   81r s    3  r s
                                                                                                                                      11 0
                                                                                                                                                                                       –3 7  ×  4
                                                                                                                                                                                                  3 –9
                                                                                                                                                                                     5r  s
                                12
                      13.  (a)  2  ÷ 8 = 2 n                                                                                     Answer: B                                          5 184
                              2  ÷ 2  = 2 n                                                                                                                                       =    5  r   3 + 3 – (–3)   × S 2 – 9 – 7
                               12
                                    3
                                2 12 – 3  = 2 n
                                   2  = 2 n                                                                                                                                          5 184  9 –14
                                    9
                                    n = 9                                                                                                                                         =    5   r s
                                                                    3                          © Penerbitan Pelangi Sdn. Bhd.  © Penerbitan Pelangi Sdn. Bhd.              4
                                                                                                                      A3
                BOOKLET ANS MATH F3.indd   3                                                                                                                                                                     03/01/2020   10:20 AM
   7   8   9   10   11   12   13   14   15   16   17