Page 79 - classs 6 a_Neat
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ADDITION OF FRACTIONS
        EXAMPLE 4.                   Find the sum   7   +  11  +   9
 ADDITION OF LIKE FRACTIONS                        12      16     24
 (Sum of their numerators)  SOLUTION:    LCM of 12, 16, 24 = ( 4 × 3 × 2 × 2) = 48.
 RULE   Sum of like fractions =
 (Common denominator)    7      11      9       28 + 33 +18            [ 48 ÷ 12 = 4, 4 × 7 = 28] and

 EXAMPLE 1.    Find the sum:  12  +  16  +  24  =    48          {     [ 48 ÷ 16 = 3, 3 × 11 = 33]    }
 4  2  3  1  5                                                         [ 48 ÷ 24 = 2, 2 × 9 = 18]
 (i)  +  (ii)  +  +      =  79  =1   31
 9  9  8  8  8              48       48
 SOLUTION:    We have:
 4  2  (4 + 2)  6 2  2                       5      7      11
 (i)  +  =  =  =  EXAMPLE 5.    find the sum    +       +
 9  9  9  9  3                               6      8      12
 3
 3  1  5  (3 + 2 + 5)  9  1  SOLUTION:    we have,
 (ii)  +  +  =  =  = 1
 8  8  8  8  8  8             LCM of 6, 8,1 2 = (2 × 3 × 2 × 2) = 24.
 ADDITION OF UNLIKE FRACTIONS
                          5   +  7   +  11  =    20 + 21 + 22          [ 24 ÷ 6 = 4, 4 × 5 = 20] and

 RULE     Change the givenfractions into equivalent like fractions and then add.  6  8  12  24  {  [ 24 ÷ 8 = 3, 3 × 7 = 21]  }
                             63 21     21       5                      [ 48 ÷ 24 = 2, 2 × 11 = 22]
                          =         =      =2
 8  5                         24        8       8
 EXAMPLE 2.    Find the sum:    +  8
 9  12

 SOLUTION:    LCM of 9 and 12 = (3 × 3 × 4) = 36.
        EXAMPLE 6.            Find the sum:  2  4  +1   3   +3   1
 Now,  8  +  8 × 5  =  32  and  5  =  5 × 3  =  15  .  5  10    15
 9  9 × 4   36  12  12 × 3   36  SOLUTION:    LCM of 5, 10, 15 = (5 × 2 × 3) = 30.
 8  +  5  =  32  +  15  =  (32 + 15)  =  47  = 1  11  .  2  4  +1  3  +3  1  [ 30 ÷ 5 = 6, 6 × 14 = 84] ,

 9  12  36  36  36  36  36          5       10      15           {     [ 30 ÷ 10 = 3, 3 × 13 = 39]    }
                                    14       13      46                [ 30 ÷ 15 = 2, 2 × 46 = 92]
 Short Method  LCM of 9 and 12 = (3 × 3 × 4) = 36.  =  +  +
                                     5       10      15
                                       84 + 39 + 92
                                       84 + 39 + 92       215 43     43         1
 8  +  5  =  32 + 5  {  [ 36 ÷ 9 = 4, 4 × 8 = 32] and  }  = =  30  +  30  =  6  = 7  6
                                           30
 9  12  36  [ 36 ÷ 12 = 3, 3 × 5 = 15]                        6

 47  11  EXAMPLE 7.           Tanvi bought a notebookjor ₹ 84 and a penjor ₹ 105• How much money should
 =  =1
 36  36                       she pay to the shopkeeper?
        SOLUTION                                      3        35

 5  3                         Cost of a notebook ₹ 8      = ₹
 EXAMPLE 3:    Find the sum:  +  .                    4        4
 6  18                                                 2        52
 SOLUTION:    LCM of 6 and 8 = (2 × 3 × 4 )= 24.           Cost of a pen   ₹ 10  5  = ₹  5

 5  +  3  =  20 + 9  {  [ 24 ÷ 6 = 4, 4 × 5 = 20] and  }      (  35      52  )          (   175 + 208   )
 6  8  24  [ 24 ÷ 8 = 3, 3 × 3 = 9]           Total cost of both the articles= ₹  4  +  5  = ₹  20
                                      =  ₹  383   = ₹19   3
 29  5                                       20          20
 =  =1
 24  24                                                      3
                              Hence, Tanvi should pay  ₹19      to the shopkeeper.
                                                            20
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