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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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