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(b)  The denominators of the given fractions are 3 and 5.

                          So, LCM (3, 5) = 15
                                                                                                    +
                                      ×
                                                         ×
                          Now,    2  =  25  =  10  and  4  =  43  =  12             ∴    2  +  4  =  10 12  =  22
                                                         ×
                                      ×
                                3   35     15      5    53     15                        3   5     15      15
                     (c)  The denominators of the given fractions are 4, 3 and 5.
                          So, LCM (4, 3, 5) = 60
                                                      ×
                                                                       ×
                                3   315      45 1   120      20 1    112      12          3  1   1   45 +  20 12   77
                                      ×
                                                                                                             +
                          Now,    =       =    ;  =        =    ;  =        =        ∴     ++      =             =
                                      ×
                                4   415      60 3    320     60 5    512      60          4  3   5        60       60
                                                      ×
                                                                       ×
                Ex 2.  Solve the following using Brahmagupta’s method:
                           1    2                3  1                    2   3   1               1   7    1
                     (a)  1 + 3             (b)    −                 (c)   +   +             (d)   −+    5
                           3    3                4   3                   3   4   2               2   2    6
                                              +
                           1    2   4 11    411     15
                Sol. (a)  1 + 3 =     +   =        =    = 5
                           3    3   3   3      3     3
                     (b)  Since the denominators of the given fractions are 4 and 3, the LCM of 4 and 3 is 12.
                                                              ×
                                                       ×
                                                                               −
                                              3   1   33     14     9    4   94      5
                          ∴                     −=         −      =    −   =       =
                                                              ×
                                                       ×
                                              4   3   43     34     12 12     12     12
                     (c)  Since the denominators of the given fractions are 3, 4 and 2, the LCM of 3, 4, and 2 is 12.
                                                               ×
                                                                      ×
                                                                                             ++ 6
                          ∴               2  +  3  +  1  =  2 ×  4  +  33  +  16  =  8  +  9  +  6    =  89  =  23  = 1 11
                                                       ×
                                                               ×
                                                                      ×
                                          3   4   2   34     43      26     12 12 12          12      12    12
                                  1
                                          7
                             7
                     (d)   1  −+ 5 =   1  − +  31         (After converting the mixed fraction into an improper fraction)
                          2  2     6   2  2    6
                                               ×
                                        ×
                                    =  13   −  73 +  31                                            ( LCM (2, 6) = 6)
                                               ×
                                       23     23     6
                                        ×
                                       3   21 31    32131        13    1
                                                      −
                                                          +
                                     =  −    +    =            =    =  2
                                       6   6    6        6       6     6
                                                  4       5
                Ex 3.  Find the difference between    and   .
                                                  15     18
                Sol.  To find the difference, it is important to know which one is greater. Then we will subtract the smaller
                      one from the greater one. LCM (18, 15) = 3 × 3 × 2 × 5 = 90.
                                                        ×
                                   ×
                             5    55      25      4    46      24                                            3 18, 15
                      Now,     =       =     and    =        =   .
                                                         ×
                            18   18 5     90     15   15 6     90                                            3 6, 5
                                    ×
                             5    4                                                                          2 2, 5
                      ∴        >     ( 25 > 24)                                                             5 1, 5
                            18   15                                                                             1, 1
                                                                                  −
                      ∴  Difference between    4   and   5  =  5  −  4  =  25  −  24  =  25 24  =  1
                                              15      18   18 15     90   90     90      90
                                    2                            3
                Ex 4.  Rahim mixes   3   litres of yellow paint with   4   litres of blue paint to make green paint. What is the
                      volume of green paint he has made?                                                      (NCERT)
                                              2     3
                Sol.  Volume of green paint =   L +   L. The LCM of 3 and 4 is 12.
                                              3     4
                                            ×
                                                   +
                          2   3   2 ×  4  33     89     17           5
                      So,   +   =       +      =       =    litres = 1  litres
                                            ×
                          3   4   34      43      12    12          12
                                    ×
              208     Mathematics-6
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