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              \\November 22, 2023 3:12 PM   Surender Prajapati   Proof 5   Reader _________________________   Date: ___________________74





                                                   1                                                    2
              example 3: Siya’s father filled  8  litres of petrol in his car and he used up  7  litres on a
                                                   3                                                    3
              visit to the zoo with her. How much petrol is left in his car?
                                                           1
              Solution:     Petrol filled in the car =  8  litres.
                                                           3
                                                                        2
                            Petrol used on visiting to the zoo =  7  litres.
                                                                        3
                                                                          1
                                                                                 2
                                                                                                               2
                            Therefore, the petrol left in the car =  8  –  7  =       25  –  23  =  25 – 23  =   litre.
                                                                          3
                                                                                 3
                                                                                                      3
                                                                                                               3
                                                                                             3
                                                                                      3
                                   2
                            Thus,  litre of petrol has left in the car.
                                   3
              addition and subtraction of unlike Fractions
              To add or subtract  unlike fractions,  first  convert  them  into like fractions  and  then  add  or
              subtract as like fractions.
              example 4: Solve the following.
                                                                                                    4
                                                 8
                        7
                    (a)   +   3             (b)   –   5              (c)  7  2   +  2  7      (d)  7  –  3 5
                        9    8                   9    6                    15       12              9      4
              Solution:
                    (a)  Convert  the  given  fractions  into their  respective equivalent  fractions  having  the
                        same denominator equal to the LCM of the denominators of the given fractions.

                         Here, LCM of 9 and 8 is 72.
                                          3
                                     7
                         Therefore,   +   =    7 × 8  +  3 × 9
                                     9    8    9 × 8    8 × 9

                                              =   56  +   27  =  56 + 27  =   83  =  1  11
                                               72    72        72        72      72

                    (b)  Convert  the  given  fractions  into their  respective equivalent  fractions  having  the
                        same denominator equal to the LCM of the denominators of the given fractions.

                         Here, LCM of 9 and 6 is 18.

                                          5
                                     8
                         Therefore,   –   =    8 × 2  –  5 × 3  =   16  –   15  =  16 – 15  =   1
                                     9    6    9 × 2    6 × 3     18     18       18        18
                    (c)  First convert mixed fractions into improper fractions and then convert the unlike
                        fractions  into  their  respective equivalent fractions having  the same denominator
                        equal to the LCM of the denominators of the given fractions.

                         Here, LCM of 15 and 12 is 60.
                                        2       7     107     31    107 × 4      31 × 5
                         Therefore,  7     +  2    =       +      =           +
                                       15      12     15      12     15 × 4      12 × 5


                                                          =  428  +  155  =  428 + 155   =  583  =  9 43
                                                             60      60         60          60       60


              Mathematics-5                                                                                          91
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