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               \ 08-Oct-2025  Bharat Arora   Proof-9                                 Reader’s Sign _______________________ Date __________





                                                   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 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 is 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
                    (  a)   +     7  3      (  b)   –     8  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, the 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, the LCM of 9 and 6 is 18.

                                     8
                                          5
                         Therefore,   –   =    8 × 2  –  5 × 3  =   16  –   15  =  16 – 15  =   1
                                     9    6    9 × 2    6 × 3     18     18       18        18
                    (  c)  First convert the 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, the 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       43
                                                          =       +      =               =      =  9
                                                             60      60         60          60       60


              Mathematics-5                                                                                          89
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