Page 125 - Math Skill - 5
P. 125

Measurement                                                                                            123


        Basic Operations on Length, Mass (Weight) and Capacity
        The basic operations we perform on the measures of length, mass (weight) and capacity are:
        Addition, Subtraction, Multiplication and Division.

        Let’s learn through examples.


              Let’s Attempt

        Example 1:  Add the following.

                        (a)  6 kg 675 g, 9 kg 85 g and 10 kg 9 g       (b) 7 m 60 cm, 3 m 7 cm and 90 cm
                        (c)  5 L 200 mL, 3 L 30 mL and 6 mL


        Solution:       (a)  Adding column-wise, we get:             (b)  Adding column-wise, we get:
                                   kg           g                                   m        cm
                                      6    6    7    5                                 1

                                      9    0    8    5                                 7    6    0
                             + 1      0    0    0    9                                 3    0    7
                                 2    5    7    6    9                        +        0    9    0
                                                                                  1    1    5    7
                             Thus, required sum = 25 kg 769 g
                                                                             Thus, required sum = 11 m 57 cm

                           (c)   Adding column-wise, we get:
                                    L       mL
                                    5    2    0    0
                                    3    0    3    0
                                + 0      0    0    6
                                    8    2    3    6          Thus, required sum = 8 L 236 mL

        Example 2:   Subtract the following.

                        (a)  89 L 678 mL from 103 L 250 mL             (b) 7 kg 865 g from 10 kg
                        (c)   9 km 75 m from 16 km 20 m

        Solution:       (a)  Subtracting column-wise we get:             (b)  Subtracting column-wise we get:

                                      L             mL                               kg           g
                                  1    0    3    2    5    0                       1    0    0    0    0
                             –         8    9    6    7    8                   –        7    8    6    5

                                       1    3    5    7    2                            2    1    3    5
                           Thus, required difference = 13 L 572 mL          Thus, required difference

                                                                                 = 2 kg 135 g
                        (c)  Subtracting column-wise we get:
                                      km           m
                                    1    6    0    2     0
                                –        9    0    7     5
                                         6    9    4     5      Thus, required difference = 6 km 945 m
   120   121   122   123   124   125   126   127   128   129   130