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But if you change the position of the brackets:

                                      (12 + 8) ÷ 2 × 3 = 20 ÷ 2 × 3 = 10 × 3 = 30
                                   3 × (3 – 3) ÷ 3 + 3 = 3 × 0 ÷ 3 + 3 = 3 × 0 + 3 = 0 + 3 = 3
                        The result is different based on where the brackets are placed.
                    4.  Complex Example:
                        When dealing with multiple brackets and operations, always solve the innermost bracket first.

                       For example: 10 + {12 ÷ (3 × 2)} = 10 + {12 ÷ 6} = 10 + 2 = 12
                        Having  learnt  the  BODMAS rule,  let  us do some  examples  on  BODMAS and  the  addition  of
                       number patterns.
                    Ex 1.  Write the expression for each of the following using brackets:
                         (a)  Four multiplied by the sum of nine and two.
                         (b)  Divide the difference of eighteen and six by four.
                         (c)  Forty-five divided by three times the sum of three and two.
                    Sol. (a)  Sum of nine and two = 9 + 2

                              So, four multiplied by the sum of nine and two = 4 × (9 + 2).
                         (b)  Difference of eighteen and six = 18 – 6
                              So, division of 18 – 6 by 4 = (18 – 6) ÷ 4.
                         (c)  The required expression is 45 ÷ {3(3 + 2)}.

                    Ex 2.  Simplify the following using the BODMAS rule.
                         (a)  16 + 3(91 ÷ 7)    (b)  {5(8 + 2) – 5} × 2   (c)  80 + [190 – {8 × 9 + (110 – 50)}]
                    Sol: (a)  16 + 3(91 ÷ 7) = 16 + 3 × 13                                                  [  91 ÷ 7 = 13]
                                           = 16 + 39 = 55
                         (b)  {5(8 + 2) – 5} × 2 = {5 × 10 – 5} × 2                                          [  8 + 2 = 10]

                                               = {50 – 5} × 2                                               [  5 × 10 = 50]
                                               = 45 × 2 = 90
                         (c)  80 + [190 – {8 × 9 + (110 – 50)}] = 80 + [190 – {8 × 9 + 60}]              [  110 – 50 = 60]

                                                             = 80 + [190 – {72 + 60}] = 80 + [190 – 132]
                                                             = 80 + 58 = 138
                    Ex 3.  Evaluate 104 × 105 using brackets.
                    Sol.  104 × 105 = (100 + 4) × (100 + 5)  = (100 + 4) × 100 + (100 + 4) × 5
                                      = 100 × 100 + 4 × 100 + 100 × 5 + 4 × 5
                                      = 10,000 + 400 + 500 + 20 = 10,000 + 900 + 20 = 10,920

                    Ex 4.  In the figure (a), some numbers are placed in a pattern. Find out   40   40      40     40
                          the sum of the numbers in the pattern. Should we add them one   50    50      50      50     50
                          by one, or can we use a quicker way?               (NCERT)

                    Sol.  Obviously, we can add these numbers in Figure (a), one by one,     40     40      40     40
                          but that won’t be an intelligent way of doing it. It will also make   50  50  50      50     50
                          our calculations lengthy and boring.
                                                                                             40     40      40     40
                                                                                                      Figure (a)


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