Page 39 - Start Up Mathematics_6
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Solution:     (a)  297 × 17 + 297 × 3 = 297 × (17 + 3) = 297 × 20 = 5,940

                          (b)  38,443 × 94 + 6 × 38,443 = 38,443 × 94 + 38,443 × 6 = 38,443 × (94 + 6)
                                                          = 38,443 × 100 = 38,44,300
                          (c)  81,265 × 169 – 81,265 × 69 = 81,265 × (169 – 69) = 81,265 × 100 = 81,26,500
            Example 8:  Find the products using suitable properties.

                          (a)  8,431 × 110             (b)  854 × 99                (c)  1,007 × 168
            Solution:     (a)  8,431 × 110 = 8,431 × (100 + 10) = 8,431 × 100 + 8,431 × 10
                                            = 8,43,100 + 84,310 = 9,27,410

                          (b)  854 × 99     = 854 × (100 – 1) = 854 × 100 – 854 × 1
                                            = 85,400 – 854 = 84,546

                          (c)  1,007 × 168 = (1,000 + 7) × (100 + 68)
                                            = 1,000 × 100 + 1,000 × 68 + 7 × 100 + 7 × 68
                                            = 1,00,000 + 68,000 + 700 + 476 = 1,69,176

            Example 9:  A bus driver filled the bus tank with 33 L of diesel. Two days later he filled the tank
                          with 17 L of diesel. If the diesel costs ` 28 per litre, how much did he spend in all
                          on diesel?
            Solution:         Diesel filled on first day  = 33 L
                           Diesel filled two days later  = 17 L
                                         Cost of diesel  = ` 28 per litre

                            Total expenditure on fuel  = 33 × 28 + 17 × 28
                                                         = (33 + 17) × 28 = 50 × 28 = ` 1,400
            Example 10: Match the following:

                          (a)  325 × 136 = 325 × (6 + 30 + 100)  (i)  Commutativity under multiplication
                          (b)  3 × 49 × 50 = 3 × 50 × 49           (ii)  Commutativity under addition
                          (c)  80 + 740 + 20 = 80 + 20 + 740       (iii)  Distributivity of multiplication over
                                                                         addition
            Solution:     (a) (iii), (b) (i), (c) (ii)
            Example 11: If the product of two whole numbers is zero, can we say that one or both of them
                          will be zero? Justify.                                                        (NCERT)
            Solution:     (a)  Let’s assume that the two whole numbers ‘a’ and ‘b’ are such that neither ‘a’
                              nor ‘b’ is zero. The product of any two non-zero whole numbers is a non-zero
                              whole number. But it is given that a × b = 0, which is not possible. Hence, both
                              ‘a’ and ‘b’ cannot be non-zero.

                          (b)  Let’s assume that any one of the two whole numbers ‘a’ and ‘b’ is zero and the
                              other is non-zero. Therefore, the product of ‘a’ and ‘b’ would be zero.

                          (c)  Now let’s assume that both the whole numbers ‘a’ and ‘b’ are zeros. Then their
                              product will also be zero.

                          Therefore, we can say that if the product of two whole numbers is zero, either one
                          or both of them are zeros.


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