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| Approximation |
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| We know that the digits of a number, one by one in order from left to right decrease in value rapidly. |
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| Let us illustrate it by examples. |
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| Mass of a gold ornament is 23.473 g. Round off to |
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| (i) two decimal places, (ii) one decimal place. |
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23.473 g = (2 x 10) + (3 x 1) +  |
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| The successive digits from left to right decrease in value rapidly. |
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| (i) 3rd decimal place is less than 5, |
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 We merely omit it. |
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 23.473 g = 23.47g corrected to two decimal places |
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| (ii) 2nd decimal place is 7. |
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 If it is 5 or more than 5, we increase 1st decimal place by one. |
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 23.473 g = 23.5 g corrected to one decimal place. |
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 |
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| Write 9847.982 corrected to |
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| (i) one decimal place (ii) integer (iii) hundred (iv) thousand |
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(i) 9848.0  2nd d.p. is 5 or more than 5 [ 8 ] |
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 1st d.p. is increased by 1 |
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(ii) 9848  1st d.p. is 5 more than 5 [ 9 ] |
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 unit's place is increased by 1. |
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(iii) 9800  ten's place is less than 5 [4] |
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 we write 0 at unit's and ten's place. |
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(iv) 10,000  hundred's place is 5 or more than 5 [ 8 ] |
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 thousands' place is increased by 1. |
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 9 increases to 10 and in unit's, ten's and hundred's place we write 0. |
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| [ i.e., 0 in the place of all digits to the right ] |
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