If $0.2 ¯ ¯¯¯¯ ¯ 41$ is expressed as a rational number in reduced form, then the sum of the values of the numerator and denominator is:

1223

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1227

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1229

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1232

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Solution

The correct option is D $1229$
Let $x$ be $0.24141...$
$=> 10 x = 2.4141........ (i)$
$=> 10 x \times 100 = 241.4141..$
$=> 1000 x = 239 + 2.4141$
$=> 1000 x = 239 + 10 x$ (from $i$ )
$=> 1000 x - 10 x = 239$
$=> x = \frac{239}{990}$
Therefore, the sum of the numerator and denominator is
$239 + 990 = 1229$

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If 0.2¯¯¯¯¯¯41 is expressed as a rational number in reduced form, then the sum of the values of the numerator and denominator is:

The correct option is D 1229Let x be 0.24141...=>10x=2.4141........(i)=>10x×100=241.4141..=>1000x=239+2.4141=>1000x=239+10x (from i)=>1000x−10x=239=>x=239990Therefore, the sum of the numerator and denominator is239+990=1229

If $0.2 ¯ ¯¯¯¯ ¯ 41$ is expressed as a rational number in reduced form, then the sum of the values of the numerator and denominator is: