Aptitude Question ID : 94704

Simplify :
$latex \left [ \left ( 1+\frac{1}{10+\frac{1}{10}} \right )\times \left ( 1+\frac{1}{10+\frac{1}{10}} \right )- \left ( 1-\frac{1}{10+\frac{1}{10}} \right )\times \left ( 1-\frac{1}{10+\frac{1}{10}} \right ) \right ]\div \left [ \left ( 1+\frac{1}{10+\frac{1}{10}} \right )+\left ( 1-\frac{1}{10+\frac{1}{10}} \right ) \right ]&s=1$
[A]$latex \frac{101}{100}&s=1$
[B]$latex \frac{20}{101}&s=1$
[C]$latex \frac{100}{101}&s=1$
[D]$latex \frac{90}{101}&s=1$

$latex \mathbf{\frac{20}{101}}&s=1$
Suppose that,
$latex 1+\frac{1}{10+\frac{1}{10}} = \frac{111}{101} = a&s=1$
$latex and, 1-\frac{1}{10+\frac{1}{10}} = \frac{91}{101} = b&s=1$
$latex \therefore \frac{a^{2}-b^{2}}{(a+b)} = \frac{(a+b)(a-b)}{(a+b)}&s=1$
$latex = (a-b)$
$latex = \frac{111}{101} – \frac{91}{101} = \frac{20}{101}&s=1$
Hence option [B] is right answer.


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