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Simplified value of $$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, - \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$ $$ ÷ $$ $$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, + \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$ = ?
Answer & Solution
Correct Answer:
Option
A
$$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, - \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$ $$ ÷ $$ $$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, + \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$
$$ {\text{Let }}\left[ {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right] = a,$$ $${\text{ }}\left[ {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right] = b $$
$$\eqalign{ & \Rightarrow \left( {{a^2} - {b^2}} \right) \div \left( {a + b} \right) = a - b = ? \cr & \Rightarrow a = 1 + \frac{{10}}{{101}} \cr & \Rightarrow a = \frac{{111}}{{101}} \cr & \Rightarrow b = 1 - \frac{{10}}{{101}} \cr & \Rightarrow b = \frac{{91}}{{101}} \cr & \Rightarrow a - b = \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr & \Rightarrow a - b = \frac{{20}}{{101}} \cr} $$
$$ {\text{Let }}\left[ {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right] = a,$$ $${\text{ }}\left[ {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right] = b $$
$$\eqalign{ & \Rightarrow \left( {{a^2} - {b^2}} \right) \div \left( {a + b} \right) = a - b = ? \cr & \Rightarrow a = 1 + \frac{{10}}{{101}} \cr & \Rightarrow a = \frac{{111}}{{101}} \cr & \Rightarrow b = 1 - \frac{{10}}{{101}} \cr & \Rightarrow b = \frac{{91}}{{101}} \cr & \Rightarrow a - b = \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr & \Rightarrow a - b = \frac{{20}}{{101}} \cr} $$
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