The value of $$\frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \div \frac{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}}{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}}{\text{is:}}$$

Solution (By Examveda Team)

$$\eqalign{ & \frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \div \frac{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}}{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}} \cr & = \frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \times \frac{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}}{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}} \cr & = \frac{{203 \times 292}}{{7 \times 365 \times 29}} \times \frac{{0.25\left( {0.25 + 19.99} \right)}}{{\left( {20.24 \times 4} \right)}} \cr & = \frac{{29 \times 292}}{{365 \times 29}} \times \frac{{0.25 \times 2024}}{{20.24 \times 4}} \cr & = \frac{{292}}{{365}} \times \frac{1}{{16}} \cr & = 0.05 \cr} $$