The value of $$\frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \div \frac{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}}{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}}{\text{is:}}$$
A. 0.05
B. 0.5
C. 0.01
D. 0.1
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \div \frac{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}}{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}} \cr & = \frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \times \frac{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}}{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}} \cr & = \frac{{203 \times 292}}{{7 \times 365 \times 29}} \times \frac{{0.25\left( {0.25 + 19.99} \right)}}{{\left( {20.24 \times 4} \right)}} \cr & = \frac{{29 \times 292}}{{365 \times 29}} \times \frac{{0.25 \times 2024}}{{20.24 \times 4}} \cr & = \frac{{292}}{{365}} \times \frac{1}{{16}} \cr & = 0.05 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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