The value of $$\frac{1}{4} + \frac{{\left[ {{{\left( {20.35} \right)}^2} - {{\left( {8.35} \right)}^2}} \right] \times 0.0175}}{{{{\left( {1.05} \right)}^2} + \left( {1.05} \right)\left( {27.65} \right)}}{\text{is:}}$$
A. $$\frac{7}{{20}}$$
B. $$\frac{3}{{20}}$$
C. $$\frac{3}{{10}}$$
D. $$\frac{9}{{20}}$$
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \frac{1}{4} + \frac{{\left[ {{{\left( {20.35} \right)}^2} - {{\left( {8.35} \right)}^2}} \right] \times 0.0175}}{{{{\left( {1.05} \right)}^2} + \left( {1.05} \right)\left( {27.65} \right)}} \cr & = \frac{1}{4} + \frac{{\left[ {\left( {20.35 + 8.35} \right)\left( {20.35 - 8.35} \right)} \right] \times 0.0175}}{{\left( {1.05} \right) + \left[ {1.05 + 27.65} \right]}} \cr & = \frac{1}{4} + \frac{{\left[ {\left( {28.70} \right)\left( {12} \right) \times 0.0175} \right]}}{{\left( {1.05} \right)\left[ {28.70} \right]}} \cr & = \frac{1}{4} + \frac{{12 \times 5}}{{300}} \cr & = \frac{1}{4} + \frac{1}{5} \cr & = \frac{9}{{20}} \cr} $$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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