A) 10
B) 100
C) 1000
D) None of these
Correct Answer: B
Solution :
Let a = 0.05, b = 0.4 and c = 0.073 \[\therefore \]\[\frac{{{(0.05)}^{2}}+{{(0.41)}^{2}}+{{(0.073)}^{2}}}{{{(0.005)}^{2}}+{{(0.041)}^{2}}+{{(0.0073)}^{2}}}\] \[=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{{{\left( \frac{a}{10} \right)}^{2}}{{\left( \frac{b}{10} \right)}^{2}}+{{\left( \frac{c}{10} \right)}^{2}}}=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{\frac{1}{100}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\] \[\frac{({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\times 100}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=100\]You need to login to perform this action.
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