A) 25 : 16
B) 5 : 3
C) 16 : 1
D) 25 : 9
Correct Answer: C
Solution :
\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{5}\] \[\therefore \frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}=\frac{{{(3+5)}^{2}}}{{{(3-5)}^{2}}}=\frac{16}{1}\]
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