| I. \[(b-a)\,\,(a+b)=40%\]of \[60-120%\] of 20 |
| II. \[a<b\] |
A) Only I
B) only II
C) both I and II
D) Neither 1 nor li
Correct Answer: C
Solution :
I. \[(b-a)(b+a)\] \[=\left( \frac{40}{100}\times 60 \right)-\left( \frac{120}{100}\times 20 \right)=24-24=0\] \[\Rightarrow \]\[{{b}^{2}}-{{a}^{2}}=0\Rightarrow {{b}^{2}}={{a}^{2}}\] \[\Rightarrow \]\[\frac{{{a}^{2}}}{{{b}^{2}}}=1.\]So, \[{{a}^{2}}:{{b}^{2}}\measuredangle 1.\] II. \[a<b\Rightarrow {{a}^{2}}<{{b}^{2}}\] \[\Rightarrow \]\[\frac{{{a}^{2}}}{{{b}^{2}}}<1\Rightarrow {{a}^{2}}:{{b}^{2}}<1.\]
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