A) 9
B) 16
C) 25
D) None of these
Correct Answer: B
Solution :
Let the required number is x So, \[x=\sqrt{x}+12\,\,\,\Rightarrow x-12=\sqrt{x}\] Þ \[{{x}^{2}}-25x+144=0\] Þ \[{{x}^{2}}-16x-9x+144=0\]\[\Rightarrow x=16\] Since \[x=9\] does not hold the condition. Trick: By inspection, since 16 exceeds its positive square root i.e.,4 by 12.
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