| \[\sqrt{{{13}^{2}}+28\div 4-{{(3)}^{2}}+107}={{(?)}^{2}}\] [IBPS (SO) 2012] |
A) 2
B) 16
C) 256
D) 4
E) \[{{(256)}^{2}}\]
Correct Answer: D
Solution :
| \[{{(?)}^{2}}=\sqrt{{{13}^{2}}+28\div 4-{{(3)}^{3}}+107}\] |
| \[\Rightarrow \]\[?=\sqrt{169+7-27+107}=\sqrt{256}=16\] |
| \[\Rightarrow \]\[?=\sqrt{16}=4\] |
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