Question: $ \sqrt{13^{2}+28\div 4-{{(3)}^{2}}+107}={{(?)}^{2}} $ [IBPS (SO) 2012]
Options:
A) 2
B) 16
C) 256
D) 4
E) $ {{(256)}^{2}} $
Show Answer
Answer:
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 $