A) \[1\frac{7}{12}\]
B) \[1\frac{1}{12}\]
C) \[5\frac{5}{12}\]
D) \[7\frac{1}{12}\]
Correct Answer: C
Solution :
| [c] Given expression \[=\frac{({{a}^{4}}-{{b}^{4}})}{({{a}^{2}}-{{b}^{2}})},\] where \[a=3\frac{1}{4}=\frac{13}{4}\] and \[b=4\frac{1}{3}=\frac{13}{3}\] \[=({{a}^{2}}+{{b}^{2}})=\left( \frac{169}{16}+\frac{169}{9} \right)\] \[=169\times \left( \frac{1}{16}+\frac{1}{9} \right)=\frac{169\times 25}{144}\] \[\therefore \] Required square root \[=\sqrt{\frac{169\times 25}{144}}=\frac{13\times 5}{12}\] \[=\frac{65}{12}=5\frac{5}{12}\] |
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