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