Bissoy.com
Doctors Medicines MCQs Q&A

If $$\sqrt {{\text{4096}}} $$  = 64, then the value of $$\sqrt {{\text{40}}{\text{.96}}} $$   $$ + $$ $$\sqrt {{\text{0}}{\text{.4096}}} $$   $$ + $$ $$\sqrt {{\text{0}}{\text{.004096}}} $$    $$ + $$ $$\sqrt {{\text{0}}{\text{.00004096}}} $$     up to two place of decimals is = ?

If $$\sqrt {{\text{4096}}} $$  = 64, then the value of $$\sqrt {{\text{40}}{\text{.96}}} $$   $$ + $$ $$\sqrt {{\text{0}}{\text{.4096}}} $$   $$ + $$ $$\sqrt {{\text{0}}{\text{.004096}}} $$    $$ + $$ $$\sqrt {{\text{0}}{\text{.00004096}}} $$     up to two place of decimals is = ? Correct Answer 7.11

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {40.96} {\text{ + }}\sqrt {0.4096} {\text{ + }}\sqrt {0.004096} {\text{ + }}\sqrt {0.00004096} \cr & \Rightarrow \sqrt {\frac{{4096}}{{100}}} + \sqrt {\frac{{4096}}{{10000}}} + \sqrt {\frac{{4096}}{{1000000}}} + \sqrt {\frac{{4096}}{{100000000}}} \cr & \Rightarrow \frac{{64}}{{10}} + \frac{{64}}{{100}} + \frac{{64}}{{1000}} + \frac{{64}}{{10000}} \cr & \Rightarrow 6.4 + 0.64 + 0.064 + 0.0064 \cr & \Rightarrow 7.11 \cr} $$

Related Questions

Which of the following is TRUE? \({\rm{I}}.{\rm{\;}}\sqrt[3]{{11}} > \sqrt 7 > \sqrt[4]{{45}}\) \({\rm{II}}.{\rm{\;}}\sqrt 7 > \sqrt[3]{{11}} > \sqrt[4]{{45}}\) \({\rm{III}}.{\rm{\;}}\sqrt 7 > \sqrt[4]{{45}} > \sqrt[3]{{11}}\) \({\rm{IV}}.{\rm{\;}}\sqrt[4]{{45}} > \sqrt 7 > \sqrt[3]{{11}}\)
Which of the following is TRUE? \({\rm{I}}.{\rm{\;}}\frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt 5 }}\) \({\rm{II}}.\;\frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt 5 }}\) \({\rm{III}}.\;\frac{1}{{\sqrt 5 }} > \frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt[4]{{29}}}}\) \({\rm{IV}}.{\rm{\;}}\frac{1}{{\sqrt 5 }} > \frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt[3]{{12}}}}\)
Simplify : $$\frac{1}{{\sqrt 3 + \sqrt 4 }} \,+ $$   $$\frac{1}{{\sqrt 4 + \sqrt 5 }} \,+ $$   $$\frac{1}{{\sqrt 5 + \sqrt 6 }} \,+ $$   $$\frac{1}{{\sqrt 6 + \sqrt 7 }} \,+ $$   $$\frac{1}{{\sqrt 7 + \sqrt 8 }}\, + $$   $$\frac{1}{{\sqrt 8 + \sqrt 9 }} = ?$$
If the value of \(\frac{{3x\sqrt y + 2y\sqrt x }}{{3x\sqrt y - 2y\sqrt x }} - \frac{{3x\sqrt y - 2y\sqrt x }}{{3x\sqrt y + 2y\sqrt x }}\) is same as that of \(\sqrt x\sqrt y\), then which of the following relations between x and y is correct?
Which of the following statement(s) is/are TRUE? I. \(\sqrt{1}+ \sqrt{2} + \sqrt{3} + \sqrt{4}+ \sqrt{5} + \sqrt{6} > 10\) II. \(\sqrt{(10)} + \sqrt{(12)} + \sqrt{(14)} > 3 \sqrt{(12)}\)
$$\frac{1}{{\left( {\sqrt 9 - \sqrt 8 } \right)}} \, - $$   $$\frac{1}{{\left( {\sqrt 8 - \sqrt 7 } \right)}} \, + $$   $$\frac{1}{{\left( {\sqrt 7 - \sqrt 6 } \right)}} \, - $$   $$\frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}} \, + $$   $$\frac{1}{{\left( {\sqrt 5 - \sqrt 4 } \right)}}$$   is equal to ?
If $$a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }},$$   $$b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$$   then the value of $${a^2} + {b^2}$$   would be = ?
If N = $$\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - $$     $$\sqrt {3 - 2\sqrt 2 } {\text{,}}$$   then the value of N is = ?
If $$x = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}$$   and $$y = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}$$   then the value of $$\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = ?$$
If $$a = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$$    and $$b = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}{\text{,}}$$    then the value of $$\frac{{{a^2}}}{b}$$  + $$\frac{{{b^2}}}{a}$$  = ?