Answer: Option D. -> 3600
L.C.M. of 3, 4, 5, 6, 8 is 120
Now 120 = 2 × 2 × 2 × 3 × 5
To make it a perfect square, it must be multiplied by 2 × 3 × 5
So, required number
$$\eqalign{
& = {2^2} \times {2^2} \times {3^2} \times {5^2} \cr
& = 3600 \cr} $$

Answer: Option D. -> 13
$$\eqalign{
& {\text{Clearly, }}a*b = \sqrt {{a^2} + {b^2}} \cr
& \therefore 5*12 \cr
& = \sqrt {{5^2} + {{12}^2}} \cr
& = \sqrt {25 + 144} \cr
& = \sqrt {169} \cr
& = 13 \cr} $$

Answer: Option C. -> 1.9841
$$\eqalign{
& \Rightarrow \frac{1}{{\sqrt 5 - \sqrt 3 }} \cr
& = \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)}} \cr
& = \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} \cr
& = \frac{{\left( {2.2361 + 1.7321} \right)}}{2} \cr
& = \frac{{3.9682}}{2} \cr
& = 1.9841{\text{ }} \cr} $$

Answer: Option B. -> 1.732
$$\eqalign{
& {\text{Given expression,}} \cr
& = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 2\sqrt {12} - \sqrt {32} + \sqrt {50} }} \cr
& = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 4\sqrt 3 - 4\sqrt 2 + 5\sqrt 2 }} \cr
& = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \cr
& = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \times \frac{{\left( {\sqrt 3 - \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)}} \cr
& = \frac{{3\sqrt 3 - 3\sqrt 2 + 3\sqrt 2 - 2\sqrt 3 }}{{\left( {3 - 2} \right)}} \cr
& = \sqrt 3 \cr
& = 1.732 \cr} $$

Answer: Option D. -> 5
Given expression,
$$ = \frac{1}{{\left( {\sqrt 9 - \sqrt 8 } \right)}} \times \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {\sqrt 9 + \sqrt 8 } \right)}}$$     $$ - \frac{1}{{\left( {\sqrt 8 - \sqrt 7 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {\sqrt 8 + \sqrt 7 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 7 - \sqrt 6 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {\sqrt 7 + \sqrt 6 } \right)}}$$   $$ - \frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 + \sqrt 5 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 5 - \sqrt 4 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {\sqrt 5 + \sqrt 4 } \right)}}$$
$$ = \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {9 - 8} \right)}} - \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {8 - 7} \right)}}$$     $$ + \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {7 - 6} \right)}}$$   $$ - \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {6 - 5} \right)}}$$   $$ + \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {5 - 4} \right)}}$$
$$ = \left( {\sqrt 9 + \sqrt 8 } \right) - \left( {\sqrt 8 + \sqrt 7 } \right)$$     $$ + \left( {\sqrt 7 + \sqrt 6 } \right)$$   $$ - \left( {\sqrt 6 + \sqrt 5 } \right)$$   $$ + \left( {\sqrt 5 + \sqrt 4 } \right)$$
$$ = \left( {\sqrt 9 + \sqrt 4 } \right)$$
$$ = 3 + 2$$
$$ = 5$$

Answer: Option B. -> 0.414
$$\eqalign{
& = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \cr
& = \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \times \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}} \cr
& = {\left( {\sqrt 2 - 1} \right)^2} \cr
& \therefore \sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \cr
& = \left( {\sqrt 2 - 1} \right) \cr
& = \left( {1.414 - 1} \right) \cr
& = 0.414 \cr} $$

Answer: Option B. -> 10
Given expressing,
$$ = \frac{1}{{\sqrt 1 + \sqrt 2 }}{\text{ + }}\frac{1}{{\sqrt 2 + \sqrt 3 }}$$     $$ + \frac{1}{{\sqrt 3 + \sqrt 4 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {120} + \sqrt {121} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }}{\text{ + }}\frac{1}{{\sqrt 3 + \sqrt 2 }}$$     $$ + \frac{1}{{\sqrt 4 + \sqrt 3 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }} \times $$   $$\frac{{\sqrt 2 - \sqrt 1 }}{{\sqrt 2 - \sqrt 1 }}{\text{ + }}$$   $$\frac{1}{{\sqrt 3 + \sqrt 2 }} \times $$   $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} + $$   $$\frac{1}{{\sqrt 4 + \sqrt 3 }} \times $$   $$\frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} + $$   $$...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }} \times $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{\sqrt {121} - \sqrt {120} }}$$
$$ = \frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}$$     $$ + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}}$$   $$ + ...... + $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{121 - 120}}$$
$$ = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 $$     $$ + \sqrt 4 - \sqrt 3 $$   $$ + ...... + $$   $$\sqrt {121} - \sqrt {120} $$
$$ = - 1 + \sqrt {121} $$
$$ = - 1 + 11$$
$$ = 10$$

Answer: Option A. -> $$16 - \sqrt 3 $$
Given expression,
$$ = \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}}$$     $$ + \frac{{\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}}$$   $$ \times \frac{{\left( {2 - \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)}}$$   $$ + \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}}$$   $$ \times \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}}$$
$$ = \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{\left( {4 - 3} \right)}} + \frac{{{{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {4 - 3} \right)}}$$     $$ + \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {3 - 1} \right)}}$$
$$ = \left[ {{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}} \right]$$     $$ + \frac{{4 - 2\sqrt 3 }}{2}$$
$$ = 2\left( {4 + 3} \right) + 2 - \sqrt 3 $$
$$ = 16 - \sqrt 3 $$

Answer: Option B. -> 98
$$\eqalign{
& \because a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr
& = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr
& = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr
& = \frac{{3 + 2 + 2\sqrt 6 }}{{3 - 2}} \cr
& = 5 + 2\sqrt 6 \cr
& \because b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr
& {\text{ = }}\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr
& {\text{ = }}\frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr
& = \frac{{3 + 2 - 2\sqrt 6 }}{{3 - 2}} \cr
& = 5 - 2\sqrt 6 \cr
& \therefore {\text{ }}{a^2} + {b^2} \cr
& = {\left( {5 + 2\sqrt 6 } \right)^2} + {\left( {5 - 2\sqrt 6 } \right)^2} \cr
& = 2\left[ {{{\left( 5 \right)}^2} + {{\left( {2\sqrt 6 } \right)}^2}} \right] \cr
& = 2\left( {25 + 24} \right) \cr
& = 2 \times 49 \cr
& = 98 \cr} $$

Answer: Option B. -> 34
$$\eqalign{
& \because x = 3 + \sqrt 8 \cr
& \Rightarrow {x^2} = {\left( {3 + \sqrt 8 } \right)^2} \cr
& \Rightarrow {x^2} = {3^2} + {\left( {\sqrt 8 } \right)^2} + 2 \times 3 \times \sqrt 8 \cr
& \Rightarrow {x^2} = 9 + 8 + 6\sqrt 8 \cr
& \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr} $$
$$\therefore {x^2} + \frac{1}{{{x^2}}}$$
$$ = \left( {17 + 12\sqrt 2 } \right)$$   $$ + \frac{1}{{\left( {17 + 12\sqrt 2 } \right)}}$$   $$ \times \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{\left( {17 - 12\sqrt 2 } \right)}}$$
$$\eqalign{
& = \left( {17 + 12\sqrt 2 } \right) + \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{289 - 288}} \cr
& = \left( {17 + 12\sqrt 2 } \right) + \left( {17 - 12\sqrt 2 } \right) \cr
& = 17 + 17 \cr
& = 34 \cr} $$