$$\eqalign{ & = \frac{{3\sqrt 2 }}{{2\sqrt 3 }} \cr & = \frac{{3\sqrt 2 }}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{3\sqrt 6 }}{{2 \times 3}} \cr & = \frac{{\sqrt 6 }}{2} \cr & = \frac{{2.449}}{2} \cr & = 1.2245 \cr} $$

$$\eqalign{ & = \frac{{\sqrt 5 }}{2} - \frac{{10}}{{\sqrt 5 }} + \sqrt {125} \cr & = \frac{{{{\left( {\sqrt 5 } \right)}^2} - 20 + 2\sqrt 5 \times 5\sqrt 5 }}{{2\sqrt 5 }} \cr & = \frac{{5 - 20 + 50}}{{2\sqrt 5 }} \cr & = \frac{{35}}{{2\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} \cr & = \frac{{35\sqrt 5 }}{{10}} \cr & = \frac{7}{2} \times 2.236 \cr & = 7 \times 1.118 \cr & = 7.826 \cr} $$

$$21600 = {2^5} \times {3^3} \times {5^2}$$
To make it a perfect cube, it must be multiplied by (2 × 5), i.e., 10

$$3600 = {2^3} \times {5^2} \times {3^2} \times 2$$
To make it a perfect cube, it must be divided by   $${5^2} \times {3^2} \times 2,i.e.,450$$

$$\eqalign{ & = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \cr & = \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \times \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}} \cr & = {\left( {\sqrt 2 - 1} \right)^2} \cr & \therefore \sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \cr & = \left( {\sqrt 2 - 1} \right) \cr & = \left( {1.414 - 1} \right) \cr & = 0.414 \cr} $$

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

$$\eqalign{ & {\text{Let,}} \cr & {\text{ }}99 \times 21 - \root 3 \of x = 1968 \cr & {\text{Then,}} \cr & \Leftrightarrow 2079 - \root 3 \of x = 1968 \cr & \Leftrightarrow \root 3 \of x = 2079 - 1968 \cr & \Leftrightarrow \root 3 \of x = 111 \cr & \Leftrightarrow x = {\left( {111} \right)^3} \cr & \Leftrightarrow x = 1367631 \cr} $$

Let the total number of camels be x
Then,
$$\eqalign{ & \Leftrightarrow x - \left( {\frac{x}{4} + 2\sqrt x } \right) = 15 \cr & \Leftrightarrow \frac{{3x}}{4} - 2\sqrt x = 15 \cr & \Leftrightarrow 3x - 8\sqrt x = 60 \cr & \Leftrightarrow 8\sqrt x = 3x - 60 \cr & \Leftrightarrow 64x = {\left( {3x - 60} \right)^2} \cr & \Leftrightarrow 64x = 9{x^2} + 3600 - 360x \cr & \Leftrightarrow 9{x^2} - 424x + 3600 = 0 \cr & \Leftrightarrow 9{x^2} - 324x - 100x + 3600 = 0 \cr & \Leftrightarrow 9x\left( {x - 36} \right) - 100\left( {x - 36} \right) = 0 \cr & \Leftrightarrow \left( {x - 36} \right)\left( {9x - 100} \right) = 0 \cr & \Leftrightarrow x = 36\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because x \ne \frac{{100}}{9}} \right] \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,9|\overline {99} \,\,\overline {99} \,\,(99 \cr & \,\,\,\,\,\,\,\,\,\,|\,81 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,189|\,18\,99 \cr & \,\,\,\,\,\,\,\,\,\,|\,17\,01 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,1\,98 \cr & \therefore {\text{Required number}} \cr & = \left( {9999 - 198} \right) \cr & = 9801 \cr} $$

Least number of 4 digits is 1000
$$\eqalign{ & \,\,\,\,3|\overline {10} \,\,\overline {00} \,\,(31 \cr & \,\,\,\,\,\,\,|\,\,\,\,9 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & 61|\,\,\,\,\,1\,00 \cr & \,\,\,\,\,\,\,|\,\,\,\,\,\,61 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,39 \cr & \therefore {\left( {31} \right)^2} < 1000 < {\left( {32} \right)^2} \cr & {\text{Hence, required number}} \cr & = {\left( {32} \right)^2} \cr & = 1024 \cr} $$

$$\eqalign{ & a + b + c = 27\sqrt {29} \cr & \Rightarrow 2c + \frac{3}{2}c + c = 27\sqrt {29} \cr & \Rightarrow \frac{9}{2}c = 27\sqrt {29} \cr & \Rightarrow c = 6\sqrt {29} \cr & \therefore \sqrt {{a^2} + {b^2} + {c^2}} \cr & = \sqrt {{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right)} \cr & = \sqrt {{{\left( {27\sqrt {29} } \right)}^2} - 2\left( {2c \times \frac{3}{2}c + \frac{3}{2}c \times c + c \times 2c} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2\left( {3{c^2} + \frac{3}{2}{c^2} + 2{c^2}} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2 \times \frac{{13}}{2}{c^2}} \cr & = \sqrt {\left( {729 \times 29} \right) - 13{{\left( {6\sqrt {29} } \right)}^2}} \cr & = \sqrt {29\left( {729 - 468} \right)} \cr & = \sqrt {29 \times 261} \cr & = \sqrt {29 \times 29 \times 9} \cr & = 29 \times 3 \cr & = 87 \cr} $$

$$\eqalign{ & {\text{Let the missing digit be x}} \cr & \,\,\,\,\,\,\,1|\overline 1 \,\,\overline {53} \,\,\overline {7{\text{x}}} \,\,(124 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,1 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,22|\,\,\,\,\,\,\,53 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,44 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 244\,|\,\,\,\,\,\,\,\,\,\,\,\,\,97{\text{x}} \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,976 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x}} \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & {\text{Then, x}} = 6 \cr} $$

$$\eqalign{ & = \frac{1}{{\sqrt 5 }} \cr & = \frac{1}{{\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} \cr & = \frac{{\sqrt 5 }}{5} \cr & = \frac{{2.236}}{5} \cr & = 0.447 \cr} $$

$$\eqalign{ & = \sqrt {\frac{8}{3}} \cr & = \sqrt {\frac{{8 \times 3}}{{3 \times 3}}} \cr & = \frac{{\sqrt {24} }}{3} \cr & = \frac{{4.899}}{3} \cr & = 1.633 \cr} $$

$$\eqalign{ & = \sqrt {{\text{0}}{\text{.}}\overline {\text{4}} {\text{ }}} \cr & = \sqrt {\frac{4}{9}} \cr & = \frac{2}{3} \cr & = 0.666...... \cr & = 0.\overline 6 \cr} $$

$$\eqalign{ & \,\,\,\,\,4|0\,.\,\overline {20} \,\,\overline {00} \,\,\overline {00} \,\,(.447 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,16 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - - - \cr & \,\,\,84|\,\,\,\,\,\,\,\,\,\,4\,00 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,3\,36 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - - - \cr & \,887|\,\,\,\,\,\,\,\,\,\,\,\,\,\,64\,00 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,62\,09 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - - - \cr & \therefore \sqrt {0.2} = 0.447 \cr} $$

$$\eqalign{ & \,\,\,\,\,3|0\,.\,\overline {12} \,\,\overline {10} \,\,\overline {00} \,\,(0.347 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,9 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - - - \cr & \,\,\,64|\,\,\,\,\,\,\,\,\,\,3\,10 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,2\,56 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - - - \cr & \,687|\,\,\,\,\,\,\,\,\,\,\,\,\,\,54\,00 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,48\,00 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - - - \cr & \therefore \sqrt {0.121} = 0.347 \cr} $$

$$\eqalign{ & = \sqrt {\frac{{0.16}}{{0.4}}} \cr & = \sqrt {\frac{{0.16}}{{0.40}}} \cr & = \sqrt {\frac{{16}}{{40}}} \cr & = \sqrt {\frac{4}{{10}}} \cr & = \sqrt {0.4} \cr & = 0.63 \cr} $$

$$\eqalign{ & {\text{Give}}\,{\text{Expression}} \cr & = \frac{{25}}{{11}} \times \frac{{14}}{5} \times \frac{{11}}{{14}} \cr & = 5 \cr} $$

Sum of decimal places in the numerator and denominator under the radical sign being the same, we remove the decimal.
∴ Given expression,
$$\eqalign{ & = \sqrt {\frac{{0.081 \times 0.484}}{{0.0064 \times 6.25}}} \cr & = \sqrt {\frac{{81 \times 484}}{{64 \times 625}}} \cr & = \frac{{9 \times 22}}{{8 \times 25}} \cr & = 0.99 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & = \sqrt {\frac{{{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}}}{{{{\left( {\frac{{0.03}}{{10}}} \right)}^2} + {{\left( {\frac{{0.21}}{{10}}} \right)}^2} + {{\left( {\frac{{0.065}}{{10}}} \right)}^2}}}} \cr & = \sqrt {\frac{{100\left[ {{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}} \right]}}{{{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}}}} \cr & = \sqrt {100} \cr & = 10 \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,1|\overline 2 \,.\,\,\overline {00} \,\,\overline {00} \,\,\overline {00} \,(1.414 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,24|\,\,1\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,96 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,281\,|\,\,\,\,\,\,\,\,\,400 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,281 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & 2824\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11900 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11296 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \therefore \sqrt 2 = 1.414 \cr} $$

$$\eqalign{ & \Rightarrow 3\sqrt 5 + \sqrt {125} = 17.88 \cr & \Rightarrow {\text{ }}3\sqrt 5 + \sqrt {25 \times 5} = 17.88 \cr & \Rightarrow {\text{ }}3\sqrt 5 + 5\sqrt 5 = 17.88 \cr & \Rightarrow {\text{ }}8\sqrt 5 = 17.88 \cr & \Rightarrow \sqrt 5 = 2.235 \cr & \therefore \sqrt {80} + 6\sqrt 5 \cr & = \sqrt {16 \times 5} + 6\sqrt 5 \cr & = 4\sqrt 5 + 6\sqrt 5 \cr & = 10\sqrt 5 \cr & = \left( {10 \times 2.235} \right) \cr & = 22.35 \cr} $$

$$\eqalign{ & {\text{Let ,}} \cr & \sqrt {\frac{{0.0196}}{x}} = 0.2 \cr & {\text{Then,}} \cr & \Leftrightarrow \frac{{0.0196}}{x} = 0.04 \cr & \Leftrightarrow x = \frac{{0.0196}}{{0.04}} \cr & \Leftrightarrow x = \frac{{1.96}}{4} \cr & \Leftrightarrow x = 0.49 \cr} $$

$$\eqalign{ & \Leftrightarrow 37 + \sqrt {0.0615 + x} = 37.25 \cr & \Leftrightarrow \sqrt {0.0615 + x} = 0.25 \cr & \Leftrightarrow 0.0615 + x = {\left( {0.25} \right)^2} = 0.0625 \cr & \Leftrightarrow x = 0.001 \cr & \Leftrightarrow x = \frac{1}{{{{10}^3}}} \cr & \Leftrightarrow x = {10^{ - 3}} \cr} $$

$$\eqalign{ & = \sqrt {\left( {7 + 3\sqrt 5 } \right)\left( {7 - 3\sqrt 5 } \right)} \cr & = \sqrt {{{\left( 7 \right)}^2} - {{\left( {3\sqrt 5 } \right)}^2}} \cr & = \sqrt {49 - 45} \cr & = \sqrt 4 \cr & = 2 \cr} $$

Given expression,
$$\sqrt {4 \times 2} + 2\sqrt {16 \times 2} - 3\sqrt {64 \times 2} $$       $$ + $$ $$4\sqrt {25 \times 2} $$
$$\eqalign{ & = 2\sqrt 2 + 8\sqrt 2 - 24\sqrt 2 + 20\sqrt 2 \cr & = 6\sqrt 2 \cr & = 6 \times 1.414 \cr & = 8.484 \cr} $$

$$\eqalign{ & {\text{Let}}\sqrt {0.0169 \times x} = 1.3 \cr & {\text{Then}},\,0.0169x = {\left( {1.3} \right)^2} = 1.69 \cr & \Rightarrow x = \frac{{1.69}}{{0.0169}} = 100 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & = \frac{{3\sqrt {12} }}{{2\sqrt {28} }} \times \frac{{\sqrt {98} }}{{2\sqrt {21} }} \cr & = \frac{{3\sqrt {4 \times 3} }}{{2\sqrt {4 \times 7} }} \times \frac{{\sqrt {49 \times 2} }}{{2\sqrt {21} }} \cr & = \frac{{6\sqrt 3 }}{{4\sqrt 7 }} \times \frac{{7\sqrt 2 }}{{2\sqrt {21} }} \cr & = \frac{{21\sqrt 6 }}{{4\sqrt {7 \times 21} }} \cr & = \frac{{21\sqrt 6 }}{{28\sqrt 3 }} \cr & = \frac{3}{4}\sqrt 2 \cr & = \frac{3}{4} \times 1.414 \cr & = 3 \times 0.3535 \cr & = 1.0605 \cr} $$

$$\eqalign{ & \Leftrightarrow \sqrt {\left( {x - 1} \right)\left( {y + 2} \right)} = 7 \cr & \Leftrightarrow \left( {x - 1} \right)\left( {y + 2} \right) = {\left( 7 \right)^2} \cr & \Leftrightarrow \left( {x - 1} \right) = 7{\text{ and}}\left( {y + 2} \right) = 7 \cr & \Leftrightarrow x = 8{\text{ and }}y = 5 \cr} $$