4 x 162 = 648. Sum of decimal places = 6.
So,
0.04 x 0.0162
= 0.000648
= 6.48 x 10^-4

$$\eqalign{ & \frac{{144}}{{0.144}} = \frac{{14.4}}{x} \cr & \Rightarrow \frac{{144 \times 1000}}{{144}} = \frac{{14.4}}{x} \cr & \Rightarrow x = \frac{{14.4}}{{1000}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.0144 \cr} $$

2 x 5 = 10.
Sum of decimal places = 4
∴ 0.002 x 0.5 = 0.001

Given expression :
58.621 - (13.829 + 7.302 + 1.214)
= 58.621 - 22.345
= 36.276

Let original fraction be $$\frac{a}{b}$$
Now, according to the question,
$$\eqalign{ & \Leftrightarrow \frac{{a - a \times \frac{{25}}{{100}}}}{{b + b \times \frac{{250}}{{100}}}} = \frac{6}{5} \cr & \Rightarrow \frac{{0.75a}}{{3.50b}} = \frac{6}{5} \cr & \Rightarrow \frac{a}{b} = \frac{6}{5} \times \frac{{3.50}}{{0.75}} \cr & \Rightarrow \frac{a}{b} = \frac{{6 \times 350 \times 100}}{{5 \times 75 \times 100}} \cr & \Rightarrow \frac{a}{b} = \frac{{28}}{5} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & \frac{{52416}}{{312}} = 168 \cr & \Leftrightarrow \frac{{52416}}{{168}} = 312 \cr & {\text{Now,}} \cr & \frac{{52.416}}{{0.0168}} \cr & = \frac{{524160}}{{168}} \cr & = \frac{{52416}}{{168}} \times 10 \cr & = 312 \times 10 \cr & = 3120 \cr} $$

Given expression :
(1.25)³ - (0.75)³ - 3 × (1.25)² × 0.75 + 3 × 1.25 × (0.75)²
= (1.25 - 0.75)³
= (0.5)³
= $${\left( {\frac{1}{2}} \right)^3}$$
= $$\frac{1}{8}$$
[∵ (a - b)³ = a³ - b³ - 3a² b + 3ab²]

383 × 38 × 38 = 553052
Number of decimal place = 1
∴ 383 × 38 × 3.8 = 55305.2

Converting each of the given fractions into decimal form, we get :
$$\frac{2}{3}$$ = 0.66
$$\frac{3}{4}$$ = 0.75
$$\frac{4}{5}$$ = 0.8
$$\frac{5}{6}$$ = 0.833
Since 0.833 > 0.8 > 0.75 > 0.66
So, $$\frac{5}{6}$$ > $$\frac{4}{5}$$ > $$\frac{3}{4}$$ > $$\frac{2}{3}$$
∴ Required difference = $$\left( {\frac{5}{6} - \frac{2}{3}} \right) = \frac{1}{6}$$

324 × 115 × 85 = 3167100
Sum of decimal place = 3
∴ 32.4 × 11.5 × 8.5 = 3167.1

$$\eqalign{ & = \frac{{3.6 \times 0.48 \times 2.50}}{{0.12 \times 0.09 \times 0.5}} \cr & = \frac{{36 \times 48 \times 250}}{{12 \times 9 \times 5}} \cr & = 800 \cr} $$

$$\frac{- 7}{10}$$ = - 0.7
$$\frac{5}{- 8}$$ = - $$\frac{5}{8}$$ = - 0.625,
$$\frac{2}{- 3}$$ = - $$\frac{2}{3}$$ = - 0.66
Since, -0.7 < -0.66 < -0.625
So, $$\frac{- 7}{10}$$ < $$\frac{2}{- 3}$$ < $$\frac{5}{- 8}$$

Given expression :
$$ = \frac{8 - 2.8}{1.3}$$
$$ = \frac{5.2}{1.3}$$
$$ = \frac{52}{13}$$
= 4

Let the missing number be x
$$\eqalign{ & \frac{{21.5}}{5} + \frac{{21}}{6} - \frac{{13.5}}{{15}} = \frac{{{{\left( x \right)}^{\frac{1}{3}}}}}{4} + \frac{{17}}{{30}} \cr & \frac{{21.5}}{5} + \frac{{21}}{6} - \frac{{13.5}}{{15}} - \frac{{17}}{{30}} = \frac{{{{\left( x \right)}^{\frac{1}{3}}}}}{4} \cr} $$
L.C.M of 5, 6, 15 and 30 is 30
$$\eqalign{ & \frac{{129 + 105 - 27 - 17}}{{30}} = \frac{{{{\left( x \right)}^{\frac{1}{3}}}}}{4} \cr & \root 3 \of x = \frac{{190 \times 4}}{{30}} \cr & \root 3 \of x = 25.33 \approx 25 \cr & x = {25^3} \cr & x = 15625 \cr} $$
Hence, the numbers 15625

Converting each of the given fractions into decimal form, we get:
$$\frac{9}{31}$$ = 0.29
$$\frac{3}{17}$$ = 0.176
$$\frac{6}{23}$$ = 0.26
$$\frac{4}{11}$$ = 0.363
$$\frac{7}{25}$$ = 0.28
Clearly, 0.363 > 0.29 > 0.28 > 0.26 > 0.176
So, $$\frac{4}{11}$$ > $$\frac{9}{31}$$ > $$\frac{7}{25}$$ > $$\frac{6}{23}$$ > $$\frac{3}{17}$$

$$\eqalign{ & 0.121212...... \cr & = 0.\overline {12} \cr & = \frac{{12}}{{99}} \cr & = \frac{4}{{33}} \cr} $$

Given expression :
$$\eqalign{ & \left[ {1 + \frac{1}{{1 \times 2}} + \frac{1}{{1 \times 2 \times 4}} + \frac{1}{{1 \times 2 \times 4 \times 8}} + \frac{1}{{1 \times 2 \times 4 \times 8 \times 16}}} \right] \cr & = \frac{{2 \times 4 \times 8 \times 16 + 4 \times 8 \times 16 + 8 \times 16 + 16 + 1}}{{2 \times 4 \times 8 \times 16}} \cr & = \frac{{1024 + 512 + 128 + 16 + 1}}{{1024}} \cr & = \frac{{1681}}{{1024}} \cr & = 1.6416 \cr} $$

Converting each of the given fractions in to decimal form, we get
$$\frac{5}{9}$$ = 0.55
$$\frac{7}{11}$$ = 0.63
$$\frac{8}{15}$$ = 0.533
$$\frac{11}{17}$$ = 0.647
Clearly, 0.647 > 0.63 > 0.55 > 0.533
So, $$\frac{11}{17}$$ > $$\frac{7}{11}$$ > $$\frac{5}{9}$$ > $$\frac{8}{15}$$

Given expression :
$$\eqalign{ & = 3927 + \frac{{5526}}{{12.5}} \cr & = 3927 + \frac{{55260}}{{125}} \cr} $$
= 3927 + 442.08
= 4369.08

$$\eqalign{ & = \frac{{3.157 \times 4126 \times 3.198}}{{63.972 \times 2835.121}} \cr & \approx \frac{{3.2 \times 4126 \times 3.2}}{{64 \times 2835}} \cr & = \frac{{32 \times 4126 \times 32}}{{64 \times 2835}} \times \frac{1}{{100}} \cr & = \frac{{66016}}{{2835}} \times \frac{1}{{100}} \cr & = \frac{{23.28}}{{100}} \cr & = 0.23 \cr & \approx 0.2 \cr} $$

477 × 124 × 86 = 5086728
Sum of decimal place = 3
∴ 47.7 × 12.4 × 8.6 = 5086.728

Given expression :
$$\eqalign{ & = \frac{{{{\left( {0.943} \right)}^2} - \left( {0.943 \times 0.057} \right) + {{\left( {0.057} \right)}^2}}}{{{{\left( {0.943} \right)}^3} + {{\left( {0.057} \right)}^3}}} \cr & = \frac{{{a^2} - ab + {b^2}}}{{{a^3} + {b^3}}} \cr & = \frac{1}{{a + b}} \cr & = \frac{1}{{0.943 + 0.057}} \cr & = 1 \cr} $$

Let, $$\frac{{{x^2} + {{\left( {18} \right)}^2}}}{{125}} = 3.56$$
Then,
$$\eqalign{ & {x^2} + 324 = 125 \times 3.56 = 445 \cr & \Rightarrow {x^2} = 121 \cr & \Rightarrow x = 11 \cr} $$

$$\eqalign{ & 0.125125.... \cr & = 0.\overline {125} \cr & = \frac{{125}}{{999}} \cr} $$

$$\eqalign{ & \frac{2}{3} = 0.666 \cr & \frac{3}{5} = 0.6 \cr & \frac{2}{5} = 0.4 \cr & \frac{1}{3} = 0.333 \cr & \frac{1}{{15}} = 0.066 \cr & \frac{{31}}{{50}} = 0.62 \cr} $$
Clearly, 0.62 lies between 0.6 and 0.666
So, $$\frac{31}{50}$$ lies between $$\frac{2}{3}$$ and $$\frac{3}{5}$$

$$ = 7\frac{1}{2} - $$  $$\left[ {2\frac{1}{4} \div \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {1\frac{1}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right]$$
$$ = \frac{{15}}{2} - $$   $$\left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{3}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right]$$
$$ = \frac{{15}}{2} - $$   $$\left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{{9 - 2 - 1}}{6}} \right)} \right\}} \right]$$
$$\eqalign{ & = \frac{{15}}{2} - \left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2}} \right\}} \right] \cr & = \frac{{15}}{2} - \left[ {\frac{9}{4} \div \left\{ {\frac{{5 - 2}}{4}} \right\}} \right] \cr & = \frac{{15}}{2} - \left[ {\frac{9}{4} \div \frac{3}{4}} \right] \cr & = \frac{{15}}{2} - \left[ {\frac{9}{4} \times \frac{4}{3}} \right] \cr & = \frac{{15}}{2} - 3 \cr & = \frac{{15 - 6}}{2} \cr & = \frac{9}{2} \cr & = 4\frac{1}{2} \cr} $$

Clearly,
$$\eqalign{ & \frac{{11}}{{14}} = 0.785 \cr & \frac{{16}}{{19}} = 0.842 \cr & \frac{{19}}{{21}} = 0.904 \cr} $$
Now, 0.785 < 0.842 < 0.904
So, $$\frac{11}{14}$$ < $$\frac{16}{19}$$ < $$\frac{19}{21}$$

$$\eqalign{ & 0.\overline {142857} \div 0.\overline {285714} \cr & = \frac{{142857}}{{999999}} \div \frac{{285714}}{{999999}} \cr & = \left( {\frac{{142857}}{{999999}} \times \frac{{999999}}{{285714}}} \right) \cr & = \frac{1}{2} \cr} $$

$$\eqalign{ & \text{Let } a = 4.2 \text{ and } b = 1.9 \cr & {\text{Given Expression}} \cr & = \frac{{ {{a^2} - {b^2}} }}{{\left( {a + b} \right)\left( {a - b} \right)}} \cr & = \frac{{ {{a^2} - {b^2}} }}{{ {{a^2} - {b^2}} }} \cr & = 1 \cr} $$

Given expression :
$$\eqalign{ & \frac{{{{\left( {0.013} \right)}^3} + 0.000000343}}{{{{\left( {0.013} \right)}^2} - 0.000091 + 0.000049}} \cr & a = 0.013 \text{ and } b = 0.007 \cr & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = a + b \cr & = 0.013 + 0.007 \cr & = 0.020 \cr} $$