$$\frac{4}{5}$$ = 0.8
$$\frac{7}{13}$$ = 0.53
$$\frac{1}{2}$$ = 0.5
$$\frac{2}{3}$$ = 0.66
$$\frac{3}{4}$$ = 0.75
$$\frac{5}{7}$$ = 0.714
Clearly, 0.5 does not lie between 0.53 and 0.8
∴ $$\frac{1}{2}$$ does not lie between $$\frac{4}{5}$$ and $$\frac{7}{13}$$

Given, 168 × 32 = 5376 or 5376 ÷ 168 = 32
Now,
$$\eqalign{ & = \frac{{5.376}}{{16.8}} \cr & = \frac{{53.76}}{{168}} \cr & = \left( {\frac{{5376}}{{168}} \times \frac{1}{{100}}} \right) \cr & = \frac{{32}}{{100}} \cr & = 0.32 \cr} $$

Converting each of the given fractions into decimal form, we get :
$$\frac{2}{3}$$ = 0.66
$$\frac{3}{5}$$ = 0.6
$$\frac{7}{9}$$ = 0.77
$$\frac{9}{11}$$ = 0.81
$$\frac{8}{9}$$ = 0.88
Clearly, 0.6 < 0.66 < 0.77 < 0.81 < 0.88
So, $$\frac{3}{5}$$ < $$\frac{2}{3}$$ < $$\frac{7}{9}$$ < $$\frac{9}{11}$$ < $$\frac{8}{9}$$

Let 534.596 + 61.472 - 496.708 = x + 27.271
Then, x = (534.596 + 61.472) - (496.708 + 27.271)
= 596.068 - 523.979
= 72.089

$$\eqalign{ & {\left( {\frac{{18}}{4}} \right)^2} \times \left( {\frac{{455}}{{19}}} \right) \div \left( {\frac{{61}}{{799}}} \right) \cr & = \frac{{324}}{{16}} \times \frac{{455}}{{19}} \times \frac{{799}}{{61}} \cr & = 6350 \cr} $$

Let the missing number be x
$$\eqalign{ & \frac{{294 \div 14 \times 5 + 11}}{x} = {8^2} \div 5 + 1.7 \cr & \Rightarrow \frac{{\frac{{294}}{{14}} \times 5 + 11}}{x} = \frac{{64}}{5} + 1.7 \cr & \Rightarrow \frac{{21 \times 5 + 11}}{x} = 12.8 + 1.7 \cr & \Rightarrow \frac{{105 + 11}}{x} = 12.8 + 1.7 \cr & \Rightarrow \frac{{116}}{x} = 14.5 \cr & \Rightarrow x = \frac{{116}}{{14.5}} \cr & \Rightarrow x = \frac{{116 \times 10}}{{145}} \cr & \Rightarrow x = 8 \cr} $$
Hence, the number is 8

Let the third number be 2960
∵ Second number = $$\frac{1}{4}$$ of the third number = $$\frac{1}{4}$$ × 2960 = 740
$$\frac{5}{9}$$ of first number = 25% of second number
$$\frac{5}{9}$$ first number = $$\frac{25 × 740}{100}$$  = 185
⇒ First number = $$\frac{185 × 9}{5}$$  = 333
∴ 30% of 333 = $$\frac{30}{100}$$ × 333 = 99.9

Given expression
= $$\frac{35 × 1.4}{7}$$
= 5 × 1.4
= 7

$$\eqalign{ & = \frac{{4.036}}{{0.04}} \cr & = \frac{{403.6}}{4} \cr & = 100.9 \cr} $$

9 is at thousand^ths place in 0.06945
So, place value of 9 = $$\frac{9}{1000}$$

$$\eqalign{ & \frac{1}{3} = 0.333 \cr & \frac{3}{4} = 0.75 \cr & \frac{{117}}{{300}} = 0.39 \cr & \frac{{287}}{{400}} = 0.7175 \cr & \frac{{95}}{{300}} = 0.316 \cr & \frac{{301}}{{400}} = 0.7525 \cr & \frac{{99}}{{300}} = 0.33 \cr & \frac{{97}}{{300}} = 0.323 \cr & \frac{{299}}{{500}} = 0.598 \cr} $$
Clearly, each one of 0.39 and 0.7175 lies between 0.333 and 0.75
So, $$\frac{117}{300}$$ and $$\frac{287}{400}$$ lie between $$\frac{1}{3}$$ and $$\frac{3}{4}$$

Given expression :
$$\eqalign{ & = \frac{{{{\left( {0.1} \right)}^3} + {{\left( {0.02} \right)}^3}}}{{{2^3}\left[ {{{\left( {0.1} \right)}^3} + {{\left( {0.02} \right)}^3}} \right]}} \cr & = \frac{1}{8} \cr & = 0.125 \cr} $$

$$\eqalign{ & 617.00 \cr & \,\,\,\,\,\,\,6.017 \cr & \,\,\,\,\,\,\,0.617 \cr & + \,\,\,6.0017 \cr & - - - - - - \cr & \,\,\,629.6357 \cr & - - - - - - \cr} $$

Let 3889 + 12.952 - x = 3854.002.
Then x = (3889 + 12.952) - 3854.002
    = 3901.952 - 3854.002
    = 47.95

Sum of decimal places = 7.
Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal point.

$$\therefore \frac{{1999}}{{2111}} = 0.946$$

$$\eqalign{ & {\text{Given Expression}} \cr & = \frac{{8 - 2.8}}{{1.3}} \cr & = \frac{{5.2}}{{1.3}} \cr & = \frac{{52}}{{13}} \cr & = 4 \cr} $$

$$\eqalign{ & 6.\overline {46} \cr & = 6 + 0.\overline {46} \cr & = 6 + \frac{{46}}{{99}} \cr & = \frac{{594 + 46}}{{99}} \cr & = \frac{{640}}{{99}} \cr} $$

$$\eqalign{ & 101\frac{{27}}{{100000}} \cr & = 101 + \frac{{27}}{{100000}} \cr & = 101 + .00027 \cr & = 101.00027 \cr} $$

$$\eqalign{ & \frac{{0.0203 \times 2.92}}{{0.0073 \times 14.5 \times 0.7}} \cr & = \frac{{203 \times 292}}{{73 \times 145 \times 7}} \cr & = \frac{4}{5} \cr & = 0.8 \cr} $$

$$\eqalign{ & \,\,\,\,34.95 \cr & \,240.016 \cr & + 23.98 \cr & - - - - - \cr & 298.946 \cr & - - - - - \cr} $$

3.14 x 10⁶
= 3.14 x 1000000
= 3140000

$$\eqalign{ & \frac{3}{4} = 0.75,{\kern 1pt} \cr & \frac{5}{6} = 0.833,{\kern 1pt} \cr & \frac{1}{2} = 0.5, \cr & \frac{2}{3} = 0.66, \cr & {\kern 1pt} \frac{4}{5} = 0.8, \cr & \frac{9}{{10}} = 0.9 \cr} $$
Clearly, 0.8 lies between 0.75 and 0.833
$$\therefore \frac{4}{5}{\text{lies between}}\frac{3}{4}{\text{and}}\frac{5}{6}$$

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

$$0.232323... = 0.\overline {23} = \frac{{23}}{{99}}$$

$$\eqalign{ & {\text{Let}}\,\frac{{.009}}{x} = .01; \cr & {\text{Then}}\,x = \frac{{.009}}{{.01}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{.9}}{1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = .9 \cr} $$

$$\eqalign{ & \frac{{4.036}}{{0.04}} \cr & = \frac{{403.6}}{4} \cr & = 100.9 \cr} $$

$$\eqalign{ & 3.\overline {87} - 2.\overline {59} \cr & = \left( {3 + 0.\overline {87} } \right) - \left( {2 + 0.\overline {59} } \right) \cr & = \left( {3 + \frac{{87}}{{99}}} \right) - \left( {2 + \frac{{59}}{{99}}} \right) \cr & = 1 + \left( {\frac{{87}}{{99}} - \frac{{59}}{{99}}} \right) \cr & = 1 + \frac{{28}}{{99}} \cr & = 1.\overline {28} \cr} $$

$$\eqalign{ & 1 \div 0.2 = \frac{1}{{0.2}} = \frac{{10}}{2} = 5; \cr & 0.\overline 2 = 0.222...; \cr & {\left( {0.2} \right)^2} = 0.04 \cr & 0.04 < 0.2 < 0.22.... < 5 \cr & {\text{Since}}\,0.04\,{\text{is}}\,{\text{the}}\,{\text{least,}}\,{\text{so}}\,{\left( {0.2} \right)^2}\,{\text{is}}\,{\text{the}}\,{\text{least}}. \cr} $$