(a) (0.09)² = 0.0081
(b) 0.09
(c) (1 - 0.9)² = (0.1)² = 0.01
(d) 1 - (0.9)² = 1 - 0.81 = 0.19
Clearly, 0.0081 < 0.01 < 0.09 < 0.19
So, 0.0081 is closest to zero.

Given $$\sqrt {197} $$ × 6.99 + 626.96
≈ $$\sqrt {196} $$ × 7 + 627
= 14 × 7 + 627
= 98 + 627
= 725

Let, the missing number be x,
Then,
$$\frac{{{{\left( {36.54} \right)}^2} - {{\left( {3.46} \right)}^2}}}{x} = 40$$
x = $$\frac{{{{\left( {36.54} \right)}^2} - {{\left( {3.46} \right)}^2}}}{40} $$
x = $$\frac{{{{\left( {36.54} \right)}^2} - {{\left( {3.46} \right)}^2}}}{{36.54 + 3.46}}$$
Let 36.54 = $$a$$ and 3.46 = $$b$$
x = $$\frac{{{a^2} - {b^2}}}{{a + b}}$$
x = (a - b)
x = (36.54 - 3.46)
x = 33.08

Given = 1599 ÷ 39.99 + $$\frac{4}{5}$$ × 2449 - 120.05
= 1600 ÷ 40 + $$\frac{4}{5}$$ × 2450 - 120
= 1600 ÷ 40 + 1960 - 120
= 40 + 1960 - 120
= 1880

$$\frac{144}{0.144}$$ = $$\frac{14.4}{x}$$
⇒ $$\frac{144 × 1000}{144}$$   = $$\frac{14.4}{x}$$
⇒ x = $$\frac{14.4}{1000}$$
⇒ x = 0.0144

$$\eqalign{ & = 0.4\overline {23} \cr & = \frac{{423 - 4}}{{990}} \cr & = \frac{{419}}{{990}} \cr} $$

Given expression :
= 293.48 - 141.27
= 152.21

Each fraction is equivalent to decimal
$$\frac{3}{2}$$ = 1.5
$$\frac{7}{3}$$ = 2.3
$$\frac{5}{4}$$ = 1.25
$$\frac{7}{2}$$ = 3.5
Hence, $$\frac{7}{2}$$ is largest fraction.

Given expression :
$$\eqalign{ & = \frac{{5.32 \times \left( {56 + 44} \right)}}{{\left( {7.66 + 2.34} \right)\left( {7.66 - 2.34} \right)}} \cr & = \frac{{5.32 \times 100}}{{10 \times 5.32}} \cr & = 10 \cr} $$

$$\eqalign{ & = 0.29\overline {56} \cr & = \frac{{2956 - 29}}{{9900}} \cr & = \frac{{2927}}{{9900}} \cr} $$

$$\eqalign{ & \left( {8.3\bar 1 + 0.\bar 6 + 0.00\bar 2} \right) \cr & = 8 + \frac{{31 - 3}}{{90}} + \frac{6}{9} + \frac{2}{{900}} \cr & = \frac{{7200 + 280 + 600 + 2}}{{900}} \cr & = \frac{{8082}}{{900}} \cr & = 8\frac{{882}}{{900}} \cr & = 8 + \frac{{979 - 97}}{{900}} \cr & = 8.97\bar 9 \cr} $$

Given = 1599 ÷ 39.99 + $$\frac{4}{5}$$ × 2449 - 120.05
≈ 1600 ÷ 40 + $$\frac{4}{5}$$ × 2450 - 120
= 1600 ÷ 40 + 1960 - 120
= 40 + 1960 - 120
= 1880

Given expression :
= (11.71 + 1.78) - (0.86 + 9.20)
= 13.49 - 10.06
= 3.43

9 × 7 = 63
Sum of decimal places = 5
∴ 0.09 × 0.007 = 0.00063

Given expression :
71.808 + 95.76
= 167.568

$$\eqalign{ & = 1.\overline {27} \cr & = 1 + 0.\overline {27} \cr & = 1 + \frac{{27}}{{99}} \cr & = 1 + \frac{3}{{11}} \cr & = \frac{{11 + 3}}{{11}} \cr & = \frac{{14}}{{11}} \cr} $$

For the expressions to be equivalent, the difference between the sum of the decimal places in the numerator and that in the denominator must be equal.
This difference is 1 in the given expression and 1 in (D).
So, (D) is the answer.

Given expression :
$$\eqalign{ & \frac{{0.125 + 0.027}}{{0.5 \times 0.5 + 0.09 - 0.15}} \cr & = \frac{{{{\left( {0.5} \right)}^3} + {{\left( {0.3} \right)}^3}}}{{{{\left( {0.5} \right)}^2} + {{\left( {0.3} \right)}^2} - \left( {0.5 \times 0.3} \right)}} \cr & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}}} \right) \cr & = \frac{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}{{\left( {{a^2} - ab + {b^2}} \right)}} \cr & = (a + b) \cr & = \left( {0.5 + 0.3} \right) \cr & = 0.8 \cr} $$

$$\frac{2}{5}$$%
= $$\frac{2}{5}$$ × $$\frac{1}{100}$$
= $$\frac{1}{250}$$

Given expression :
$$\eqalign{ & \left[ {\frac{{8{{\left( {3.75} \right)}^3} + 1}}{{{{\left( {7.5} \right)}^2} - 6.5}}} \right] \cr & = \frac{{{{\left( {2 \times 3.75} \right)}^3} + {1^3}}}{{{{\left( {7.5} \right)}^2} - \left( {7.5 \times 1} \right) + {1^2}}} \cr & = \frac{{{{\left( {7.5} \right)}^3} + {1^3}}}{{{{\left( {7.5} \right)}^2} - \left( {7.5 \times 1} \right) + {1^2}}} \cr & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = \frac{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}{{\left( {{a^2} - ab + {b^2}} \right)}} \cr & = \left( {a + b} \right) \cr & = \left( {7.5 + 1} \right) \cr & = 8.5 \cr} $$

Given expression :
$$\eqalign{ & = 35.7 - \left( {3 + \frac{1}{{\frac{{10}}{3}}}} \right) - \left( {2 + \frac{1}{{\frac{5}{2}}}} \right) \cr & = 35.7 - \left( {3 + \frac{3}{{10}}} \right) - \left( {2 + \frac{2}{5}} \right) \cr & = 35.7 - \frac{{33}}{{10}} - \frac{{12}}{5} \cr & = 35.7 - \left( {\frac{{33}}{{10}} + \frac{{12}}{5}} \right) \cr & = 35.7 - \frac{{57}}{{10}} \cr & = 35.7 - 5.7 \cr & = 30 \cr} $$

Given expression :
$$ = \frac{{{a^2} + {b^2} + {c^2}}}{{{{\left( {\frac{a}{{10}}} \right)}^2} + {{\left( {\frac{b}{{10}}} \right)}^2} + {{\left( {\frac{c}{{10}}} \right)}^2}}}$$
Where a = 0.06, b = 0.47 and c = 0.079
$$\eqalign{ & = \frac{{100\left( {{a^2} + {b^2} + {c^2}} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}} \cr & = 100 \cr} $$

The given expression can be written in this form also
N = $${\text{ 0}}{\text{.39}}\overline {{\text{39}}} $$ ..... (i)
Multiply equation (i) with 100 on both sides.
100N = $${\text{39}}{\text{.}}\overline {{\text{39}}} $$ ..... (ii)
Subtracting equation (i) from (ii) we get,
⇒ 100N - N = $${\text{39}}\overline {{\text{.39}}} $$  - $${\text{0}}\overline {{\text{.39}}} $$
⇒ 99N = 39
⇒ N = $$\frac{39}{99}$$ = $$\frac{13}{33}$$

Given expression :
$$ = \frac{{{{\left( {0.051} \right)}^3} + {{\left( {0.041} \right)}^3}}}{{{{\left( {0.051} \right)}^2} - \left( {0.051 \times 0.041} \right) + {{\left( {0.041} \right)}^2}}}$$
Let 0.051 = $$a$$ and 0.041 = $$b$$
$$\eqalign{ & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = (a + b) \cr & = \left( {0.051 + 0.041} \right) \cr & = 0.092 \cr} $$

Given expression :
$$\eqalign{ & = \frac{{{{\left( {10.3} \right)}^3} + {1^3}}}{{{{\left( {10.3} \right)}^2} - \left( {10.3 \times 1} \right) + {1^2}}} \cr & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = (a + b) \cr & = \left( {10.3 + 1} \right) \cr & = 11.3 \cr} $$

Given expression :
$$ = \frac{{{{\left( {4.53 - 3.07} \right)}^3} + {{\left( {3.07 - 2.15} \right)}^3} + {{\left( {2.15 - 4.53} \right)}^3}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}}$$
Let (4.53 - 3.07) = $$a$$, (3.07 - 2.15) = $$b$$ and (2.15 - 4.53) = $$c$$
$$\eqalign{ & = \frac{{{a^3} + {b^3} + {c^3}}}{{abc}} \cr & = \frac{{3abc}}{{abc}} \cr & = 3 \cr} $$
[∵ If a + b + c = 0, a ³ + b³ + c³ = 3abc]