Solution:
According to the question,
$$\eqalign{ \frac{{25 - 5\left[ {2 + 3\left\{ {2 - 2\left( {5 - 3} \right) + 5} \right\} - 10} \right]}}{4} \cr = \frac{{25 - 5\left[ {2 + 3\left\{ {2 - 2 * 2 + 5} \right\} - 10} \right]}}{4} \cr = \frac{{25 - 5\left[ {2 + 3\left\{ {2 - 4 + 5} \right\} - 10} \right]}}{4} \cr = \frac{{25 - 5\left[ {2 + 3 * 3 - 10} \right]}}{4} \cr = \frac{{25 - 5\left[ {11 - 10} \right]}}{4} \cr = 25 - \frac{5}{4} \cr = \frac{{100 - 5}}{4} \cr = \frac{{95}}{4} \cr = 23.75 \cr} $$

Solution:
[7 + 7 × (7 + 7 ÷ 7)] + 7 ÷ 7
= [7 + 7 × (7 + 1)] + 1
= [7 + 56] + 1
= 64

Solution:
$$\eqalign{ \frac{a}{{b + c}} = k \cr \Rightarrow a = k\left( {b + c} \right)\,....(i) \cr \frac{b}{{c + a}} = k \cr \Rightarrow b = k\left( {c + a} \right)\,....(ii) \cr \frac{c}{{a + b}} = k \cr \Rightarrow c = k\left( {a + b} \right)\,....(iii) \cr {\text{Adding (i), (ii) and (iii) we get,}} \cr a + b + c \cr = k\left( {2a + 2b + 2c} \right) \cr \Rightarrow k = \frac{{a + b + c}}{{2(a + b + c)}} \cr \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$

Solution:
$$\eqalign{ \,a = \left( {4 \div 3} \right) \div {\text{3}} \div {\text{4}} \cr \,\,\,\,\,\,{\text{ = }}\frac{4}{3} \times \frac{1}{3} \times \frac{1}{4} \cr \,\,\,\,\,\, = \frac{1}{9} \cr b = 4 \div \left( {3 \div 3} \right) \div {\text{4}} \cr \,\,\,\,\,{\text{ = 4}} \div 1 \div 4 \cr \,\,\,\,\, = 1 \cr c = 4 \div 3 \div \left( {3 \div 4} \right) \cr \,\,\,\,\, = 4 \div 3 \div \frac{3}{4} \cr \,\,\,\,\, = \frac{4}{3} \times \frac{4}{3} \cr \,\,\,\,\, = \frac{{16}}{9} \cr {\text{Clearly, c is the greatest}}{\text{.}} \cr} $$

Solution:
Clearly, $$\frac{{x + 1}}{x}$$ is the only fraction in which the numerator is greater then the denominator. So, it is greatest fraction.

Solution:
$$\eqalign{ {\text{Given, }} \cr x + y = 2a \cr \therefore \frac{a}{{x - a}} + \frac{a}{{y - a}} \cr = \frac{{a\left( {y - a} \right) + a\left( {x - a} \right)}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr = \frac{{ay - {a^2} + ax - {a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr = \frac{{a\left( {x + y} \right) - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr = \frac{{a.2a - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr = \frac{{2{a^2} - 2{a^2}}}{{\left( {x - a} \right)\left( {y - a} \right)}} \cr = 0 \cr} $$

Solution:
$$\eqalign{ {\text{According to question,}} \cr \,\,\,\left[ {4 - \frac{5}{{1 + \frac{1}{{3 + \frac{1}{{2 + \frac{1}{4}}}}}}}} \right] \cr = 4 - \frac{5}{{1 + \frac{1}{{3 + \frac{1}{{\frac{9}{4}}}}}}} \cr = 4 - \frac{5}{{1 + \frac{1}{{3 + \frac{4}{9}}}}} \cr = 4 - \frac{5}{{1 + \frac{1}{{\frac{{31}}{9}}}}} \cr = 4 - \frac{5}{{1 + \frac{9}{{31}}}} \cr = 4 - \frac{5}{{\frac{{40}}{{31}}}} \cr = 4 - \frac{{5 \times 31}}{{40}} \cr = 4 - \frac{{31}}{8} \cr = \frac{1}{8} \cr {\text{According to question}} \cr \frac{1}{8}{\text{part = 10 minutes}} \cr {\text{1 part = 10 minutes}} \cr \frac{3}{5}{\text{ part = 80}} \times \frac{3}{5} \cr {\text{ = 48 minutes}} \cr} $$