Simplification

$${\text{Given equation is :}}$$
$$\frac{{\left( {5x + 1} \right)}}{x} \times \frac{{\left( {4y + 3} \right)}}{4} = 20$$
$$ ⇒ \left( {5x + 1} \right)\left( {4y + 3} \right) = 80x$$
$${\text{Clearly, }}x = 3{\text{ and}}$$
$${\text{ }}y = 3{\text{ satisfy}}$$

$${\text{According to question,}}$$
$$\sqrt {248 + \sqrt {52 + \sqrt {144} } } $$
$$ ⇒ \sqrt {248 + \sqrt {52 + 12} } $$
$$ ⇒ \sqrt {248 + \sqrt {64} } $$
$$ ⇒ \sqrt {256} $$
$$ ⇒ \pm 16$$

$${\text{According to question,}}$$
$$\sqrt {40.96} {\text{ + }}\sqrt {0.4096} {\text{ + }}\sqrt {0.004096} {\text{ + }}\sqrt {0.00004096} $$
$$ ⇒ \sqrt {\frac{{4096}}{{100}}} + \sqrt {\frac{{4096}}{{10000}}} + \sqrt {\frac{{4096}}{{1000000}}} + \sqrt {\frac{{4096}}{{100000000}}} $$
$$ ⇒ \frac{{64}}{{10}} + \frac{{64}}{{100}} + \frac{{64}}{{1000}} + \frac{{64}}{{10000}}$$
$$ ⇒ 6.4 + 0.64 + 0.064 + 0.0064$$
$$ ⇒ 7.11$$

$${\text{Required differnce}}$$
$${\text{ = }}\frac{{19}}{{16}} - \frac{{16}}{{19}} = \frac{{{{19}^2} - {{16}^2}}}{{304}}$$
$$ = \frac{{\left( {19 + 16} \right)\left( {19 - 16} \right)}}{{304}}$$
$$ = \frac{{35 \times 3}}{{304}}$$
$$ = \frac{{105}}{{304}}$$

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

$${\text{Let}}\,C's\,{\text{share}} = ₹\,x$$
$$\text{Then,}\,B's\,{\text{share}} = \,₹\,\frac{x}{4},$$
$$A's\,{\text{share}} = ₹\,\left( {\frac{2}{3} \times \frac{x}{4}} \right) = ₹\,\frac{x}{6}$$
$$\therefore \frac{x}{6} + \frac{x}{4} +×= 1360$$
$$ ⇒ \frac{{17x}}{{12}} = 1360$$
$$ ⇒ x = \frac{{1360 \times 12}}{{17}} = ₹\,960$$
$${\text{Hence}},\,B's\,{\text{share}} = ₹\,\left( {\frac{{960}}{4}} \right)$$
$$ = ₹\,240$$

$${\text{Let}}\,{\text{saving}}\,{\text{in}}\,{\text{N}}{\text{.S}}{\text{.C}}\,{\text{and}}\,{\text{P}}{\text{.P}}{\text{.F}}{\text{.}}\,{\text{be}}\,$$
$$₹\,x\,{\text{and}}\,₹\,\left( {150000 - x} \right){\text{respectively}}.$$
$${\text{Then}},\,\frac{1}{3}x = \frac{1}{2}\left( {150000 - x} \right)$$
$$ ⇒ \frac{x}{3} + \frac{x}{2} = 75000$$
$$ ⇒ \frac{{5x}}{6} = 75000$$
$$ ⇒ x = \frac{{75000 \times 6}}{5} = 90000$$
$$\therefore {\text{Savings}}\,{\text{in}}\,{\text{Public}}\,{\text{Provident}}\,{\text{Fund}}$$
$$ = ₹\,\left( {150000 - 90000} \right)$$
$$ = ₹\,60000$$

$${\text{Let}}\,{\text{the}}\,{\text{total}}\,{\text{number}}\,{\text{of}}\,{\text{shots}}\,{\text{be}}\,x.$$
$${\text{Then,}}\,{\text{Shots}}\,{\text{fired}}\,{\text{by}}\,{\text{A}} = \frac{5}{8}x$$
$${\text{Shots}}\,{\text{fired}}\,{\text{by}}\,{\text{B}} = \frac{3}{8}x$$
$${\text{Killing}}\,{\text{shots}}\,{\text{by}}\,{\text{A}}$$
$$ = \frac{1}{3}{\text{of}}\frac{5}{8}x = \frac{5}{{24}}x$$
$${\text{Shots}}\,{\text{missed}}\,{\text{by}}\,{\text{B}}$$
$$ = \frac{1}{2}{\text{of}}\,\frac{3}{8}x = \frac{3}{{16}}x$$
$$\therefore \frac{{3x}}{{16}} = 27\,{\text{or}}\,x = {\frac{{27 \times 16}}{3}} = 144$$
$${\text{Birds}}\,{\text{killed}}\,{\text{by}}\,{\text{A}}$$
$$ = \frac{{5x}}{{24}} = {\frac{5}{{24}} \times 144} = 30$$

$${\text{Original}}\,{\text{share}}\,{\text{of}}\,{\text{1}}\,{\text{person}} = \frac{1}{8}$$
$${\text{New}}\,{\text{share}}\,{\text{of}}\,{\text{1}}\,{\text{person}} = \frac{1}{7}$$
$${\text{Increase}} = {\frac{1}{7} \times \frac{1}{8}} = \frac{1}{{56}}$$
$$\therefore {\text{Required}}\,{\text{fraction}}$$
$$ = \frac{{ {1/56} }}{{ {1/8} }} = {\frac{1}{{56}} \times \frac{8}{1}} = \frac{1}{7}$$

$${\text{Let}}\,{\text{the}}\,{\text{capacity}}\,{\text{of}}\,{\text{1}}\,{\text{bucket}} =× $$
$$ {\text{Then,}}\,{\text{the}}\,{\text{capacity}}\,{\text{of}}\,{\text{tank}} = 25x$$
$${\text{New}}\,{\text{capacity}}\,{\text{of}}\,{\text{bucket}} = \frac{2}{5}x$$
$$\therefore {\text{Required}}\,{\text{number}}\,{\text{of}}\,{\text{buckets}}$$
$$ = \frac{{25x}}{{\left( {2x/5} \right)}}$$
$$ = {25x \times \frac{5}{{2x}}} $$
$$ = \frac{{125}}{2}$$
$$ = 62.5$$