$$\eqalign{ & \left( {{\text{A + B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{6} \cr & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{8} \cr & \left( {{\text{B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{12}} \cr & \refore \left( {{\text{A + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \left( {2 \times \frac{1}{6}} \right) - \left( {\frac{1}{8} + \frac{1}{{12}}} \right) \cr & = {\frac{1}{3} - \frac{5}{{24}}} \cr & = \frac{3}{{24}} \cr & = \frac{1}{8} \cr }$$
So, A and C toger will do work in 8 days

Number of pages typed by Ronald in 1 hour
$$\eqalign{ & = \frac{{32}}{6} \cr & = \frac{{16}}{3} \cr} $$
Number of pages typed by Elan in 1 hour
$$\eqalign{ & = \frac{{40}}{5} \cr & = 8 \cr} $$
Number of pages typed by both in 1 hour
$$\eqalign{ & = \left( {\frac{{16}}{3} + 8} \right) \cr & = \frac{{40}}{3} \cr} $$
∴Time taken by both to type 110 pages
$$\eqalign{ & {\text{ = }}\left( {100 \times \frac{3}{{40}}} \right){\text{hours}} \cr & = 8\frac{1}{4}{\text{hours}} \cr & = {\text{8 hours 15 minutes}} \cr} $$

$$\eqalign{ & \frac{3}{{10}}\left( {{\text{B}} + {\text{C }}} \right) = 2{\text{ days}} \cr & \left( {{\text{B}} + {\text{C }}} \right) = 2 \times \frac{{10}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{3}{\text{ days}} \cr & \frac{1}{2}\left( {{\text{A}} + {\text{B}} + {\text{C }}} \right) = 2{\text{ days}} \cr & {\text{A}} + {\text{B}} + {\text{C}} = {\text{4 days}} \cr} $$
L.C.M of total work = 20
One day work of A + B + C = $$\frac{{20}}{4}$$ = 5 unit/day
One day work of B + C = $$\frac{{20}}{{\frac{{20}}{3}}}$$ = 3 unit/day
A = 5 - 3 = 2
A alone will complete work
$$\eqalign{ & = \frac{{20}}{2}{\text{days}} \cr & = {\text{10 days}} \cr} $$

$$\eqalign{ & {\text{Ratio}}\,{\text{of}}\,{\text{times}}\,{\text{taken}}\,{\text{by}}\,{\text{A}}\,{\text{and}}\,{\text{B}} \cr & = 100:130 = 10:13 \cr & {\text{Suppose}}\,{\text{B}}\,{\text{takes}}\,x\,{\text{days}}\,{\text{to}}\,{\text{do}}\,{\text{}}\,{\text{work}} \cr & {\text{Then}},10:13::23:x \cr & \Rightarrow x = {\frac{{23 \times 13}}{{10}}} \cr & \Rightarrow x = \frac{{299}}{{10}} \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{23}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{{10}}{{299}} \cr & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{23}} + \frac{{10}}{{299}}} \cr & = \frac{{23}}{{299}} \cr & = \frac{1}{{13}} \cr & \refore A\,{\text{and}}\,{\text{B}}\,{\text{toger}}\,{\text{can}}\,{\text{complete}} \cr & \,{\text{}}\,{\text{work}}\,{\text{in}}\,{\text{13}}\,{\text{days}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{A's 1 hour's work}} = \frac{1}{4} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 hour's work}} = \frac{1}{3} \cr & \left( {{\text{A}} + {\text{C}}} \right){\text{'s 1 hour's work}} = \frac{1}{2} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 hour's work}} \cr & = \frac{1}{4} + \frac{1}{3} \cr & = \frac{7}{{12}} \cr & \refore {\text{B's 1 hour's work}} \cr} $$
= (A + B + C)'s 1 hour's work - (A + C)'s 1 hour's work
$$\eqalign{ & = \frac{7}{{12}} - \frac{1}{2} \cr & = \frac{1}{{12}} \cr} $$
So, B alone can complete work in 12 hours.

$$\eqalign{ & {\text{C's 1 day's work}} \cr & = \frac{1}{7} \cr & {\text{B's 1 day's work}} \cr & = \left( {\frac{1}{2} \times \frac{1}{7}} \right) \cr & = \frac{1}{{14}} \cr & {\text{A's 1 day's work}} \cr & = \left( {\frac{1}{2} \times \frac{1}{{14}}} \right) \cr & = \frac{1}{{28}} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{28}} + \frac{1}{{14}} + \frac{1}{7}} \right) \cr & = \frac{7}{{28}} \cr} $$
Hence, A, B and C toger can complete work in 4 days.

Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y
Then, 4x + 6y = $$\frac{1}{8}$$ and 3x + 7y = $$\frac{1}{{10}}$$
Solving two equations, we get
$$x = \frac{{11}}{{400}},\,\,\,y = \frac{1}{{400}}$$
∴ 1 woman's 1 day's work = $$\frac{1}{{400}}$$
⇒ 10 women's 1 day's work = $$ {\frac{1}{{400}} \times 10} $$   = $$\frac{1}{{40}}$$
Hence, 10 women will complete work in 40 days

$$\eqalign{ & {\text{1 men's 1 day's work}} \cr & = \frac{1}{{99 \times 2}} \cr & = \frac{1}{{198}} \cr & {\text{1 women's 1 day's work}} \cr & = \frac{1}{{99 \times 6}}. \cr & = \frac{1}{{594}} \cr & {\text{1 boy's 1 day's work}} \cr & = \frac{1}{{99 \times 4}} \cr & = \frac{1}{{396}} \cr} $$
(1 men + 1 women + boy)'s 1 day's work
$$\eqalign{ & = \frac{1}{{198}} + \frac{1}{{594}} + \frac{1}{{396}} \cr & = \frac{{11}}{{1188}} \cr & = \frac{1}{{108}} \cr} $$
Hence, 1 mens , 1 women and 1 boys toger take 108 days to finish work.

Number of pairs knit by A and B toger in 1 day
$$\eqalign{ & = \left( {\frac{1}{3} + \frac{1}{9}} \right) \cr & = \frac{4}{9}{\text{ }} \cr} $$
∴ Required number of days,
$$\eqalign{ & = \left( {2 \div \frac{4}{9}} \right) \cr & = \left( {2 \times \frac{9}{4}} \right) \cr & = \frac{9}{2} \cr & = 4\frac{1}{2} \text{ days} \cr} $$

$$\eqalign{ & {\text{From, }} \cr & \frac{{{{\text{m}}_1} \times {{\text{d}}_1} \times {{\text{t}}_1}}}{{{{\text{w}}_1}}} = \frac{{{{\text{m}}_2} \times {{\text{d}}_2} \times {{\text{t}}_2}}}{{{{\text{w}}_2}}} \cr & {\text{Let extra workers be x}} \cr & \Rightarrow \frac{{20 \times 12}}{{\frac{5}{8}}} = \frac{{\left( {20 + x} \right) \times 4}}{{\frac{3}{8}}} \cr & \Rightarrow 4 \times 12 = \frac{{\left( {20 + x} \right) \times 4}}{3} \cr & \Rightarrow 36 = 20 + x \cr & \Rightarrow x = 16 \cr & \Rightarrow {\text{Extra workers }} = {\text{16}} \cr} $$