Number of rosogollas eaten by Rosa in 1 minute = $$\frac{{32}}{{60}}$$
Number of rosogollas eaten by Lila in 1 minute =$$\frac{{32}}{{180}}$$
Number of rosogollas eaten by Rosa and Lila together in 1 minute
$$\eqalign{ & {\text{ = }}\left( {\frac{{32}}{{60}} + \frac{{32}}{{180}}} \right) \cr & {\text{ = }}\frac{{128}}{{180}} \cr & \therefore {\text{Required time}} \cr & = \left( {32 \div \frac{{128}}{{180}}} \right){\text{minutes}} \cr & = \left( {\frac{{32 \times 180}}{{128}}} \right){\text{minutes}} \cr & = {\text{ 45 minutes}} \cr} $$

Length of cloth produced by A and B in 10 hours
$$ = 300000{\text{ metres}}$$
Length of cloth produced by B in 10 hours
$$\eqalign{ & = \left( {\frac{{300000}}{{15}} \times 10} \right) \cr & = 200000{\text{ metres}} \cr} $$
∴ Length of cloth produced by A in 10 hours
$$\eqalign{ & = \left( {300000 - 200000} \right) \cr & = 100000{\text{ metres}} \cr} $$

(X + Y)'s 1 day's work
$$\eqalign{ & = \left( {\frac{1}{{16}} + \frac{1}{{16}}} \right) \cr & = \frac{2}{{16}} \cr & = \frac{1}{8} \cr} $$
∴ Z's 1 day's work
(X + Y + Z)'s 1 day's work - (X + Y)'s 1 day's work
$$\eqalign{ & = \frac{1}{6} - \frac{1}{8} \cr & = \frac{1}{{24}} \cr} $$
∴ Z alone can finish the work in 24 days.

x does $$\frac{1}{4}$$ work in 6 days.
∴ x does complete work in 6 × 4 = 24 days
y does complete the $$\frac{3}{4}$$ work in 12 days.
∴ y does complete work in 12 × $$\frac{4}{3}$$ = 16 days
x and y together can complete a work in
$$\eqalign{ & = \frac{{16 \times 24}}{{16 + 24}} \cr & = \frac{{48}}{5} \cr & = 9\frac{3}{5}\,{\text{days}} \cr} $$

$$\eqalign{ & {\text{Reena's 1 hour's work}} = \frac{1}{6}{\text{ }} \cr & {\text{Aastha's 1 hour's work}} = \frac{1}{4}{\text{ }} \cr & {\text{Shloka's 1 hour's work}} = \frac{1}{{12}}{\text{ }} \cr} $$

( Reena + Aastha + Shloka )'s 1 hour's work
$$\eqalign{ & = \frac{1}{4} + \frac{1}{6}{\text{ + }}\frac{1}{{12}} \cr & {\text{ = }}\frac{6}{{12}} \cr & = \frac{1}{2} \cr} $$
Hence, Reema, Aastha and Shloka together take 2 hours to complete the work.

$$\eqalign{ & {\text{A's 1 day's work}} = \frac{1}{{18}} \cr & {\text{And}} \cr & {\text{B's 1 day's work}} = \frac{1}{9} \cr & \therefore \left( {{\text{A}} + {\text{B}}} \right)'{\text{s 1 day's work}} \cr & = \left( {\frac{1}{{18}} + \frac{1}{9}} \right) \cr & = \frac{1}{6} \cr} $$

$$\eqalign{ & \frac{3}{{10}}\left( {{\text{B}} + {\text{C }}} \right) = 2{\text{ days}} \cr & \left( {{\text{B}} + {\text{C }}} \right) = 2 \times \frac{{10}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{3}{\text{ days}} \cr & \frac{1}{2}\left( {{\text{A}} + {\text{B}} + {\text{C }}} \right) = 2{\text{ days}} \cr & {\text{A}} + {\text{B}} + {\text{C}} = {\text{4 days}} \cr} $$
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 the work
$$\eqalign{ & = \frac{{20}}{2}{\text{days}} \cr & = {\text{10 days}} \cr} $$

$$\eqalign{ & {\text{Amit's 1 day's work }} \cr & = \left( {\frac{1}{4} - \frac{1}{6}} \right) \cr & = \frac{1}{{12}} \cr} $$
∴ Amit alone can plough the field in 12 days.

P completes $$\frac{1}{4}$$ of work in 10 days
P completes full of work in
$$\eqalign{ & = \frac{{10}}{1} \times 4 \cr & = 40{\text{ days}} \cr} $$
Q completes 40% of work in 145 days
Q completes full 100% of work in
$$\eqalign{ & = \frac{{145}}{{40}} \times 100 \cr & = 362.5{\text{ days}} \cr} $$
R completes $$\frac{1}{3}$$ of work in 13 days
R completes full of work in
$$\eqalign{ & = \frac{{13}}{1} \times 3 \cr & = 39{\text{ days}} \cr} $$
S completes $$\frac{1}{6}$$ of work in 7 days
S completes full of work in
$$\eqalign{ & = \frac{7}{1} \times 6 \cr & = 42{\text{ days}} \cr} $$
Clearly, we can see R completes the work first

Number of pages typed by Gorge in 1 hour
$$\eqalign{ & = \frac{{50}}{8} \cr & = \frac{{25}}{4} \cr} $$
Number of pages typed by Sonia in 1 hour
$$\eqalign{ & = \frac{{50}}{6} \cr & = \frac{{25}}{3} \cr} $$
Number of pages typed by Gorge and Sonia together in 1 hour
$$\eqalign{ & = \left( {\frac{{25}}{4} + \frac{{25}}{3}} \right) \cr & = \left( {\frac{{75 + 100}}{{12}}} \right) \cr & = \frac{{175}}{{12}} \cr & \therefore {\text{Required time}} \cr & = \left( {100 \div \frac{{175}}{{12}}} \right){\text{hours}} \cr & = \left( {\frac{{100 \times 12}}{{175}}} \right){\text{ hours}} \cr & = \frac{{48}}{7}{\text{ hours}} \cr & = 6\frac{6}{7}{\text{ hours}} \cr} $$

$$\eqalign{ & {\text{Required average,}} \cr & = \frac{{ {\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{{10}} + \frac{1}{{12}}} }}{5} \cr & = {\frac{{48}}{{60}} \times \frac{1}{5}} \cr & = \frac{4}{{25}} \cr & = 0.16 \cr} $$

Try to solve these kind of question by option
Because of his being absent he was paid Rs. 750 less
Now check with option
Since maximum possible daily wages is asked so it will be Rs. 250
OR :
Maximum possible daily wage
= HCF of Rs. 5750 and Rs. 5000
= Rs. 250

In these kind of questions, always take total work as L.C.M. of number of days
Here L.C.M. of Total Work = 120
One day work of A + B = $$\frac{{120}}{8}$$ = 15 unit/day
One day work of B + C = $$\frac{{120}}{10}$$ = 12 unit/day
One day work of C + A = $$\frac{{120}}{6}$$ = 20 unit/day
∴ Total units per day = 15 + 12 + 20 = 47
∴ Efficiency of :
$$\eqalign{ & 2\left( {{\text{A}} + {\text{B}} + {\text{C}}} \right) = 47 \cr & {\text{A}} + {\text{B}} + {\text{C}} = \frac{{47}}{2} \cr} $$
(A + B + C) will complete the whole work in
$$\eqalign{ & = \frac{{120}}{{\frac{{47}}{2}}} \cr & = \frac{{240}}{{47}} \cr & = 5\frac{5}{{47}}{\text{days}} \cr} $$

Work done by X in 8 days = $$ {\frac{1}{{40}} \times 8} $$   = $$\frac{1}{5}$$
Remaining work = $$ {1 - \frac{1}{5}} $$   = $$\frac{4}{5}$$
Now, $$\frac{4}{5}$$ work is done by Y in 16 days
Whole work will be done by Y in = $$ {16 \times \frac{5}{4}} $$   = 20 days
∴ X's 1 day's work = $$\frac{1}{{40}}$$
∴ Y's 1 day's work = $$\frac{1}{{20}}$$
(X + Y)'s 1 day's work
$$\eqalign{ & = {\frac{1}{{40}} + \frac{1}{{20}}} \cr & = \frac{3}{{40}} \cr} $$
Hence, X and Y will together complete the work in
$$\eqalign{ & = {\frac{{{\text{40}}}}{{\text{3}}}} \cr & {\text{ = 13}}\frac{{\text{1}}}{{\text{3}}}\,\,{\text{days}} \cr} $$

$$\eqalign{ & \left( {{\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}} \right){\text{:}}\left( {{\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}} \right) \cr & = \frac{7}{4}:1 = 7:4 \cr & {\text{Let}}\,{\text{A's}}\,{\text{and}}\,{\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}\,{\text{be}} \cr & 7x\,{\text{and}}\,4x\,{\text{respectively}} \cr & {\text{Then}},\,7x + 4x = \frac{1}{7} \cr & \Rightarrow 11x = \frac{1}{7} \cr & \Rightarrow x = \frac{1}{{77}} \cr & \therefore {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{77}} \times {\text{7}}} = \frac{1}{{11}} \cr} $$
So, A will do the work in 11 days

$$\eqalign{ & {\text{A's 1 day's work}} \cr & = \frac{1}{{{\text{T}} + 3}} \cr & {\text{B's 1 day's work}} \cr & = \frac{1}{{{\text{T}} + 12}} \cr & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{{\text{T}}} \cr & \therefore \frac{1}{{{\text{T}} + 3}} + \frac{1}{{{\text{T}} + 12}} = \frac{1}{{\text{T}}} \cr & \Rightarrow \frac{{2{\text{T}} + 15}}{{\left( {{\text{T}} + 3} \right)\left( {{\text{T}} + 12} \right)}} = \frac{1}{{\text{T}}} \cr & \Rightarrow 2{{\text{T}}^2} + 15{\text{T}} = {{\text{T}}^2} + 15{\text{T}} + 36 \cr & \Rightarrow {{\text{T}}^2} = 36 \cr & \Rightarrow {\text{T}} = 6 \cr} $$

$$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{pages}}\,{\text{typed}}\,{\text{by}}\,{\text{Ravi}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = \frac{{32}}{6} = \frac{{16}}{3} \cr & {\text{Number}}\,{\text{of}}\,{\text{pages}}\,{\text{typed}}\,{\text{by}}\,{\text{Kumar}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = \frac{{40}}{5} = 8 \cr & {\text{Number}}\,{\text{of}}\,{\text{pages}}\,{\text{typed}}\,{\text{by}}\,{\text{both}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = {\frac{{16}}{3} + 8} = \frac{{40}}{3} \cr & \therefore {\text{Time}}\,{\text{taken}}\,{\text{by}}\,{\text{both}}\,{\text{to}}\,{\text{type}}\,{\text{110}}\,{\text{pages}} \cr & = {110 \times \frac{3}{{40}}} {\text{hours}} \cr & = 8\frac{1}{4}\,{\text{hours}}\,{\text{(or)}}\,{\text{8}}\,{\text{hours}}\,{\text{15}}\,{\text{minutes}} \cr} $$

L.C.M. of total work = 24
One day work of A + B = $$\frac{{24}}{8}$$ = 3 unit/day
B's 1 day work = $$\frac{{24}}{12}$$   = 2 units/days
A's 1 day work = 3 - 2 = 1 units/days
4 days work of B = 4 × 2 = 8 units/days
Work left = 24 - 8 = 16 units
A will complete the remaining work in
$$\eqalign{ & = \frac{{{\text{16 units}}}}{{{\text{1 unit/day}}}} \cr & = {\text{16 days}} \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = x\,{\text{and}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = y \cr & {\text{Then}},\,x + y = \frac{1}{{30}} \cr & 16x + 44y = 1 \cr & {\text{Solving}}\,{\text{these}}\,{\text{two}}\,{\text{equations,}}\,{\text{we}}\,{\text{get}} \cr & x = \frac{1}{{60}}\,{\text{and}}\,y = \frac{1}{{60}} \cr & \therefore {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{60}} \cr & {\text{Hence,}}\,{\text{B}}\,{\text{alone}}\,{\text{shall}}\,{\text{finish}}\,{\text{the}} \cr & {\text{whole}}\,{\text{work}}\,{\text{in}}\,{\text{60}}\,{\text{days}} \cr} $$

So let's solve now,
A's 1 day work = $$\frac{1}{{18}}$$
B's 1 day work = $$\frac{1}{9}$$ [because B take half time than A]
(A + B)'s one day work
$$\eqalign{ & = {\frac{1}{{18}} + \frac{1}{9}} \cr & = {\frac{{1 + 2}}{{18}}} \cr & = \frac{1}{6} \cr} $$

Formula: If A can do a piece of work in n days, then A's 1 day's work = $$\frac{1}{{\text{n}}}$$
$$\eqalign{ & (A + B + C)'s\,1\,{\text{day's work}} \cr & = {\frac{1}{{24}} + \frac{1}{6} + \frac{1}{{12}}} = \frac{7}{{24}} \cr} $$
Formula: If A's 1 day's work = $$\frac{1}{{\text{n}}}$$ , then A can finish the work in n days
So, all the three together will complete the job in
$$ {\frac{{24}}{7}} {\text{ days}} = 3\frac{3}{7}{\text{ days}}$$

Ratio of times taken by Sakshi and Tanya
= 125 : 100
= 5 : 4
Suppose Tanya takes x days to do the work
5 : 4 :: 20 : x
⇒ $$x = {\frac{{4 \times 20}}{5}} $$
⇒ x = 16 days
Hence, Tanya takes 16 days to complete the work

Assume work to be done 100%. First case, A + B = $$\frac{{100}}{5}$$ = 20% work done per day -------- (1) Second case, 2A + $$\frac{{\text{B}}}{2}$$ = $$\frac{{100}}{4}$$ = 25% work done per day ------ (2) On solving equation (1) and (2), we get A = 10 days

Let D days required to complete the remaining work
$${40_{{\text{men}}}} \times {18_{{\text{days}}}}$$    = $$\left( {{{40}_{{\text{men}}}} \times {8_{{\text{days}}}}} \right)$$    + $$\left( {{{50}_{{\text{men}}}} \times {\text{D}}} \right)$$
$$\eqalign{ & \Rightarrow 720 - 320 = 50{\text{D}} \cr & \Rightarrow {\text{D}} = 8 \cr} $$

$$\eqalign{ & {\text{2(A + B + C)'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{30}} + \frac{1}{{24}} + \frac{1}{{20}}} \cr & = \frac{{15}}{{120}} = \frac{1}{8} \cr & \therefore \left( {{\text{A + B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \frac{1}{{2 \times 8}} = \frac{1}{{16}} \cr & {\text{Work}}\,{\text{done}}\,{\text{by}}\,{\text{A,}}\,{\text{B,}}\,{\text{C}}\,{\text{in}}\,{\text{10}}\,{\text{days}} \cr & = \frac{{10}}{{16}} = \frac{5}{8} \cr & {\text{Remaining}}\,{\text{work}} \cr & = {1 - \frac{5}{8}} = \frac{3}{8} \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{16}} - \frac{1}{{24}}} = \frac{1}{{48}} \cr & {\text{Now}},\,\frac{1}{{48}}\,{\text{work}}\,{\text{isdone}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{1}}\,{\text{day}} \cr & {\text{So}},\,\frac{3}{8}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}} \cr & {48 \times \frac{3}{8}} = 18\,{\text{days}} \cr} $$

In these kind of questions, always take total work as L.C.M. of number of days.
Here L.C.M. of Total Work = 360
One day work of A + B = $$\frac{{360}}{{72}}$$  = 5 unit/day
One day work of B + C = $$\frac{{360}}{{120}}$$  = 3 unit/day
One day work of C + A = $$\frac{{360}}{{90}}$$  = 4 unit/day
∴ Total units/day = 5 + 3 + 4 = 12
(Here , 12 unit represents twice of the work done by A, B and C. So we will divide it by 2)
Work done by (A + B + C) per day = $$\frac{{12}}{2}$$ = 6 units/day
∴Total time taken by (A + B + C)
$$\eqalign{ & = \frac{{360}}{6} \cr & = 60{\text{ days}} \cr} $$

$$\eqalign{ & {\text{Suppose}}\,{\text{A,}}\,{\text{B}}\,{\text{and}}\,{\text{C}}\,{\text{take}} \cr & x,\,\frac{x}{2},\,\frac{x}{3}\,{\text{days}}\,{\text{respectively}}\,{\text{to}}\,{\text{finish}}\,{\text{the}}\,{\text{work}} \cr & {\text{Then}},\, {\frac{1}{x} + \frac{2}{x} + \frac{3}{x}} = \frac{1}{2} \cr & \Rightarrow \frac{6}{x} = \frac{1}{2} \cr & \Rightarrow x = 12 \cr & {\text{So,}}\,{\text{B}}\,{\text{takes}}\, {\frac{{12}}{2}} \cr & = 6\,{\text{days}}\,{\text{to}}\,{\text{finish}}\,{\text{the}}\,{\text{work}} \cr} $$

$$\eqalign{ & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{15}} + \frac{1}{{10}}} = \frac{1}{6} \cr & {\text{Work}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{and}}\,{\text{B}}\,{\text{in}}\,{\text{2}}\,{\text{days}} \cr & = {\frac{1}{6} \times 2} = \frac{1}{3} \cr & {\text{Remaining}}\,{\text{work}} \cr & = {1 - \frac{1}{3}} = \frac{2}{3} \cr & {\text{Now}},\,\frac{1}{{15}}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{1}}\,{\text{day}} \cr & \therefore \frac{2}{3}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{done}}\,{\text{by}}\,{\text{a}}\,{\text{in}} \cr & {15 \times \frac{2}{3}} = 10\,{\text{days}} \cr & {\text{Hence,}}\,{\text{the}}\,{\text{total}}\,{\text{time}}\,{\text{taken}} \cr & = {10 + 2} = 12\,{\text{days}} \cr} $$

$$\eqalign{ & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{20}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{30}} \times 20} = \frac{2}{3} \cr & {\text{Remaining}}\,{\text{work}} \cr & = {1 - \frac{2}{3}} = \frac{1}{3} \cr & {\text{Now}},\frac{1}{3}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{20}}\,{\text{days}} \cr & \therefore {\text{The}}\,{\text{whole}}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}} \cr & {20 \times 3} = 60\,days \cr} $$

$$\eqalign{ & {\text{P}}\,{\text{can}}\,{\text{complete}}\,{\text{the}}\,{\text{work}} \cr & = \,\left( {12 \times 8} \right){\text{hrs}}{\text{.}} = 96\,{\text{hrs}}{\text{.}} \cr & {\text{Q}}\,{\text{can}}\,{\text{complete}}\,{\text{the}}\,{\text{work}} \cr & = \left( {8 \times 10} \right){\text{hrs}}{\text{.}} = 80\,{\text{hrs}}{\text{.}} \cr & \therefore {\text{P's}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} = \frac{1}{{96}}\,{\text{and}} \cr & \therefore {\text{Q's}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} = \frac{1}{{80}} \cr & \left( {{\text{P + Q}}} \right){\text{'s}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} \cr & = {\frac{1}{{96}} + \frac{1}{{80}}} = \frac{{11}}{{480}} \cr & {\text{So,}}\,{\text{both}}\,{\text{P}}\,{\text{and}}\,{\text{Q}}\,{\text{will}}\,{\text{finish}}\,{\text{the}}\,{\text{work}} \cr & = {\frac{{480}}{{11}}} {\text{ hrs}}{\text{.}} \cr & \therefore {\text{Number}}\,{\text{of}}\,{\text{days}}\,{\text{of}}\,{\text{8}}\,{\text{hours}}\,{\text{each}} \cr & {\frac{{480}}{{11}} \times \frac{1}{8}} = \frac{{60}}{{11}}{\text{days}} = 5\frac{5}{{11}}{\text{days}} \cr} $$