Time and Work Questions in English

(A) Let the work done by $10$ men be represented by $X = 15$ days.
Let the work done by $15$ women be represented by $Y = 12$ days.
Since the group of $10$ men and the group of $15$ women are working together,we use the formula for combined work time: $\text{Time} = \frac{X \times Y}{X + Y}$.
Substituting the values: $\text{Time} = \frac{15 \times 12}{15 + 12}$.
$\text{Time} = \frac{180}{27}$.
Dividing both numerator and denominator by $9$,we get $\frac{20}{3}$ days.
Converting to a mixed fraction,we get $6 \frac{2}{3}$ days.

(B) Let the time taken by $A$ be $X = 30 \text{ days}$ and the time taken by $B$ be $Y = 40 \text{ days}$.
The formula for the time taken by two people working together is $\frac{X \times Y}{X + Y} \text{ days}$.
Substituting the values: $\frac{30 \times 40}{30 + 40} = \frac{1200}{70} \text{ days}$.
Simplifying the fraction: $\frac{120}{7} \text{ days}$.
Converting to a mixed fraction: $17 \frac{1}{7} \text{ days}$.

(B) can do $\frac{1}{3}$ of the work in $5 \text{ days}$. Therefore,$A$ can complete the whole work in $5 \times 3 = 15 \text{ days}$.
$B$ can do $\frac{2}{5}$ of the work in $10 \text{ days}$. Therefore,$B$ can complete the whole work in $10 \times \frac{5}{2} = 25 \text{ days}$.
Let $X = 15$ and $Y = 25$.
The time taken by $A$ and $B$ working together to complete the work is given by the formula $\frac{X \times Y}{X + Y}$.
Substituting the values: $\frac{15 \times 25}{15 + 25} = \frac{375}{40} = \frac{75}{8} \text{ days}$.
Converting to a mixed fraction: $\frac{75}{8} = 9 \frac{3}{8} \text{ days}$.

(C) Let the work done by $A, B,$ and $C$ in one day be $\frac{1}{6}, \frac{1}{12},$ and $\frac{1}{24}$ respectively.
The total work done by $A, B,$ and $C$ together in one day is:
$\frac{1}{6} + \frac{1}{12} + \frac{1}{24} = \frac{4 + 2 + 1}{24} = \frac{7}{24}$ units of work per day.
Therefore,the total time taken by them to complete the work is the reciprocal of the work done in one day:
Time $= \frac{24}{7}$ days.
Converting this into a mixed fraction:
$\frac{24}{7} = 3 \frac{3}{7}$ days.

(B) Let $B$ take $x$ days to do the work.
Then,$A$ takes $(x - 60)$ days to complete the work.
Since the ratio of efficiency (work done per day) of $A$ and $B$ is $3:1$,the ratio of time taken by $A$ and $B$ is $1:3$.
We have,$\frac{x - 60}{x} = \frac{1}{3}$.
$\Rightarrow 3(x - 60) = x \Rightarrow 3x - 180 = x \Rightarrow 2x = 180 \Rightarrow x = 90$.
Therefore,time taken by $B = 90 \text{ days}$ and time taken by $A = 90 - 60 = 30 \text{ days}$.
Working together,they will complete the work in $\frac{A \times B}{A + B} = \frac{30 \times 90}{30 + 90} = \frac{2700}{120} = \frac{270}{12} = 22.5 \text{ days}$.
Thus,the time taken is $22 \frac{1}{2} \text{ days}$.

(A) Let the time taken by Ramesh to complete the work be $x$ days.
Since Ramesh takes twice as much time as Mahesh,Mahesh takes $x/2$ days.
Since Ramesh takes thrice as much time as Suresh,Suresh takes $x/3$ days.
The work done by Ramesh in $1$ day is $1/x$.
The work done by Mahesh in $1$ day is $2/x$.
The work done by Suresh in $1$ day is $3/x$.
Together,they complete the work in $4$ days,so their combined $1$-day work is $1/4$.
Therefore,$1/x + 2/x + 3/x = 1/4$.
$6/x = 1/4$.
$x = 24$.
Thus,Ramesh takes $24$ days,Mahesh takes $24/2 = 12$ days,and Suresh takes $24/3 = 8$ days.

(A) Let Gita take $x$ days to complete the work. Then,Sita takes $2x$ days to complete the same work.
Rita takes the same time as Sita and Gita together. The time taken by Sita and Gita together is $\frac{x \times 2x}{x + 2x} = \frac{2x^2}{3x} = \frac{2x}{3}$ days.
So,Rita takes $\frac{2x}{3}$ days to complete the work.
When all three work together,the time taken is given by the formula $\frac{1}{T} = \frac{1}{x} + \frac{1}{2x} + \frac{1}{2x/3}$.
$\frac{1}{6} = \frac{1}{x} + \frac{1}{2x} + \frac{3}{2x}$.
$\frac{1}{6} = \frac{2 + 1 + 3}{2x} = \frac{6}{2x} = \frac{3}{x}$.
Therefore,$x = 18$ days.
Thus,Gita takes $18$ days,Sita takes $2x = 36$ days,and Rita takes $\frac{2x}{3} = 12$ days.

(B) The work done by $1$ man in $1$ day is $1/10$ of the total work (since $5$ men take $2$ days,$1$ man takes $5 \times 2 = 10$ days).
The work done by $1$ woman in $1$ day is $1/12$ of the total work (since $4$ women take $3$ days,$1$ woman takes $4 \times 3 = 12$ days).
The work done by $1$ child in $1$ day is $1/15$ of the total work (since $5$ children take $3$ days,$1$ child takes $5 \times 3 = 15$ days).
Combined work done by $1$ man,$1$ woman,and $1$ child in $1$ day = $\frac{1}{10} + \frac{1}{12} + \frac{1}{15}$.
Finding the common denominator $(60)$: $\frac{6+5+4}{60} = \frac{15}{60} = \frac{1}{4}$.
Therefore,they can complete the work together in $4$ days.

(B) Let the time taken by $A$ and $B$ together be $T_{AB} = 6$ days.
Let the time taken by $A$ alone be $T_A = 9$ days.
The work done by $A$ and $B$ in $1$ day is $\frac{1}{6}$.
The work done by $A$ in $1$ day is $\frac{1}{9}$.
Therefore,the work done by $B$ in $1$ day is $\frac{1}{6} - \frac{1}{9} = \frac{3-2}{18} = \frac{1}{18}$.
Thus,$B$ alone will complete the work in $18$ days.

(C) Let the work done by $(A+B)$ in $1 \text{ day} = \frac{1}{30}$.
Let the work done by $(B+C)$ in $1 \text{ day} = \frac{1}{40}$.
Let the work done by $(C+A)$ in $1 \text{ day} = \frac{1}{60}$.
Adding these,we get $2(A+B+C)$'s $1 \text{ day}$ work:
$2(A+B+C) = \frac{1}{30} + \frac{1}{40} + \frac{1}{60} = \frac{4+3+2}{120} = \frac{9}{120} = \frac{3}{40}$.
Therefore,$(A+B+C)$'s $1 \text{ day}$ work $= \frac{3}{40} \times \frac{1}{2} = \frac{3}{80}$.
Thus,$A, B, C$ together can complete the work in $\frac{80}{3} = 26 \frac{2}{3} \text{ days}$.

(A) Let the work done by $A, B,$ and $C$ in one day be $a, b,$ and $c$ respectively.
Given:
$a + b = 1/18$
$b + c = 1/24$
$c + a = 1/36$
Adding these three equations:
$2(a + b + c) = 1/18 + 1/24 + 1/36$
$2(a + b + c) = (4 + 3 + 2) / 72 = 9/72 = 1/8$
$a + b + c = 1/16$
To find the time taken by $A$ alone,we subtract the work done by $B$ and $C$ from the total:
$a = (a + b + c) - (b + c) = 1/16 - 1/24$
$a = (3 - 2) / 48 = 1/48$
Thus,$A$ alone can complete the work in $48$ days.

(B) Let the work done by Ajay,Sunil,and Sanjay in one day be $A, S,$ and $J$ respectively.
Given:
$A + S = 1/10$
$S + J = 1/15$
$J + A = 1/20$
Adding these,$2(A + S + J) = 1/10 + 1/15 + 1/20 = (6+4+3)/60 = 13/60$.
So,$A + S + J = 13/120$.
Work done by all three in $6$ days $= 6 \times (13/120) = 13/20$.
Remaining work $= 1 - 13/20 = 7/20$.
Sunil and Sanjay work for $4$ more days. Their rate is $S + J = 1/15$.
Work done by them in $4$ days $= 4 \times (1/15) = 4/15$.
Remaining work for Sanjay $= 7/20 - 4/15 = (21-16)/60 = 5/60 = 1/12$.
Sanjay's individual rate $J = (A+S+J) - (A+S) = 13/120 - 1/10 = (13-12)/120 = 1/120$.
Time taken by Sanjay to complete $1/12$ of the work $= (1/12) / (1/120) = 10$ days.

(C) Anu completes the work in $10$ days.
Manu is $25\%$ more efficient than Anu,so Manu's time taken $= 10 \times \frac{100}{125} = 8$ days.
Sonu is $60\%$ more efficient than Manu,so Sonu's time taken $= 8 \times \frac{100}{160} = 5$ days.
Let the total work be $40$ units ($LCM$ of $10, 8, 5$).
Efficiency of Anu $= \frac{40}{10} = 4$ units/day.
Efficiency of Manu $= \frac{40}{8} = 5$ units/day.
Efficiency of Sonu $= \frac{40}{5} = 8$ units/day.
Combined efficiency $= 4 + 5 + 8 = 17$ units/day.
Time taken by all three working together $= \frac{40}{17} = 2 \frac{6}{17}$ days.

(A) Let the work done by $(A+B)$ in $1 \text{ day} = \frac{1}{12}$.
Let the work done by $(A+B+C)$ in $1 \text{ day} = \frac{1}{8}$.
Work done by $C$ in $1 \text{ day} = (\text{Work of } A+B+C) - (\text{Work of } A+B)$.
Work done by $C$ in $1 \text{ day} = \frac{1}{8} - \frac{1}{12} = \frac{3-2}{24} = \frac{1}{24}$.
Therefore,$C$ alone will finish the job in $24 \text{ days}$.

(B) Let the work done by Bansal,Gupta,and Singhal in one day be $B, G,$ and $S$ respectively.
Given: $B + G + S = \frac{1}{4}$ (Equation $1$)
$B + G = \frac{1}{4.8} = \frac{1}{24/5} = \frac{5}{24}$ (Equation $2$)
$G + S = \frac{1}{8}$ (Equation $3$)
From Equation $1$ and Equation $2$,we get $S = \frac{1}{4} - \frac{5}{24} = \frac{6-5}{24} = \frac{1}{24}$. So,Singhal takes $24 \, \text{days}$.
From Equation $1$ and Equation $3$,we get $B = \frac{1}{4} - \frac{1}{8} = \frac{2-1}{8} = \frac{1}{8}$. So,Bansal takes $8 \, \text{days}$.
Now,substitute $B$ in Equation $2$: $\frac{1}{8} + G = \frac{5}{24} \implies G = \frac{5}{24} - \frac{1}{8} = \frac{5-3}{24} = \frac{2}{24} = \frac{1}{12}$.
Therefore,Gupta alone can complete the work in $12 \, \text{days}$.

(C) Let the work done by Nikita,Nishita,and Kavita be $A, B,$ and $C$ respectively.
Given: $(A+B+C)$ can complete the work in $2 \frac{2}{3} = \frac{8}{3}$ days.
$(B+C)$ can complete the work in $4$ days.
$B$ alone can complete the work in $6$ days.
First,find the time taken by $C$ (Kavita) alone:
Since $(B+C)$ takes $4$ days and $B$ takes $6$ days,the work done by $C$ in $1$ day is $\frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$.
So,$C$ alone takes $12$ days.
Next,find the time taken by $A$ (Nikita) alone:
Since $(A+B+C)$ takes $\frac{8}{3}$ days,their $1$-day work is $\frac{3}{8}$.
$A$'s $1$-day work $= \frac{3}{8} - (B+C)$'s $1$-day work $= \frac{3}{8} - \frac{1}{4} = \frac{3-2}{8} = \frac{1}{8}$.
So,$A$ alone takes $8$ days.
Now,find the time taken by $A$ and $B$ (Nikita and Nishita) together:
Their combined $1$-day work $= \frac{1}{8} + \frac{1}{6} = \frac{3+4}{24} = \frac{7}{24}$.
Therefore,they can complete the work in $\frac{24}{7} = 3 \frac{3}{7}$ days.

(A) Let the efficiency of $B$ be $1$ unit/day. Then the efficiency of $A$ is $2$ units/day.
Combined efficiency of $A$ and $B$ is $1 + 2 = 3$ units/day.
Total work = (Combined efficiency) $\times$ (Time taken) = $3 \times 1.5 = 4.5$ units.
Time taken by $A$ alone = $\frac{\text{Total work}}{\text{Efficiency of } A} = \frac{4.5}{2} = 2.25$ days.

(A) Let the total work be $1$ unit.
Bindal's efficiency $= \frac{1}{10}$ work per day.
Since Jindal is twice as efficient as Bindal,Jindal's efficiency $= 2 \times \frac{1}{10} = \frac{1}{5}$ work per day.
When they work together,their combined efficiency $= \frac{1}{10} + \frac{1}{5} = \frac{1+2}{10} = \frac{3}{10}$ work per day.
Time taken to complete the work together $= \frac{1}{3/10} = \frac{10}{3} = 3\frac{1}{3}$ days.

(B) Let the time taken by $A$ and $B$ together to complete the job be $x \, \text{days}$.
Then, $A$ alone takes $(x + 27) \, \text{days}$ and $B$ alone takes $(x + 3) \, \text{days}$.
The work done by $A$ and $B$ in one day is $\frac{1}{x + 27} + \frac{1}{x + 3} = \frac{1}{x}$.
Solving for $x$: $\frac{(x + 3) + (x + 27)}{(x + 27)(x + 3)} = \frac{1}{x}$.
$\frac{2x + 30}{x^2 + 30x + 81} = \frac{1}{x}$.
$2x^2 + 30x = x^2 + 30x + 81$.
$x^2 = 81$, which gives $x = 9$.
Thus, working together, they take $9 \, \text{days}$.

(A) Let the time taken by $B$ to complete the work be $x$ days.
Since $A$ is $4$ times as fast as $B$,the time taken by $A$ to complete the same work is $\frac{x}{4}$ days.
According to the problem,$A$ takes $45$ days less than $B$,so $x - \frac{x}{4} = 45$.
Solving for $x$: $\frac{3x}{4} = 45$,which gives $x = 60$ days.
Thus,$B$ takes $60$ days and $A$ takes $\frac{60}{4} = 15$ days.
The work done by $A$ in $1$ day is $\frac{1}{15}$ and by $B$ in $1$ day is $\frac{1}{60}$.
Working together,their $1$ day's work is $\frac{1}{15} + \frac{1}{60} = \frac{4+1}{60} = \frac{5}{60} = \frac{1}{12}$.
Therefore,$A$ and $B$ together can complete the work in $12$ days.

(C) Given that $A$ completes the whole work in $16$ days.
Therefore,the time taken to complete $\frac{3}{4}$ of the work is calculated as:
$\text{Time} = \frac{3}{4} \times 16 \text{ days}$.
$\text{Time} = 3 \times 4 = 12 \text{ days}$.
Thus,$A$ can complete $\frac{3}{4}$ of the work in $12$ days.

(A) We use the formula: $\frac{M_{1} \times D_{1} \times t_{1}}{W_{1}} = \frac{M_{2} \times D_{2} \times t_{2}}{W_{2}}$
Given:
$M_{1} = 24, D_{1} = 27, t_{1} = 7, W_{1} = 1$
$M_{2} = 14, D_{2} = ?, t_{2} = 9, W_{2} = 1$
Substituting the values:
$24 \times 27 \times 7 = 14 \times D_{2} \times 9$
$D_{2} = \frac{24 \times 27 \times 7}{14 \times 9}$
$D_{2} = \frac{24 \times 3 \times 7}{14}$
$D_{2} = \frac{24 \times 3}{2} = 12 \times 3 = 36$
Therefore,$14$ men will complete the work in $36$ $\text{days}$.

(B) We use the formula for work done: $\frac{M_1 \times D_1}{W_1} = \frac{M_2 \times D_2}{W_2}$,where $M$ is the number of men,$D$ is the time (duration),and $W$ is the work done.
Given: $M_1 = 10$,$D_1 = 2$ hours,$W_1 = 15$ trees.
After $2$ men leave,$M_2 = 10 - 2 = 8$ men.
We need to find $W_2$ for $D_2 = 3$ hours.
Substituting the values into the formula:
$\frac{10 \times 2}{15} = \frac{8 \times 3}{W_2}$
$\frac{20}{15} = \frac{24}{W_2}$
$\frac{4}{3} = \frac{24}{W_2}$
$W_2 = \frac{24 \times 3}{4} = 6 \times 3 = 18$ trees.
Thus,$18$ trees will be cut in $3$ hours.

(A) We use the formula: $\frac{M_{1} \times D_{1} \times t_{1}}{W_{1}} = \frac{M_{2} \times D_{2} \times t_{2}}{W_{2}}$
Given:
$M_{1} = 45, D_{1} = 30, t_{1} = 12$
$M_{2} = 60, D_{2} = ?, t_{2} = 10$
Since the work $W$ is the same,$W_{1} = W_{2} = 1$.
Substituting the values:
$45 \times 30 \times 12 = 60 \times D_{2} \times 10$
$16200 = 600 \times D_{2}$
$D_{2} = \frac{16200}{600} = 27$ days.
Therefore,$60$ men will complete the work in $27$ days.

(B) Given that the work done by $1$ woman in $8\, \text{hours} = 1$ man in $6\, \text{hours} = 1$ boy in $12\, \text{hours}$.
Let the work rate of a man be $M$,a woman be $W$,and a boy be $B$.
$8W = 6M = 12B$.
From $6M = 12B$,we get $M = 2B$.
From $8W = 6M$,we get $8W = 6(2B) = 12B$,so $W = 1.5B$.
We need to find the work done by $12$ men,$12$ women,and $12$ boys in terms of men:
$12$ men $+ 12$ women $+ 12$ boys $= 12$ men $+ (12 \times \frac{6}{8})$ men $+ (12 \times \frac{6}{12})$ men.
$= 12 + 9 + 6 = 27$ men.
Using the formula $M_1 D_1 t_1 = M_2 D_2 t_2$:
$9 \times 6 \times 6 = 27 \times D_2 \times 8$.
$324 = 216 \times D_2$.
$D_2 = \frac{324}{216} = \frac{3}{2} = 1 \frac{1}{2}$ days.

(B) Given that $4$ men = $6$ women,so $1$ man = $\frac{6}{4} = 1.5$ women.
Total work can be expressed in terms of women: $6$ women can finish the work in $20$ days.
We need to find the time taken by $6$ men and $11$ women.
Since $1$ man = $1.5$ women,$6$ men = $6 \times 1.5 = 9$ women.
Total workers = $9$ women + $11$ women = $20$ women.
Using the formula $M_1 D_1 = M_2 D_2$:
$6 \times 20 = 20 \times D_2$
$D_2 = \frac{6 \times 20}{20} = 6$ days.
Therefore,$6$ men and $11$ women can finish the work in $6$ days.

(B) Let the work be $W$.
$10$ men finish the work in $10$ days,so $1$ man's $1$ day work $= W / (10 \times 10) = W / 100$.
$12$ women finish the work in $10$ days,so $1$ woman's $1$ day work $= W / (12 \times 10) = W / 120$.
Now,$15$ men and $6$ women's $1$ day work $= 15 \times (W / 100) + 6 \times (W / 120)$.
$= (3W / 20) + (W / 20) = 4W / 20 = W / 5$.
Therefore,the time taken to complete the work is $5$ days.

(B) Work done by $A$ in $1$ day $= 1/10$.
Work done by $B$ in $1$ day $= 1/15$.
Work done by $(A+B)$ in $5$ days $= 5 \times (1/10 + 1/15) = 5 \times (3+2)/30 = 5 \times (5/30) = 5/6$.
Remaining work $= 1 - 5/6 = 1/6$.
This remaining work is done by $C$ in $2$ days.
Since wages are distributed in the ratio of work done,we calculate the work done by each person:
Work done by $A = 5/10 = 1/2 = 3/6$.
Work done by $B = 5/15 = 1/3 = 2/6$.
Work done by $C = 1/6$.
The ratio of work done by $A:B:C = 3/6 : 2/6 : 1/6 = 3:2:1$.
Total parts $= 3+2+1 = 6$.
$A$'s share $= (3/6) \times 450 = Rs. 225$.
$B$'s share $= (2/6) \times 450 = Rs. 150$.
$C$'s share $= (1/6) \times 450 = Rs. 75$.

(B) Wages are distributed in proportion to the work done by each individual.
Work done by the first man in $3$ days $= 3 \times (1/7) = 3/7$.
Wages of the first man $= (3/7) \times 1400 = Rs. 600$.
Work done by the second man in $3$ days $= 3 \times (1/8) = 3/8$.
Wages of the second man $= (3/8) \times 1400 = Rs. 525$.
Work done by the boy $= 1 - (3/7 + 3/8) = 1 - (24+21)/56 = 1 - 45/56 = 11/56$.
Wages of the boy $= (11/56) \times 1400 = Rs. 275$.
Therefore,the shares are $Rs. 600, Rs. 525,$ and $Rs. 275$ respectively.

(A) Let the time taken by $B$ to complete the work be $x$ days.
Since $A$ does half as much work as $B$ in $\frac{3}{4}$ of the time,the efficiency of $A$ relative to $B$ can be calculated.
If $B$ does $1$ unit of work in $x$ days,$A$ does $0.5$ units of work in $\frac{3}{4}x$ days.
To compare their rates,we find the time $A$ would take to do the same amount of work as $B$ ($1$ unit).
Time taken by $A = \frac{3}{4}x \times 2 = \frac{3x}{2}$ days.
Given that $(A + B)$ together take $18$ days,their combined $1$-day work is $\frac{1}{18}$.
So,$\frac{1}{x} + \frac{1}{3x/2} = \frac{1}{18}$.
$\frac{1}{x} + \frac{2}{3x} = \frac{1}{18}$.
Taking the common denominator: $\frac{3 + 2}{3x} = \frac{1}{18}$.
$\frac{5}{3x} = \frac{1}{18}$.
$3x = 90$,which gives $x = 30$ days.

(A) The efficiency of a person is inversely proportional to the time taken to complete the work.
$A$'s $1$ day's work $= \frac{1}{30}$
$B$'s $1$ day's work $= \frac{1}{40}$
Therefore,the ratio of the shares of $A$ and $B$ is equal to the ratio of their work efficiencies:
Ratio $= \frac{1}{30} : \frac{1}{40} = 4 : 3$
Total parts $= 4 + 3 = 7$
$B$'s share $= \frac{3}{7} \times 2100 = 3 \times 300 = Rs. 900$

(C) The first man's $3$ days' work $= \frac{3}{6} = \frac{1}{2}$.
The second man's $3$ days' work $= \frac{3}{8}$.
The boy's $3$ days' work $= 1 - (\frac{1}{2} + \frac{3}{8}) = 1 - (\frac{4+3}{8}) = 1 - \frac{7}{8} = \frac{1}{8}$.
Since the total work is completed in $3$ days,the ratio of work done by the first man,the second man,and the boy is $\frac{1}{2} : \frac{3}{8} : \frac{1}{8}$.
Multiplying by $8$,the ratio becomes $4 : 3 : 1$.
The boy's share $= \frac{1}{4+3+1} \times 600 = \frac{1}{8} \times 600 = Rs. 75$.

(A) 's $1$ day's work $= \frac{1}{8}$.
$(A+B)$'s $1$ day's work $= \frac{1}{6}$.
$B$'s $1$ day's work $= (A+B)$'s $1$ day's work $- A$'s $1$ day's work $= \frac{1}{6} - \frac{1}{8} = \frac{4-3}{24} = \frac{1}{24}$.
The ratio of work done by $A$ and $B$ is the ratio of their $1$ day's work,which is $\frac{1}{8} : \frac{1}{24} = 3 : 1$.
Total amount $= Rs. 320$.
$B$'s share $= \frac{1}{3+1} \times 320 = \frac{1}{4} \times 320 = Rs. 80$.

(B) Let the work done by a man in one hour be $M$ and the work done by a boy in one hour be $B$.
According to the problem,the work done by $5$ men and $2$ boys in one hour is $4$ times the work done by $1$ man and $1$ boy in one hour.
So,the equation is: $(5M + 2B) = 4(1M + 1B)$.
Expanding the equation: $5M + 2B = 4M + 4B$.
Rearranging the terms to solve for $M$ and $B$: $5M - 4M = 4B - 2B$.
This simplifies to: $M = 2B$.
Therefore,the ratio of the work done by a man to that of a boy is $M/B = 2/1$,which is $2:1$.

(D) The $1$ day's work of $A, B$ and $C$ is $\frac{1}{16}, \frac{1}{32}$ and $\frac{1}{48}$ respectively.
$(A+B+C)$'s $1$ day's work $= \frac{1}{16} + \frac{1}{32} + \frac{1}{48} = \frac{6+3+2}{96} = \frac{11}{96}$.
Work done by $A, B$ and $C$ in the first $4$ days $= 4 \times \frac{11}{96} = \frac{11}{24}$.
Remaining work $= 1 - \frac{11}{24} = \frac{13}{24}$.
Let the total time taken be $x$ days. $B$ left $2$ days before completion,so $B$ worked for $(x-2)$ days. $A$ worked for the entire $x$ days. $C$ worked for $4$ days.
Since $C$ left after $4$ days,the remaining work $\frac{13}{24}$ was done by $A$ and $B$.
Work done by $A$ in $x$ days + Work done by $B$ in $(x-2)$ days $= \frac{13}{24}$.
$\frac{x}{16} + \frac{x-2}{32} = \frac{13}{24}$.
Multiply by $96$: $6x + 3(x-2) = 52$.
$6x + 3x - 6 = 52 \Rightarrow 9x = 58 \Rightarrow x = \frac{58}{9} = 6\frac{4}{9}$.
Total days $= 4$ (initial) $+ 6\frac{4}{9} = 10\frac{4}{9}$ days.

(B) Work done by $A$ in $1$ day $= \frac{1}{9}$.
Work done by $B$ in $1$ day $= \frac{1}{12}$.
Work done by $(A+B)$ in $2$ days (one pair) $= \frac{1}{9} + \frac{1}{12} = \frac{4+3}{36} = \frac{7}{36}$.
In $5$ pairs of days ($10$ days),work done $= 5 \times \frac{7}{36} = \frac{35}{36}$.
Remaining work $= 1 - \frac{35}{36} = \frac{1}{36}$.
On the $11^{th}$ day,it is $A$'s turn. $A$ completes $\frac{1}{9}$ work in $1$ day.
Time taken by $A$ to complete $\frac{1}{36}$ work $= \frac{1/36}{1/9} = \frac{9}{36} = \frac{1}{4}$ day.
Total time $= 10 + \frac{1}{4} = 10 \frac{1}{4}$ days.

(C) 's $23$ days of work $= \frac{23}{40}$.
Remaining work $= 1 - \frac{23}{40} = \frac{17}{40}$.
Now,$(A+B)$'s $1$ day of work $= \frac{1}{45} + \frac{1}{40} = \frac{8+9}{360} = \frac{17}{360}$.
Since $\frac{17}{360}$ work is done by $A$ and $B$ in $1$ day,the time taken to complete $\frac{17}{40}$ work is $\frac{17}{40} \div \frac{17}{360} = \frac{17}{40} \times \frac{360}{17} = 9$ days.
Therefore,$A$ left after $9$ days.

(B) Suppose $C$ alone can do this work in $x$ days.
Therefore,$C$ will do $\frac{1}{x}$ work in $1$ day.
Now,work done by $(B + C)$ in $1$ day $= \frac{1}{16}$.
Therefore,work done by $B$ in $1$ day $= \left(\frac{1}{16} - \frac{1}{x}\right)$.
And,work done by $(A + B)$ in $1$ day $= \frac{1}{12}$.
Therefore,work done by $A$ in $1$ day $= \frac{1}{12} - \left(\frac{1}{16} - \frac{1}{x}\right) = \frac{4-3}{48} + \frac{1}{x} = \frac{1}{48} + \frac{1}{x}$.
As per the question,work done by $A$ in $5$ days $+$ work done by $B$ in $7$ days $+$ work done by $C$ in $13$ days $= 1$ (whole work).
Therefore,$5\left(\frac{1}{48} + \frac{1}{x}\right) + 7\left(\frac{1}{16} - \frac{1}{x}\right) + \frac{13}{x} = 1$.
$\frac{5}{48} + \frac{5}{x} + \frac{7}{16} - \frac{7}{x} + \frac{13}{x} = 1$.
$\frac{5}{48} + \frac{21}{48} + \frac{11}{x} = 1$.
$\frac{26}{48} + \frac{11}{x} = 1$.
$\frac{11}{x} = 1 - \frac{13}{24} = \frac{11}{24}$.
Thus,$x = 24$.
$C$ alone would complete this work in $24$ days.

(A) Work done by $A$ in $5$ days $= \frac{5}{40} = \frac{1}{8}$.
Remaining work $= 1 - \frac{1}{8} = \frac{7}{8}$.
$B$ completes $\frac{7}{8}$ of the work in $21$ days.
Therefore,$B$ would complete the whole work in $\frac{21 \times 8}{7} = 24$ days.
Now,$A$ takes $40$ days and $B$ takes $24$ days to complete the work individually.
Working together,$A$ and $B$ would complete the work in $\frac{x \times y}{x + y}$ days,where $x = 40$ and $y = 24$.
Time taken $= \frac{40 \times 24}{40 + 24} = \frac{960}{64} = 15$ days.

(A) $(A+B)$'s $1$ day's work $= \frac{1}{30} + \frac{1}{50} = \frac{5+3}{150} = \frac{8}{150} = \frac{4}{75}$ units.
$(A+C)$'s $1$ day's work $= \frac{1}{30} + \frac{1}{40} = \frac{4+3}{120} = \frac{7}{120}$ units.
Work done in a cycle of $2$ days $= \frac{4}{75} + \frac{7}{120} = \frac{32+35}{600} = \frac{67}{600}$ units.
In $8$ cycles ($16$ days),work done $= 8 \times \frac{67}{600} = \frac{67}{75}$ units.
Remaining work $= 1 - \frac{67}{75} = \frac{8}{75}$ units.
On the $17^{th}$ day,$(A+B)$ works. They do $\frac{4}{75}$ units. Remaining work $= \frac{8}{75} - \frac{4}{75} = \frac{4}{75}$ units.
On the $18^{th}$ day,$(A+C)$ works. They do $\frac{7}{120}$ units per day. Time taken to finish remaining work $= \frac{4/75}{7/120} = \frac{4}{75} \times \frac{120}{7} = \frac{4 \times 8}{5 \times 7} = \frac{32}{35}$ days.
Total time $= 17 + \frac{32}{35} = 17 \frac{32}{35}$ days.

(A) Let $R, M,$ and $V$ represent the work done per day by Rohan,Mohit,and Vikas respectively.
Given:
$R + M = \frac{1}{8}$ (work per day)
$M + V = \frac{1}{10}$ (work per day)
$V + R = \frac{1}{12}$ (work per day)
Adding these equations:
$2(R + M + V) = \frac{1}{8} + \frac{1}{10} + \frac{1}{12}$
$2(R + M + V) = \frac{15 + 12 + 10}{120} = \frac{37}{120}$
$R + M + V = \frac{37}{240}$ (work per day by all three together)
Therefore,the time taken by all three to complete the wall is the reciprocal of the work done per day:
Time $= \frac{240}{37} \, days$.

(D) Let the total work be $18$ units.
Since $A$ and $B$ together complete the task in $18$ hours, their combined efficiency is $1$ unit/hour.
In $6$ hours, $A$ and $B$ together complete $6 \times 1 = 6$ units of work.
Remaining work $= 18 - 6 = 12$ units.
$B$ takes $36$ hours to complete the remaining $12$ units, so $B$'s efficiency $= \frac{12}{36} = \frac{1}{3}$ unit/hour.
$A$'s efficiency $= (\text{Combined efficiency}) - (B\text{'s efficiency}) = 1 - \frac{1}{3} = \frac{2}{3}$ unit/hour.
Time taken by $A$ to complete the total work alone $= \frac{\text{Total work}}{A\text{'s efficiency}} = \frac{18}{2/3} = 18 \times \frac{3}{2} = 27$ hours.

(C) Given:
$A$ alone takes $63 \, \text{hours}$.
$A + B$ together take $36 \, \text{hours}$.
Let the total work be the Least Common Multiple $(LCM)$ of $63$ and $36$, which is $252 \, \text{units}$.
Efficiency of $A = \frac{252}{63} = 4 \, \text{units/hour}$.
Efficiency of $(A + B) = \frac{252}{36} = 7 \, \text{units/hour}$.
Efficiency of $B = (A + B) - A = 7 - 4 = 3 \, \text{units/hour}$.
The ratio of work done (and thus the ratio of wages) by $A$ and $B$ is $4:3$.
Total parts $= 4 + 3 = 7$.
$B$'s share $= \frac{3}{7} \times 5950 = 3 \times 850 = Rs. 2550$.

(D) can do $50\%$ (or $1/2$) of the work in $16$ days,so $A$ can complete the whole work in $16 \times 2 = 32$ days.
$B$ can do $1/4$ of the work in $24$ days,so $B$ can complete the whole work in $24 \times 4 = 96$ days.
Total work is the $\operatorname{LCM}(32, 96) = 96$ units.
Efficiency of $A = 96 / 32 = 3$ units/day.
Efficiency of $B = 96 / 96 = 1$ unit/day.
Combined efficiency of $A$ and $B = 3 + 1 = 4$ units/day.
They need to complete $3/4$ of the total work,which is $(3/4) \times 96 = 72$ units.
Time taken by $A$ and $B$ together to complete $72$ units $= 72 / 4 = 18$ days.

(C) can complete the work in $30$ days,so $A$'s $1$ day work is $\frac{1}{30}$.
$B$ can complete the work in $40$ days,so $B$'s $1$ day work is $\frac{1}{40}$.
Combined $1$ day work of $A$ and $B$ is $\frac{1}{30} + \frac{1}{40} = \frac{4+3}{120} = \frac{7}{120}$.
Therefore,$A$ and $B$ together can complete the work in $\frac{120}{7}$ days.
$\frac{120}{7} = 17 \frac{1}{7}$ days.

(B) The formula for work completion is $\frac{M_1 \times D_1}{W_1} = \frac{M_2 \times D_2}{W_2}$.
Given:
$M_1 = 20$,$D_1 = 10 \text{ days}$,$W_1 = \frac{1}{4}$ of the work.
Remaining work $W_2 = 1 - \frac{1}{4} = \frac{3}{4}$.
Remaining time $D_2 = 40 - 10 = 30 \text{ days}$.
Let the total number of persons required for the remaining work be $M_2$.
Substituting the values:
$\frac{20 \times 10}{1/4} = \frac{M_2 \times 30}{3/4}$
$800 = \frac{M_2 \times 30 \times 4}{3}$
$800 = M_2 \times 40$
$M_2 = 20$.
Since $20$ persons are required in total for the remaining work and he already had $20$ persons (who left),the number of additional persons needed is $20 - 20 = 0$. However,checking the calculation: $\frac{200}{1/4} = 800$. $\frac{M_2 \times 30}{3/4} = M_2 \times 40$. Thus $M_2 = 20$. The question asks for the number of persons to employ to finish the remaining part. If $20$ are needed and none are currently working,he must employ $20$ persons.

(B) Let the total work be $30$ units.
Since $A$ and $B$ together complete the work in $30$ days,their combined efficiency is $(A+B) = \frac{30}{30} = 1$ unit/day.
In $6$ days,they complete $6 \times 1 = 6$ units of work.
Remaining work $= 30 - 6 = 24$ units.
This remaining $24$ units of work is completed by $B$ in $36$ days.
Therefore,$B$'s efficiency $= \frac{24}{36} = \frac{2}{3}$ units/day.
Since $(A+B) = 1$ unit/day,$A$'s efficiency $= 1 - \frac{2}{3} = \frac{1}{3}$ units/day.
Time taken by $A$ to complete the entire work $= \frac{\text{Total Work}}{\text{Efficiency of } A} = \frac{30}{1/3} = 90$ days.

(C) Let the total work be $10$ units (assuming the rate of $A+B$ is $1$ unit/day).
$A+B$ together complete $1$ unit per day.
In $2$ days,$A+B$ complete $2 \times 1 = 2$ units of work.
Remaining work $= 10 - 2 = 8$ units.
$B$ completes these $8$ units in $12$ days.
Therefore,$B$'s efficiency $= \frac{8}{12} = \frac{2}{3}$ units/day.
Since $A+B$ efficiency is $1$ unit/day,$A$'s efficiency $= 1 - \frac{2}{3} = \frac{1}{3}$ units/day.
Time taken by $A$ to complete the entire work $= \frac{\text{Total work}}{\text{Efficiency of } A} = \frac{10}{1/3} = 30$ days.

(C) The total time required to complete the job is $36 \text{ hours}$.
The labourer works for $9 \text{ hours}$.
The fraction of the work completed is $\frac{\text{Time worked}}{\text{Total time}} = \frac{9}{36} = \frac{1}{4}$.
The fraction of the work remaining is $1 - \text{Fraction completed} = 1 - \frac{1}{4} = \frac{3}{4}$.
Converting the fraction to a decimal,$\frac{3}{4} = 0.75$.

(C) The work done by $P, Q,$ and $R$ in one day is the reciprocal of the total days taken by each.
$P$'s one-day work $= 1/60$
$Q$'s one-day work $= 1/100$
$R$'s one-day work $= 1/80$
The ratio of their work efficiency is $1/60 : 1/100 : 1/80$.
To simplify,find the $LCM$ of $60, 100,$ and $80$,which is $1200$.
Multiplying by $1200$,we get the ratio: $(1200/60) : (1200/100) : (1200/80) = 20 : 12 : 15$.
The sum of the ratio parts $= 20 + 12 + 15 = 47$.
$P$'s share $= (20/47) \times 2256$.
$2256 / 47 = 48$.
$P$'s share $= 20 \times 48 = 960$.
Thus,$P$ will receive $Rs. 960$.