Time and Work Questions in English

(A) Work done by $A$ and $B$ in $1$ day $= \frac{1}{8} + \frac{1}{12} = \frac{3+2}{24} = \frac{5}{24}$.
Work done by $A$ and $B$ in $2$ days $= 2 \times \frac{5}{24} = \frac{10}{24} = \frac{5}{12}$.
Remaining work $= 1 - \frac{5}{12} = \frac{7}{12}$.
One day work of $B$ and $C$ together $= \frac{1}{12} + \frac{1}{15} = \frac{5+4}{60} = \frac{9}{60} = \frac{3}{20}$.
Days required by $B$ and $C$ to finish the remaining work $= \frac{7/12}{3/20} = \frac{7}{12} \times \frac{20}{3} = \frac{7 \times 5}{3 \times 3} = \frac{35}{9}$ days.
Total days to complete the work $= 2 + \frac{35}{9} = \frac{18+35}{9} = \frac{53}{9} = 5 \frac{8}{9}$ days.

(A) Let the efficiency of a man per hour be $E_m$ and that of a woman be $E_w$.
Given $E_w = \frac{2}{3} E_m$,which implies $E_m = \frac{3}{2} E_w$.
Total work done by $10$ men and $6$ women in $18$ days:
Daily work of $10$ men $= 10 \times 9 \times E_m = 90 E_m$.
Daily work of $6$ women $= 6 \times 7.5 \times E_w = 45 E_w$.
Total daily work $= 90 E_m + 45 E_w = 90(\frac{3}{2} E_w) + 45 E_w = 135 E_w + 45 E_w = 180 E_w$.
Total work $= 180 E_w \times 18 = 3240 E_w$.
Now,for $10$ men and $9$ women:
Daily work $= 10 \times 9 \times E_m + 9 \times 7.5 \times E_w = 90 E_m + 67.5 E_w$.
Substituting $E_m = \frac{3}{2} E_w$:
Daily work $= 90(\frac{3}{2} E_w) + 67.5 E_w = 135 E_w + 67.5 E_w = 202.5 E_w$.
Number of days required $= \frac{\text{Total Work}}{\text{Daily Work}} = \frac{3240 E_w}{202.5 E_w} = 16$ days.

(B) Let $M$ be the daily wage of a man and $W$ be the daily wage of a woman.
In $5$ days,$(4M + 6W) \times 5 = 1600 \Rightarrow 4M + 6W = 320$ (Equation $1$).
In $6$ days,$(3M + 7W) \times 6 = 1740 \Rightarrow 3M + 7W = 290$ (Equation $2$).
Multiplying Equation $1$ by $3$ and Equation $2$ by $4$:
$12M + 18W = 960$
$12M + 28W = 1160$
Subtracting the first from the second: $10W = 200 \Rightarrow W = 20$.
Substituting $W = 20$ into Equation $1$: $4M + 6(20) = 320 \Rightarrow 4M + 120 = 320 \Rightarrow 4M = 200 \Rightarrow M = 50$.
Daily wage of $7$ men and $6$ women $= 7(50) + 6(20) = 350 + 120 = 470$.
Number of days required to earn $Rs. \, 3760 = \frac{3760}{470} = 8$ days.

(B) Let the number of men,women,and children be $M, W,$ and $C$ respectively.
Given $M + W + C = 18$.
The ratio of total wages is $18: 10: 12$. Let the total wages be $18k, 10k,$ and $12k$ respectively,where $18k + 10k + 12k = 4000 \Rightarrow 40k = 4000 \Rightarrow k = 100$.
Total wages for men $= 1800$,women $= 1000$,children $= 1200$.
The ratio of individual wages is $6: 5: 3$. Let individual wages be $6y, 5y,$ and $3y$.
The number of people is given by $\frac{\text{Total Wages}}{\text{Individual Wage}}$.
$M = \frac{1800}{6y} = \frac{300}{y}$,$W = \frac{1000}{5y} = \frac{200}{y}$,$C = \frac{1200}{3y} = \frac{400}{y}$.
Sum of people: $\frac{300+200+400}{y} = 18 \Rightarrow \frac{900}{y} = 18 \Rightarrow y = 50$.
Individual wage of a woman $= 5y = 5 \times 50 = Rs. 250$.

(A) Let $M$ be the work done by $1$ man per day and $B$ be the work done by $1$ boy per day.
From the given data,the work done is proportional to the number of workers and time.
$(5M + 3B) \times 4 / 23 = (3M + 2B) \times 2 / 7$
$28(5M + 3B) = 46(3M + 2B)$
$140M + 84B = 138M + 92B$
$2M = 8B \Rightarrow M = 4B$.
Now,substitute $M = 4B$ into the first condition:
$5(4B) + 3B = 23B$ can reap $23 \text{ hectares}$ in $4 \text{ days}$.
This means $23B$ can reap $23 \text{ hectares}$ in $4 \text{ days}$,so $1B$ can reap $1 \text{ hectare}$ in $4 \text{ days}$.
Therefore,$1$ boy can reap $1/4 \text{ hectare}$ in $1 \text{ day}$.
We need to find the number of boys $x$ such that $(7M + xB)$ can reap $45 \text{ hectares}$ in $6 \text{ days}$.
$(7(4B) + xB) \times 6 / 45 = 1 \text{ (work unit)}$
$(28B + xB) \times 6 = 45 \times 4B$
$(28 + x) \times 6 = 180$
$28 + x = 30$
$x = 2$.
Thus,$2$ boys must assist $7$ men.

(A) Subhash's rate of copying = $\frac{50 \text{ pages}}{10 \text{ h}} = 5 \text{ pages/h}$.
Combined rate of Subhash and Prakash = $\frac{300 \text{ pages}}{40 \text{ h}} = 7.5 \text{ pages/h}$.
Prakash's rate of copying = $7.5 - 5 = 2.5 \text{ pages/h}$.
Time taken by Prakash to copy $30$ pages = $\frac{30 \text{ pages}}{2.5 \text{ pages/h}} = 12 \text{ h}$.

(D) Work done by $A$ in $1 \text{ day} = \frac{1}{15}$.
Work done by $(A + B)$ in $1 \text{ day} = \frac{1}{6 \frac{2}{3}} = \frac{1}{20/3} = \frac{3}{20}$.
Work done by $B$ in $1 \text{ day} = (A + B)'s \text{ work} - A's \text{ work}$.
Work done by $B$ in $1 \text{ day} = \frac{3}{20} - \frac{1}{15} = \frac{9 - 4}{60} = \frac{5}{60} = \frac{1}{12}$.
Therefore,$B$ can do the work alone in $12 \text{ days}$.

(B) The work done by $A, B$ and $C$ together in $1 \, h$ is $\frac{1}{2}$ of the total work.
The work done by $A$ alone in $1 \, h$ is $\frac{1}{6}$ of the total work.
The work done by $B$ alone in $1 \, h$ is $\frac{1}{5}$ of the total work.
Let the work done by $C$ alone in $1 \, h$ be $x$.
Therefore,$\frac{1}{6} + \frac{1}{5} + x = \frac{1}{2}$.
$x = \frac{1}{2} - (\frac{1}{6} + \frac{1}{5}) = \frac{1}{2} - \frac{5+6}{30} = \frac{1}{2} - \frac{11}{30}$.
$x = \frac{15-11}{30} = \frac{4}{30} = \frac{2}{15}$.
So,$C$ finishes $\frac{2}{15}$ of the work in $1 \, h$. The time taken by $C$ to finish the whole work is $\frac{15}{2} = 7.5 \, h$ or $7 \frac{1}{2} \, h$.

(A) Work done by $A$ in $1 \, day = \frac{1}{20}$.
Work done by $B$ in $1 \, day = \frac{1}{40}$.
Combined work done by $A$ and $B$ in $1 \, day = \frac{1}{20} + \frac{1}{40} = \frac{2+1}{40} = \frac{3}{40}$.
Work done by both in $5 \, days = 5 \times \frac{3}{40} = \frac{15}{40} = \frac{3}{8}$.
Remaining work = $1 - \frac{3}{8} = \frac{8-3}{8} = \frac{5}{8}$.

(C) The work done by $A$ in $1\, day = \frac{1}{10}$.
The work done by $B$ in $1\, day = \frac{1}{12}$.
The work done by $C$ in $1\, day = \frac{1}{15}$.
Part of the work done by all of them working together in $1\, day = \frac{1}{10} + \frac{1}{12} + \frac{1}{15}$.
Taking the least common multiple $(LCM)$ of $10, 12,$ and $15$,which is $60$:
$= \frac{6 + 5 + 4}{60} = \frac{15}{60} = \frac{1}{4}$.
Therefore,the number of days required to finish the work by $A, B,$ and $C$ working together is $4\, days$.

(A) Let the time required by $B$ to finish the whole work be $x\, \text{days}$.
Then, $A$ will take $(x-10)\, \text{days}$.
According to the problem, the combined work rate of $A$ and $B$ is $\frac{1}{12}$ of the work per day.
So, $\frac{1}{x} + \frac{1}{x-10} = \frac{1}{12}$.
$\frac{(x-10) + x}{x(x-10)} = \frac{1}{12} \Rightarrow \frac{2x-10}{x^2-10x} = \frac{1}{12}$.
Cross-multiplying gives $12(2x-10) = x^2-10x$, which simplifies to $24x - 120 = x^2 - 10x$.
Rearranging into a quadratic equation: $x^2 - 34x + 120 = 0$.
Factoring the equation: $(x-30)(x-4) = 0$.
This gives $x = 30$ or $x = 4$.
Since $x$ must be greater than $10$ (as $A$ takes $x-10$ days), $x = 4$ is rejected.
Therefore, the time taken by $B$ alone is $30\, \text{days}$.

(B) Let the total work be $1$ unit.
Rate of $(A+B+C) = \frac{1}{30}$ work per $min$.
Rate of $(A+B) = \frac{1}{50}$ work per $min$.
Rate of $C = \text{Rate of } (A+B+C) - \text{Rate of } (A+B) = \frac{1}{30} - \frac{1}{50}$.
Taking $LCM$ of $30$ and $50$ as $150$,we get:
Rate of $C = \frac{5-3}{150} = \frac{2}{150} = \frac{1}{75}$ work per $min$.
Therefore,$C$ alone can complete the work in $75\, min$.

(A) The work done by $A$ and $B$ together in $1$ day is $\frac{1}{15}$.
The work done by $B$ alone in $1$ day is $\frac{1}{20}$.
The work done by $A$ alone in $1$ day is $\frac{1}{15} - \frac{1}{20}$.
Taking the least common multiple $(LCM)$ of $15$ and $20$,which is $60$,we get $\frac{4-3}{60} = \frac{1}{60}$.
Therefore,$A$ alone can complete the work in $60$ days.

(C) Given that $(A + B)$'s $1$ day work $= \frac{1}{10}$.
$A$'s $1$ day work $= \frac{1}{30}$.
Therefore,$B$'s $1$ day work $= (A + B)$'s $1$ day work $- A$'s $1$ day work.
$B$'s $1$ day work $= \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30} = \frac{1}{15}$.
Thus,$B$ alone can complete the work in $15$ days.

(A) The work done by $A$ in $1 \, day = 1/6$.
The work done by $B$ in $1 \, day = 1/12$.
When working together,their combined work in $1 \, day = 1/6 + 1/12 = (2+1)/12 = 3/12 = 1/4$.
They complete the work in $4 \, days$.
In $4 \, days$,the work done by $A = 4 \times (1/6) = 4/6 = 2/3$.
Alternatively,the ratio of their work efficiency is $1/6 : 1/12 = 2 : 1$. Thus,$A$ does $2/3$ of the total work.

(A) Let the work done by $A, B,$ and $C$ in one day be $a, b,$ and $c$ respectively.
Given:
$a + b = \frac{1}{10}$ (Equation $1$)
$b + c = \frac{1}{12}$ (Equation $2$)
$a + c = \frac{1}{15}$ (Equation $3$)
Adding all three equations:
$2(a + b + c) = \frac{1}{10} + \frac{1}{12} + \frac{1}{15} = \frac{6 + 5 + 4}{60} = \frac{15}{60} = \frac{1}{4}$
Therefore,$a + b + c = \frac{1}{8}$ (Equation $4$)
To find the time taken by $A$ alone,we subtract Equation $2$ from Equation $4$:
$a = (a + b + c) - (b + c) = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$
Thus,$A$ will take $24$ days to complete the work alone.

(C) Given:
$(A+B)$'s $1$ day work $= \frac{1}{8}$
$(B+C)$'s $1$ day work $= \frac{1}{12}$
$(C+A)$'s $1$ day work $= \frac{1}{8}$
Adding these equations:
$2(A+B+C)$'s $1$ day work $= \frac{1}{8} + \frac{1}{12} + \frac{1}{8} = \frac{3+2+3}{24} = \frac{8}{24} = \frac{1}{3}$
Therefore,$(A+B+C)$'s $1$ day work $= \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$
Thus,$A, B,$ and $C$ together can complete the work in $6$ days.

(D) Let the work done by $A, B,$ and $C$ in one day be $A, B,$ and $C$ respectively.
Given:
$A + B = \frac{1}{30}$
$B + C = \frac{1}{20}$
$A + C = \frac{1}{15}$
Adding these equations: $2(A + B + C) = \frac{1}{30} + \frac{1}{20} + \frac{1}{15} = \frac{2 + 3 + 4}{60} = \frac{9}{60} = \frac{3}{20}$.
Therefore,$(A + B + C) = \frac{3}{20} \times \frac{1}{2} = \frac{3}{40}$.
The time taken by all three working together is the reciprocal of their combined one-day work: $\frac{40}{3} = 13 \frac{1}{3} \text{ days}$.

(B) The work done by $A$ in $1\, h$ is $\frac{1}{4}$.
The work done by $(B+C)$ in $1\, h$ is $\frac{1}{3}$.
The work done by $(A+C)$ in $1\, h$ is $\frac{1}{2}$.
First,find the work done by $C$ in $1\, h$ by subtracting $A$'s work from $(A+C)$'s work:
$C = (A+C) - A = \frac{1}{2} - \frac{1}{4} = \frac{2-1}{4} = \frac{1}{4}$.
Now,find the work done by $B$ in $1\, h$ by subtracting $C$'s work from $(B+C)$'s work:
$B = (B+C) - C = \frac{1}{3} - \frac{1}{4} = \frac{4-3}{12} = \frac{1}{12}$.
Therefore,$B$ alone will take $12\, h$ to complete the work.

(A) Let the total work be $1$ unit.
$A$'s $1$ day work $= \frac{1}{12}$ unit.
$B$ is $60\%$ more efficient than $A$,so $B$'s efficiency $= A$'s efficiency $\times (1 + 0.60) = \frac{1}{12} \times 1.6 = \frac{1}{12} \times \frac{16}{10} = \frac{1}{12} \times \frac{8}{5} = \frac{8}{60} = \frac{2}{15}$ unit/day.
Time taken by $B$ to complete the work $= \frac{\text{Total Work}}{\text{Efficiency of } B} = \frac{1}{2/15} = \frac{15}{2} = 7\frac{1}{2}$ days.

(B) Let the efficiency of $B$ be $1$ unit/day. Since $A$ is twice as good as $B$,the efficiency of $A$ is $2$ units/day.
Combined efficiency of $A$ and $B = 2 + 1 = 3$ units/day.
Total work = (Combined efficiency) $\times$ (Total days) = $3 \times 14 = 42$ units.
Time taken by $A$ alone = $\frac{\text{Total work}}{\text{Efficiency of } A} = \frac{42}{2} = 21$ days.

(C) Kamal's $1$ day work $= \frac{1}{15}$.
Sita is $50\%$ more efficient than Kamal,so Sita's efficiency $= 1 + \frac{50}{100} = 1.5 = \frac{3}{2}$ times Kamal's efficiency.
Sita's $1$ day work $= \frac{3}{2} \times \frac{1}{15} = \frac{1}{10}$.
Therefore,the time taken by Sita to complete the work $= 10$ days.

(A) Let the time taken by $B$ to complete the work alone be $x$ days.
Therefore,the work done by $B$ in one day is $\frac{1}{x}$.
$A$ does half the work of $B$ in $\frac{3}{4}$ of the time taken by $B$.
Let $W_B$ be the work done by $B$ in time $T$. Then $W_A = \frac{1}{2} W_B$ in time $T_A = \frac{3}{4} T$.
Efficiency of $B$ $(E_B)$ = $\frac{W_B}{T}$.
Efficiency of $A$ $(E_A)$ = $\frac{W_A}{T_A} = \frac{0.5 W_B}{0.75 T} = \frac{1/2}{3/4} E_B = \frac{2}{3} E_B$.
Given that together they take $18$ days to complete the work,their combined efficiency is $\frac{1}{18}$.
$E_A + E_B = \frac{1}{18} \Rightarrow \frac{2}{3} E_B + E_B = \frac{1}{18}$.
$\frac{5}{3} E_B = \frac{1}{18} \Rightarrow E_B = \frac{1}{18} \times \frac{3}{5} = \frac{1}{30}$.
Since $E_B = \frac{1}{x}$,we have $x = 30$ days.
Thus,$B$ takes $30$ days to complete the work alone.

(B) Let the work done by $1$ man in $1$ day be $M$ and by $1$ woman in $1$ day be $W$.
Given that $4$ men can complete the work in $20$ days,the total work is $4 \times 20 = 80$ man-days.
Thus,$1$ man's $1$ day work is $M = 1/80$.
Given that $2$ men and $3$ women can complete the work in $20$ days,we have: $2M + 3W = 1/20$.
Substituting $M = 1/80$: $2(1/80) + 3W = 1/20 \Rightarrow 1/40 + 3W = 1/20$.
$3W = 1/20 - 1/40 = 1/40 \Rightarrow W = 1/120$.
Now,we need to find the time taken by $3$ men and $3$ women to complete the work.
Total work done by $3$ men and $3$ women in $1$ day $= 3M + 3W = 3(1/80) + 3(1/120) = 3/80 + 1/40 = (3+2)/80 = 5/80 = 1/16$.
Therefore,the time required is $16$ days.

(B) The formula for work completion is given by $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$.
Here,$M_1 = 15$ (men),$D_1 = 20$ (days),$H_1 = 8$ (hours/day),and $W_1 = 1$ (total work).
For the second case,$M_2 = 20$ (men),$D_2 = 12$ (days),$H_2 = ?$ (hours/day),and $W_2 = 1$ (same work).
Substituting the values into the formula:
$15 \times 20 \times 8 = 20 \times 12 \times H_2$
$2400 = 240 \times H_2$
$H_2 = \frac{2400}{240} = 10$ hours per day.

(C) Total work $= 45 \times 16 = 720$ man-days.
Work done by $45$ men in $4$ days $= 45 \times 4 = 180$ man-days.
Remaining work $= 720 - 180 = 540$ man-days.
New number of men $= 45 + 36 = 81$ men.
Time taken to complete the remaining work $= \frac{540}{81}$ days.
$= \frac{540 \div 27}{81 \div 27} = \frac{20}{3} = 6\frac{2}{3}$ days.

(B) The relationship between the number of men $(M)$ and the number of days $(D)$ required to complete a fixed amount of work is inversely proportional,given by the formula: $M_1 \times D_1 = M_2 \times D_2$.
Given:
$M_1 = 10$ men
$D_1 = 12$ days
$M_2 = 12$ men
$D_2 = ?$
Substituting the values into the formula:
$10 \times 12 = 12 \times D_2$
$120 = 12 \times D_2$
$D_2 = \frac{120}{12}$
$D_2 = 10$ days.
Therefore,$12$ men will take $10$ days to complete the same work.

(A) The formula for work completion is $M_1 \times D_1 \times H_1 = M_2 \times D_2 \times H_2$, where $M$ is the number of men, $D$ is the number of days, and $H$ is the number of hours per day.
Given:
$M_1 = 10, D_1 = 18, H_1 = 6$
$M_2 = 15, D_2 = 12, H_2 = ?$
Substituting the values into the formula:
$10 \times 18 \times 6 = 15 \times 12 \times H_2$
$1080 = 180 \times H_2$
$H_2 = \frac{1080}{180} = 6\, \text{hours per day}$.

(B) Let the original number of persons be $x$.
According to the problem,the total work remains constant.
Work done = $\text{Number of persons} \times \text{Time taken}$.
Original work = $x \times 55$.
If there were $6$ more persons,the number of persons becomes $(x + 6)$ and the time taken is $(55 - 11) = 44 \text{ days}$.
Equating the work: $55x = 44(x + 6)$.
Dividing both sides by $11$: $5x = 4(x + 6)$.
Expanding the equation: $5x = 4x + 24$.
Subtracting $4x$ from both sides: $x = 24$.
Therefore,the original number of persons was $24$.

(C) Given: $6$ men = $12$ women,so $1$ man = $2$ women.
Work done by $6$ men in $20$ days = $1$ unit of work.
Total work capacity of $6$ men = $6 \times 20 = 120$ man-days.
Total work capacity of $12$ women = $12 \times 20 = 240$ woman-days.
We need to complete $2$ units of work.
Combined efficiency of $8$ men and $16$ women:
$8$ men = $8 \times 2 = 16$ women.
Total workers = $16$ women + $16$ women = $32$ women.
Since $12$ women take $240$ days for $1$ unit of work,$1$ woman takes $240 \times 12 = 2880$ days for $1$ unit.
For $32$ women,time taken for $1$ unit = $2880 / 32 = 90$ days.
For $2$ units of work,time taken = $90 \times 2 = 180$ days.
Wait,re-evaluating: $6$ men = $12$ women $\Rightarrow 1$ man = $2$ women.
$8$ men + $16$ women = $8(2) + 16 = 32$ women.
$12$ women do $1$ work in $20$ days.
$32$ women do $1$ work in $(12 \times 20) / 32 = 240 / 32 = 7.5$ days.
For $2$ times the work,time = $7.5 \times 2 = 15$ days.

(A) 's one day work $= \frac{1}{12}$.
In $3$ days,the work done by $A = 3 \times \frac{1}{12} = \frac{1}{4}$.
Remaining work $= 1 - \frac{1}{4} = \frac{3}{4}$.
Let $B$ alone finish the work in $x$ days,so $B$'s one day work $= \frac{1}{x}$.
Together,$A$ and $B$ complete the remaining work in $3$ days: $3 \times (\frac{1}{12} + \frac{1}{x}) = \frac{3}{4}$.
Dividing by $3$: $\frac{1}{12} + \frac{1}{x} = \frac{1}{4}$.
$\frac{1}{x} = \frac{1}{4} - \frac{1}{12} = \frac{3-1}{12} = \frac{2}{12} = \frac{1}{6}$.
Therefore,$x = 6$ days.

(C) Time taken by $B$ to complete the work $= 12$ days.
Time taken by $A$ and $B$ together to complete the work $= 8$ days.
Efficiency of $A$ and $B$ together $= \frac{1}{8}$ work/day.
Efficiency of $B$ alone $= \frac{1}{12}$ work/day.
Efficiency of $A$ alone $= \frac{1}{8} - \frac{1}{12} = \frac{3-2}{24} = \frac{1}{24}$ work/day.
Thus,$A$ alone can complete the work in $24$ days.
Work done by $B$ in $4$ days $= 4 \times \frac{1}{12} = \frac{1}{3}$.
Remaining work $= 1 - \frac{1}{3} = \frac{2}{3}$.
Time taken by $A$ to complete the remaining work $= \frac{\text{Remaining work}}{\text{Efficiency of } A} = \frac{2/3}{1/24} = \frac{2}{3} \times 24 = 16$ days.

(C) The $1$ day work of $B$ and $C$ is $\frac{1}{9} + \frac{1}{12} = \frac{4+3}{36} = \frac{7}{36}$.
The work done by $B$ and $C$ in $3$ days is $3 \times \frac{7}{36} = \frac{21}{36} = \frac{7}{12}$.
The remaining work is $1 - \frac{7}{12} = \frac{5}{12}$.
Since $A$ can finish the whole work in $24$ days,$A$'s $1$ day work is $\frac{1}{24}$.
The time taken by $A$ to complete the remaining work is $\frac{5/12}{1/24} = \frac{5}{12} \times 24 = 10$ days.

(A) Total work hours for $A = 7 \times 6 = 42$ hours.
Total work hours for $B = 8 \times 7 = 56$ hours.
$A$'s $1$ hour work $= \frac{1}{42}$ of the total work.
$B$'s $1$ hour work $= \frac{1}{56}$ of the total work.
Combined $1$ hour work of $A$ and $B = \frac{1}{42} + \frac{1}{56} = \frac{4 + 3}{168} = \frac{7}{168} = \frac{1}{24}$ of the total work.
If they work $8$ hours a day together,their daily work $= 8 \times \frac{1}{24} = \frac{1}{3}$ of the total work.
Therefore,the time required to complete the work $= \frac{1}{1/3} = 3$ days.

(C) The work done by $A$ in $1 \text{ day} = \frac{1}{10}$.
The work done by $B$ in $1 \text{ day} = \frac{1}{15}$.
Since they work on alternate days starting with $A$,the work done in a cycle of $2 \text{ days}$ is: $\frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6}$.
This means in $2 \text{ days}$,$\frac{1}{6}$ of the work is completed.
To complete the whole work ($1$ unit),we need $6$ such cycles.
Total days $= 6 \times 2 = 12 \text{ days}$.

(B) 's $1$ day work $= 1/9$.
$B$'s $1$ day work $= 1/15$.
$(A+B)$'s $2$ days work (alternate) $= 1/9 + 1/15 = (5+3)/45 = 8/45$.
In $10$ days ($5$ pairs of $A$ and $B$),work done $= 5 \times (8/45) = 40/45 = 8/9$.
Remaining work $= 1 - 8/9 = 1/9$.
On the $11$th day,it is $A$'s turn. $A$ does $1/9$ of the work in $(1/9) / (1/9) = 1$ day.
Total time $= 10 + 1 = 11 \, \text{days}$.

(C) $(A+B)$'s $1 \, \text{day}$ work $= \frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$.
$(A+B)$'s $7 \, \text{days}$ work $= 7 \times \frac{1}{12} = \frac{7}{12}$.
Remaining work $= 1 - \frac{7}{12} = \frac{5}{12}$.
$C$ finishes this remaining $\frac{5}{12}$ part of the work in $10 \, \text{days}$.
Therefore, the time taken by $C$ to finish the full work $= 10 \div \frac{5}{12} = 10 \times \frac{12}{5} = 24 \, \text{days}$.

(C) The work done by $A$ in $1 \, \text{day}$ is $1/12$ and by $B$ in $1 \, \text{day}$ is $1/15$.
Together,their $1 \, \text{day}$ work is $(1/12 + 1/15) = (5+4)/60 = 9/60 = 3/20$.
They worked together for $4 \, \text{days}$,so the work completed is $4 \times (3/20) = 12/20 = 3/5$.
The remaining work is $1 - 3/5 = 2/5$.
$B$ completes the remaining work at a rate of $1/15$ per day.
Time taken by $B$ to complete the remaining work $= (2/5) / (1/15) = (2/5) \times 15 = 6 \, \text{days}$.

(B) Let $A$ leave after $x$ days.
$A$'s one day work = $\frac{1}{45}$
$B$'s one day work = $\frac{1}{40}$
Since $A$ and $B$ worked together for $x$ days,and $B$ worked alone for $23$ days,the total work done is:
$x \left( \frac{1}{45} + \frac{1}{40} \right) + 23 \left( \frac{1}{40} \right) = 1$
$\frac{x}{45} + \frac{x}{40} + \frac{23}{40} = 1$
$\frac{x}{45} + \frac{x+23}{40} = 1$
Taking the $LCM$ of $45$ and $40$,which is $360$:
$\frac{8x + 9(x+23)}{360} = 1$
$8x + 9x + 207 = 360$
$17x = 360 - 207$
$17x = 153$
$x = 9$
Therefore,$A$ left after $9$ days.

(C) The work rates of $A, B$ and $C$ are $\frac{1}{6}, \frac{1}{12}$ and $\frac{1}{15}$ units per day respectively.
After $\frac{1}{8}$ of the work is completed,the remaining work is $1 - \frac{1}{8} = \frac{7}{8}$ units.
The combined work rate of $A$ and $B$ is $\frac{1}{6} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12} = \frac{1}{4}$ units per day.
Time taken by $A$ and $B$ to complete the remaining work $= \frac{\text{Remaining Work}}{\text{Combined Rate of } A+B} = \frac{7/8}{1/4} = \frac{7}{8} \times 4 = \frac{7}{2} = 3 \frac{1}{2}$ days.

(B) Given that $4$ men,$6$ women,or $10$ children can paint a house in $5$ days.
Work done by $1$ man in $1$ day $= \frac{1}{4 \times 5} = \frac{1}{20}$ of the house.
Work done by $1$ woman in $1$ day $= \frac{1}{6 \times 5} = \frac{1}{30}$ of the house.
Work done by $1$ child in $1$ day $= \frac{1}{10 \times 5} = \frac{1}{50}$ of the house.
The group consists of a couple (a man and a woman) and $5$ sons (children).
Total work done by them in $1$ day $= 1 \times (\frac{1}{20}) + 1 \times (\frac{1}{30}) + 5 \times (\frac{1}{50})$
$= \frac{1}{20} + \frac{1}{30} + \frac{1}{10} = \frac{3 + 2 + 6}{60} = \frac{11}{60}$.
Time required to finish the work $= \frac{1}{\frac{11}{60}} = \frac{60}{11} = 5 \frac{5}{11}$ days.

(B) The formula for work completion is $M_{1} \times D_{1} \times H_{1} = M_{2} \times D_{2} \times H_{2}$,where $M$ is the number of persons,$D$ is the number of days,and $H$ is the hours worked per day.
Given:
$M_{1} = 39, D_{1} = 12, H_{1} = 5$
$M_{2} = 30, H_{2} = 6, D_{2} = ?$
Substituting the values into the formula:
$39 \times 12 \times 5 = 30 \times D_{2} \times 6$
$2340 = 180 \times D_{2}$
$D_{2} = \frac{2340}{180} = 13$
Therefore,$30$ persons working $6$ hours per day will complete the work in $13$ days.

(B) Let the original number of carpenters be $x$.
Since the total work remains the same,we use the formula: $\text{Number of men} \times \text{Time} = \text{Constant Work}$.
According to the problem,$x$ carpenters can complete the work in $9$ days,so total work $= 9x$.
If $5$ carpenters are absent,the number of remaining carpenters is $(x - 5)$,and they complete the work in $12$ days,so total work $= 12(x - 5)$.
Equating the two expressions for total work:
$9x = 12(x - 5)$
$9x = 12x - 60$
$60 = 12x - 9x$
$60 = 3x$
$x = 20$.
Thus,the original number of carpenters was $20$.

(B) Let the work done by one woman in one day be $W$ and by one man be $M$.
Total work done by $20$ women in $16$ days $= 20 \times 16 \times W = 320W$.
Total work done by $16$ men in $15$ days $= 16 \times 15 \times M = 240M$.
Since the work is the same,$320W = 240M$.
Therefore,the ratio of the capacity of a man to a woman is $\frac{M}{W} = \frac{320}{240} = \frac{4}{3}$.
Thus,the ratio is $4:3$.

(B) Total work = $60 \times 250 = 15000$ man-days.
Work completed in $200$ days = $60 \times 200 = 12000$ man-days.
Remaining work = $15000 - 12000 = 3000$ man-days.
Total time allotted = $250$ days.
Time already spent = $200$ days (work) + $10$ days (stoppage) = $210$ days.
Remaining time = $250 - 210 = 40$ days.
Let the total number of men required to finish the remaining work in $40$ days be $x$.
$x \times 40 = 3000$.
$x = \frac{3000}{40} = 75$ men.
Additional men required = $75 - 60 = 15$ men.

(D) Let the total work be $1$ unit.
Work done by $P$ and $Q$ in $1 \, \text{day} = \frac{1}{6}$.
Work done by $Q$ in $1 \, \text{day} = \frac{1}{8}$.
Work done by $P$ in $1 \, \text{day} = (\text{Work done by } P+Q) - (\text{Work done by } Q) = \frac{1}{6} - \frac{1}{8}$.
Taking the least common multiple of $6$ and $8$,which is $24$,we get: $\frac{4-3}{24} = \frac{1}{24}$.
Therefore,$P$ alone can finish the work in $24 \, \text{days}$.

(B) Let the total sum of money be $S$.
Daily wage of $X = \frac{S}{21}$.
Daily wage of $Y = \frac{S}{28}$.
Combined daily wage of $X$ and $Y = \frac{S}{21} + \frac{S}{28} = \frac{4S + 3S}{84} = \frac{7S}{84} = \frac{S}{12}$.
Number of days for which the money is sufficient for both = $\frac{\text{Total Money}}{\text{Combined Daily Wage}} = \frac{S}{S/12} = 12$ days.
Alternatively,using the formula: $\frac{x \times y}{x + y} = \frac{21 \times 28}{21 + 28} = \frac{588}{49} = 12$ days.

(C) Given that $3$ men = $5$ women.
Therefore,$1$ man = $\frac{5}{3}$ women.
So,$5$ men = $5 \times \frac{5}{3} = \frac{25}{3}$ women.
Total work force in terms of women = $5$ men + $6$ women = $\frac{25}{3} + 6 = \frac{25 + 18}{3} = \frac{43}{3}$ women.
We know that $M_1 D_1 = M_2 D_2$,where $M$ is the number of workers and $D$ is the number of days.
Given $M_1 = 5$ women,$D_1 = 43$ days.
We need to find $D_2$ for $M_2 = \frac{43}{3}$ women.
$5 \times 43 = \frac{43}{3} \times D_2$.
$D_2 = \frac{5 \times 43 \times 3}{43} = 5 \times 3 = 15$ days.

(D) Let the work done by $1$ man in $1$ day be $M$ and by $1$ boy in $1$ day be $B$.
According to the problem:
$8(3M + 4B) = 1$ (Total work)
$24M + 32B = 1$ .....$(1)$
$6(4M + 4B) = 1$
$24M + 24B = 1$ .....$(2)$
Subtracting equation $(2)$ from $(1)$:
$(24M + 32B) - (24M + 24B) = 1 - 1$
$8B = 0$,which implies $B = 0$ (The boys contribute no work).
Substituting $B = 0$ in equation $(2)$:
$24M = 1 \Rightarrow M = 1/24$.
So,$1$ man completes the work in $24$ days.
We need to find the time taken by $2$ men and $4$ boys:
Work done by $(2M + 4B)$ in $1$ day $= 2(1/24) + 4(0) = 2/24 = 1/12$.
Therefore,$2$ men and $4$ boys will finish the work in $12$ days.

(B) Let $M$ be the work done by $1$ man in $1$ day and $W$ be the work done by $1$ woman in $1$ day.
From the given information:
$10(2M + 3W) = 1$ (Total work)
$10(4M) = 1$ (Total work)
Equating the two: $10(4M) = 10(2M + 3W) \Rightarrow 4M = 2M + 3W \Rightarrow 2M = 3W$.
Thus,$1$ man's work is equivalent to $1.5$ women's work.
From $4M$ in $10$ days,total work $= 40$ man-days.
We need to find the time for $3$ men and $3$ women.
Since $2M = 3W$,$3$ men $+ 3$ women $= 3$ men $+ 2$ men $= 5$ men.
Using the formula $M_1 D_1 = M_2 D_2$:
$4 \text{ men} \times 10 \text{ days} = 5 \text{ men} \times D_2 \text{ days}$.
$D_2 = \frac{40}{5} = 8$ days.