Time and Work Questions in English

(D) Let the time taken by $1$ man to complete the work be $x$ days. Then,the time taken by $1$ child is $2x$ days.
Since work is inversely proportional to time,the efficiency of $1$ child is half the efficiency of $1$ man.
Therefore,$1$ man $= 2$ children,or $1$ child $= 0.5$ men.
Given that $8$ children and $12$ men complete the work in $9$ days.
Converting children to men: $8$ children $= 8 \times 0.5 = 4$ men.
Total men $= 4 + 12 = 16$ men.
So,$16$ men complete the work in $9$ days.
Using the formula $M_1 D_1 = M_2 D_2$:
$16 \times 9 = 12 \times D_2$
$D_2 = \frac{16 \times 9}{12} = \frac{144}{12} = 12$ days.
Thus,$12$ men will finish the work in $12$ days.

(A) Daily earning of $(A+B+C) = 2700 / 18 = Rs. 150$ ... $(i)$
Daily earning of $(A+C) = 940 / 10 = Rs. 94$ ... $(ii)$
Daily earning of $(B+C) = 1520 / 20 = Rs. 76$ ... $(iii)$
From $(i)$ and $(ii)$,we get $B = (A+B+C) - (A+C) = 150 - 94 = Rs. 56$.
Substitute $B = 56$ into $(iii)$:
$56 + C = 76$
$C = 76 - 56 = Rs. 20$.
Therefore,the daily earning of $C$ is $Rs. 20$.

(A) Sita's work rate is $\frac{1}{8}$ field/hr and Gita's work rate is $\frac{1}{12}$ field/hr.
In the first $2 \, hrs$ (one hour each),the work done is $\frac{1}{8} + \frac{1}{12} = \frac{3+2}{24} = \frac{5}{24}$ of the field.
In $8 \, hrs$ ($4$ cycles of $2 \, hrs$),the work done is $\frac{5}{24} \times 4 = \frac{5}{6}$ of the field.
Remaining work $= 1 - \frac{5}{6} = \frac{1}{6}$.
In the $9^{th} \, hr$,Sita works and completes $\frac{1}{8}$ of the field.
Remaining work after $9 \, hrs = \frac{1}{6} - \frac{1}{8} = \frac{4-3}{24} = \frac{1}{24}$.
In the $10^{th} \, hr$,Gita works. She completes $\frac{1}{12}$ of the field in $1 \, hr$,so she will complete $\frac{1}{24}$ of the field in $\frac{1}{24} \div \frac{1}{12} = 0.5 \, hr$.
Total time $= 9.5 \, hrs$.
Starting at $9:00 \, a.m.$,$9.5 \, hrs$ later is $6:30 \, p.m.$

(B) Let the daily wage of a woman be $w$. Then,the daily wage of a man is $2w$.
Total wages for $45$ women in $48$ days: $45 \times 48 \times w = 46575$.
Therefore,$w = \frac{46575}{45 \times 48} = \frac{1035}{48} = Rs. 21.5625$.
Let the number of men be $x$. The total wages for $x$ men in $16$ days is $x \times 16 \times (2w) = 17250$.
Substituting $w = \frac{46575}{2160}$:
$x \times 32 \times \frac{46575}{2160} = 17250$.
$x = \frac{17250 \times 2160}{32 \times 46575} = \frac{37260000}{1490400} = 25$.
Thus,$25$ men are required.

(D) Given that $5$ men $= 7$ women earn $Rs. 5,250$ per day.
Therefore,the daily earning of $1$ woman $= \frac{5250}{7} = Rs. 750$.
Since $5$ men $= 7$ women,the daily earning of $1$ man $= \frac{7}{5} \times 750 = Rs. 1,050$.
We need to find the earning of $7$ men and $13$ women.
Total earning $= (7 \times 1050) + (13 \times 750)$.
Total earning $= 7350 + 9750 = Rs. 17,100$.

(D) The work done by $A, B$ and $C$ in $1$ day is $\frac{1}{4}$.
The work done by $A$ in $1$ day is $\frac{1}{12}$.
The work done by $B$ in $1$ day is $\frac{1}{18}$.
Let the work done by $C$ in $1$ day be $x$.
Then,$\frac{1}{12} + \frac{1}{18} + x = \frac{1}{4}$.
$x = \frac{1}{4} - (\frac{1}{12} + \frac{1}{18})$.
$x = \frac{1}{4} - (\frac{3+2}{36}) = \frac{1}{4} - \frac{5}{36}$.
$x = \frac{9-5}{36} = \frac{4}{36} = \frac{1}{9}$.
Therefore,$C$ can complete the work alone in $9$ days.

(A) 's $1$ day's work $= \frac{1}{20}$.
$A$'s $4$ days' work $= \frac{4}{20} = \frac{1}{5}$.
Remaining work $= 1 - \frac{1}{5} = \frac{4}{5}$.
This remaining work is completed by $A$ and $B$ together.
$(A+B)$'s $1$ day's work $= \frac{1}{20} + \frac{1}{12} = \frac{3+5}{60} = \frac{8}{60} = \frac{2}{15}$.
Time taken by $(A+B)$ to complete $\frac{4}{5}$ of the work $= \frac{4/5}{2/15} = \frac{4}{5} \times \frac{15}{2} = 6$ days.
Total time for which the work lasted $= 4 + 6 = 10$ days.

(A) Let the work done by a man,a woman,and a boy in $1$ day be $M, W,$ and $B$ respectively.
Given that $(M + W + B) = \frac{1}{3}$ of the work per day.
Given that $M = \frac{1}{6}$ of the work per day.
Given that $B = \frac{1}{18}$ of the work per day.
To find the work done by a woman in $1$ day $(W)$:
$W = (M + W + B) - M - B$
$W = \frac{1}{3} - \frac{1}{6} - \frac{1}{18}$
Taking the least common multiple $(LCM)$ of $3, 6,$ and $18$,which is $18$:
$W = \frac{6 - 3 - 1}{18} = \frac{2}{18} = \frac{1}{9}$
Therefore,a woman alone will take $9$ days to complete the work.

(A) Let the original number of men be $x$.
According to the problem,$x$ men can complete the job in $20$ days.
Total work = $20x$ man-days.
When $12$ men did not turn up,the number of men remaining is $(x - 12)$.
These $(x - 12)$ men completed the job in $32$ days.
Total work = $32(x - 12)$ man-days.
Since the total work remains the same,we equate the two expressions:
$20x = 32(x - 12)$
$20x = 32x - 384$
$384 = 32x - 20x$
$384 = 12x$
$x = \frac{384}{12} = 32$.
Therefore,the original number of men was $32$.

(D) Given that $16$ men = $20$ women,which implies $4$ men = $5$ women.
$16$ men can complete the work in $25$ days.
Total work = $16 \times 25 = 400$ man-days.
We need to find the time taken by $28$ men and $15$ women.
Convert the group into equivalent men: $15$ women = $15 \times (4/5)$ men = $12$ men.
Total men = $28 + 12 = 40$ men.
Time taken = $\frac{\text{Total work}}{\text{Total men}} = \frac{400}{40} = 10$ days.

(D) Let the original number of men be $x$.
According to the formula for work done,$M_1 \times D_1 = M_2 \times D_2$,where $M$ is the number of men and $D$ is the number of days.
Given: $M_1 = x$,$D_1 = 40$,$M_2 = (x + 45)$,and $D_2 = 25$.
Substituting the values into the equation:
$x \times 40 = (x + 45) \times 25$
Divide both sides by $5$:
$x \times 8 = (x + 45) \times 5$
$8x = 5x + 225$
$8x - 5x = 225$
$3x = 225$
$x = 75$
Therefore,the original number of men employed was $75$.

(C) Given: $M_1 = 7$,$D_1 = 12$,$W_1 = 1$.
Let the total number of men required for the second task be $M_2$. We are given $D_2 = 8$ and $W_2 = 2$ (double the work).
Using the formula $\frac{M_1 D_1}{W_1} = \frac{M_2 D_2}{W_2}$:
$\frac{7 \times 12}{1} = \frac{M_2 \times 8}{2}$
$84 = 4 M_2$
$M_2 = \frac{84}{4} = 21$.
The number of additional men required is $M_2 - M_1 = 21 - 7 = 14$.

(C) $6$ men $= 12$ women,which implies $1$ man $= 2$ women.
Now,$8$ men $+ 16$ women $= (8 \times 2 + 16)$ women $= 32$ women.
Let the initial work be $W_1 = 1$ unit and the new work be $W_2 = 2$ units.
Using the formula $\frac{M_1 D_1}{W_1} = \frac{M_2 D_2}{W_2}$:
$\frac{12 \times 20}{1} = \frac{32 \times D_2}{2}$
$240 = 16 \times D_2$
$D_2 = \frac{240}{16} = 15$ days.

(C) Work done by $A$ in $1$ day $= \frac{1}{9}$.
Work done by $B$ in $1$ day $= \frac{1}{12}$.
Work done in a cycle of $2$ days (starting with $A$) $= \frac{1}{9} + \frac{1}{12} = \frac{4+3}{36} = \frac{7}{36}$.
In $10$ days ($5$ cycles),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. $A$ does $\frac{1}{9}$ 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.

(A) Given: $10$ men = $18$ boys,which implies $5$ men = $9$ boys.
To convert $15$ men into boys: $15$ men = $3 \times (5$ men) = $3 \times 9$ boys = $27$ boys.
Total workers = $15$ men + $33$ boys = $27$ boys + $33$ boys = $60$ boys.
Using the formula $\frac{M_1 D_1}{W_1} = \frac{M_2 D_2}{W_2}$:
Here,$M_1 = 18$ boys,$D_1 = 15$ days,$W_1 = 1$ unit of work.
$M_2 = 60$ boys,$D_2 = x$ days,$W_2 = 2$ units of work.
$\frac{18 \times 15}{1} = \frac{60 \times x}{2}$.
$270 = 30x$.
$x = \frac{270}{30} = 9$ days.

(B) Let the efficiency of one woman be $W$ and 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,we equate the two:
$320W = 240M$.
Dividing both sides by $80$,we get $4W = 3M$.
Therefore,the ratio of the working capacity of a man to that of a woman is $\frac{M}{W} = \frac{4}{3}$,which is $4:3$.

(C) First,calculate the total time taken by each person in hours:
Ram's total time $= 15 \text{ days} \times 8 \text{ hrs/day} = 120 \text{ hrs}$.
Fari's total time $= \frac{20}{3} \text{ days} \times 9 \text{ hrs/day} = 60 \text{ hrs}$.
Now,calculate the combined rate of work per hour:
Combined work per hour $= \frac{1}{120} + \frac{1}{60} = \frac{1+2}{120} = \frac{3}{120} = \frac{1}{40} \text{ of the work per hour}$.
So,they finish the work together in $40 \text{ hrs}$.
Given that they work $10 \text{ hrs/day}$,the number of days required is:
$\text{Days} = \frac{40 \text{ hrs}}{10 \text{ hrs/day}} = 4 \text{ days}$.

(C) Let $M$ be the work done by $1$ man per day and $W$ be the work done by $1$ woman per day.
From the given information:
$(3M + 2W) \times 15 = (2M + 3W) \times 18$
Dividing by $3$:
$(3M + 2W) \times 5 = (2M + 3W) \times 6$
$15M + 10W = 12M + 18W$
$3M = 8W \Rightarrow M = \frac{8}{3}W$
Total work = $(3M + 2W) \times 15 = (3(\frac{8}{3}W) + 2W) \times 15 = (8W + 2W) \times 15 = 10W \times 15 = 150W$
We need to find the time taken by $1$ man and $1$ woman $(M + W)$:
$M + W = \frac{8}{3}W + W = \frac{11}{3}W$
Time taken = $\frac{\text{Total Work}}{\text{Work per day}} = \frac{150W}{\frac{11}{3}W} = \frac{150 \times 3}{11} = \frac{450}{11} = 40 \frac{10}{11}$ days.

(D) Let $B$ take $x$ days to finish the work.
Then,$A$ takes $2x$ days and $C$ takes $3x$ days to complete the same work.
The work done by $A, B,$ and $C$ in one day is $\frac{1}{2x}, \frac{1}{x},$ and $\frac{1}{3x}$ respectively.
Given that working together they finish the work in $12$ days,their combined one-day work is $\frac{1}{12}$.
Therefore,$\frac{1}{2x} + \frac{1}{x} + \frac{1}{3x} = \frac{1}{12}$.
Finding a common denominator $(6x)$: $\frac{3 + 6 + 2}{6x} = \frac{1}{12}$.
$\frac{11}{6x} = \frac{1}{12}$.
Cross-multiplying: $6x = 11 \times 12 = 132$.
$x = \frac{132}{6} = 22$ days.
Since $A$ takes $2x$ days,the time taken by $A$ alone is $2 \times 22 = 44$ days.

(A) Let the total work be completed by $100$ men in $x$ days.
Work done by $100$ men in $35$ days + work done by $200$ men in $5$ days = $1$ (total work).
Using the formula $\frac{M_1 D_1}{W_1} = \frac{M_2 D_2}{W_2}$:
$\frac{100 \times 35}{W} + \frac{200 \times 5}{W} = 1$,where $W$ is the total work in man-days.
$W = (100 \times 35) + (200 \times 5) = 3500 + 1000 = 4500$ man-days.
If only $100$ men were employed,the time taken would be $\frac{4500}{100} = 45$ days.
The stipulated time was $40$ days.
Therefore,the delay would be $45 - 40 = 5$ days.

(C) The work done by $P$,$Q$,and $R$ in one day is proportional to their efficiency.
Efficiency of $P = 1/6$,$Q = 1/10$,$R = 1/12$.
Ratio of efficiencies $= 1/6 : 1/10 : 1/12$.
To simplify,multiply by the $LCM$ of $6, 10, 12$,which is $60$.
Ratio $= (60/6) : (60/10) : (60/12) = 10 : 6 : 5$.
Total parts $= 10 + 6 + 5 = 21$.
Share of $R = (5 / 21) \times 4200 = 5 \times 200 = Rs. 1000$.

(D) Given that $8$ men $= 10$ women,which implies $4$ men $= 5$ women.
Since $8$ men can complete the work in $50$ days,the total work is $8 \times 50 = 400$ man-days.
We need to find the time taken by $28$ men and $15$ women.
Convert the group into a single unit (women):
$28$ men $= (28 / 4) \times 5 = 7 \times 5 = 35$ women.
Total group $= 35$ women $+ 15$ women $= 50$ women.
Since $10$ women take $50$ days,$50$ women will take $D$ days.
Using the formula $M_1 D_1 = M_2 D_2$:
$10 \times 50 = 50 \times D$
$D = (10 \times 50) / 50 = 10$ days.

(A) Let the daily wage of $1$ man be $m$ and the daily wage of $1$ boy be $b$.
From the first condition:
$(3m + 4b) \times 7 = 2100 \Rightarrow 3m + 4b = 300$ .....$(i)$
From the second condition:
$(11m + 13b) \times 8 = 8300 \Rightarrow 11m + 13b = 1037.5$ .....$(ii)$
Multiply equation $(i)$ by $11$ and equation $(ii)$ by $3$:
$33m + 44b = 3300$ .....$(iii)$
$33m + 39b = 3112.5$ .....$(iv)$
Subtracting $(iv)$ from $(iii)$:
$5b = 187.5 \Rightarrow b = 37.5$
Substitute $b = 37.5$ into $(i)$:
$3m + 4(37.5) = 300 \Rightarrow 3m + 150 = 300 \Rightarrow 3m = 150 \Rightarrow m = 50$
Daily earnings of $7$ men and $9$ boys $= 7(50) + 9(37.5) = 350 + 337.5 = 687.5$.
Number of days required $= \frac{11000}{687.5} = 16$ days.

(B) Total food available is for $400$ men for $31$ days.
After $28$ days,the food consumed is for $400$ men for $28$ days.
The remaining food is enough for $400$ men for $(31 - 28) = 3$ days.
Number of men remaining $= 400 - 280 = 120$.
Let the remaining food last for $x$ days for $120$ men.
Since the total amount of food is constant,we use the relation: $M_1 \times D_1 = M_2 \times D_2$.
$400 \times 3 = 120 \times x$.
$1200 = 120x$.
$x = 1200 / 120 = 10$ days.

(C) We use the formula $\frac{M_1 D_1}{W_1} = \frac{M_2 D_2}{W_2}$,where $M$ is the number of men,$D$ is the number of days (or weeks),and $W$ is the work done.
Given: $M_1 = 28$,$D_1 = 1$ week,$W_1 = \frac{7}{8}$.
Remaining work $W_2 = 1 - \frac{7}{8} = \frac{1}{8}$.
Time for remaining work $D_2 = 1$ week.
Let $x$ be the number of men required.
Substituting the values: $\frac{28 \times 1}{7/8} = \frac{x \times 1}{1/8}$.
$\frac{28 \times 8}{7} = x \times 8$.
$4 \times 8 = x \times 8$.
$x = 4$.
Thus,$4$ men are required to complete the remaining work.

(B) The formula for work completion is $\frac{M_1 D_1}{W_1} = \frac{M_2 D_2}{W_2}$.
Given:
$M_1 = 45$ men,
$D_1 = 200$ days,
$W_1 = 4.5 \, km$.
Remaining work $W_2 = 12 - 4.5 = 7.5 \, km$.
Remaining time $D_2 = 350 - 200 = 150$ days.
Let $M_2$ be the total number of men required to complete the remaining work in time.
Substituting the values:
$\frac{45 \times 200}{4.5} = \frac{M_2 \times 150}{7.5}$
$M_2 = \frac{45 \times 200 \times 7.5}{4.5 \times 150}$
$M_2 = \frac{45 \times 200 \times 7.5}{675}$
$M_2 = 100$ men.
Extra men required $= M_2 - M_1 = 100 - 45 = 55$ men.

(C) Total money $= Rs. 4500$.
$A$'s one day's work $= \frac{1}{8}$.
$B$'s one day's work $= \frac{1}{12}$.
$(A+B+C)$'s one day's work $= \frac{1}{4}$.
$C$'s one day's work $= \frac{1}{4} - (\frac{1}{8} + \frac{1}{12}) = \frac{1}{4} - (\frac{3+2}{24}) = \frac{1}{4} - \frac{5}{24} = \frac{6-5}{24} = \frac{1}{24}$.
Ratio of their one day's work $A:B:C = \frac{1}{8} : \frac{1}{12} : \frac{1}{24} = 3:2:1$ (by taking $LCM$ as $24$).
Sum of the ratio $= 3+2+1 = 6$.
$C$'s share $= (\frac{1}{6} \times 4500) = Rs. 750$.

(A) Work done by $A$ in $1$ day $= 1/16$.
Work done by $B$ in $1$ day $= 1/24$.
Work done by $A, B,$ and $C$ in $1$ day $= 1/6$.
Work done by $C$ in $1$ day $= (A+B+C)$'s $1$ day work $- (A+B)$'s $1$ day work.
$1/C = 1/6 - (1/16 + 1/24) = 1/6 - (3+2)/48 = 1/6 - 5/48 = (8-5)/48 = 3/48 = 1/16$.
The ratio of work done by $A, B,$ and $C$ is $1/16 : 1/24 : 1/16$.
To simplify,multiply by $48$: $48/16 : 48/24 : 48/16 = 3 : 2 : 3$.
Total parts $= 3 + 2 + 3 = 8$.
Share of $A = (3/8) \times 400 = Rs. 150$.
Share of $B = (2/8) \times 400 = Rs. 100$.
Share of $C = (3/8) \times 400 = Rs. 150$.

(B) Total money $= Rs. 1800$.
The ratio of daily wages of $A, B$ and $C$ is $5: 6: 4$.
The total work done by each person is proportional to the product of their daily wage and the number of days worked.
Ratio of total earnings $= (5 \times 6) : (6 \times 4) : (4 \times 9) = 30 : 24 : 36$.
Simplifying the ratio by dividing by $6$,we get $5 : 4 : 6$.
Sum of ratio parts $= 5 + 4 + 6 = 15$.
Amount received by $A = \frac{5}{15} \times 1800 = \frac{1}{3} \times 1800 = Rs. 600$.

(A) man can complete a work in $10\, days$.
(Man + boy) can complete the same work in $6\, days$.
Let the total work be the Least Common Multiple of $10$ and $6$,which is $30\, units$.
Efficiency of the man $= 30 / 10 = 3\, units/day$.
Efficiency of (man + boy) $= 30 / 6 = 5\, units/day$.
Efficiency of the boy $= 5 - 3 = 2\, units/day$.
Since wages are distributed in proportion to the work done,the share of the boy is calculated based on his efficiency relative to the total efficiency.
Share of the boy $= (2 / 5) \times 50 = Rs.\, 20$.

(D) Let the efficiency of a man, a woman, and a child be $5k, 4k,$ and $2k$ respectively.
Total work done by $(2\, \text{men} + 3\, \text{women} + 4\, \text{children})$ in $10\, \text{days}$ is:
$\text{Work} = (2 \times 5k + 3 \times 4k + 4 \times 2k) \times 10 = (10k + 12k + 8k) \times 10 = 30k \times 10 = 300k$.
Given that this work corresponds to $10\, \text{hectares},$ so $10\, \text{hectares} = 300k,$ which means $1\, \text{hectare} = 30k$.
Now, we need to reap $16\, \text{hectares},$ which is equivalent to $16 \times 30k = 480k$ units of work.
The group of $(6\, \text{men} + 4\, \text{women} + 7\, \text{children})$ has a daily efficiency of:
$\text{Efficiency} = (6 \times 5k + 4 \times 4k + 7 \times 2k) = (30k + 16k + 14k) = 60k$.
Let the number of days required be $x$.
$\text{Total Work} = \text{Efficiency} \times \text{Days} \rightarrow 480k = 60k \times x$.
$x = \frac{480k}{60k} = 8\, \text{days}$.

(C) Let the work done by $A$ in $1$ day be $a$,by $B$ be $b$,and by $C$ be $c$.
Given that $A$ does the work in the same time as $B$ and $C$ together,so $a = b + c$.
Given $(A+B)$'s $1$ day work is $\frac{1}{10}$,so $a + b = \frac{1}{10}$.
Given $C$'s $1$ day work is $c = \frac{1}{50}$.
Since $a = b + c$,we substitute $b = a - c$ into the equation $a + b = \frac{1}{10}$:
$a + (a - c) = \frac{1}{10}$
$2a - c = \frac{1}{10}$
$2a - \frac{1}{50} = \frac{1}{10}$
$2a = \frac{1}{10} + \frac{1}{50} = \frac{5+1}{50} = \frac{6}{50} = \frac{3}{25}$
$a = \frac{3}{50}$.
Now,find $b$ using $a + b = \frac{1}{10}$:
$b = \frac{1}{10} - a = \frac{1}{10} - \frac{3}{50} = \frac{5-3}{50} = \frac{2}{50} = \frac{1}{25}$.
Thus,$B$ alone can complete the work in $25$ days.

(B) Let $M, W, B$ be the work done by $1$ man,$1$ woman,and $1$ boy in $1$ hour respectively.
From the given data:
$1M + 3W + 4B = \frac{1}{96}$ ...$(i)$
$2M + 8B = \frac{1}{80} \implies 1M + 4B = \frac{1}{160}$ ...$(ii)$
$2M + 3W = \frac{1}{120}$ ...$(iii)$
Subtracting $(ii)$ from $(i)$:
$(1M + 3W + 4B) - (1M + 4B) = \frac{1}{96} - \frac{1}{160}$
$3W = \frac{5-3}{480} = \frac{2}{480} = \frac{1}{240}$
So,$3$ women can do the work in $240$ hours.
From $(iii)$,$2M = \frac{1}{120} - 3W = \frac{1}{120} - \frac{1}{240} = \frac{2-1}{240} = \frac{1}{240}$.
So,$2$ men can do the work in $240$ hours,which means $1$ man can do it in $480$ hours.
From $(ii)$,$4B = \frac{1}{160} - 1M = \frac{1}{160} - \frac{1}{480} = \frac{3-1}{480} = \frac{2}{480} = \frac{1}{240}$.
So,$4$ boys can do the work in $240$ hours,which means $1$ boy can do it in $960$ hours.
Now,for $5$ men and $12$ boys:
Work per hour $= 5 \times (\frac{1}{480}) + 12 \times (\frac{1}{960}) = \frac{5}{480} + \frac{12}{960} = \frac{10}{960} + \frac{12}{960} = \frac{22}{960} = \frac{11}{480}$.
Time taken $= \frac{480}{11} = 43 \frac{7}{11}$ hours.

(A) Total work = $40 \times 40 = 1600$ man-days.
Work done in first $10$ days by $40$ men = $40 \times 10 = 400$ units.
Work done in next $10$ days by $35$ men = $35 \times 10 = 350$ units.
Work done in next $10$ days by $30$ men = $30 \times 10 = 300$ units.
Work done in next $10$ days by $25$ men = $25 \times 10 = 250$ units.
Work done in next $10$ days by $20$ men = $20 \times 10 = 200$ units.
Total work done in $50$ days = $400 + 350 + 300 + 250 + 200 = 1500$ units.
Remaining work = $1600 - 1500 = 100$ units.
Now,$15$ men are left. Time taken to complete the remaining work = $\frac{100}{15} = \frac{20}{3} = 6 \frac{2}{3}$ days.
Total time = $50 + 6 \frac{2}{3} = 56 \frac{2}{3}$ days.

(D) Let the work done by $1$ woman in $1$ day be $W$ and by $1$ child in $1$ day be $C$.
Given that $6$ women complete the work in $3$ days,the total work is $6 \times 3 = 18$ units (in terms of woman-days).
Thus,$1$ woman's $1$ day work $= 1/18$ of the total work.
Given that $3$ women and $18$ children complete the work in $2$ days,their combined $1$ day work is $1/2$.
So,$3(1/18) + 18C = 1/2$.
$1/6 + 18C = 1/2$.
$18C = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3$.
$C = 1/(3 \times 18) = 1/54$.
This means $1$ child completes $1/54$ of the work in $1$ day.
Therefore,$9$ children will complete $9 \times (1/54) = 9/54 = 1/6$ of the work in $1$ day.
To complete the whole work,$9$ children will take $1 / (1/6) = 6$ days.

(C) 's $1$ day's work $= \frac{1}{120}$.
$B$'s $1$ day's work $= \frac{1}{150}$.
$(A+B)$'s $1$ day's work $= \frac{1}{120} + \frac{1}{150} = \frac{5+4}{600} = \frac{9}{600} = \frac{3}{200}$.
$(A+B)$ work together for $20$ days,so work done $= 20 \times \frac{3}{200} = \frac{3}{10}$.
After $B$ leaves,$A$ works alone for $12$ days. $A$'s $12$ day's work $= 12 \times \frac{1}{120} = \frac{1}{10}$.
Then $C$ joins $A$ and they work for $48$ days to complete the remaining work.
$A$'s work during these $48$ days $= 48 \times \frac{1}{120} = \frac{2}{5}$.
Total work done by $A$ and $B$ before $C$ joined $= \frac{3}{10} + \frac{1}{10} = \frac{4}{10} = \frac{2}{5}$.
Total work done by $A$ in the final $48$ days $= \frac{2}{5}$.
Total work done by $A$ and $B$ throughout $= \frac{3}{10} + \frac{1}{10} + \frac{2}{5} = \frac{4}{10} + \frac{4}{10} = \frac{8}{10} = \frac{4}{5}$.
Remaining work $= 1 - \frac{4}{5} = \frac{1}{5}$.
This remaining work is done by $C$ in $48$ days.
$C$'s $1$ day's work $= \frac{1/5}{48} = \frac{1}{240}$.
Therefore,$C$ alone can finish the work in $240$ days.

(B) Let the efficiency of $1$ man be $m$ and $1$ woman be $w$ units of work per day.
According to the problem:
$14(2m + 1w) = 1$ (Total work) ...$(i)$
$8(2m + 4w) = 1$ (Total work) ...$(ii)$
Equating $(i)$ and $(ii)$:
$14(2m + w) = 8(2m + 4w)$
$28m + 14w = 16m + 32w$
$28m - 16m = 32w - 14w$
$12m = 18w$
$\frac{m}{w} = \frac{18}{12} = \frac{3}{2}$
The ratio of the efficiency of a man to a woman is $3:2$. Since wages are proportional to the work done (efficiency),the ratio of their daily wages will also be $3:2$.
Let the daily wage of a woman be $x$.
$\frac{90}{x} = \frac{3}{2}$
$3x = 180$
$x = 60$
Therefore,the daily wage of a woman is $Rs. 60$.

(C) Efficiency of Heena $= 1/20$ of the work per day.
Efficiency of Himani $= 1/25$ of the work per day.
Since Heena and Himani worked for the entire $10$ days,the work done by them is:
Work done by Heena $= 10 \times (1/20) = 1/2 = 50\%$ of the total work.
Work done by Himani $= 10 \times (1/25) = 10/25 = 2/5 = 40\%$ of the total work.
Total work done by Heena and Himani $= 50\% + 40\% = 90\%$.
Remaining work done by Mayuri $= 100\% - 90\% = 10\%$.
Since wages are distributed based on the work done,Mayuri's share $= 10\%$ of $Rs. 700 = Rs. 70$.

(A) 's $1 \text{ day}$'s work $= \frac{1}{12}$.
$B$'s $1 \text{ day}$'s work $= \frac{1}{18}$.
Part of work done by $A$ and $B$ in the first $2 \text{ days} = \frac{1}{12} + \frac{1}{18} = \frac{3+2}{36} = \frac{5}{36}$.
In $14 \text{ days}$ ($7$ cycles of $2 \text{ days}$),the work done $= 7 \times \frac{5}{36} = \frac{35}{36}$.
Remaining work $= 1 - \frac{35}{36} = \frac{1}{36}$.
On the $15^{\text{th}}$ day,$A$ works. Since $A$ completes $\frac{1}{12}$ work in $1 \text{ day}$,$A$ will complete $\frac{1}{36}$ work in $\frac{1}{36} \times 12 = \frac{1}{3} \text{ day}$.
Total time taken $= 14 + \frac{1}{3} = 14\frac{1}{3} \text{ days}$.

(D) $(A + B)$'s $2$ days' work $= 2 \times (\frac{1}{8} + \frac{1}{10}) = 2 \times \frac{5+4}{40} = 2 \times \frac{9}{40} = \frac{9}{20}$.
Remaining work $= 1 - \frac{9}{20} = \frac{11}{20}$.
$(B + C)$'s $1$ day's work $= \frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6}$.
Days taken by $B$ and $C$ to complete the remaining work $= \frac{\text{Remaining work}}{\text{Efficiency of } (B+C)} = \frac{11/20}{1/6} = \frac{11}{20} \times 6 = \frac{33}{10}$ days.
Total days to complete the work $= 2 + \frac{33}{10} = \frac{20+33}{10} = \frac{53}{10}$ days.

(B) 's work for the first day $= 1/10$.
$B$'s work for the second day $= 1/20$.
$C$'s work for the third day $= 1/40$.
Work done in $3$ days by them $= 1/10 + 1/20 + 1/40 = (4+2+1)/40 = 7/40$.
Hence,$7/40$ part of the work is completed in $3$ days.
In $15$ days ($5$ cycles of $3$ days),the work completed $= 5 \times (7/40) = 35/40 = 7/8$.
Remaining work $= 1 - 7/8 = 1/8$.
On the $16^{th}$ day,$A$ works. $A$ completes $1/10$ of the work.
Remaining work after $16$ days $= 1/8 - 1/10 = (5-4)/40 = 1/40$.
On the $17^{th}$ day,$B$ works. $B$ completes $1/20$ of the work in $1$ day.
Time taken by $B$ to complete $1/40$ of the work $= 20 \times (1/40) = 1/2$ day.
Total time $= 15 + 1 + 1/2 = 16.5$ days.

(A) Let the total work be completed in $x$ days.
The work done by $A$ in $2$ days is $2 \times \frac{1}{10} = \frac{1}{5}$.
$B$ left $3$ days before the completion,so $B$ worked for $(x - 3)$ days. The work done by $B$ is $(x - 3) \times \frac{1}{12}$.
$C$ worked for the entire duration of $x$ days. The work done by $C$ is $x \times \frac{1}{15}$.
The sum of the work done by $A, B,$ and $C$ equals $1$ (the total work).
$\frac{1}{5} + \frac{x-3}{12} + \frac{x}{15} = 1$
Multiply the entire equation by the least common multiple of $5, 12,$ and $15$,which is $60$:
$12 + 5(x - 3) + 4x = 60$
$12 + 5x - 15 + 4x = 60$
$9x - 3 = 60$
$9x = 63$
$x = 7$ days.
Thus,the work lasted for $7$ days.

(B) Let $x$ be the number of men employed at first.
$x$ men complete $\frac{1}{2}$ of the work in $24$ days.
Therefore,$1$ man would complete the whole work in $24 \times 2 \times x = 48x$ days.
Now,$(x + 16)$ men do the remaining work $\left(1 - \frac{1}{2} = \frac{1}{2}\right)$ in $(40 - 24 = 16)$ days.
Therefore,$1$ man would complete the whole work in $16 \times 2 \times (x + 16) = 32(x + 16)$ days.
Since the total work is the same,we equate the two expressions:
$48x = 32(x + 16)$
$48x = 32x + 512$
$16x = 512$
$x = 32$.
Thus,$32$ men were employed at first.

(C) Using the formula: $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$
Given:
$M_1 = 105, D_1 = 25, H_1 = 8, W_1 = \frac{2}{5}$
Let the additional men be $x$.
Then,$M_2 = 105 + x, D_2 = 50 - 25 = 25, H_2 = 9, W_2 = 1 - \frac{2}{5} = \frac{3}{5}$
Substituting the values into the formula:
$\frac{105 \times 25 \times 8}{2/5} = \frac{(105 + x) \times 25 \times 9}{3/5}$
$\frac{105 \times 25 \times 8 \times 5}{2} = \frac{(105 + x) \times 25 \times 9 \times 5}{3}$
$105 \times 4 = (105 + x) \times 3$
$420 = 315 + 3x$
$3x = 105$
$x = 35$
Therefore,$35$ additional men must be employed.

(C) Let the time taken by $B$ to complete the work be $x \, \text{days}$.
Then, the time taken by $A$ to complete the work is $(x-5) \, \text{days}$.
Given that both together take $11 \frac{1}{9} = \frac{100}{9} \, \text{days}$.
The work done by $A$ and $B$ in one day is $\frac{1}{x-5} + \frac{1}{x} = \frac{9}{100}$.
Solving the equation: $\frac{x + x - 5}{x(x-5)} = \frac{9}{100}$.
$\frac{2x - 5}{x^2 - 5x} = \frac{9}{100}$.
$100(2x - 5) = 9(x^2 - 5x)$.
$200x - 500 = 9x^2 - 45x$.
$9x^2 - 245x + 500 = 0$.
Factoring the quadratic equation: $9x^2 - 225x - 20x + 500 = 0$.
$9x(x - 25) - 20(x - 25) = 0$.
$(9x - 20)(x - 25) = 0$.
So, $x = \frac{20}{9}$ or $x = 25$.
Since $x$ must be greater than $5$ (as $A$ takes $x-5$ days), we have $x = 25$.
Therefore, the time taken by $B$ alone is $25 \, \text{days}$.

(D) Let $A$ work for $x$ days.
$A$'s $1$ day's work $= \frac{1}{45}$.
$B$'s $1$ day's work $= \frac{1}{40}$.
Both $A$ and $B$ worked together for $x$ days.
Work done by $(A+B)$ in $x$ days $= x \times (\frac{1}{45} + \frac{1}{40}) = x \times (\frac{8+9}{360}) = \frac{17x}{360}$.
Remaining work $= 1 - \frac{17x}{360} = \frac{360-17x}{360}$.
This remaining work is completed by $B$ in $23$ days.
So,$23 \times (\text{B's } 1 \text{ day's work}) = \frac{360-17x}{360}$.
$23 \times \frac{1}{40} = \frac{360-17x}{360}$.
$\frac{23}{40} = \frac{360-17x}{360}$.
$23 \times 9 = 360 - 17x$.
$207 = 360 - 17x$.
$17x = 360 - 207 = 153$.
$x = \frac{153}{17} = 9$ days.
Therefore,$A$ worked for $9$ days.

(D) Let $m$ be the work done by one man per day and $w$ be the work done by one woman per day.
According to the problem:
$8(4m + 6w) = 10(3m + 7w)$
$32m + 48w = 30m + 70w$
$2m = 22w \Rightarrow m = 11w$
Now,calculate the total work in terms of women:
Total work = $8(4m + 6w) = 8(4(11w) + 6w) = 8(44w + 6w) = 8(50w) = 400w$
If $20$ women work together,the time taken $D_2$ is:
$D_2 = \frac{\text{Total work}}{20w} = \frac{400w}{20w} = 20$ days.

(B) Let the efficiency of $A$ be $a$ and $B$ be $b$ units per day.
Total work $= 5(a + b)$.
According to the second condition,the work is completed in $3$ days with efficiencies $2a$ and $\frac{b}{3}$.
So,$3(2a + \frac{b}{3}) = 5(a + b)$.
$6a + b = 5a + 5b$.
$a = 4b$,which means $\frac{a}{b} = \frac{4}{1}$.
Total work $= 5(4 + 1) = 25$ units.
Time taken by $A$ alone $= \frac{\text{Total Work}}{\text{Efficiency of } A} = \frac{25}{4} = 6 \frac{1}{4}$ days.

(A) Remaining work $= (1 - \frac{4}{7}) = \frac{3}{7}$.
Remaining period $= (92 - 66) = 26$ days.
Let the number of additional men be $x$.
Using the formula $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$:
$M_1 = 234, D_1 = 66, H_1 = 16, W_1 = \frac{4}{7}$.
$M_2 = (234 + x), D_2 = 26, H_2 = 18, W_2 = \frac{3}{7}$.
Substituting the values:
$\frac{234 \times 66 \times 16}{4/7} = \frac{(234 + x) \times 26 \times 18}{3/7}$.
$234 \times 66 \times 16 \times 3 = (234 + x) \times 26 \times 18 \times 4$.
$234 + x = \frac{234 \times 66 \times 16 \times 3}{26 \times 18 \times 4}$.
$234 + x = 9 \times 11 \times 4 = 396$.
$x = 396 - 234 = 162$.
Therefore,$162$ additional men are required.

(B) $1$. Calculate the efficiency of each person:
Efficiency of $A = 120 / 15 = 8$ units/day.
Efficiency of $B = 120 / 30 = 4$ units/day.
Efficiency of $C = 120 / 40 = 3$ units/day.
$2$. Assume $A$ and $B$ did not leave the work. Add the work they would have done if they had stayed until the end:
Work done by $A$ in $2$ days $= 2 \times 8 = 16$ units.
Work done by $B$ in $4$ days $= 4 \times 4 = 16$ units.
$3$. Calculate the new total work:
Total work $= 120 + 16 + 16 = 152$ units.
$4$. Calculate the total time taken by $A, B$ and $C$ working together:
Combined efficiency $= 8 + 4 + 3 = 15$ units/day.
Total time $= 152 / 15 = 10\frac{2}{15}$ days.