Time and Work Questions in English

(C) Work done by $Q$ alone in $12$ days $= \frac{12}{40} = \frac{3}{10}$ of the total work.
Remaining work to be done by $P$ and $Q$ together $= 1 - \frac{3}{10} = \frac{7}{10}$ of the total work.
$P$'s one day work $= \frac{1}{30}$ and $Q$'s one day work $= \frac{1}{40}$.
Combined one day work of $P$ and $Q = \frac{1}{30} + \frac{1}{40} = \frac{4+3}{120} = \frac{7}{120}$.
Time taken by $P$ and $Q$ to complete $\frac{7}{10}$ of the work $= \frac{7/10}{7/120} = \frac{7}{10} \times \frac{120}{7} = 12$ days.
Therefore,$P$ left after $12$ days.

(B) Let the efficiency of $Q$ be $1$ unit per day.
Since $P$ is twice as good as $Q,$ the efficiency of $P$ is $2$ units per day.
Combined efficiency of $P$ and $Q = 1 + 2 = 3$ units per day.
Total work = $\text{Combined efficiency} \times \text{Total time} = 3 \times 16 = 48$ units.
Time taken by $P$ alone = $\frac{\text{Total work}}{\text{Efficiency of } P} = \frac{48}{2} = 24$ days.
Time taken by $Q$ alone = $\frac{\text{Total work}}{\text{Efficiency of } Q} = \frac{48}{1} = 48$ days.
Therefore,$P$ and $Q$ can finish the work in $24$ days and $48$ days respectively.

(C) The work done by Ram in $1$ day $= \frac{1}{6}$.
The work done by Rajan in $1$ day $= \frac{1}{12}$.
The work done by Rangan in $1$ day $= \frac{1}{24}$.
Total work done by all three in $1$ day $= \frac{1}{6} + \frac{1}{12} + \frac{1}{24} = \frac{4 + 2 + 1}{24} = \frac{7}{24}$.
Therefore,the time taken by them to complete the work together $= \frac{1}{7/24} = \frac{24}{7}$ days.

(B) Let the work done by $(P+Q)$,$(Q+R)$,and $(R+P)$ in one day be $\frac{1}{30}$,$\frac{1}{40}$,and $\frac{1}{60}$ respectively.
Adding these,we get $2(P+Q+R) = \frac{1}{30} + \frac{1}{40} + \frac{1}{60}$.
$2(P+Q+R) = \frac{4+3+2}{120} = \frac{9}{120} = \frac{3}{40}$.
Therefore,$(P+Q+R)$ together can do the work in $2 \times \frac{40}{3} = \frac{80}{3} \, days$.

(C) The work done by $P, Q,$ and $R$ in $1$ day is $\frac{1}{10}, \frac{1}{12},$ and $\frac{1}{15}$ respectively.
In $2$ days,the work completed by all three is $2 \times (\frac{1}{10} + \frac{1}{12} + \frac{1}{15}) = 2 \times (\frac{6+5+4}{60}) = 2 \times \frac{15}{60} = 2 \times \frac{1}{4} = \frac{1}{2}$ of the total work.
Remaining work $= 1 - \frac{1}{2} = \frac{1}{2}$.
Now,$P$ and $R$ work together. Their combined $1$-day work is $\frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6}$.
Time taken by $P$ and $R$ to complete the remaining $\frac{1}{2}$ work $= \frac{1/2}{1/6} = \frac{1}{2} \times 6 = 3$ days.

(A) Let the work done by $1$ man in $1$ day be $x$ and the work done by $1$ boy in $1$ day be $y$.
According to the problem:
$10(6x + 8y) = 1 \implies 60x + 80y = 1$ .....$(1)$
$2(26x + 48y) = 1 \implies 52x + 96y = 1$ .....$(2)$
Multiply equation $(1)$ by $6$ and equation $(2)$ by $5$ to eliminate $y$:
$360x + 480y = 6$
$260x + 480y = 5$
Subtracting these equations: $100x = 1 \implies x = \frac{1}{100}$.
Substitute $x$ in equation $(1)$:
$60(\frac{1}{100}) + 80y = 1 \implies 0.6 + 80y = 1 \implies 80y = 0.4 \implies y = \frac{0.4}{80} = \frac{1}{200}$.
Now,the work done by $15$ men and $20$ boys in $1$ day is:
$15x + 20y = 15(\frac{1}{100}) + 20(\frac{1}{200}) = \frac{15}{100} + \frac{10}{100} = \frac{25}{100} = \frac{1}{4}$.
Therefore,the time taken to complete the work is $4$ days.

(B) $1$ man's $1$ day's work $= \frac{1}{24 \times 16} = \frac{1}{384}$ units.
$1$ woman's $1$ day's work $= \frac{1}{32 \times 24} = \frac{1}{768}$ units.
$16$ men and $16$ women's $1$ day's work $= 16 \times \frac{1}{384} + 16 \times \frac{1}{768} = \frac{1}{24} + \frac{1}{48} = \frac{2+1}{48} = \frac{3}{48} = \frac{1}{16}$ units.
Work done in $12$ days $= 12 \times \frac{1}{16} = \frac{3}{4}$ units.
Remaining work $= 1 - \frac{3}{4} = \frac{1}{4}$ units.
Let $x$ be the number of additional men required. Total men $= 16 + x$.
Work done by $(16+x)$ men and $16$ women in $2$ days $= (16+x) \times 2 \times \frac{1}{384} + 16 \times 2 \times \frac{1}{768} = \frac{1}{4}$.
$\frac{16+x}{192} + \frac{32}{768} = \frac{1}{4}$.
$\frac{16+x}{192} + \frac{1}{24} = \frac{1}{4}$.
$\frac{16+x}{192} = \frac{1}{4} - \frac{1}{24} = \frac{6-1}{24} = \frac{5}{24}$.
$16+x = \frac{5}{24} \times 192 = 5 \times 8 = 40$.
$x = 40 - 16 = 24$.

(C) The work done by $1$ woman in $1$ day $= \frac{1}{10 \times 7} = \frac{1}{70}$.
The work done by $1$ child in $1$ day $= \frac{1}{10 \times 14} = \frac{1}{140}$.
Now,the work done by $5$ women and $10$ children in $1$ day is:
$= 5 \times \frac{1}{70} + 10 \times \frac{1}{140}$
$= \frac{5}{70} + \frac{10}{140}$
$= \frac{1}{14} + \frac{1}{14}$
$= \frac{2}{14} = \frac{1}{7}$.
Since they complete $\frac{1}{7}$ of the work in $1$ day,the total time required to complete the work is $7$ days.

(B) Let the work done by $1$ man in $1$ day be $x$ and by $1$ woman in $1$ day be $y$.
According to the problem:
$4x + 6y = \frac{1}{8}$ ... $(1)$
$3x + 7y = \frac{1}{10}$ ... $(2)$
To solve for $y$,multiply equation $(1)$ by $3$ and equation $(2)$ by $4$:
$12x + 18y = \frac{3}{8}$
$12x + 28y = \frac{4}{10} = \frac{2}{5}$
Subtracting the first from the second:
$(28y - 18y) = \frac{2}{5} - \frac{3}{8}$
$10y = \frac{16 - 15}{40} = \frac{1}{40}$
$y = \frac{1}{400}$
This means $1$ woman completes $\frac{1}{400}$ of the work in $1$ day.
Therefore,$10$ women will complete $10 \times \frac{1}{400} = \frac{1}{40}$ of the work in $1$ day.
Thus,$10$ women will complete the work in $40$ days.

(A) Let $x$ be the work done by a man in one day and $y$ be the work done by a boy in one day.
According to the problem:
$5(12x + 16y) = 1 \implies 60x + 80y = 1$ ....$(1)$
$4(13x + 24y) = 1 \implies 52x + 96y = 1$ ....$(2)$
To solve for $x$ and $y$,multiply equation $(1)$ by $6$ and equation $(2)$ by $5$:
$360x + 480y = 6$ ....$(3)$
$260x + 480y = 5$ ....$(4)$
Subtracting $(4)$ from $(3)$:
$100x = 1 \implies x = \frac{1}{100}$
Substitute $x = \frac{1}{100}$ into equation $(1)$:
$60(\frac{1}{100}) + 80y = 1$
$0.6 + 80y = 1$
$80y = 0.4$
$y = \frac{0.4}{80} = \frac{4}{800} = \frac{1}{200}$
The ratio of work done by a man to that of a boy is $x:y = \frac{1}{100} : \frac{1}{200} = 2:1$.

(A) Let the work done by a man in one day be $x$.
Since a woman does double the work of a man,the work done by a woman in one day is $2x$.
Since a child does half the work of a man,the work done by a child in one day is $\frac{x}{2}$.
The total work done by $3$ men,$4$ women,and $6$ children in one day is $3x + 4(2x) + 6(\frac{x}{2}) = 3x + 8x + 3x = 14x$.
Given that they complete the work in $7$ days,the total work is $14x \times 7 = 98x$.
We need to find the number of women $(n)$ required to complete the same work in $7$ days.
The work done by $n$ women in $7$ days is $n \times (2x) \times 7 = 14nx$.
Equating the total work: $14nx = 98x$.
Dividing both sides by $14x$,we get $n = 7$.
Therefore,$7$ women are required to complete the work in $7$ days.

(B) Work done by $A$ in $1$ day $= \frac{1}{20}$.
Work done by $B$ in $1$ day $= \frac{1}{30}$.
Work done by $C$ in $1$ day $= \frac{1}{60}$.
$A$ works every day,while $B$ and $C$ assist him only on every $3^{rd}$ day.
Work done in a $3$-day cycle $= (A+A+A+B+C) = 3 \times (\text{Work of } A) + (\text{Work of } B) + (\text{Work of } C)$.
Work done in $3$ days $= 3 \times \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{9}{60} + \frac{2}{60} + \frac{1}{60} = \frac{12}{60} = \frac{1}{5}$.
Since $\frac{1}{5}$ of the work is completed in $3$ days,the total time required to complete the whole work $= 3 \times 5 = 15$ days.

(B) Let the capacity of one woman be $W$ and one man be $M$ units of work per day.
Total work $= 20 \times 16 \times W = 320W$.
Also,total work $= 16 \times 15 \times M = 240M$.
Since the work is the same,$320W = 240M$.
Therefore,the ratio of the capacity of a man to a woman is $M/W = 320/240$.
Simplifying the fraction,$M/W = 32/24 = 4/3$.
Thus,the ratio is $4:3$.

(A) $1$ day's work by a man $= \frac{1}{12 \times 4} = \frac{1}{48}$.
$1$ day's work by a woman $= \frac{1}{15 \times 4} = \frac{1}{60}$.
Part of work done by $6$ men in $2$ days $= (6 \times \frac{1}{48}) \times 2 = \frac{12}{48} = \frac{1}{4}$.
Remaining work $= 1 - \frac{1}{4} = \frac{3}{4}$.
Let the number of women required to complete the remaining work in $3$ days be $x$.
Then,$x \times (\text{work of } 1 \text{ woman per day}) \times 3 = \text{Remaining work}$.
$x \times \frac{1}{60} \times 3 = \frac{3}{4}$.
$x \times \frac{1}{20} = \frac{3}{4}$.
$x = \frac{3}{4} \times 20 = 15$.
Therefore,$15$ women are required.

(D) Let the work done by $1$ man in $1$ day be $x$ and by $1$ woman in $1$ day be $y$.
Given that $10$ men and $15$ women complete the work in $6$ days,so their $1$ day's work is $10x + 15y = \frac{1}{6}$ ....$(1)$
Given that $1$ man alone completes the work in $100$ days,so $x = \frac{1}{100}$ ....$(2)$
Substitute $x = \frac{1}{100}$ in equation $(1)$:
$10 \times \frac{1}{100} + 15y = \frac{1}{6}$
$\frac{1}{10} + 15y = \frac{1}{6}$
$15y = \frac{1}{6} - \frac{1}{10}$
$15y = \frac{5 - 3}{30} = \frac{2}{30} = \frac{1}{15}$
$y = \frac{1}{15 \times 15} = \frac{1}{225}$
Thus,one woman alone can complete the work in $225$ days.

(D) Let the work done by a man,a woman,and a boy in one day be $x, y,$ and $z$ respectively.
$x = \frac{1}{3}, y = \frac{1}{4}, z = \frac{1}{12}$.
Let $N$ be the number of boys required to assist $1$ man and $1$ woman to complete the work in $\frac{1}{4}$ of a day.
The total work done by $1$ man,$1$ woman,and $N$ boys in $\frac{1}{4}$ of a day is $1$ (the whole job).
Therefore,$(1 \times x + 1 \times y + N \times z) \times \frac{1}{4} = 1$.
Substituting the values: $(1 \times \frac{1}{3} + 1 \times \frac{1}{4} + N \times \frac{1}{12}) \times \frac{1}{4} = 1$.
$(\frac{4 + 3}{12} + \frac{N}{12}) \times \frac{1}{4} = 1$.
$(\frac{7 + N}{12}) \times \frac{1}{4} = 1$.
$\frac{7 + N}{48} = 1$.
$7 + N = 48$.
$N = 48 - 7 = 41$.

(A) Total work $= 12 \times 9 = 108$ man-days.
Work done in $6$ days by $12$ men $= 12 \times 6 = 72$ man-days.
Remaining work $= 108 - 72 = 36$ man-days.
After $6$ days,the number of men $= 12 + 6 = 18$ men.
Let the remaining work be completed in $x$ days.
$18 \times x = 36$
$x = \frac{36}{18} = 2$ days.
Therefore,the remaining work will be completed in $2$ days.

(C) Work done by $10$ men in $1$ day $= \frac{1}{15}$ of the total work.
Work done by $15$ women in $1$ day $= \frac{1}{12}$ of the total work.
When $10$ men and $15$ women work together,the work done in $1$ day is $\frac{1}{15} + \frac{1}{12}$.
Calculating the sum: $\frac{4 + 5}{60} = \frac{9}{60} = \frac{3}{20}$ of the work per day.
Therefore,the total number of days required to complete the work is the reciprocal of the work done per day.
Days $= \frac{20}{3} = 6\frac{2}{3}$ days.

(D) Work done by $A$ in $1$ day $= \frac{1}{16}$.
Work done by $B$ in $1$ day $= \frac{1}{12}$.
Work done in a cycle of $2$ days (starting with $A$) $= \frac{1}{16} + \frac{1}{12} = \frac{3+4}{48} = \frac{7}{48}$.
In $6$ cycles ($12$ days),work done $= 6 \times \frac{7}{48} = \frac{42}{48} = \frac{7}{8}$.
Remaining work $= 1 - \frac{7}{8} = \frac{1}{8}$.
On the $13$th day,$A$ works. Work done by $A$ in $1$ day is $\frac{1}{16}$.
Remaining work after $13$ days $= \frac{1}{8} - \frac{1}{16} = \frac{1}{16}$.
On the $14$th day,$B$ works. $B$ completes $\frac{1}{12}$ work in $1$ day,so $B$ completes $\frac{1}{16}$ work in $\frac{1/16}{1/12} = \frac{12}{16} = \frac{3}{4}$ days.
Total time $= 13 + \frac{3}{4} = 13\frac{3}{4}$ days.

(B) The total work is considered as $1$ unit.
Given that $A+B$ do $\frac{19}{23}$ of the work.
Therefore,the work done by $C$ alone $= 1 - \frac{19}{23} = \frac{4}{23}$.
Given that $B+C$ do $\frac{8}{23}$ of the work.
Therefore,the work done by $A$ alone $= 1 - \frac{8}{23} = \frac{15}{23}$.
The wages are distributed in proportion to the work done.
Amount to be paid to $A = \text{Total amount} \times (\text{Work done by } A)$
Amount to be paid to $A = 529 \times \frac{15}{23} = 23 \times 15 = Rs. 345$.

(C) Part of work done by $A$ and $B$ per day $= \frac{1}{12}$.
Part of work done by $A$ alone per day $= \frac{1}{20}$.
Therefore,part of work done by $B$ alone per day $= \frac{1}{12} - \frac{1}{20} = \frac{5 - 3}{60} = \frac{1}{30}$.
Since $B$ works only for half a day,the work done by $B$ per day $= \frac{1}{2} \times \frac{1}{30} = \frac{1}{60}$.
Combined work done by $A$ (full day) and $B$ (half day) per day $= \frac{1}{20} + \frac{1}{60} = \frac{3 + 1}{60} = \frac{4}{60} = \frac{1}{15}$.
Let the total number of days required to complete the work be $x$.
$\frac{1}{15} \times x = 1$.
Therefore,$x = 15$ days.

(C) Work done by $A$ in $1$ day $= \frac{1}{14}$.
Work done by $B$ in $1$ day $= \frac{1}{21}$.
Let the total number of days to complete the work be $x$.
Since $A$ leaves $3$ days before completion,$A$ worked for $(x-3)$ days and $B$ worked for $x$ days.
The total work done is $1$:
$(x-3) \times \frac{1}{14} + x \times \frac{1}{21} = 1$.
Taking the $LCM$ of $14$ and $21$,which is $42$:
$\frac{3(x-3) + 2x}{42} = 1$.
$3x - 9 + 2x = 42$.
$5x = 51$.
$x = \frac{51}{5} = 10 \frac{1}{5}$ days.

(B) Work done by $A$ in $1$ day $= \frac{1}{6}$.
Work done by $B$ in $1$ day $= \frac{1}{8}$.
Work done by $(A+B+C)$ in $1$ day $= \frac{1}{3}$.
Work done by $C$ in $1$ day $= \text{Work}(A+B+C) - \text{Work}(A+B) = \frac{1}{3} - (\frac{1}{6} + \frac{1}{8}) = \frac{1}{3} - (\frac{4+3}{24}) = \frac{1}{3} - \frac{7}{24} = \frac{8-7}{24} = \frac{1}{24}$.
The ratio of work done by $A, B,$ and $C$ is $\frac{1}{6} : \frac{1}{8} : \frac{1}{24}$.
Multiplying by $24$,the ratio is $4 : 3 : 1$.
Total share $= 4+3+1 = 8$ units.
Total amount $= Rs. 3200$.
Share of $C = \frac{1}{8} \times 3200 = Rs. 400$.

(C) Work done by $A$ in $1$ day $= \frac{1}{45}$.
Work done by $B$ in $1$ day $= \frac{1}{40}$.
Let $A$ work for $x$ days. Both $A$ and $B$ worked together for $x$ days.
$B$ worked alone for $23$ days.
Total work done $= 1$.
Equation: $x \left( \frac{1}{45} + \frac{1}{40} \right) + 23 \left( \frac{1}{40} \right) = 1$.
$x \left( \frac{8 + 9}{360} \right) + \frac{23}{40} = 1$.
$x \left( \frac{17}{360} \right) = 1 - \frac{23}{40} = \frac{17}{40}$.
$x = \frac{17}{40} \times \frac{360}{17} = 9$ days.
Thus,$A$ left the work after $9$ days.

(B) The work done by Kim in $1 \, \text{day}$ is $1/3$ of the total work.
The work done by David in $1 \, \text{day}$ is $1/2$ of the total work.
The ratio of their work efficiency is $1/3 : 1/2$, which simplifies to $2 : 3$.
Since the wages are distributed in the ratio of their work efficiency, the share of Kim is calculated as:
$\text{Share of Kim} = \frac{2}{2+3} \times 150 = \frac{2}{5} \times 150 = 2 \times 30 = Rs. 60$.

(C) Let the work done by $A$ in one day be $a$ and by $B$ in one day be $b$.
Given that $(a + b) = 1/30$.
$A$ works for $16$ days and $B$ works for $44$ days,completing the work: $16a + 44b = 1$.
We can rewrite this as $16(a + b) + 28b = 1$.
Substituting $(a + b) = 1/30$,we get: $16(1/30) + 28b = 1$.
$16/30 + 28b = 1$.
$8/15 + 28b = 1$.
$28b = 1 - 8/15 = 7/15$.
$b = 7 / (15 \times 28) = 1 / (15 \times 4) = 1/60$.
Since $B$ does $1/60$ of the work in one day,$B$ will finish the whole work in $60$ days.

(C) Work done by $A$ in $1 \, day = \frac{1}{24}$.
Work done by $B$ in $1 \, day = \frac{1}{9}$.
Work done by $C$ in $1 \, day = \frac{1}{12}$.
Work done by $B$ and $C$ together in $1 \, day = \frac{1}{9} + \frac{1}{12} = \frac{4+3}{36} = \frac{7}{36}$.
Work done by $B$ and $C$ in $3 \, days = 3 \times \frac{7}{36} = \frac{7}{12}$.
Remaining work $= 1 - \frac{7}{12} = \frac{5}{12}$.
Let $A$ take $x$ days to complete the remaining work.
$x \times \frac{1}{24} = \frac{5}{12}$.
$x = \frac{5}{12} \times 24 = 10 \, days$.

(C) completes $\frac{4}{5}$ of the work in $20 \, \text{days}$.
Therefore,the rate of work of $A$ is $\frac{4/5}{20} = \frac{1}{25}$ of the work per day.
Remaining work $= 1 - \frac{4}{5} = \frac{1}{5}$.
$A$ and $B$ together finish the remaining $\frac{1}{5}$ work in $3 \, \text{days}$.
Therefore,their combined rate of work is $\frac{1/5}{3} = \frac{1}{15}$ of the work per day.
Let $x$ be the rate of work of $B$ per day.
Then,$\frac{1}{25} + x = \frac{1}{15}$.
$x = \frac{1}{15} - \frac{1}{25} = \frac{5 - 3}{75} = \frac{2}{75}$.
Time taken by $B$ alone to complete the whole work $= \frac{1}{2/75} = \frac{75}{2} = 37\frac{1}{2} \, \text{days}$.

(D) Let the total work be $1$ unit.
Part of work done by $A$ and $B$ in $1$ day is $\frac{1}{30}$.
Work done by $A$ and $B$ in $20$ days is $\frac{20}{30} = \frac{2}{3}$.
Remaining work = $1 - \frac{2}{3} = \frac{1}{3}$.
This remaining work is finished by $A$ in $20$ days.
Therefore,$A$'s $1$ day work = $\frac{1/3}{20} = \frac{1}{60}$.
Thus,$A$ alone can finish the work in $60$ days.

(C) Part of work done per day by $A = \frac{1}{18}$.
Part of work done per day by $B = \frac{1}{15}$.
Work done by $B$ in $10$ days $= 10 \times \frac{1}{15} = \frac{10}{15} = \frac{2}{3}$.
Remaining work $= 1 - \frac{2}{3} = \frac{1}{3}$.
Let the number of days taken by $A$ to finish the remaining work be $x$.
Since $A$ does $\frac{1}{18}$ of the work in $1$ day,the time taken to finish $\frac{1}{3}$ of the work is:
$x = \frac{1/3}{1/18} = \frac{1}{3} \times 18 = 6$ days.
Therefore,$A$ will finish the remaining work in $6$ days.

(D) Work done per hour by machine $P = \frac{1}{8}$.
Work done per hour by machine $Q = \frac{1}{10}$.
Work done per hour by machine $R = \frac{1}{12}$.
All three machines work together from $9.00$ a.m. to $11.00$ a.m. ($2$ hours).
Work done in $2$ hours $= 2 \times (\frac{1}{8} + \frac{1}{10} + \frac{1}{12}) = 2 \times (\frac{15+12+10}{120}) = 2 \times \frac{37}{120} = \frac{37}{60}$.
Remaining work $= 1 - \frac{37}{60} = \frac{23}{60}$.
Machines $Q$ and $R$ work together to complete the remaining work in $x$ hours.
Work done per hour by $Q$ and $R = \frac{1}{10} + \frac{1}{12} = \frac{6+5}{60} = \frac{11}{60}$.
$x = \frac{23}{60} \div \frac{11}{60} = \frac{23}{11} = 2 \frac{1}{11}$ hours.
$2 \frac{1}{11}$ hours is approximately $2$ hours and $5$ minutes.
Starting from $11.00$ a.m.,adding $2$ hours and $5$ minutes gives $1.05$ p.m.,which is approximately $1.00$ p.m.

(B) Efficiency of $X = \frac{1}{20}$ work/day.
Efficiency of $Y = \frac{1}{12}$ work/day.
$X$ worked alone for $4$ days,so work done by $X = 4 \times \frac{1}{20} = \frac{1}{5}$ of the total work.
Remaining work = $1 - \frac{1}{5} = \frac{4}{5}$.
After $4$ days,$X$ and $Y$ worked together. Their combined efficiency = $\frac{1}{20} + \frac{1}{12} = \frac{3+5}{60} = \frac{8}{60} = \frac{2}{15}$ work/day.
Time taken to complete the remaining work = $\frac{4/5}{2/15} = \frac{4}{5} \times \frac{15}{2} = 6$ days.
Total time for the work = $4$ days (initial) + $6$ days (together) = $10$ days.

(A) Given: $(A + B) = 1/30$,$(B + C) = 1/24$,$(C + A) = 1/20$ units per day.
Adding these,we get $2(A + B + C) = 1/30 + 1/24 + 1/20 = (4 + 5 + 6)/120 = 15/120 = 1/8$.
So,$(A + B + C) = 1/16$ units per day.
Work done by $A, B, C$ in $10$ days $= 10 \times (1/16) = 10/16 = 5/8$.
Remaining work $= 1 - 5/8 = 3/8$.
Efficiency of $A = (A + B + C) - (B + C) = 1/16 - 1/24 = (3 - 2)/48 = 1/48$.
Time taken by $A$ to finish the remaining work $= (3/8) / (1/48) = (3/8) \times 48 = 18$ days.

(C) Work done by $A$ in $1$ day = $1/15$.
Work done by $B$ in $1$ day = $1/10$.
Work done by $(A+B)$ in $2$ days = $2 \times (1/15 + 1/10) = 2 \times (2/30 + 3/30) = 2 \times (5/30) = 2 \times (1/6) = 1/3$.
Remaining work = $1 - 1/3 = 2/3$.
Time taken by $A$ to complete the remaining work = $(2/3) / (1/15) = (2/3) \times 15 = 10$ days.
Total time taken = $2$ days (together) + $10$ days ($A$ alone) = $12$ days.

(B) Let the efficiency of $B$ be $x$ units per day.
Since $A$ is $1\frac{3}{4} = \frac{7}{4}$ times as efficient as $B$,the efficiency of $A$ is $\frac{7}{4}x$ units per day.
Together,their combined efficiency is $x + \frac{7}{4}x = \frac{11}{4}x$ units per day.
Given that they complete the job in $7$ days,the total work is $7 \times \frac{11}{4}x = \frac{77}{4}x$ units.
Time taken by $A$ alone to complete the work is $\frac{\text{Total Work}}{\text{Efficiency of } A} = \frac{77x/4}{7x/4} = \frac{77}{7} = 11$ days.

(C) Let the work done by $A, B,$ and $C$ in one day be $a, b,$ and $c$ respectively.
Given that $A$ can do the work in the same time as $B$ and $C$ together,so $a = b + c$.
Given that $(A + B)$ can do the work in $10 \text{ days}$,so $a + b = \frac{1}{10}$.
Given that $C$ can do the work in $50 \text{ days}$,so $c = \frac{1}{50}$.
From $a = b + c$,we substitute $a$ in the equation $a + b = \frac{1}{10}$:
$(b + c) + b = \frac{1}{10} \implies 2b + c = \frac{1}{10}$.
Substituting $c = \frac{1}{50}$:
$2b + \frac{1}{50} = \frac{1}{10} \implies 2b = \frac{1}{10} - \frac{1}{50}$.
$2b = \frac{5 - 1}{50} = \frac{4}{50} = \frac{2}{25}$.
$b = \frac{1}{25}$.
Thus,$B$ alone can do the work in $25 \text{ days}$.

(B) Let the number of days taken by $B$ to complete the work be $x$.
Since $A$ is thrice as efficient as $B$,$A$ takes one-third the time taken by $B$.
Therefore,the time taken by $A = \frac{x}{3}$.
According to the problem,$A$ finishes the work $60\, days$ earlier than $B$,so $x - \frac{x}{3} = 60$.
$\frac{2x}{3} = 60 \implies x = \frac{60 \times 3}{2} = 90\, days$.
Thus,$B$ takes $90\, days$ and $A$ takes $\frac{90}{3} = 30\, days$.
Working together,the time taken by $A$ and $B$ is $\frac{A \times B}{A + B} = \frac{30 \times 90}{30 + 90} = \frac{2700}{120} = 22.5\, days$ or $22\frac{1}{2}\, days$.

(D) The work done by $A$ in $1\, day = 1/15$.
The work done by $B$ in $1\, day = 1/20$.
Work done by $A$ and $B$ together in $1\, day = (1/15 + 1/20) = (4 + 3) / 60 = 7/60$.
Work done by them in $4\, days = (7/60) \times 4 = 7/15$.
Fraction of the work left $= 1 - (\text{Work done}) = 1 - 7/15 = 8/15$.

(B) Let the efficiency of $B$ be $100$ units/day.
Since $A$ is $30\%$ more efficient than $B$,the efficiency of $A$ is $130$ units/day.
Total work = Efficiency of $A \times$ Time taken by $A = 130 \times 23 = 2990$ units.
Combined efficiency of $A$ and $B = 130 + 100 = 230$ units/day.
Time taken by $A$ and $B$ together = $\frac{\text{Total work}}{\text{Combined efficiency}} = \frac{2990}{230} = 13$ days.

(B) Sakshi's work rate per day $= \frac{1}{20}$.
Since Tanya is $25\%$ more efficient than Sakshi,her efficiency is $125\%$ of Sakshi's efficiency.
Tanya's work rate per day $= \frac{125}{100} \times \frac{1}{20} = \frac{5}{4} \times \frac{1}{20} = \frac{1}{16}$.
Therefore,the number of days taken by Tanya to complete the work $= 16$ days.

(C) Work done per day by $(A + B) = \frac{1}{8}$.
Work done per day by $(B + C) = \frac{1}{12}$.
Work done per day by $(A + B + C) = \frac{1}{6}$.
To find the work done by $(A + C)$,we first find the work done by $B$ alone:
Work done by $B = (A + B + C) - (A + C)$ is not direct,so we use:
Work done by $B = (A + B) + (B + C) - (A + B + C) = \frac{1}{8} + \frac{1}{12} - \frac{1}{6} = \frac{3 + 2 - 4}{24} = \frac{1}{24}$.
Now,work done by $A = (A + B) - B = \frac{1}{8} - \frac{1}{24} = \frac{3 - 1}{24} = \frac{2}{24} = \frac{1}{12}$.
Work done by $C = (B + C) - B = \frac{1}{12} - \frac{1}{24} = \frac{2 - 1}{24} = \frac{1}{24}$.
Work done by $(A + C)$ per day $= \frac{1}{12} + \frac{1}{24} = \frac{2 + 1}{24} = \frac{3}{24} = \frac{1}{8}$.
Therefore,$A$ and $C$ together will complete the work in $8$ days.

(A) Total work done by $P$ in terms of hours = $12 \times 8 = 96$ hours.
Total work done by $Q$ in terms of hours = $8 \times 10 = 80$ hours.
Work done by $P$ in $1$ hour = $\frac{1}{96}$.
Work done by $Q$ in $1$ hour = $\frac{1}{80}$.
Combined work done by $P$ and $Q$ in $1$ hour = $\frac{1}{96} + \frac{1}{80} = \frac{5 + 6}{480} = \frac{11}{480}$.
If they work together for $8$ hours a day,the work done per day = $8 \times \frac{11}{480} = \frac{11}{60}$.
Let the number of days be $x$. Then,$x \times \frac{11}{60} = 1$.
$x = \frac{60}{11} = 5 \frac{5}{11}$ days.

(B) Let the time taken by $A$ to complete the work be $x$ days.
Since $A$ takes twice as much time as $B$,the time taken by $B$ is $\frac{x}{2}$ days.
Since $A$ takes thrice as much time as $C$,the time taken by $C$ is $\frac{x}{3}$ days.
The work done by $A, B,$ and $C$ in one day is $\frac{1}{x} + \frac{2}{x} + \frac{3}{x} = \frac{6}{x}$.
Given that they finish the work together in $2$ days,the total work done in one day is $\frac{1}{2}$.
Therefore,$\frac{6}{x} = \frac{1}{2}$.
Solving for $x$,we get $x = 12$ days.
Thus,the time taken by $B$ to complete the work alone is $\frac{x}{2} = \frac{12}{2} = 6$ days.

(C) Given that:
$(A + B)$'s one day work $= \frac{1}{12}$
$(B + C)$'s one day work $= \frac{1}{15}$
$(C + A)$'s one day work $= \frac{1}{20}$
Adding these three equations:
$2(A + B + C)$'s one day work $= \frac{1}{12} + \frac{1}{15} + \frac{1}{20}$
$= \frac{5 + 4 + 3}{60} = \frac{12}{60} = \frac{1}{5}$
Therefore,$(A + B + C)$'s one day work $= \frac{1}{2} \times \frac{1}{5} = \frac{1}{10}$
Thus,$A, B,$ and $C$ together will complete the work in $10$ days.

(C) Part of work done by $A$ in $1 \, day = \frac{1}{16}$.
Part of work done by $B$ in $1 \, day = \frac{1}{12}$.
Part of work done by $A, B$ and $C$ together in $1 \, day = \frac{1}{4}$.
To find the work done by $C$ alone in $1 \, day$,we subtract the work done by $A$ and $B$ from the total work done by all three:
Work done by $C$ in $1 \, day = \frac{1}{4} - (\frac{1}{16} + \frac{1}{12})$.
Finding a common denominator for $4, 16,$ and $12$,which is $48$:
Work done by $C$ in $1 \, day = \frac{12}{48} - (\frac{3}{48} + \frac{4}{48}) = \frac{12 - 7}{48} = \frac{5}{48}$.
Therefore,$C$ alone can complete the job in $\frac{48}{5} = 9\frac{3}{5} \, days$.

(C) Let the number of days required by the son to do the work alone be $x$.
The man's one-day work is $\frac{1}{5}$.
The combined one-day work of the man and his son is $\frac{1}{3}$.
Therefore,the son's one-day work is $\frac{1}{x} = \frac{1}{3} - \frac{1}{5}$.
$\frac{1}{x} = \frac{5-3}{15} = \frac{2}{15}$.
Thus,$x = \frac{15}{2} = 7 \frac{1}{2}$ days.

(A) Work done by $A$ in $1$ day $= 1/18$.
Since $B$ takes half the time taken by $A$,$B$ takes $18/2 = 9$ days to finish the work.
Work done by $B$ in $1$ day $= 1/9$.
When working together,the part of the work finished in $1$ day $= 1/18 + 1/9$.
$= (1 + 2) / 18 = 3 / 18 = 1/6$.

(C) The work done by $A$ in $1 \text{ day} = \frac{1}{24}$.
The work done by $B$ in $1 \text{ day} = \frac{1}{6}$.
The work done by $C$ in $1 \text{ day} = \frac{1}{12}$.
Work done by $A, B$ and $C$ together in $1 \text{ day} = \frac{1}{24} + \frac{1}{6} + \frac{1}{12}$.
Taking the least common multiple $(LCM)$ of $24, 6$ and $12$,which is $24$:
$= \frac{1 + 4 + 2}{24} = \frac{7}{24}$.
Therefore,the total time taken to complete the work together $= \frac{1}{7/24} = \frac{24}{7} = 3\frac{3}{7} \text{ days}$.

(B) The work done by $A$ in $1$ $day$ is $1/10$.
The work done by $B$ in $1$ $day$ is $1/15$.
Together,the work done by $A$ and $B$ in $1$ $day$ is $(1/10 + 1/15) = (3+2)/30 = 5/30 = 1/6$.
Therefore,they will complete the work together in $6$ $days$.
Alternatively,using the formula: $\text{Days} = (A \times B) / (A + B) = (10 \times 15) / (10 + 15) = 150 / 25 = 6$ $days$.

(C) Let the total capacity of air in the tyre be $1$ unit.
Rate of air leakage by the first puncture $= \frac{1}{9}$ unit/minute.
Rate of air leakage by the second puncture $= \frac{1}{6}$ unit/minute.
Combined rate of air leakage $= \frac{1}{9} + \frac{1}{6} = \frac{2+3}{18} = \frac{5}{18}$ unit/minute.
Time taken to make the tyre flat when both punctures are active $= \frac{1}{5/18} = \frac{18}{5} = 3.6$ minutes.
Converting $0.6$ minutes to fractions: $0.6 = \frac{6}{10} = \frac{3}{5}$.
Thus,the total time is $3 \frac{3}{5}$ minutes.