Time and Work Questions in English

(D) Efficiency of $S = 240/40 = 6$ units/day.
Efficiency of $T = 240/48 = 5$ units/day.
Efficiency of $U = 240/60 = 4$ units/day.
Let the total work be $240$ units.
Let the total time taken be $x$ days.
$S$ worked for $x$ days,$T$ worked for $(x-2)$ days,and $U$ worked for $(x-5)$ days.
Total work: $6x + 5(x-2) + 4(x-5) = 240$
$6x + 5x - 10 + 4x - 20 = 240$
$15x - 30 = 240$
$15x = 270 \Rightarrow x = 18$ days.
Work done by $S = 6 \times 18 = 108$ units.
Share of $S = (108/240) \times 10800 = 0.45 \times 10800 = Rs. 4860$.

(D) Let the efficiency of Aman be $100$ units/day.
Since Raman is $25\%$ more efficient than Aman,the efficiency of Raman is $100 + 25 = 125$ units/day.
The ratio of efficiency of Raman to Aman is $125 : 100 = 5 : 4$.
Since time taken is inversely proportional to efficiency,the ratio of time taken by Raman to Aman is $4 : 5$.
Given that Aman takes $25$ days to complete the work,we have $5$ units of time $= 25$ days.
Therefore,$1$ unit of time $= 5$ days.
The time taken by Raman is $4$ units of time $= 4 \times 5 = 20$ days.

(A) Efficiency ratio of $A$ to $B$ is $120:100$,which simplifies to $6:5$.
Since time taken is inversely proportional to efficiency,the ratio of time taken by $A$ to $B$ is $5:6$.
Given that $B$ takes $12$ days to complete the work,we have $6$ units $= 12$ days.
Therefore,$1$ unit $= 2$ days.
Time taken by $A = 5$ units $= 5 \times 2 = 10$ days.
Thus,$A$ can complete the work in $10$ days.

(B) Sister's rate of baking: $50 \text{ cakes} / 25 \text{ hours} = 2 \text{ cakes/hour}$.
Combined rate of Sister and Mummy: $75 \text{ cakes} / 15 \text{ hours} = 5 \text{ cakes/hour}$.
Mummy's rate of baking: $5 \text{ cakes/hour} - 2 \text{ cakes/hour} = 3 \text{ cakes/hour}$.
Number of cakes Mummy can bake in $15$ hours: $3 \text{ cakes/hour} \times 15 \text{ hours} = 45 \text{ cakes}$.

(C) To find who completes the work first,we calculate the total time taken by each person to complete $1$ full unit of work:
For $P$: $\frac{1}{4}$ of the work is done in $10$ days. Therefore,the total time taken by $P = 10 \times 4 = 40$ days.
For $Q$: $40\%$ (or $\frac{40}{100} = \frac{2}{5}$) of the work is done in $40$ days. Therefore,the total time taken by $Q = 40 \times \frac{5}{2} = 100$ days.
For $R$: $\frac{1}{3}$ of the work is done in $13$ days. Therefore,the total time taken by $R = 13 \times 3 = 39$ days.
Comparing the total time taken: $P = 40$ days,$Q = 100$ days,and $R = 39$ days.
Since $39 < 40 < 100$,$R$ will complete the work first.

(C) The total time required to build the cupboard is $48 \, \text{hours}$.
In $1 \, \text{hour}$, the carpenter completes $\frac{1}{48}$ of the work.
In $12 \, \text{hours}$, the work completed is $12 \times \frac{1}{48} = \frac{12}{48} = \frac{1}{4}$ of the cupboard.
The remaining work to be done is $1 - \frac{1}{4} = \frac{3}{4}$.
Converting the fraction $\frac{3}{4}$ to a decimal gives $0.75$.

$A$ completes the work in $12$ days, and $B$ completes it in $20$ days.
Taking the Least Common Multiple $(LCM)$ of $12$ and $20$, the total work is $60$ units.
Efficiency of $A = 60 / 12 = 5$ units/day.
Efficiency of $B = 60 / 20 = 3$ units/day.
Combined efficiency of $(A + B) = 5 + 3 = 8$ units/day.
Work done by $(A + B)$ in $5$ days $= 8 \times 5 = 40$ units.
Remaining work $= 60 - 40 = 20$ units.
$C$ completes the remaining $20$ units in $3$ days, so the efficiency of $C = 20 / 3$ units/day.
Time taken by $C$ to complete the whole work alone $= \text{Total Work} / \text{Efficiency of } C = 60 / (20 / 3) = 60 \times (3 / 20) = 9$ days.

(D) Let the efficiency of $B$ be $1$ unit/day.
Since $A$ does thrice the work of $B$,the efficiency of $A$ is $3$ units/day.
Combined efficiency of $A$ and $B$ = $3 + 1 = 4$ units/day.
Total work = (Combined efficiency) $\times$ (Total days) = $4 \times 9 = 36$ units.
Time taken by $A$ alone = $\frac{\text{Total work}}{\text{Efficiency of } A} = \frac{36}{3} = 12$ days.

(D) Let $R$ and $H$ be the daily work rates of Ram and Hari in $kg/day$.
Given: $(R + H) \times 2 = 12 \, kg \implies R + H = 6 \, kg/day$.
In $5$ days, they cut: $5 \times 6 = 30 \, kg$.
Remaining nuts: $58 - 30 = 28 \, kg$.
Ram cut these $28 \, kg$ in $8$ days, so Ram's rate $R = \frac{28}{8} = 3.5 \, kg/day$.
Since $R + H = 6$, then $H = 6 - 3.5 = 2.5 \, kg/day$.
Time taken by Hari to cut $10 \, kg$ of nuts $= \frac{10}{2.5} = 4 \, days$.

(A) Let the total work be the least common multiple of $5, 15,$ and $3$,which is $15$ units.
Efficiency of $A = 15 / 5 = 3$ units/day.
Efficiency of $B = 15 / 15 = 1$ unit/day.
Efficiency of $(A + B + C) = 15 / 3 = 5$ units/day.
Efficiency of $C = (A + B + C) - (A + B) = 5 - (3 + 1) = 1$ unit/day.
The ratio of work done by $A : B : C$ is $3 : 1 : 1$.
The total share is $Rs. 250$.
$C$'s share $= (1 / (3 + 1 + 1)) \times 250 = (1 / 5) \times 250 = Rs. 50$.

(B) Let the efficiency of Shashi be $100$ units/day.
Since Tanya is $25\%$ more efficient than Shashi,the efficiency of Tanya is $125$ units/day.
The ratio of efficiencies of Tanya to Shashi is $125:100$,which simplifies to $5:4$.
Since time taken is inversely proportional to efficiency,the ratio of time taken by Tanya to Shashi is $4:5$.
Given that Shashi takes $20$ days to complete the work,we have $5$ units of time $= 20$ days.
Therefore,$1$ unit of time $= 4$ days.
The time taken by Tanya is $4$ units of time $= 4 \times 4 = 16$ days.

(A) Let the work done by $1$ man in $1$ day be $M$ and the work done by $1$ boy in $1$ day be $B$.
According to the problem,the total work done is the same in both cases.
$(12M + 16B) \times 5 = (13M + 24B) \times 4$
$60M + 80B = 52M + 96B$
Rearranging the terms to group $M$ and $B$:
$60M - 52M = 96B - 80B$
$8M = 16B$
$\frac{M}{B} = \frac{16}{8} = \frac{2}{1}$
Therefore,the ratio of the daily work done by a man to that of a boy is $2:1$.

(C) Let the total work be $50$ units.
Since one worker completes the work in $50$ days,the efficiency of one worker is $1$ unit per day.
On day $1$,the number of workers is $1$,so work done = $1$ unit.
On day $2$,the number of workers is $2$,so work done = $2$ units.
On day $3$,the number of workers is $3$,so work done = $3$ units.
This follows an arithmetic progression where the work done on day $n$ is $n$ units.
The total work done in $n$ days is the sum of the first $n$ natural numbers: $S_n = \frac{n(n+1)}{2}$.
We need to find $n$ such that $S_n \ge 50$.
$\frac{n(n+1)}{2} \ge 50 \implies n(n+1) \ge 100$.
For $n = 9$,$9 \times 10 = 90 < 100$.
For $n = 10$,$10 \times 11 = 110 \ge 100$.
Thus,the work will be completed on the $10^{th}$ day.

(D) Step $1$: Calculate the work done by $A$ in $9$ days.
$A$ does $\frac{2}{5}$ of the work in $9$ days.
So,$A$'s efficiency $= \frac{2/5}{9} = \frac{2}{45}$ work per day.
Step $2$: Calculate the remaining work.
Remaining work $= 1 - \frac{2}{5} = \frac{3}{5}$ of the work.
Step $3$: Calculate the combined efficiency of $A$ and $B$.
They complete $\frac{3}{5}$ of the work in $6$ days.
Combined efficiency $(A+B) = \frac{3/5}{6} = \frac{3}{30} = \frac{1}{10}$ work per day.
Step $4$: Calculate $B$'s efficiency.
$B$'s efficiency $= (A+B) - A = \frac{1}{10} - \frac{2}{45}$.
Taking $LCM$ of $10$ and $45$ as $90$:
$B$'s efficiency $= \frac{9-4}{90} = \frac{5}{90} = \frac{1}{18}$ work per day.
Step $5$: Calculate the time taken by $B$ to finish the whole work.
Time taken by $B = \frac{1}{1/18} = 18$ days.

(B) $1$. Calculate Akka's rate of baking: Akka bakes $100$ cakes in $20$ hours,so her rate is $100 / 20 = 5$ cakes per hour.
$2$. Calculate the combined rate of Akka and Tai: They bake $75$ cakes in $10$ hours,so their combined rate is $75 / 10 = 7.5$ cakes per hour.
$3$. Calculate Tai's rate of baking: Tai's rate = (Combined rate) - (Akka's rate) = $7.5 - 5 = 2.5$ cakes per hour.
$4$. Calculate the number of cakes Tai bakes in $20$ hours: $2.5 \text{ cakes/hour} \times 20 \text{ hours} = 50$ cakes.

(B) Let the time taken by $A$ to complete the task be $x \, \text{hours}$.
Then,the time taken by $B$ to complete the task is $(x + 10) \, \text{hours}$.
Working together,their combined rate of work is $\frac{1}{x} + \frac{1}{x + 10} = \frac{1}{12}$.
Solving for $x$: $\frac{x + 10 + x}{x(x + 10)} = \frac{1}{12} \implies 12(2x + 10) = x^2 + 10x$.
$24x + 120 = x^2 + 10x \implies x^2 - 14x - 120 = 0$.
Factoring the quadratic equation: $(x - 20)(x + 6) = 0$.
Since time cannot be negative,$x = 20 \, \text{hours}$.
Thus,$B$ takes $x + 10 = 30 \, \text{hours}$ to complete the whole task.
To complete $50 \%$ of the task,$B$ will take $50 \% \text{ of } 30 \, \text{hours} = 15 \, \text{hours}$.

(D) Let $1$ man's $1 \, \text{day}$ work be $M$ and $1$ woman's $1 \, \text{day}$ work be $W$.
Total work = $28 \times 15 \times M = 15 \times 24 \times W$.
$28 M = 24 W \implies \frac{M}{W} = \frac{24}{28} = \frac{6}{7}$.
Work done by $30$ men in $1 \, \text{day} = 30 M$.
Work done by $18$ women in $1 \, \text{day} = 18 W$.
Ratio = $\frac{30 M}{18 W} = \frac{30}{18} \times \frac{M}{W} = \frac{5}{3} \times \frac{6}{7} = \frac{5 \times 2}{7} = \frac{10}{7}$.
Thus,the ratio is $10:7$.

(C) Total work = $20 \times 16 = 320$ man-days.
Work done in $5$ days = $20 \times 5 = 100$ man-days.
Remaining work = $320 - 100 = 220$ man-days.
Let $x$ be the number of men who left. The remaining men = $(20 - x)$.
The remaining work is completed in $18 \frac{1}{3} = \frac{55}{3}$ days.
Using the formula $\text{Work} = \text{Men} \times \text{Days}$:
$220 = (20 - x) \times \frac{55}{3}$.
$220 \times \frac{3}{55} = 20 - x$.
$4 \times 3 = 20 - x$.
$12 = 20 - x$.
$x = 20 - 12 = 8$.
Therefore,$8$ men left the work.

(C) Efficiency of $A = 1/30$ units per day.
Efficiency of $B = 1/36$ units per day.
Let $A$ and $B$ work together for $x$ days.
Work done by $B$ alone in $25$ days $= 25 \times (1/36) = 25/36$.
Remaining work done by $A$ and $B$ together $= 1 - 25/36 = 11/36$.
Combined efficiency of $A$ and $B = 1/30 + 1/36 = (6+5)/180 = 11/180$.
Time taken by $A$ and $B$ to complete $11/36$ of the work $= (11/36) / (11/180) = (11/36) \times (180/11) = 5$ days.
Therefore,$A$ left the work after $5$ days.

(B) Let the total time taken to complete the work be $x$ days.
Since $A$ worked for the entire duration,$B$ left $8$ days before completion,and $C$ left $6$ days before completion,their individual work contributions are:
Work done by $A = \frac{x}{16}$
Work done by $B = \frac{x-8}{32}$
Work done by $C = \frac{x-6}{48}$
The sum of their work equals $1$ (the whole work):
$\frac{x}{16} + \frac{x-8}{32} + \frac{x-6}{48} = 1$
Taking the least common multiple $(LCM)$ of $16, 32,$ and $48$,which is $96$:
$\frac{6x + 3(x-8) + 2(x-6)}{96} = 1$
$6x + 3x - 24 + 2x - 12 = 96$
$11x - 36 = 96$
$11x = 132$
$x = \frac{132}{11} = 12$ days.
Thus,the work is finished in $12$ days.

(B) $(P+Q)$'s $1$ day work $= 1/6$.
$(Q+R)$'s $1$ day work $= 7/60$.
Let $P$ alone take $x$ days to finish the work.
According to the question,$P$ worked for $3$ days and $(Q+R)$ worked for $6$ days to complete the total work ($1$ unit).
So,$3/x + 6 \times (7/60) = 1$.
$3/x + 42/60 = 1$.
$3/x + 7/10 = 1$.
$3/x = 1 - 7/10 = 3/10$.
$x = 10$ days.
$Q$'s $1$ day work $= (P+Q) - P = 1/6 - 1/10 = (5-3)/30 = 2/30 = 1/15$.
$R$'s $1$ day work $= (Q+R) - Q = 7/60 - 1/15 = (7-4)/60 = 3/60 = 1/20$.
$R$ alone takes $20$ days to finish the work.
The difference in days taken by $R$ and $P$ is $20 - 10 = 10$ days.

(C) Let the total work be the Least Common Multiple $(LCM)$ of $8, 12,$ and $6$,which is $24$ units.
Efficiency of $(A+B) = 24 / 8 = 3$ units/day.
Efficiency of $(B+C) = 24 / 12 = 2$ units/day.
Efficiency of $(A+B+C) = 24 / 6 = 4$ units/day.
To find the efficiency of $B$,we use the equation: $(A+B) + (B+C) - (A+B+C) = B$.
Efficiency of $B = 3 + 2 - 4 = 1$ unit/day.
Now,find the efficiency of $(A+C)$:
Efficiency of $(A+C) = (A+B+C) - B = 4 - 1 = 3$ units/day.
Time taken by $A$ and $C$ to complete the work = Total work / Efficiency of $(A+C) = 24 / 3 = 8$ days.

(D) Efficiency of $A = \frac{1}{24}$ units/day.
Efficiency of $B = \frac{1}{30}$ units/day.
Efficiency of $C = \frac{1}{40}$ units/day.
Total work ($LCM$ of $24, 30, 40$) $= 120$ units.
Efficiency of $A = 5$ units/day,$B = 4$ units/day,$C = 3$ units/day.
Combined efficiency of $(A+B+C) = 5+4+3 = 12$ units/day.
Let the total time taken to complete the work be $x$ days.
$C$ left $4$ days before completion,so $C$ worked for $(x-4)$ days,while $A$ and $B$ worked for $x$ days.
Equation: $5x + 4x + 3(x-4) = 120$.
$9x + 3x - 12 = 120$.
$12x = 132$.
$x = 11$ days.
Thus,the total work was completed in $11$ days.

(B) Step $1$: Calculate the work done per day by $A$ and $B$.
$A$ completes the work in $10 \text{ days}$,so $A$'s $1 \text{ day}$ work $= \frac{1}{10}$.
$B$ completes the work in $20 \text{ days}$,so $B$'s $1 \text{ day}$ work $= \frac{1}{20}$.
Step $2$: Calculate the combined work done by $A$ and $B$ in $1 \text{ day}$.
$(A + B)$'s $1 \text{ day}$ work $= \frac{1}{10} + \frac{1}{20} = \frac{2+1}{20} = \frac{3}{20}$.
Step $3$: Calculate the work done by them in $5 \text{ days}$.
Work done in $5 \text{ days} = 5 \times \frac{3}{20} = \frac{15}{20} = \frac{3}{4}$.
Step $4$: Calculate the remaining work.
Remaining work $= 1 - \frac{3}{4} = \frac{1}{4}$.

(A) Let the total work be the $LCM$ of $30$ and $20$,which is $60$ units.
Efficiency of $(A + B) = \frac{60}{30} = 2$ units/day.
Efficiency of $(B + C) = \frac{60}{20} = 3$ units/day.
Let $A, B, C$ be the efficiencies of $A, B, C$ respectively.
$A + B = 2$ $(i)$
$B + C = 3$ (ii)
Total work done = $5A + 15B + 18C = 60$.
We can rewrite this as $5(A + B) + 10B + 18C = 60$.
Substitute $(A + B) = 2$:
$5(2) + 10B + 18C = 60 \Rightarrow 10 + 10B + 18C = 60 \Rightarrow 10B + 18C = 50 \Rightarrow 5B + 9C = 25$ (iii).
From (ii),$B = 3 - C$. Substitute this into (iii):
$5(3 - C) + 9C = 25 \Rightarrow 15 - 5C + 9C = 25 \Rightarrow 4C = 10 \Rightarrow C = 2.5$ units/day.
Time taken by $C$ alone = $\frac{\text{Total Work}}{\text{Efficiency of } C} = \frac{60}{2.5} = 24$ days.

(B) Let the efficiency of $C$ be $k$. Since $A$ is thrice as fast as $C$,the efficiency of $A$ is $3k$.
Since $A$ is twice as fast as $B$,the efficiency of $B$ is $\frac{3k}{2} = 1.5k$.
The total efficiency of $A, B,$ and $C$ is $3k + 1.5k + k = 5.5k$.
Given that they complete the work in $6$ days,the total work is $6 \times 5.5k = 33k$.
To find the time taken by $C$ alone,we divide the total work by the efficiency of $C$:
Time taken by $C = \frac{33k}{k} = 33$ days.

(D) Given that $A$'s $2$ days work is equal to $B$'s $3$ days work.
Let the efficiency of $A$ be $E_A$ and $B$ be $E_B$.
So,$2 \times E_A = 3 \times E_B$.
This implies the ratio of their efficiencies is $\frac{E_A}{E_B} = \frac{3}{2}$.
Since time taken is inversely proportional to efficiency,the ratio of time taken by $A$ and $B$ to complete the same work is $\frac{T_A}{T_B} = \frac{E_B}{E_A} = \frac{2}{3}$.
Given that $A$ completes the work in $8$ days $(T_A = 8)$.
Substituting the values: $\frac{8}{T_B} = \frac{2}{3}$.
$2 \times T_B = 8 \times 3$.
$2 \times T_B = 24$.
$T_B = 12$ days.
Therefore,$B$ will take $12$ days to complete the work.

(D) Let $M$ be the work done by a man in one day and $W$ be the work done by a woman in one day.
According to the problem:
$(4M + 6W) \times 8 = (2M + 9W) \times 8$
Dividing both sides by $8$,we get:
$4M + 6W = 2M + 9W$
$2M = 3W$
This means $2$ men are equivalent to $3$ women.
Now,substitute $2M = 3W$ into the first case:
$4M + 6W = 2(2M) + 6W = 2(3W) + 6W = 6W + 6W = 12W$.
So,$12$ women can complete the work in $8$ days.
To find the time taken by $18$ women,we use the formula $W_1 \times D_1 = W_2 \times D_2$:
$12 \times 8 = 18 \times D_2$
$D_2 = \frac{12 \times 8}{18} = \frac{96}{18} = \frac{16}{3} = 5 \frac{1}{3}$ days.

(A) Given that $4$ men can do the work in $15$ days,so $1$ man can do the work in $15 \times 4 = 60$ days.
Therefore,the work done by $1$ man in $1$ day is $\frac{1}{60}$.
Similarly,$8$ women can do the work in $15$ days,so $1$ woman can do the work in $15 \times 8 = 120$ days.
Therefore,the work done by $1$ woman in $1$ day is $\frac{1}{120}$.
Now,we need to find the work done by $6$ men and $12$ women in $1$ day:
Work done by $6$ men $= 6 \times \frac{1}{60} = \frac{1}{10}$.
Work done by $12$ women $= 12 \times \frac{1}{120} = \frac{1}{10}$.
Total work done by $6$ men and $12$ women in $1$ day $= \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$.
Thus,the total time taken to complete the work is $5$ days.

(B) The total work is considered as $1$ unit.
Given that $A$ and $C$ together finish $\frac{19}{23}$ of the work.
Therefore,the work done by $B$ is $1 - \frac{19}{23} = \frac{23 - 19}{23} = \frac{4}{23}$.
The wages are distributed in proportion to the work done.
Thus,the share of $B$ is $\frac{4}{23}$ of the total amount.
$B$'s share $= \frac{4}{23} \times 575 = 4 \times 25 = Rs. 100$.

(A) Let the efficiency of a boy be $1$ unit/day.
Since a woman is twice as fast as a boy, the efficiency of a woman is $2$ units/day.
Since a man is twice as fast as a woman, the efficiency of a man is $2 \times 2 = 4$ units/day.
Total efficiency of a man, a woman, and a boy together $= 4 + 2 + 1 = 7$ units/day.
Total work $= \text{Total efficiency} \times \text{Total days} = 7 \text{ units/day} \times 7 \text{ days} = 49$ units.
Time taken by a boy to complete the work alone $= \frac{\text{Total work}}{\text{Efficiency of a boy}} = \frac{49}{1} = 49$ days.

(C) Step $1$: Calculate the work done by $C$ before leaving.
$C$ leaves after $\frac{1}{8}$ of the work is completed.
Step $2$: Calculate the remaining work.
Remaining work $= 1 - \frac{1}{8} = \frac{7}{8}$.
Step $3$: Calculate the combined efficiency of $A$ and $B$.
$A$'s $1$ day work $= \frac{1}{6}$.
$B$'s $1$ day work $= \frac{1}{12}$.
$(A + B)$'s $1$ day work $= \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{3}{12} = \frac{1}{4}$.
Step $4$: Calculate the time taken by $A$ and $B$ to complete the remaining work.
Time $= \frac{\text{Remaining Work}}{\text{Combined Efficiency}} = \frac{7/8}{1/4} = \frac{7}{8} \times 4 = \frac{7}{2} = 3 \frac{1}{2}$ days.

(B) We use the formula for work equivalence: $M_{1} \times D_{1} \times T_{1} = M_{2} \times D_{2} \times T_{2}$
Given:
$M_{1} = 15$ men,$D_{1} = 20$ days,$T_{1} = 8$ hours/day
$M_{2} = 20$ men,$D_{2} = 12$ days,$T_{2} = ?$
Substituting the values into the formula:
$15 \times 20 \times 8 = 20 \times 12 \times T_{2}$
Solving for $T_{2}$:
$T_{2} = \frac{15 \times 20 \times 8}{20 \times 12}$
$T_{2} = \frac{15 \times 8}{12}$
$T_{2} = \frac{120}{12} = 10$ hours/day.

(D) The work done by $(Raj + Ram)$ in $1 \text{ day} = \frac{1}{10}$.
The work done by $Raj$ in $1 \text{ day} = \frac{1}{12}$.
Therefore,the work done by $Ram$ in $1 \text{ day} = \frac{1}{10} - \frac{1}{12}$.
Taking the least common multiple $(LCM)$ of $10$ and $12$,which is $60$:
$Ram$'s $1 \text{ day work} = \frac{6 - 5}{60} = \frac{1}{60}$.
Thus,$Ram$ alone will complete the work in $60 \text{ days}$.

(B) Let the time spent by each person be $T$ minutes.
The rate of typing for $A$ is $1/12$ pages per minute.
The rate of typing for $B$ is $1/15$ pages per minute.
The rate of typing for $C$ is $1/24$ pages per minute.
Since they spend equal time $T$,the number of pages typed by $A, B,$ and $C$ are $T/12, T/15,$ and $T/24$ respectively.
The total number of pages is $506$,so:
$T/12 + T/15 + T/24 = 506$
To solve for $T$,find the least common multiple of $12, 15,$ and $24$,which is $120$.
$(10T + 8T + 5T) / 120 = 506$
$23T / 120 = 506$
$T = (506 \times 120) / 23 = 22 \times 120 = 2640$ minutes.
The number of pages typed by $B$ is $T/15 = 2640 / 15 = 176$ pages.

(D) Work done by $A$ in $1$ day $= \frac{1}{m}$.
Work done by $B$ in $1$ day $= \frac{1}{n}$.
Work done by both $A$ and $B$ together in $1$ day $= \frac{1}{m} + \frac{1}{n} = \frac{n+m}{mn}$.
Therefore,the total time taken by $A$ and $B$ to complete the work together is the reciprocal of the work done in $1$ day.
Time taken $= \frac{mn}{m+n}$ days.

(D) Let the time taken by $A, B, C$ working together be $t$ hours.
Then $A$ takes $(t+6)$ hours,$B$ takes $(t+1)$ hours,and $C$ takes $2t$ hours.
The combined work rate is $\frac{1}{t+6} + \frac{1}{t+1} + \frac{1}{2t} = \frac{1}{t}$.
Subtracting $\frac{1}{2t}$ from both sides: $\frac{1}{t+6} + \frac{1}{t+1} = \frac{1}{2t}$.
$\frac{(t+1) + (t+6)}{(t+6)(t+1)} = \frac{1}{2t} \Rightarrow \frac{2t+7}{t^2+7t+6} = \frac{1}{2t}$.
$4t^2 + 14t = t^2 + 7t + 6 \Rightarrow 3t^2 + 7t - 6 = 0$.
$(3t-2)(t+3) = 0$. Since time must be positive,$t = 2/3$ hours.
Now,$A$ takes $2/3 + 6 = 20/3$ hours and $B$ takes $2/3 + 1 = 5/3$ hours.
Combined rate of $A$ and $B = \frac{3}{20} + \frac{3}{5} = \frac{3+12}{20} = \frac{15}{20} = 3/4$ work per hour.
Time taken by $A$ and $B$ together = $4/3$ hours.

(B) Let the time taken by $(B+C)$ be $x$ days. Then,the time taken by $A$ is $3x$ days.
Therefore,the work done by $(A+B+C)$ in $1$ day is $\frac{1}{3x} + \frac{1}{x} = \frac{1+3}{3x} = \frac{4}{3x}$.
Given that $(A+B+C)$ complete the work in $24$ days,so $\frac{4}{3x} = \frac{1}{24}$.
$3x = 4 \times 24 = 96$,so $x = 32$ days.
Thus,$A$ takes $3x = 3 \times 32 = 96$ days to finish the job alone.

(A) The efficiency of $P$ is taken as $100\%$. Since $Q$ is $50\%$ more efficient than $P$,the efficiency of $Q$ is $150\%$.
The ratio of efficiency of $Q$ to $P$ is $150 : 100$,which simplifies to $3 : 2$.
Since time taken is inversely proportional to efficiency,the ratio of time taken by $Q$ to $P$ is $2 : 3$.
Given that $P$ takes $9$ days,which corresponds to $3$ parts in the ratio.
Therefore,$1$ part corresponds to $\frac{9}{3} = 3$ days.
Since $Q$ takes $2$ parts,the time taken by $Q$ is $2 \times 3 = 6$ days.

(A) Given that $16$ men can complete the work in $15$ days.
Therefore,$8$ men can complete the same work in $16 \times 15 / 8 = 30$ days.
Given that $24$ children can complete the work in $20$ days.
Therefore,$8$ children can complete the same work in $24 \times 20 / 8 = 60$ days.
Let the total work be the $LCM$ of $30$ and $60$,which is $60$ units.
Efficiency of $8$ men per day $= 60 / 30 = 2$ units/day.
Efficiency of $8$ children per day $= 60 / 60 = 1$ unit/day.
Combined efficiency of $8$ men and $8$ children $= 2 + 1 = 3$ units/day.
Time taken to complete the work $= 60 / 3 = 20$ days.

(B) Given that $A$ completes the work in $4 \, days$ and $B$ completes it in $12 \, days$.
Total work is taken as the $LCM$ of $4$ and $12$,which is $12 \, units$.
Efficiency of $A = \frac{12 \, units}{4 \, days} = 3 \, units/day$.
Efficiency of $B = \frac{12 \, units}{12 \, days} = 1 \, unit/day$.
Combined efficiency of $A$ and $B = 3 + 1 = 4 \, units/day$.
Time taken by both working together $= \frac{\text{Total work}}{\text{Combined efficiency}} = \frac{12}{4} = 3 \, days$.

(A) If $A$ can do $\frac{1}{4}$ of the work in $10$ days,then $A$ can complete the whole work in $10 \times 4 = 40$ days.
If $B$ can do $\frac{1}{3}$ of the work in $20$ days,then $B$ can complete the whole work in $20 \times 3 = 60$ days.
Let the total work be the Least Common Multiple $(LCM)$ of $40$ and $60$,which is $120$ units.
Efficiency of $A = \frac{120}{40} = 3$ units/day.
Efficiency of $B = \frac{120}{60} = 2$ units/day.
Combined efficiency of $A$ and $B = 3 + 2 = 5$ units/day.
Time taken by $A$ and $B$ together to complete the work $= \frac{\text{Total work}}{\text{Combined efficiency}} = \frac{120}{5} = 24$ days.

(C) Let the total work be $W = 1$ unit.
Combined efficiency of $A$ and $B$ is $(A+B) = \frac{1}{30}$ units/day.
Work done by $A$ and $B$ in $20$ days $= 20 \times \frac{1}{30} = \frac{2}{3}$ units.
Remaining work $= 1 - \frac{2}{3} = \frac{1}{3}$ units.
$A$ completes this $\frac{1}{3}$ units of work in $20$ days.
Therefore,efficiency of $A = \frac{1/3}{20} = \frac{1}{60}$ units/day.
Time taken by $A$ alone to finish the total work $= \frac{1}{1/60} = 60$ days.

(C) According to the formula for work,$M_{1} \times D_{1} = M_{2} \times D_{2}$,where $M$ is the number of men and $D$ is the number of days.
Given: $M_{1} = x$,$D_{1} = x$,$M_{2} = y$,and we need to find $D_{2}$.
Substituting the values: $x \times x = y \times D_{2}$.
Therefore,$D_{2} = \frac{x^{2}}{y}$ days.

(C) Let the work done by the $first$,$second$,$third$,and $fourth$ person in $1$ day be $W_1, W_2, W_3$,and $W_4$ respectively.
$W_1 = 1/8$,$W_2 = 1/12$,$W_3 = 1/16$.
Let the $fourth$ person complete the work in $x$ days,so $W_4 = 1/x$.
Since they complete the work together in $3$ days:
$3(1/8 + 1/12 + 1/16 + 1/x) = 1$
$1/8 + 1/12 + 1/16 + 1/x = 1/3$
$1/x = 1/3 - (6+4+3)/48 = 1/3 - 13/48 = (16-13)/48 = 3/48 = 1/16$.
Thus,the $fourth$ person's work share in $3$ days is $3 \times (1/16) = 3/16$.
The wages are distributed in proportion to the work done by each person.
$Fourth$ person's share $= (3/16) \times 1200 = 3 \times 75 = 225$ $Rs.$

(B) Let $A$ alone do the work in $x$ days and $B$ alone do the work in $y$ days.
According to the problem,the work done by $A$ and $B$ in one day is $1/x$ and $1/y$ respectively.
Given that $A$ and $B$ together complete the job in $5$ days,we have: $1/x + 1/y = 1/5$ .....$(1)$
If $A$ works with double efficiency $(2/x)$ and $B$ works with $1/3$ efficiency $(1/(3y))$,they complete the job in $3$ days: $2/x + 1/(3y) = 1/3$ .....$(2)$
To solve for $x$,multiply equation $(2)$ by $3$: $6/x + 1/y = 1$ .....$(3)$
Subtract equation $(1)$ from equation $(3)$: $(6/x + 1/y) - (1/x + 1/y) = 1 - 1/5$
$5/x = 4/5$
$x = 25/4 = 6 \frac{1}{4}$ days.
Thus,$A$ alone would require $6 \frac{1}{4}$ days to complete the job.

(C) Work done by $A$ in one day $= \frac{1}{20}$.
Work done by $B$ in one day $= \frac{1}{30}$.
Combined work of $A$ and $B$ in one day $= \frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$.
Work done by $A$ and $B$ in $7$ days $= 7 \times \frac{1}{12} = \frac{7}{12}$.
Remaining work $= 1 - \frac{7}{12} = \frac{5}{12}$.
$C$ finishes the remaining $\frac{5}{12}$ work in $10$ days.
Therefore,$C$ will finish the full work in $10 \times \frac{12}{5} = 24$ days.

(D) Sunil's $1$ day work $= \frac{1}{4}$.
Dinesh's $1$ day work $= \frac{1}{6}$.
Ramesh is $1 \frac{1}{2} = \frac{3}{2}$ times as fast as Sunil.
Therefore,Ramesh's $1$ day work $= \frac{3}{2} \times \frac{1}{4} = \frac{3}{8}$.
Work done by all three together in $1$ day $= \frac{1}{4} + \frac{1}{6} + \frac{3}{8}$.
Taking the least common multiple $(LCM)$ of $4, 6, 8$ as $24$:
$= \frac{6 + 4 + 9}{24} = \frac{19}{24}$.
Time taken by all three to complete the work $= \frac{24}{19} = 1 \frac{5}{19}$ days.

(B) The total work done is constant. Let $D_1$ and $T_1$ be the days and hours per day for the first scenario,and $D_2$ and $T_2$ be the days and hours per day for the second scenario.
The formula for total work is $D_1 \times T_1 = D_2 \times T_2$.
Given:
$D_1 = 18$ days
$T_1 = 6$ hours/day
$D_2 = 12$ days
Substituting the values:
$18 \times 6 = 12 \times T_2$
$108 = 12 \times T_2$
$T_2 = \frac{108}{12} = 9$ hours/day.
Therefore,the farmer must work $9$ hours per day.

(A) The work done by $2$ men in $x$ days is $1$ unit of work.
Therefore,the work done by $1$ man in $1$ day is $\frac{1}{2x}$.
The work done by $y$ women in $3$ days is $1$ unit of work.
Therefore,the work done by $1$ woman in $1$ day is $\frac{1}{3y}$.
The ratio of the work done by $1$ man to that of $1$ woman is $\frac{1}{2x} : \frac{1}{3y}$.
Multiplying both sides by $6xy$,we get $3y : 2x$.