Solution:

Let Hasan's 1-day work be $h$ and Kamal's 1-day work be $k$.
From the first condition: $h + k = \frac{1}{12}$
From the second condition: $\frac{h}{2} + 3k = \frac{1}{9}$
Multiply the first equation by 3 to match the $k$ terms: $3h + 3k = \frac{3}{12} = \frac{1}{4}$
Now, subtract the second equation from this new equation:
$(3h + 3k) - (\frac{h}{2} + 3k) = \frac{1}{4} - \frac{1}{9}$
$\frac{5h}{2} = \frac{5}{36}$
$h = \frac{5}{36} \times \frac{2}{5} = \frac{1}{18}$
Since Hasan's 1-day work is $\frac{1}{18}$, he takes 18 days to complete the work alone.

So, the answer is B

Solution:

Net work done in a 2-day cycle = $\frac{1}{30} - \frac{1}{40} = \frac{4 - 3}{120} = \frac{1}{120}$.
A common trap is to assume it takes $120 \times 2 = 240$ days. However, on the final step, the mason will finish building the wall, and the laborer will not have a chance to demolish it.
The mason's work on his final day is $\frac{1}{30}$. We must calculate how long it takes to reach the point just before this final day.
Target work before the final day = $1 - \frac{1}{30} = \frac{29}{30}$.
Let $N$ be the number of 2-day cycles needed to reach this target: $N \times \frac{1}{120} = \frac{29}{30}$.
$N = \frac{29 \times 120}{30} = 29 \times 4 = 116$ cycles.
116 cycles $\times 2$ = 232 days. In 232 days, exactly $\frac{29}{30}$ of the wall is built.
On the 233rd day, the mason comes in and completes the remaining $\frac{1}{30}$ of the work. The wall is now fully built.

So, the answer is B

Solution:

Let Arif's 1-day work be $a$ and Babu's 1-day work be $b$.
Scenario 1: $4a + 18b = 1$ (Whole work complete)
Scenario 2: $6a + 12b = 1$
Equate the two expressions since both equal 1:
$4a + 18b = 6a + 12b \Rightarrow 2a = 6b \Rightarrow a = 3b$. (Arif is 3 times as efficient as Babu).
Substitute $a = 3b$ back into Scenario 1: $4(3b) + 18b = 1 \Rightarrow 12b + 18b = 1 \Rightarrow 30b = 1 \Rightarrow b = \frac{1}{30}$. Babu takes 30 days.
Since $a = 3b$, $a = 3(\frac{1}{30}) = \frac{1}{10}$. Arif takes 10 days.
Working together, their 1-day work is $a + b = \frac{1}{10} + \frac{1}{30} = \frac{3+1}{30} = \frac{4}{30} = \frac{2}{15}$.
Total time taken together = $\frac{15}{2} = 7.5$ days.

So, the answer is B

Solution:

Let the waste pipe take $x$ minutes to empty the full tank alone. Its 1-minute work is $-\frac{1}{x}$.
Work of Pipe A in 1 min = $\frac{1}{20}$. Work of Pipe B in 1 min = $\frac{1}{24}$.
Net work of all three pipes in 1 min = $\frac{1}{15}$.
Equation: $\frac{1}{20} + \frac{1}{24} - \frac{1}{x} = \frac{1}{15}$
$\frac{6 + 5}{120} - \frac{1}{x} = \frac{8}{120}$ (converting all fractions to a common denominator of 120)
$\frac{11}{120} - \frac{8}{120} = \frac{1}{x}$
$\frac{3}{120} = \frac{1}{x} \Rightarrow \frac{1}{40} = \frac{1}{x} \Rightarrow x = 40$.
The waste pipe takes 40 minutes to empty the tank. Since it empties at a rate of 3 gallons per minute, the Capacity = $40 \text{ mins} \times 3 \text{ gallons/min} = 120$ gallons.

So, the answer is C

Solution:

We use the concept of equating total man-days of work.
$(2M + 3B) \times 10 = (3M + 2B) \times 8$
$20M + 30B = 24M + 16B$
$14B = 4M \Rightarrow 2M = 7B$. (The work of 2 Men is equivalent to 7 Boys).
Substitute this into the first team's equation to find the work in terms of purely boys:
$(2M + 3B) \rightarrow (7B + 3B) = 10B$. So, 10 Boys can do the work in 10 days. Total work = $10 \times 10 = 100$ Boy-days.
Now, find the equivalent of the target team (2 Men and 1 Boy):
$2M + 1B = 7B + 1B = 8B$.
Time taken by 8 Boys = $\frac{\text{Total Work}}{\text{Workers}} = \frac{100}{8} = 12.5$ days.

So, the answer is B

Solution:

Use the formula: $\frac{M_1 \times D_1}{W_1} = \frac{M_2 \times D_2}{W_2}$
Initial condition: $M_1 = 30$ men, $D_1 = 25$ days, $W_1 = 20\%$ (or $0.2$).
Target condition: The remaining time is $100 - 25 = 75$ days ($D_2$). The remaining work is $100\% - 20\% = 80\%$ (or $0.8$) ($W_2$). We need to find $M_2$ (total men needed).
$\frac{30 \times 25}{0.2} = \frac{M_2 \times 75}{0.8}$
$3750 = \frac{M_2 \times 75}{4}$ (Simplifying the denominators by multiplying both sides by 0.8)
$M_2 = \frac{3750 \times 4}{75} = 50 \times 4 = 200$? Let's re-calculate.
$\frac{750}{0.2} = \frac{75M_2}{0.8} \Rightarrow 3750 = 93.75 \times M_2 \Rightarrow M_2 = \frac{3750}{93.75} = 40$ men.
Total men required to finish on time is 40. Since 30 men are already working, the extra men required = $40 - 30 = 10$.

So, the answer is A

Solution:

Wages are strictly distributed in proportion to the fraction of total work completed by each person.
X's 1-day work is $\frac{1}{10}$. In 5 days, X completes $5 \times \frac{1}{10} = \frac{1}{2}$ of the work.
Y's 1-day work is $\frac{1}{15}$. In 5 days, Y completes $5 \times \frac{1}{15} = \frac{1}{3}$ of the work.
Total work done by X and Y together = $\frac{1}{2} + \frac{1}{3} = \frac{3+2}{6} = \frac{5}{6}$.
The remaining work that Z had to complete is $1 - \frac{5}{6} = \frac{1}{6}$.
Therefore, Z's share is $\frac{1}{6}$ of the total contract amount.
Z's Share = $\frac{1}{6} \times 4500 = 750$.

So, the answer is B

Solution:

First, calculate the work B did alone in his final 23 days.
B's 1-day work is $\frac{1}{40}$. Work done by B in 23 days = $23 \times \frac{1}{40} = \frac{23}{40}$.
This means the work done by A and B together before A left was $1 - \frac{23}{40} = \frac{17}{40}$.
Now, find their combined 1-day work: $\frac{1}{45} + \frac{1}{40} = \frac{8+9}{360} = \frac{17}{360}$.
Let $d$ be the number of days they worked together.
$d \times (\frac{17}{360}) = \frac{17}{40}$
$d = \frac{17}{40} \times \frac{360}{17} = \frac{360}{40} = 9$ days.

So, the answer is C

Solution:

Let the supply tap take $x$ hours to fill the tank alone. Its rate is $+\frac{1}{x}$.
The leak empties the tank in 8 hours. Its rate is $-\frac{1}{8}$.
Together, the net result is emptying the tank in 12 hours. So the net rate is $-\frac{1}{12}$.
$\frac{1}{x} - \frac{1}{8} = -\frac{1}{12}$
$\frac{1}{x} = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$.
This means the supply tap alone could fill the tank in 24 hours.
Convert the time into minutes because the flow rate is in liters per minute: $24 \text{ hours} \times 60 \text{ mins} = 1440 \text{ minutes}$.
Capacity = $1440 \text{ mins} \times 6 \text{ liters/min} = 8640$ liters.

So, the answer is B

Solution:

Let's calculate the work done in one 3-day cycle.
Day 1: P works alone $\Rightarrow \frac{1}{10}$ of work.
Day 2: P works alone $\Rightarrow \frac{1}{10}$ of work.
Day 3: P, Q, and R work together $\Rightarrow \frac{1}{10} + \frac{1}{15} + \frac{1}{30} = \frac{3+2+1}{30} = \frac{6}{30} = \frac{1}{5}$ of work.
Total work done in one 3-day cycle = $\frac{1}{10} + \frac{1}{10} + \frac{1}{5} = \frac{2}{10} + \frac{2}{10} = \frac{4}{10} = \frac{2}{5}$.
In 2 cycles (6 days), the work done is $2 \times \frac{2}{5} = \frac{4}{5}$.
The remaining work is $1 - \frac{4}{5} = \frac{1}{5}$.
Now the 3rd cycle begins.
On Day 7, P works alone and does $\frac{1}{10}$. Remaining work = $\frac{1}{5} - \frac{1}{10} = \frac{1}{10}$.
On Day 8, P works alone again and does exactly the remaining $\frac{1}{10}$ of the work.
The task is completely finished exactly at the end of the 8th day.

So, the answer is C