Time & Work

A can do a piece of work in 15 days,

⇒ A's 1 day's work = $\frac{1}{15}$.

B can do a piece of work in 10 days,

⇒ B's 1 day's work = $\frac{1}{10}$.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work }= \Big(\frac{1}{15} + \frac{1}{10}\Big)\\[1em] = \frac{2 + 3}{30}\\[1em] = \frac{5}{30}\\[1em] = \frac{1}{6}\end{aligned}$$

∴ Both can finish the work in 6 days.

Hence, A and B together will take 6 days to do the work.

A can do a piece of work in 12 days,

⇒ A's 1 day's work = $\frac{1}{12}$.

B can do a piece of work in 15 days,

⇒ B's 1 day's work = $\frac{1}{15}$.

C can do a piece of work in 10 days,

⇒ C's 1 day's work = $\frac{1}{10}$.

$$\begin{aligned}\Rightarrow \text{ (A + B + C)'s 1 day's work }= \Big(\frac{1}{12} + \frac{1}{15} + \frac{1}{10}\Big)\\[1em] = \frac{5 + 4 + 6}{60}\\[1em] = \frac{15}{60}\\[1em] = \frac{1}{4}\end{aligned}$$

∴ All the three together can finish the work in 4 days.

Hence, A, B and C together will finish the work in 4 days.

A and B together can do a piece of work in 35 days,

⇒ (A + B)'s 1 day's work = $\frac{1}{35}$.

A alone can do the work in 60 days,

⇒ A's 1 day's work = $\frac{1}{60}$.

B's 1 day's work = [(A + B)'s 1 day's work] − (A's 1 day's work)

$$\begin{aligned}\Rightarrow \text{ B's 1 day's work }= \Big(\frac{1}{35} - \frac{1}{60}\Big)\\[1em] = \frac{12 - 7}{420}\\[1em] = \frac{5}{420}\\[1em] = \frac{1}{84}\end{aligned}$$

∴ B alone can finish the work in 84 days.

Hence, B alone would take 84 days to do the work.

A can do a piece of work in 20 days,

⇒ A's 1 day's work = $\frac{1}{20}$.

B can do a piece of work in 15 days,

⇒ B's 1 day's work = $\frac{1}{15}$.

A, B and C together finish the work in 5 days,

⇒ (A + B + C)'s 1 day's work = $\frac{1}{5}$.

C's 1 day's work = [(A + B + C)'s 1 day's work] − [(A's 1 day's work) + (B's 1 day's work)]

$$\begin{aligned}\Rightarrow \text{ C's 1 day's work }= \frac{1}{5} - \Big(\frac{1}{20} + \frac{1}{15}\Big)\\[1em] = \frac{12 - 3 - 4}{60}\\[1em] = \frac{5}{60}\\[1em] = \frac{1}{12}\end{aligned}$$

∴ C alone can finish the work in 12 days.

Hence, C alone would do the work in 12 days.

A can do a piece of work in 12 days,

⇒ A's 1 day's work = $\frac{1}{12}$.

B alone can do the work in 16 days,

⇒ B's 1 day's work = $\frac{1}{16}$.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work }= \Big(\frac{1}{12} + \frac{1}{16}\Big)\\[1em] = \frac{4 + 3}{48}\\[1em] = \frac{7}{48}\end{aligned}$$

They worked together for 3 days,

⇒ (A + B)'s 3 days' work = $\Big(\frac{7}{48} \times 3\Big) = \frac{21}{48} = \frac{7}{16}$.

Remaining work = $\Big(1 - \frac{7}{16}\Big) = \frac{16 - 7}{16} = \frac{9}{16}$.

As A left after 3 days, this work is being done by B alone.

Now, $\frac{1}{16}$ work is done by B in 1 day.

Whole work will be done by B in $\Big(1 \times \frac{16}{1}\Big)$ days = 16 days.

∴ $\frac{9}{16}$ work will be done by B in $\Big(16 \times \frac{9}{16}\Big)$ days = 9 days.

Hence, B took 9 days to finish the remaining work.

$\frac{1}{4}$ of the work is done by A in 5 days,

⇒ Whole work will be done by A in $\Big(5 \times \frac{4}{1}\Big)$ days = 20 days.

$\frac{1}{5}$ of the work is done by B in 6 days,

⇒ Whole work will be done by B in $\Big(6 \times \frac{5}{1}\Big)$ days = 30 days.

A can do the whole work in 20 days,

⇒ A's 1 day's work = $\frac{1}{20}$.

B can do the whole work in 30 days,

⇒ B's 1 day's work = $\frac{1}{30}$.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work }= \Big(\frac{1}{20} + \frac{1}{30}\Big)\\[1em] = \frac{3 + 2}{60}\\[1em] = \frac{5}{60}\\[1em] = \frac{1}{12}\end{aligned}$$

∴ Both together can do the work in 12 days.

Hence, both A and B together can do the work in 12 days.

A can dig the trench in 6 days,

⇒ A's 1 day's work = $\frac{1}{6}$.

B can dig the trench in 8 days,

⇒ B's 1 day's work = $\frac{1}{8}$.

A and B worked together for the same number of days, so the money is shared in the ratio of the work done by them in 1 day.

∴ Ratio of the shares of A and B = Ratio of their 1 day's work = $\frac{1}{6} : \frac{1}{8}$.

To write this ratio in whole numbers, we multiply each term by the L.C.M. of the denominators 6 and 8.

Calculating L.C.M. of 6 and 8 :

$$\begin{array}{l|rr} 2 & 6, & 8 \\ \hline 2 & 3, & 4 \\ \hline 2 & 3, & 2 \\ \hline 3 & 3, & 1 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. = 2 × 2 × 2 × 3 = 24.

$$\begin{aligned}\Rightarrow \frac{1}{6} : \frac{1}{8} = \Big(\frac{1}{6} \times 24\Big) : \Big(\frac{1}{8} \times 24\Big)\\[1em] = 4 : 3\end{aligned}$$

Sum of the ratio terms = (4 + 3) = 7.

∴ A's share = ₹ $\Big(1{,}120 \times \frac{4}{7}\Big)$ = ₹ (160 × 4) = ₹ 640.

B's share = ₹ $\Big(1{,}120 \times \frac{3}{7}\Big)$ = ₹ (160 × 3) = ₹ 480.

Hence, A's share is ₹ 640 and B's share is ₹ 480.

A can mow the field in 9 days,

⇒ A's 1 day's work = $\frac{1}{9}$.

B can mow the field in 12 days,

⇒ B's 1 day's work = $\frac{1}{12}$.

C can mow the field in 8 days,

⇒ C's 1 day's work = $\frac{1}{8}$.

A, B and C worked together for the same number of days, so the money is shared in the ratio of the work done by them in 1 day.

∴ Ratio of the shares of A, B, C = Ratio of their 1 day's work = $\frac{1}{9} : \frac{1}{12} : \frac{1}{8}$.

To write this ratio in whole numbers, we multiply each term by the L.C.M. of the denominators 9, 12 and 8.

Calculating L.C.M. of 9, 12 and 8 :

$$\begin{array}{l|rrr} 2 & 9, & 12, & 8 \\ \hline 2 & 9, & 6, & 4 \\ \hline 2 & 9, & 3, & 2 \\ \hline 3 & 9, & 3, & 1 \\ \hline 3 & 3, & 1, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. = 2 × 2 × 2 × 3 × 3 = 72.

$$\begin{aligned}\Rightarrow \frac{1}{9} : \frac{1}{12} : \frac{1}{8} = \Big(\frac{1}{9} \times 72\Big) : \Big(\frac{1}{12} \times 72\Big) : \Big(\frac{1}{8} \times 72\Big)\\[1em] = 8 : 6 : 9\end{aligned}$$

Sum of the ratio terms = (8 + 6 + 9) = 23.

∴ A's share = ₹ $\Big(1{,}610 \times \frac{8}{23}\Big)$ = ₹ (70 × 8) = ₹ 560.

B's share = ₹ $\Big(1{,}610 \times \frac{6}{23}\Big)$ = ₹ (70 × 6) = ₹ 420.

C's share = ₹ $\Big(1{,}610 \times \frac{9}{23}\Big)$ = ₹ (70 × 9) = ₹ 630.

Hence, A gets ₹ 560, B gets ₹ 420 and C gets ₹ 630.

A and B can do a piece of work in 30 days,

⇒ (A + B)'s 1 day's work = $\frac{1}{30} \quad$ ....(1)

B and C can do the work in 24 days,

⇒ (B + C)'s 1 day's work = $\frac{1}{24} \quad$ ....(2)

C and A can do the work in 40 days,

⇒ (C + A)'s 1 day's work = $\frac{1}{40} \quad$ ....(3)

Adding equations (1), (2) and (3), we get :

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work + (B + C)'s 1 day's work + (C + A)'s 1 day's work }= \frac{1}{30} + \frac{1}{24} + \frac{1}{40}\\[1em] \Rightarrow \text{ (A + B + B + C + C + A)'s 1 day's work }= \frac{1}{30} + \frac{1}{24} + \frac{1}{40}\\[1em] \Rightarrow \text{ (2A + 2B + 2C)'s 1 day's work }= \frac{1}{30} + \frac{1}{24} + \frac{1}{40}\\[1em] \Rightarrow \text{ 2(A + B + C)'s 1 day's work }= \Big(\frac{1}{30} + \frac{1}{24} + \frac{1}{40}\Big)\\[1em] \Rightarrow \text{ 2(A + B + C)'s 1 day's work }= \frac{4 + 5 + 3}{120}\\[1em] \Rightarrow \text{ 2(A + B + C)'s 1 day's work }= \frac{12}{120}\\[1em] \Rightarrow \text{ 2(A + B + C)'s 1 day's work }= \frac{1}{10}\end{aligned}$$

Dividing both sides by 2 :

$$\begin{aligned}\Rightarrow \text{ (A + B + C)'s 1 day's work }= \frac{1}{10} \div 2\\[1em] = \frac{1}{10} \times \frac{1}{2}\\[1em] = \frac{1}{20}\end{aligned}$$

∴ A, B, C together can finish the work in 20 days.

Now, A's 1 day's work = [(A + B + C)'s 1 day's work] − [(B + C)'s 1 day's work]

$$\begin{aligned}\Rightarrow \text{ A's 1 day's work }= \Big(\frac{1}{20} - \frac{1}{24}\Big)\\[1em] = \frac{6 - 5}{120}\\[1em] = \frac{1}{120}\end{aligned}$$

∴ A alone can finish the work in 120 days.

B's 1 day's work = [(A + B + C)'s 1 day's work] − [(C + A)'s 1 day's work]

$$\begin{aligned}\Rightarrow \text{ B's 1 day's work }= \Big(\frac{1}{20} - \frac{1}{40}\Big)\\[1em] = \frac{2 - 1}{40}\\[1em] = \frac{1}{40}\end{aligned}$$

∴ B alone can finish the work in 40 days.

C's 1 day's work = [(A + B + C)'s 1 day's work] − [(A + B)'s 1 day's work]

$$\begin{aligned}\Rightarrow \text{ C's 1 day's work }= \Big(\frac{1}{20} - \frac{1}{30}\Big)\\[1em] = \frac{3 - 2}{60}\\[1em] = \frac{1}{60}\end{aligned}$$

∴ C alone can finish the work in 60 days.

Hence, working together they will take 20 days, and working alone A will take 120 days, B will take 40 days and C will take 60 days.

A can do a piece of work in 80 days,

⇒ A's 1 day's work = $\frac{1}{80}$.

A works at it for 10 days,

⇒ A's 10 days' work = $\Big(\frac{1}{80} \times 10\Big) = \frac{10}{80} = \frac{1}{8}$.

Remaining work = $\Big(1 - \frac{1}{8}\Big) = \frac{8 - 1}{8} = \frac{7}{8}$.

B alone finishes the remaining work in 42 days,

⇒ $\frac{7}{8}$ work is done by B in 42 days.

Whole work will be done by B in $\Big(42 \times \frac{8}{7}\Big)$ days = 48 days.

∴ B's 1 day's work = $\frac{1}{48}$.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work }= \Big(\frac{1}{80} + \frac{1}{48}\Big)\\[1em] = \frac{3 + 5}{240}\\[1em] = \frac{8}{240}\\[1em] = \frac{1}{30}\end{aligned}$$

∴ Both together can do the work in 30 days.

Hence, both A and B together could do the work in 30 days.

A and B can together finish the work in 30 days,

⇒ (A + B)'s 1 day's work = $\frac{1}{30}$.

They worked at it for 20 days,

⇒ (A + B)'s 20 days' work = $\Big(\frac{1}{30} \times 20\Big) = \frac{20}{30} = \frac{2}{3}$.

Remaining work = $\Big(1 - \frac{2}{3}\Big) = \frac{3 - 2}{3} = \frac{1}{3}$.

As B left after 20 days, this work was done by A alone in 20 more days,

⇒ $\frac{1}{3}$ work is done by A in 20 days.

Whole work will be done by A in $\Big(20 \times \frac{3}{1}\Big)$ days = 60 days.

Hence, A alone can do the work in 60 days.

A can do the job in 25 days,

⇒ A's 1 day's work = $\frac{1}{25}$.

B alone can do the job in 20 days,

⇒ B's 1 day's work = $\frac{1}{20}$.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work }= \Big(\frac{1}{25} + \frac{1}{20}\Big)\\[1em] = \frac{4 + 5}{100}\\[1em] = \frac{9}{100}\end{aligned}$$

A worked alone for the first 10 days,

⇒ A's 10 days' work = $\Big(\frac{1}{25} \times 10\Big) = \frac{10}{25} = \frac{2}{5}$.

Remaining work = $\Big(1 - \frac{2}{5}\Big) = \frac{5 - 2}{5} = \frac{3}{5}$.

As B joined A after 10 days, this work is done by A and B together.

Now, $\frac{9}{100}$ work is done by (A + B) in 1 day.

Whole work is done by (A + B) in $\Big(1 \times \frac{100}{9}\Big)$ days = $\frac{100}{9}$ days.

∴ $\frac{3}{5}$ work is done by (A + B) in $\Big(\frac{100}{9} \times \frac{3}{5}\Big)$ days = $\frac{300}{45}$ days = $\frac{20}{3}$ days = $6\frac{2}{3}$ days.

∴ Total time taken to complete the whole work = $\Big(10 + \frac{20}{3}\Big)$ days = $\frac{30 + 20}{3}$ days = $\frac{50}{3}$ days = $16\frac{2}{3}$ days.

Hence, the whole work was completed in $\bm{16\frac{2}{3}}$ days.

A can do a piece of work in 14 days,

⇒ A's 1 day's work = $\frac{1}{14}$.

B can do the work in 21 days,

⇒ B's 1 day's work = $\frac{1}{21}$.

Let both A and B together work for $x$ days.

A's $x$ days' work = $\Big(\frac{1}{14} \times x\Big) = \frac{x}{14}$.

B's $x$ days' work = $\Big(\frac{1}{21} \times x\Big) = \frac{x}{21}$.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s } x \text{ days' work }= \Big(\frac{x}{14} + \frac{x}{21}\Big)\\[1em] = \frac{3x + 2x}{42}\\[1em] = \frac{5x}{42}\end{aligned}$$

Remaining work = $\Big(1 - \frac{5x}{42}\Big) = \frac{42 - 5x}{42}$.

As A leaves 3 days before the completion of the work, this remaining work was done by B alone in 3 days.

B's 3 days' work = $\Big(\frac{1}{21} \times 3\Big) = \frac{3}{21} = \frac{1}{7} = \frac{6}{42}$.

∴ Remaining work = B's 3 days' work

$$\begin{aligned}\Rightarrow \frac{42 - 5x}{42} = \frac{6}{42}\\[1em] \Rightarrow 42 - 5x = 6\\[1em] \Rightarrow 5x = 42 - 6\\[1em] \Rightarrow 5x = 36\\[1em] \Rightarrow x = \frac{36}{5}\end{aligned}$$

∴ Total number of days taken to complete the work = $\Big(\frac{36}{5} + 3\Big)$ days = $\frac{36 + 15}{5}$ days = $\frac{51}{5}$ days = $10\frac{1}{5}$ days.

Hence, the total number of days taken to complete the work is $\bm{10\frac{1}{5}}$ days.

A is thrice as good a workman as B.

∴ Ratio of one day's work of A and B = 3 : 1.

B is twice as good a workman as C.

∴ Ratio of one day's work of B and C = 2 : 1.

B's term is 1 in the first ratio and 2 in the second ratio. To make B's term the same in both, we multiply both the terms of the first ratio by 2.

∴ Ratio of one day's work of A and B = 3 : 1 = (3 × 2) : (1 × 2) = 6 : 2.

Now, B's term is 2 in both the ratios.

∴ Ratio of one day's work of A, B and C = 6 : 2 : 1.

The money is divided in the ratio of their one day's work.

Sum of the ratio terms = (6 + 2 + 1) = 9.

∴ A's share = ₹ $\Big(1{,}800 \times \frac{6}{9}\Big)$ = ₹ (200 × 6) = ₹ 1,200.

B's share = ₹ $\Big(1{,}800 \times \frac{2}{9}\Big)$ = ₹ (200 × 2) = ₹ 400.

C's share = ₹ $\Big(1{,}800 \times \frac{1}{9}\Big)$ = ₹ (200 × 1) = ₹ 200.

Hence, A gets ₹ 1,200, B gets ₹ 400 and C gets ₹ 200.

A can do the job in 12 days,

⇒ A's 1 day's work = $\frac{1}{12}$.

B is 60% more efficient than A.

So, if A's efficiency is 100%, then B's efficiency is (100 + 60)% = 160% = $\frac{160}{100} = \frac{8}{5}$ of A's efficiency.

∴ A's 1 day's work : B's 1 day's work = $1 : \frac{8}{5}$.

To write this ratio in whole numbers, we multiply each term by 5.

∴ A's 1 day's work : B's 1 day's work = $(1 \times 5) : \Big(\frac{8}{5} \times 5\Big) = 5 : 8$.

Let B's 1 day's work = $\frac{1}{x}$, i.e., B finishes the job in $x$ days.

$$\begin{aligned}\Rightarrow \frac{1}{12} : \frac{1}{x} = 5 : 8\\[1em] \Rightarrow \frac{\Big(\frac{1}{12}\Big)}{\Big(\frac{1}{x}\Big)} = \frac{5}{8}\\[1em] \Rightarrow \frac{1}{12} \times \frac{x}{1} = \frac{5}{8}\\[1em] \Rightarrow \frac{x}{12} = \frac{5}{8}\\[1em] \Rightarrow x = \frac{5 \times 12}{8}\\[1em] \Rightarrow x = \frac{60}{8}\\[1em] \Rightarrow x = \frac{15}{2}\end{aligned}$$

∴ B can finish the job in $\frac{15}{2}$ days = $7\frac{1}{2}$ days.

Hence, B will take $\bm{7\frac{1}{2}}$ days to finish the job.

A is twice as good a workman as B.

Thus, (A's 1 day's work) : (B's 1 day's work) = 2 : 1.

A and B can finish the work in 14 days.

(A + B)'s 1 day's work = $\frac{1}{14}$.

Divide $\frac{1}{14}$ in the ratio 2 : 1.

Sum of the ratio terms = (2 + 1) = 3.

∴ A's 1 day's work = $\Big(\frac{1}{14} \times \frac{2}{3}\Big) = \frac{1}{21}$.

∴ A alone can finish the work in 21 days.

Hence, A alone can do the work in 21 days.

Pipe A can fill the tank in 36 minutes,

⇒ Part of the tank filled by A in 1 minute = $\frac{1}{36}$.

Pipe B can fill the tank in 45 minutes,

⇒ Part of the tank filled by B in 1 minute = $\frac{1}{45}$.

$$\begin{aligned}\Rightarrow \text{ Part of the tank filled by (A + B) in 1 minute }= \Big(\frac{1}{36} + \frac{1}{45}\Big)\\[1em] = \frac{5 + 4}{180}\\[1em] = \frac{9}{180}\\[1em] = \frac{1}{20}\end{aligned}$$

∴ Time taken by (A + B) to fill the tank = 20 minutes.

Hence, the tank will be filled in 20 minutes.

The tap can fill the cistern in 3 hours,

⇒ Work done by the tap in 1 hour = $\frac{1}{3}$.

The waste pipe can empty the full cistern in 5 hours,

⇒ Work done by the waste pipe in 1 hour = $\frac{1}{5}$.

When both are kept open together, the tap fills the cistern while the waste pipe empties it, so the work of the waste pipe is subtracted :

$$\begin{aligned}\Rightarrow \text{ Net part of the cistern filled in 1 hour }= \Big(\frac{1}{3} - \frac{1}{5}\Big)\\[1em] = \frac{5 - 3}{15}\\[1em] = \frac{2}{15}\end{aligned}$$

∴ Time taken to fill the cistern = $\frac{15}{2}$ hours = $7\frac{1}{2}$ hours = 7 hours 30 minutes.

Hence, the empty cistern will be full in $\bm{7\frac{1}{2}}$ hours, i.e., 7 hours 30 minutes.

Pipe A can fill the cistern in 20 minutes,

⇒ Part of the cistern filled by A in 1 minute = $\frac{1}{20}$.

Pipe B can fill the cistern in 30 minutes,

⇒ Part of the cistern filled by B in 1 minute = $\frac{1}{30}$.

Pipe C can empty the full cistern in 15 minutes,

⇒ Part of the cistern emptied by C in 1 minute = $\frac{1}{15}$.

When all the three pipes are opened together, A and B fill the cistern while C empties it, so the part emptied by C is subtracted :

$$\begin{aligned}\Rightarrow \text{ Net part of the cistern filled in 1 minute }= \Big(\frac{1}{20} + \frac{1}{30} - \frac{1}{15}\Big)\\[1em] = \frac{3 + 2 - 4}{60}\\[1em] = \frac{1}{60}\end{aligned}$$

∴ Time taken to fill the cistern = 60 minutes = 1 hour.

Hence, the empty cistern is filled in 60 minutes, i.e., 1 hour.

The pipe can fill the tank in 16 hours,

⇒ Part of the tank filled by the pipe in 1 hour = $\frac{1}{16}$.

Due to the leak, the tank is filled in 24 hours,

⇒ Part of the tank filled by the pipe with the leak in 1 hour = $\frac{1}{24}$.

The leak empties the difference of these two parts in 1 hour :

$$\begin{aligned}\Rightarrow \text{ Part of the tank emptied by the leak in 1 hour }= \Big(\frac{1}{16} - \frac{1}{24}\Big)\\[1em] = \frac{3 - 2}{48}\\[1em] = \frac{1}{48}\end{aligned}$$

∴ Time taken by the leak to empty the full tank = 48 hours.

Hence, the leak will take 48 hours to empty the full tank.

A can complete the work in 30 days,

⇒ A's 1 day's work = $\frac{1}{30}$.

B can complete the work in 60 days,

⇒ B's 1 day's work = $\frac{1}{60}$.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work }= \Big(\frac{1}{30} + \frac{1}{60}\Big)\\[1em] = \frac{2 + 1}{60}\\[1em] = \frac{3}{60}\\[1em] = \frac{1}{20}\end{aligned}$$

∴ Both together can finish the work in 20 days.

Hence, Option 1 is the correct option.

A can do the job in 6 days,

⇒ A's 1 day's work = $\frac{1}{6}$.

A and B together can do the job in 2 days,

⇒ (A + B)'s 1 day's work = $\frac{1}{2}$.

B's 1 day's work = [(A + B)'s 1 day's work] − (A's 1 day's work)

$$\begin{aligned}\Rightarrow \text{ B's 1 day's work }= \Big(\frac{1}{2} - \frac{1}{6}\Big)\\[1em] = \frac{3 - 1}{6}\\[1em] = \frac{2}{6}\\[1em] = \frac{1}{3}\end{aligned}$$

∴ B alone can do the job in 3 days.

Hence, Option 1 is the correct option.

A, B and C together can finish the work in 4 days,

⇒ (A + B + C)'s 1 day's work = $\frac{1}{4}$.

A alone can do the work in 12 days,

⇒ A's 1 day's work = $\frac{1}{12}$.

B alone can do the work in 18 days,

⇒ B's 1 day's work = $\frac{1}{18}$.

C's 1 day's work = [(A + B + C)'s 1 day's work] − [(A's 1 day's work) + (B's 1 day's work)]

$$\begin{aligned}\Rightarrow \text{ C's 1 day's work }= \frac{1}{4} - \Big(\frac{1}{12} + \frac{1}{18}\Big)\\[1em] = \frac{9 - 3 - 2}{36}\\[1em] = \frac{4}{36}\\[1em] = \frac{1}{9}\end{aligned}$$

∴ C alone can do the work in 9 days.

Hence, Option 1 is the correct option.

A and B can do a piece of work in 18 days,

⇒ (A + B)'s 1 day's work = $\frac{1}{18} \quad$ ....(1)

B and C can do the work in 24 days,

⇒ (B + C)'s 1 day's work = $\frac{1}{24} \quad$ ....(2)

C and A can do the work in 36 days,

⇒ (C + A)'s 1 day's work = $\frac{1}{36} \quad$ ....(3)

Adding equations (1), (2) and (3), we get :

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work + (B + C)'s 1 day's work + (C + A)'s 1 day's work }= \frac{1}{18} + \frac{1}{24} + \frac{1}{36}\\[1em] \Rightarrow \text{ (A + B + B + C + C + A)'s 1 day's work }= \frac{1}{18} + \frac{1}{24} + \frac{1}{36}\\[1em] \Rightarrow \text{ (2A + 2B + 2C)'s 1 day's work }= \frac{1}{18} + \frac{1}{24} + \frac{1}{36}\\[1em] \Rightarrow \text{ 2(A + B + C)'s 1 day's work }= \Big(\frac{1}{18} + \frac{1}{24} + \frac{1}{36}\Big)\\[1em] \Rightarrow \text{ 2(A + B + C)'s 1 day's work }= \frac{4 + 3 + 2}{72}\\[1em] \Rightarrow \text{ 2(A + B + C)'s 1 day's work }= \frac{9}{72}\\[1em] \Rightarrow \text{ 2(A + B + C)'s 1 day's work }= \frac{1}{8}\end{aligned}$$

Dividing both sides by 2 :

$$\begin{aligned}\Rightarrow \text{ (A + B + C)'s 1 day's work }= \frac{1}{8} \div 2\\[1em] = \frac{1}{8} \times \frac{1}{2}\\[1em] = \frac{1}{16}\end{aligned}$$

∴ All the three working together can do the work in 16 days.

Hence, Option 3 is the correct option.

A can finish the work in 12 days,

⇒ A's 1 day's work = $\frac{1}{12}$.

B can do the work in 15 days,

⇒ B's 1 day's work = $\frac{1}{15}$.

A worked alone for 3 days,

⇒ A's 3 days' work = $\Big(\frac{1}{12} \times 3\Big) = \frac{3}{12} = \frac{1}{4}$.

Remaining work = $\Big(1 - \frac{1}{4}\Big) = \frac{4 - 1}{4} = \frac{3}{4}$.

As B joined A after 3 days, this work is done by A and B together.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work }= \Big(\frac{1}{12} + \frac{1}{15}\Big)\\[1em] = \frac{5 + 4}{60}\\[1em] = \frac{9}{60}\\[1em] = \frac{3}{20}\end{aligned}$$

Now, $\frac{3}{20}$ work is done by (A + B) in 1 day.

Whole work is done by (A + B) in $\Big(1 \times \frac{20}{3}\Big)$ days = $\frac{20}{3}$ days.

∴ $\frac{3}{4}$ work is done by (A + B) in $\Big(\frac{20}{3} \times \frac{3}{4}\Big)$ days = $\frac{60}{12}$ days = 5 days.

Hence, Option 2 is the correct option.

A can do a piece of work in 20 days,

⇒ A's 1 day's work = $\frac{1}{20}$.

B can do the work in 12 days,

⇒ B's 1 day's work = $\frac{1}{12}$.

B worked at it for 9 days,

⇒ B's 9 days' work = $\Big(\frac{1}{12} \times 9\Big) = \frac{9}{12} = \frac{3}{4}$.

Remaining work = $\Big(1 - \frac{3}{4}\Big) = \frac{4 - 3}{4} = \frac{1}{4}$.

This remaining work is done by A alone.

Now, $\frac{1}{20}$ work is done by A in 1 day.

Whole work is done by A in 20 days.

∴ $\frac{1}{4}$ work is done by A in $\Big(20 \times \frac{1}{4}\Big)$ days = 5 days.

Hence, Option 2 is the correct option.

B can complete the work in 12 days.

A works twice as fast as B, so A takes half the time taken by B.

∴ A can complete the work in $\Big(\frac{12}{2}\Big)$ days = 6 days.

A can complete the work in 6 days,

⇒ A's 1 day's work = $\frac{1}{6}$.

B can complete the work in 12 days,

⇒ B's 1 day's work = $\frac{1}{12}$.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work }= \Big(\frac{1}{6} + \frac{1}{12}\Big)\\[1em] = \frac{2 + 1}{12}\\[1em] = \frac{3}{12}\\[1em] = \frac{1}{4}\end{aligned}$$

∴ A and B together can finish the work in 4 days.

Hence, Option 1 is the correct option.

A can do a piece of work in 80 days,

⇒ A's 1 day's work = $\frac{1}{80}$.

A works at it for 10 days,

⇒ A's 10 days' work = $\Big(\frac{1}{80} \times 10\Big) = \frac{10}{80} = \frac{1}{8}$.

Remaining work = $\Big(1 - \frac{1}{8}\Big) = \frac{8 - 1}{8} = \frac{7}{8}$.

B alone finishes the remaining work in 42 days,

⇒ $\frac{7}{8}$ work is done by B in 42 days.

Whole work will be done by B in $\Big(42 \times \frac{8}{7}\Big)$ days = 48 days.

∴ B's 1 day's work = $\frac{1}{48}$.

$$\begin{aligned}\Rightarrow \text{ (A + B)'s 1 day's work }= \Big(\frac{1}{80} + \frac{1}{48}\Big)\\[1em] = \frac{3 + 5}{240}\\[1em] = \frac{8}{240}\\[1em] = \frac{1}{30}\end{aligned}$$

∴ Both together can finish the work in 30 days.

Hence, Option 3 is the correct option.

Pipe A can fill the tank in 45 hours,

⇒ Part of the tank filled by A in 1 hour = $\frac{1}{45}$.

Pipe B can fill the tank in 36 hours,

⇒ Part of the tank filled by B in 1 hour = $\frac{1}{36}$.

Taking both the pipes to be working together:

$$\begin{aligned}\Rightarrow \text{ Part of the tank filled by (A + B) in 1 hour }= \Big(\frac{1}{45} + \frac{1}{36}\Big)\\[1em] = \frac{4 + 5}{180}\\[1em] = \frac{9}{180}\\[1em] = \frac{1}{20}\end{aligned}$$

∴ The tank will be full in 20 hours.

Hence, Option 2 is the correct option.

Pipe A can fill the cistern in 4 hours,

⇒ Part of the cistern filled by A in 1 hour = $\frac{1}{4}$.

Pipe B can empty the cistern in 6 hours,

⇒ Part of the cistern emptied by B in 1 hour = $\frac{1}{6}$.

When both are opened together, A fills the cistern while B empties it, so the part emptied by B is subtracted :

$$\begin{aligned}\Rightarrow \text{ Net part of the cistern filled in 1 hour }= \Big(\frac{1}{4} - \frac{1}{6}\Big)\\[1em] = \frac{3 - 2}{12}\\[1em] = \frac{1}{12}\end{aligned}$$

∴ The cistern will be filled in 12 hours.

Hence, Option 4 is the correct option.

Without the leak, the cistern can be filled in 12 hours,

⇒ Part of the cistern filled in 1 hour, without the leak = $\frac{1}{12}$.

With the leak, the cistern is filled in 15 hours,

⇒ Part of the cistern filled in 1 hour, with the leak = $\frac{1}{15}$.

The leak empties the difference of these two parts in 1 hour :

$$\begin{aligned}\Rightarrow \text{ Part of the cistern emptied by the leak in 1 hour }= \Big(\frac{1}{12} - \frac{1}{15}\Big)\\[1em] = \frac{5 - 4}{60}\\[1em] = \frac{1}{60}\end{aligned}$$

∴ The leak can empty the full cistern in 60 hours.

Hence, Option 4 is the correct option.

Pipe A can fill the cistern in 4 hours,

⇒ Part of the cistern filled by A in 1 hour = $\frac{1}{4}$.

Pipe B can fill the cistern in 6 hours,

⇒ Part of the cistern filled by B in 1 hour = $\frac{1}{6}$.

Pipe C can empty the full cistern in 8 hours,

⇒ Part of the cistern emptied by C in 1 hour = $\frac{1}{8}$.

When all the pipes are turned on together, A and B fill the cistern while C empties it, so the part emptied by C is subtracted :

$$\begin{aligned}\Rightarrow \text{ Net part of the cistern filled in 1 hour }= \Big(\frac{1}{4} + \frac{1}{6} - \frac{1}{8}\Big)\\[1em] = \frac{6 + 4 - 3}{24}\\[1em] = \frac{7}{24}\end{aligned}$$

∴ Time taken to fill the cistern = $\frac{24}{7}$ hours = $3\frac{3}{7}$ hours.

Now, $\frac{3}{7}$ hour = $\Big(\frac{3}{7} \times 60\Big)$ minutes = $\frac{180}{7}$ minutes = $25\frac{5}{7}$ minutes ≈ 26 minutes.

∴ The cistern will be full in about 3 hrs 26 min.

Hence, Option 2 is the correct option.

1. A can build the wall in 30 days,

⇒ Part of the wall built by A in 1 day = $\frac{1}{30}$.

B can build the wall in 60 days,

⇒ Part of the wall built by B in 1 day = $\frac{1}{60}$.

$$\begin{aligned}\Rightarrow \text{ Part of the wall built by (A + B) in 1 day }= \Big(\frac{1}{30} + \frac{1}{60}\Big)\\[1em] = \frac{2 + 1}{60}\\[1em] = \frac{3}{60}\\[1em] = \frac{1}{20}\end{aligned}$$

So, A and B together can build the wall in 20 days.

Hence, option (b) is the correct option.

2. A can build the wall in 30 days,

⇒ Part of the wall built by A in 1 day = $\frac{1}{30}$.

B can break the wall in 60 days,

⇒ Part of the wall broken by B in 1 day = $\frac{1}{60}$.

A builds the wall while B breaks it, so the part broken by B is subtracted :

$$\begin{aligned}\Rightarrow \text{ Net part of the wall built in 1 day }= \Big(\frac{1}{30} - \frac{1}{60}\Big)\\[1em] = \frac{2 - 1}{60}\\[1em] = \frac{1}{60}\end{aligned}$$

So, the wall will be built in 60 days.

Hence, option (b) is the correct option.

3. A can build the wall in 30 days,

⇒ Part of the wall built by A in 1 day = $\frac{1}{30}$.

B can build the wall in 60 days,

⇒ Part of the wall built by B in 1 day = $\frac{1}{60}$.

C can build the wall in 20 days,

⇒ Part of the wall built by C in 1 day = $\frac{1}{20}$.

$$\begin{aligned}\Rightarrow \text{ Part of the wall built by (A + B + C) in 1 day }= \Big(\frac{1}{30} + \frac{1}{60} + \frac{1}{20}\Big)\\[1em] = \frac{2 + 1 + 3}{60}\\[1em] = \frac{6}{60}\\[1em] = \frac{1}{10}\end{aligned}$$

So, all the three together can build the wall in 10 days.

Hence, option (a) is the correct option.

4. A can build the wall in 30 days,

⇒ Part of the wall built by A in 1 day = $\frac{1}{30}$.

B can build the wall in 60 days,

⇒ Part of the wall built by B in 1 day = $\frac{1}{60}$.

C can break the wall in 20 days,

⇒ Part of the wall broken by C in 1 day = $\frac{1}{20}$.

A and B build the wall while C breaks it, so the part broken by C is subtracted :

$$\begin{aligned}\Rightarrow \text{ Net part of the wall built in 1 day }= \Big(\frac{1}{30} + \frac{1}{60} - \frac{1}{20}\Big)\\[1em] = \frac{2 + 1 - 3}{60}\\[1em] = \frac{0}{60}\\[1em] = 0\end{aligned}$$

As no part of the wall is built in 1 day, the wall will never be built.

Hence, option (d) is the correct option.

If a person can complete a work in $n$ days, then in 1 day he does $\frac{1}{n}$ of the work.

∴ Reason (R) is true.

The man can complete the work in 10 days,

⇒ Man's 1 day's work = $\frac{1}{10}$.

$$\begin{aligned}\Rightarrow \text{ Man's 2 days' work }= \Big(\frac{1}{10} \times 2\Big)\\[1em] = \frac{2}{10}\\[1em] = \frac{1}{5}\end{aligned}$$

So, he will complete $\frac{1}{5}$ of the work in 2 days.

∴ Assertion (A) is true, and Reason (R) is the rule used to obtain it, so it is the correct explanation of Assertion (A).

Hence, Option 1 is the correct option.

$\frac{1}{3}$ of the work is done by Anish in 5 days,

⇒ Whole work will be done by Anish in $\Big(5 \times \frac{3}{1}\Big)$ days = 15 days.

$\frac{1}{2}$ of the work is done by Brijesh in 6 days,

⇒ Whole work will be done by Brijesh in $\Big(6 \times \frac{2}{1}\Big)$ days = 12 days.

Anish can do the whole work in 15 days,

⇒ Anish's 1 day's work = $\frac{1}{15}$.

Brijesh can do the whole work in 12 days,

⇒ Brijesh's 1 day's work = $\frac{1}{12}$.

$$\begin{aligned}\Rightarrow \text{ (Anish + Brijesh)'s 1 day's work }= \Big(\frac{1}{15} + \frac{1}{12}\Big)\\[1em] = \frac{4 + 5}{60}\\[1em] = \frac{9}{60}\\[1em] = \frac{3}{20}\end{aligned}$$

∴ Together they will complete the work in $\frac{20}{3}$ days = $6\frac{2}{3}$ days, and not in 6 days.

∴ Assertion (A) is false.

More the number of men, less is the time taken by them to complete a work, and the product of the two remains the same. So, the number of men and the time taken vary inversely with each other.

∴ Reason (R) is true.

Hence, Option 4 is the correct option.