RD Sharma solution class 8 chapter 11 Time and Work Exercise 11.1

Exercise 11.1

Page-11.10

Question 1:

Rakesh can do a piece of work in 20 days. How much work can he do in 4 days?

Answer 1:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{Rakesh} \mathrm{can} \mathrm{do} \mathrm{a} \mathrm{piece} \mathrm{of} \mathrm{work} \mathrm{in} 20 \mathrm{days}. \\ \therefore \mathrm{Rakesh}\text{'}\mathrm{s} 1 \mathrm{day}\text{'s} \mathrm{work}=\frac{1}{20} \\ \therefore \mathrm{Rakesh}\text{'}\mathrm{s} \mathrm{work}\text{ for 4 days}=\frac{4}{20}=\frac{1}{5} \\ \mathrm{Thus}, \mathrm{he} \mathrm{can} \mathrm{do} \frac{1}{5}\mathrm{th} \mathrm{of} \mathrm{the} \mathrm{work} \mathrm{in} 4 \mathrm{days}. \end{gathered}$$

Question 2:

Rohan can paint $\frac{1}{3}$ of a painting in 6 days. How many days will he take to complete the painting?

Answer 2:

$$\begin{gathered} \mathrm{Rohan} \mathrm{can} \mathrm{paint} \frac{1}{3}\text{rd }\mathrm{of} \mathrm{a} \mathrm{painting} \mathrm{in} 6 \mathrm{days}. \\ \therefore \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{Rohan} \mathrm{to} \mathrm{complete} \mathrm{the} \mathrm{painting}=\left(6\times 3\right) \mathrm{days}=18 \mathrm{days}. \end{gathered}$$

Question 3:

Anil can do a piece of work in 5 days and Ankur in 4 days. How long will they take to do the same work, if they work together?

Answer 3:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{Anil} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=5 \mathrm{days} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{Ankur} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=4 \mathrm{days} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Anil} \mathrm{in} 1 \mathrm{day}=\frac{1}{5} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Ankur} \mathrm{in} 1 \mathrm{day}=\frac{1}{4} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Anil} \mathrm{and} \mathrm{Ankur} \mathrm{in} \mathrm{one} \mathrm{day}=\frac{1}{5}+\frac{1}{4} \\ =\frac{4+5}{20}=\frac{9}{20} \\ \mathrm{Thus}, \mathrm{Anil} \mathrm{and} \mathrm{Ankur} \mathrm{can} \mathrm{do} \mathrm{the} \mathrm{work} \mathrm{in} \frac{20}{9} \mathrm{days} \mathrm{i}.\mathrm{e}. 2\frac{2}{9}\mathrm{days}. \end{gathered}$$

Question 4:

Mohan takes 9 hours to mow a large lawn. He and Sohan together can mow it in 4 hours. How long will Sohan take to mow the lawn if he works alone?

Answer 4:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{Mohan} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=9 \mathrm{hours} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{Mohan} \mathrm{and} \mathrm{Sohan} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=4 \mathrm{hours} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Mohan}=\frac{1}{9} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Mohan} \mathrm{and} \mathrm{Sohan}=\frac{1}{4} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Sohan} =\frac{1}{4}-\frac{1}{9} \\ =\frac{9-4}{36}=\frac{5}{36} \\ \mathrm{Thus}, \mathrm{Sohan} \mathrm{can} \mathrm{do} \mathrm{the} \mathrm{work} \mathrm{in} \frac{36}{5} \mathrm{hours} \mathrm{i}.\mathrm{e}. 7\frac{1}{5}\mathrm{hours}. \end{gathered}$$

Question 5:

Sita can finish typing a 100 page document in 9 hours. Mita in 6 hours and Rita in 12 hours. How long will they take to type a 100 page document if they work together?

Answer 5:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{Sita} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=9 \mathrm{hours} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{Mita} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=6 \mathrm{hours} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{Ri}\mathrm{ta} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=12 \mathrm{hours} \\ \mathrm{Now}, \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Sita}=\frac{1}{9} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Mita}=\frac{1}{6} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Rita}=\frac{1}{12} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{them} \mathrm{together} =\frac{1}{9}+\frac{1}{6}+\frac{1}{12} \\ =\frac{4+6+3}{36}=\frac{13}{36} \\ \mathrm{Thus}, \mathrm{together} \mathrm{they} \mathrm{can} \mathrm{do} \mathrm{the} \mathrm{work} \mathrm{in} \frac{36}{13} \mathrm{hours}. \end{gathered}$$

Question 6:

A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work?

Answer 6:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{A} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=20 \mathrm{hours} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{B} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=24 \mathrm{hours} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=8 \mathrm{hours} \\ \mathrm{Now}, \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{A}=\frac{1}{20} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{B}=\frac{1}{24} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)=\frac{1}{8} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{C} =\frac{1}{8}-\left(\frac{1}{20}+\frac{1}{24}\right) \\ =\frac{1}{8}-\left(\frac{6}{120}+\frac{5}{120}\right)=\frac{1}{8}-\left(\frac{11}{120}\right) \\ =\frac{15-11}{120}=\frac{4}{120}=\frac{1}{30} \\ \mathrm{Thus}, \mathrm{C} \mathrm{can} \mathrm{do} \mathrm{the} \mathrm{work} \mathrm{in} 30 \mathrm{hours}. \end{gathered}$$

Question 7:

A and B can do a piece of work in 18 days; B and C in 24 days and A and C in 36 days. In what time can they do it, all working together?

Answer 7:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=18 \mathrm{days} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{B}+\mathrm{C}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=24 \mathrm{days} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{A}+\mathrm{C}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=36 \mathrm{days} \\ \mathrm{Now}, \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right)=\frac{1}{18} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{B}+\mathrm{C}\right)=\frac{1}{24} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{C}\right)=\frac{1}{36} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{together} =\left(\mathrm{A}+\mathrm{B}\right)+\left(\mathrm{B}+\mathrm{C}\right)+\left(\mathrm{A}+\mathrm{C}\right) \\ =\frac{1}{18}+\frac{1}{24}+\frac{1}{36} \\ =\frac{4+3+2}{72}=\frac{9}{72} \\ =\frac{1}{8} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{together} =2\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)=\frac{1}{8} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)=\frac{1}{16} \\ \mathrm{Thus}, \mathrm{together} \mathrm{they} \mathrm{can} \mathrm{finish} \mathrm{the} \mathrm{work} \mathrm{in} 16 \mathrm{days}. \end{gathered}$$

Question 8:

A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. How much time will A alone take to finish the work?

Answer 8:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=12 \mathrm{days} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{B}+\mathrm{C}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=15 \mathrm{days} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{A}+\mathrm{C}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=20 \mathrm{days} \\ \mathrm{Now}, \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right)=\frac{1}{12} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{B}+\mathrm{C}\right)=\frac{1}{15} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{C}\right)=\frac{1}{20} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{together} =\left(\mathrm{A}+\mathrm{B}\right)+\left(\mathrm{B}+\mathrm{C}\right)+\left(\mathrm{A}+\mathrm{C}\right) \\ =\frac{1}{12}+\frac{1}{15}+\frac{1}{20} \\ =\frac{5+4+3}{60}=\frac{12}{60} \\ =\frac{1}{5} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{together} =2\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)=\frac{1}{5} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)=\frac{1}{10} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{alone} =\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)-\left(\mathrm{B}+\mathrm{C}\right) \\ =\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30} \\ \mathrm{Thus}, \mathrm{A} \mathrm{alone} \mathrm{can} \mathrm{do} \mathrm{the} \mathrm{work} \mathrm{in} 30 \mathrm{days}. \end{gathered}$$

Question 9:

A, B and C can reap a field in $15\frac{3}{4}$ days; B, C and D in 14 days; C, D and A in 18 days; D, A and B in 21 days. In what time can A, B, C and D together reap it?

Answer 9:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=15\frac{3}{4} \mathrm{days}=\frac{63}{4}\mathrm{days} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{B}+\mathrm{C}+\mathrm{D}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=14 \mathrm{days} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{C}+\mathrm{D}+\mathrm{A}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=18 \mathrm{days} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \left(\mathrm{D}+\mathrm{A}+\mathrm{B}\right) \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{work}=21 \mathrm{days} \\ \mathrm{Now}, \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)=\frac{4}{63} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{B}+\mathrm{C}+\mathrm{D}\right)=\frac{1}{14} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{C}+\mathrm{D}+\mathrm{A}\right)=\frac{1}{18} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{D}+\mathrm{A}+\mathrm{B}\right)=\frac{1}{21} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{working} \mathrm{together} =\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)+\left(\mathrm{B}+\mathrm{C}+\mathrm{D}\right)+\left(\mathrm{C}+\mathrm{A}+\mathrm{D}\right)+\left(\mathrm{D}+\mathrm{A}+\mathrm{B}\right) \\ =\frac{4}{63}+\frac{1}{14}+\frac{1}{18}+\frac{1}{21} \\ =\frac{4}{63}+\left(\frac{9+7+6}{126}\right)=\frac{4}{63}+\frac{22}{126} \\ =\frac{4}{63}+\frac{11}{63}=\frac{15}{63} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{working} \mathrm{together} =3\left(\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}\right)=\frac{15}{63} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}\right)=\frac{15}{63\times 3}=\frac{5}{63} \\ \mathrm{Thus}, \mathrm{together} \mathrm{they} \mathrm{can} \mathrm{do} \mathrm{the} \mathrm{work} \mathrm{in} \frac{63}{5} \mathrm{days} \mathrm{or} 12\frac{3}{5} \mathrm{days}. \end{gathered}$$

Question 10:

A and B can polish the floors of a building in 10 days. A alone can do $\frac{1}{4}$ th of it in 12 days. In how many days can B alone polish the floor?

Answer 10:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{can} \mathrm{polish} \mathrm{the} \mathrm{floors} \mathrm{of} \mathrm{the} \mathrm{building} \mathrm{in} 10 \mathrm{days}. \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{in} 1 \mathrm{day}=\frac{1}{10} \\ \mathrm{Now}, \mathrm{A} \mathrm{alone} \mathrm{can} \mathrm{do} \frac{1}{4}\text{th} \mathrm{of} \mathrm{the} \mathrm{work} \mathrm{in} 12 \mathrm{days}. \\ \therefore \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{A} \mathrm{alone} \mathrm{to} \mathrm{do} \mathrm{the} \mathrm{complete} \mathrm{work} = \left(4\times 12\right)=48 \mathrm{days} \\ \Rightarrow \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{in} 1 \mathrm{day}=\frac{1}{48} \\ \mathrm{Now}, \mathrm{work} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} 1 \mathrm{day} =\mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{in} 1 \mathrm{day} -\mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{in} 1 \mathrm{day} \\ =\frac{1}{10}-\frac{1}{48} \\ =\frac{24-5}{240}=\frac{19}{240} \\ \mathrm{Thus}, \mathrm{B} \mathrm{alone} \mathrm{can} \mathrm{polish} \mathrm{the} \mathrm{floor} \mathrm{in} \frac{240}{19}\mathrm{days} \mathrm{or} 12\frac{12}{19} \mathrm{days}. \end{gathered}$$

Question 11:

A and B can finish a work in 20 days. A alone can do $\frac{1}{5}$ th of the work in 12 days. In how many days can B alone do it?

Answer 11:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{can} \mathrm{finish} \mathrm{the} \mathrm{work} \mathrm{in} 20 \mathrm{days}. \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{in} 1 \mathrm{day} =\frac{1}{20} \\ \mathrm{Now}, \mathrm{A} \mathrm{alone} \mathrm{can} \mathrm{do} \frac{1}{5}\text{th} \mathrm{of} \mathrm{the} \mathrm{work} \mathrm{in} 12 \mathrm{days}. \\ \therefore \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{A} \mathrm{alone} \mathrm{to} \mathrm{complete} \mathrm{the} \mathrm{work}=\left(5\times 12\right)=60 \mathrm{days} \\ \Rightarrow \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{in} 1 \mathrm{day}=\frac{1}{60} \\ \mathrm{Now}, \mathrm{work} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} 1 \mathrm{day}=\mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{in} 1 \mathrm{day} \mathrm{work}-\mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{in} 1 \mathrm{day} \\ =\frac{1}{20}-\frac{1}{60} \\ =\frac{3-1}{60}=\frac{2}{60} \\ \mathrm{Thus}, \mathrm{B} \mathrm{alone} \mathrm{can} \mathrm{polish} \mathrm{the} \mathrm{floor} \mathrm{in} \frac{60}{2}\mathrm{days} \mathrm{or} 30 \mathrm{days}. \end{gathered}$$

Question 12:

A and B can do a piece of work in 20 days and B in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work?

Answer 12:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{A} \mathrm{can} \mathrm{finish} \mathrm{the} \mathrm{work} \mathrm{in} 20 \mathrm{days} \mathrm{and} \mathrm{B} \mathrm{can} \mathrm{finish} \mathrm{the} \mathrm{same} \mathrm{work} \mathrm{in} 15 \mathrm{days}. \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{in} 1 \mathrm{day}=\frac{1}{20} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} 1 \mathrm{day}=\frac{1}{15} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{in} 1 \mathrm{day} =\frac{1}{20}+\frac{1}{15} \\ =\frac{3+4}{60}=\frac{7}{60} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{in} 2 \mathrm{days}=\frac{14}{60}=\frac{7}{30} \\ \mathrm{Remaining} \mathrm{work}=1-\frac{7}{30}=\frac{23}{30} \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{remaining} \mathrm{work} \mathrm{is} \mathrm{done} \mathrm{by} \mathrm{B}. \\ \because \mathrm{Complete} \mathrm{work} \mathrm{is} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} 15 \mathrm{days}. \\ \therefore \frac{23}{30} \mathrm{of} \mathrm{the} \mathrm{work} \mathrm{will} \mathrm{be} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} \left(15\times \frac{23}{30}\right) \mathrm{days} \mathrm{or} \frac{23}{2} \mathrm{days} \mathrm{or} 11\frac{1}{2} \mathrm{days}. \\ \mathrm{Thus}, \mathrm{the} \mathrm{remaining} \mathrm{work} \mathrm{is} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} 11\frac{1}{2} \mathrm{days}. \end{gathered}$$

Question 13:

A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days will A finish the remaining work?

Answer 13:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{A} \mathrm{can} \mathrm{finish} \mathrm{the} \mathrm{work} \mathrm{in} 40 \mathrm{days} \mathrm{and} \mathrm{B} \mathrm{can} \mathrm{finish} \mathrm{the} \mathrm{same} \mathrm{work} \mathrm{in} 45 \mathrm{days}. \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{in} 1 \mathrm{day}=\frac{1}{40} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} 1 \mathrm{day}=\frac{1}{45} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{in} 1 \mathrm{day}=\frac{1}{40}+\frac{1}{45} \\ =\frac{9+8}{360}=\frac{17}{360} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right) \mathrm{in} 10 \mathrm{day}\text{s}=10\times \frac{17}{360}=\frac{17}{36} \\ \mathrm{Remaining} \mathrm{work}=1-\frac{17}{36}=\frac{19}{36} \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{remaining} \mathrm{work} \mathrm{is} \mathrm{done} \mathrm{by} \mathrm{B}. \\ \mathrm{Complete} \mathrm{work} \mathrm{is} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} 45 \mathrm{days}. \\ \therefore \frac{19}{36} \mathrm{of} \mathrm{the} \mathrm{work} \mathrm{will} \mathrm{be} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} \left(45\times \frac{19}{36}\right) \mathrm{days} \mathrm{or} 23\frac{3}{4} \mathrm{days}. \\ \mathrm{Thus}, \mathrm{the} \mathrm{remaining} \mathrm{work} \mathrm{is} \mathrm{done} \mathrm{by} \mathrm{B} \mathrm{in} 23\frac{3}{4} \mathrm{days}. \end{gathered}$$

Question 14:

Aasheesh can paint his doll in 20 minutes and his sister Chinki can do so in 25 minutes. They paint the doll together for five minutes. At this juncture they have a quarrel and Chinki withdraws from painting. In how many minutes will Aasheesh finish the painting of the remaining doll?

Answer 14:

$$\begin{gathered} \mathrm{Aasheesh} \mathrm{can} \mathrm{paint} \mathrm{a} \mathrm{doll} \mathrm{in} 20 \mathrm{minutes}, \mathrm{and} \mathrm{Chinki} \mathrm{can} \mathrm{do} \mathrm{the} \mathrm{same} \mathrm{in} 25 \mathrm{minutes}. \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Aasheesh} \mathrm{in} 1 \mathrm{minute}=\frac{1}{20} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{Chinki} \mathrm{in} 1 \mathrm{minute}=\frac{1}{25} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{them} \mathrm{together}=\frac{1}{20}+\frac{1}{25} \\ =\frac{5+4}{100}=\frac{9}{100} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{them} \mathrm{in} 5 \mathrm{minutes}=5\times \frac{9}{100}=\frac{9}{20} \\ \mathrm{Remaining} \mathrm{work}=1-\frac{9}{20}=\frac{11}{20} \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{remaining} \mathrm{work} \mathrm{is} \mathrm{done} \mathrm{by} \mathrm{Aasheesh}. \\ \text{The} \mathrm{work} \mathrm{done} \mathrm{by} \mathrm{Aasheesh} \mathrm{in} 20 \mathrm{minutes}. \\ \therefore \frac{11}{20}\text{th} \mathrm{work} \mathrm{will} \mathrm{be} \mathrm{done} \mathrm{by} \mathrm{Aasheesh} \mathrm{in} \left(20\times \frac{11}{20}\right) \mathrm{minutes} \mathrm{or} 11 \mathrm{minutes}. \\ \mathrm{Thus}, \mathrm{the} \mathrm{remaining} \mathrm{work} \mathrm{is} \mathrm{done} \mathrm{by} \mathrm{Aasheesh} \mathrm{in} 11 \mathrm{minutes}. \end{gathered}$$

Question 15:

A and B can do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined by B. Find the total time taken to complete the work.

Answer 15:

$$\begin{gathered} \mathrm{A} \mathrm{can} \mathrm{do} \mathrm{a} \mathrm{work} \mathrm{in} 6 \mathrm{days}, \mathrm{and} \mathrm{B} \mathrm{can} \mathrm{do} \mathrm{the} \mathrm{same} \mathrm{work} \mathrm{in} 4 \mathrm{days}. \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{in} 2 \mathrm{days}=\frac{2}{6}=\frac{1}{3} \\ \mathrm{Remaining} \mathrm{work}=1-\frac{1}{3}=\frac{2}{3} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \left(\mathrm{A}+\mathrm{B}\right)\mathrm{in} 1 \mathrm{day}=\left(\frac{1}{6}+\frac{1}{4}\right) \\ =\frac{2+3}{12}=\frac{5}{12} \\ \because \frac{5}{12}\text{th} \mathrm{work} \mathrm{is} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{in} 1 \mathrm{day}. \\ \therefore \frac{2}{3}\text{rd} \mathrm{work} \mathrm{will} \mathrm{be} \mathrm{done} \mathrm{by} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{in} \left(\frac{12}{5}\times \frac{2}{3}\right) \mathrm{days} \mathrm{or} \frac{8}{5} \mathrm{days}. \\ \therefore \mathrm{Total} \mathrm{time} \mathrm{taken}=\left(\frac{8}{5}+2\right) \mathrm{days}=\frac{18}{5} \mathrm{days}=3\frac{3}{5} \mathrm{days} \end{gathered}$$

Question 16:

6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job?

Answer 16:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} 6 \mathrm{men} \mathrm{can} \mathrm{complete} \mathrm{a} \mathrm{job} \mathrm{in} 7 \mathrm{days}. \\ \therefore \mathrm{Time} \mathrm{taken} \mathrm{by} 1 \mathrm{man} \mathrm{to} \mathrm{complete} \mathrm{the} \mathrm{same} \mathrm{job}=\left(6\times 7\right)=42 \mathrm{days} \\ \therefore \mathrm{Time} \mathrm{taken} \mathrm{by} 21 \mathrm{men} \mathrm{to} \mathrm{complete} \mathrm{the} \mathrm{same} \mathrm{job}=\frac{42}{21}=2 \mathrm{days} \end{gathered}$$

Question 17:

8 men can do a piece of work in 9 days. In how many days will 6 men do it?

Answer 17:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} 8 \mathrm{men} \mathrm{can} \mathrm{complete} \mathrm{a} \mathrm{job} \mathrm{in} 9 \mathrm{days}. \\ \therefore \mathrm{Time} \mathrm{taken} \mathrm{by} 1 \mathrm{man} \mathrm{to} \mathrm{complete} \mathrm{the} \mathrm{same} \mathrm{job}=\left(8\times 9\right)=72 \mathrm{days} \\ \therefore \mathrm{Time} \mathrm{taken} \mathrm{by} 6 \mathrm{men} \mathrm{to} \mathrm{complete} \mathrm{the} \mathrm{same} \mathrm{job}=\frac{72}{6}=12 \mathrm{days} \end{gathered}$$

Question 18:

Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets?

Answer 18:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{Reema} \mathrm{weaves} 35 \mathrm{baskets} \mathrm{in} 25 \mathrm{days}. \\ \therefore \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{her} \mathrm{to} \mathrm{weave} 1 \mathrm{basket}=\frac{25}{35}\mathrm{day} \\ \therefore \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{her} \mathrm{to} \mathrm{weave} 55 \mathrm{baskets}=\left(\frac{25}{35}\times 55\right) \mathrm{days} \\ =\frac{1375}{35} \mathrm{days}=\frac{275}{7} \mathrm{days}=39\frac{2}{7} \mathrm{days} \end{gathered}$$

Question 19:

Neha types 75 pages in 14 hours. How many pages will she type in 20 hours?

Answer 19:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{Neha} \mathrm{types} 75 \mathrm{pages} \mathrm{in} 14 \mathrm{hours}. \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{pages} \mathrm{typed} \mathrm{by} \mathrm{her} \mathrm{in} 1 \mathrm{hour}=\frac{75}{14} \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{pages} \mathrm{typed} \mathrm{by} \mathrm{her} \mathrm{in} 20 \mathrm{hours}= \left(\frac{75}{14}\times 20\right) \mathrm{pages} \\ =\frac{1500}{14} \mathrm{pages} \\ =\frac{750}{7} \mathrm{pages} \\ =107\frac{1}{7} \mathrm{pages} \\ \mathrm{Thus}, \mathrm{Neha} \mathrm{types} 107\frac{1}{7} \mathrm{pages} \mathrm{in} 20 \mathrm{hours}. \end{gathered}$$

Question 20:

If 12 boys earn Rs 840 in 7 days, what will 15 boys earn in 6 days?

Answer 20:

$$\begin{gathered} \mathrm{Earnings} \mathrm{of} 12 \mathrm{boys} \mathrm{in} 7 \mathrm{days}=\mathrm{Rs}. 840 \\ \therefore \mathrm{Earnings} \mathrm{of} 12 \mathrm{boys} \mathrm{in} 1 \mathrm{day}=\mathrm{Rs}. \frac{840}{7}=\mathrm{Rs}. 120 \\ \therefore \mathrm{Earnings} \mathrm{of} 1 \mathrm{boy} \mathrm{in} 1 \mathrm{day}=\mathrm{Rs}.\frac{120}{12}=\mathrm{Rs}. 10 \\ \therefore \mathrm{Earnings} \mathrm{of} 1 \mathrm{boy} \mathrm{in} 6 \mathrm{days}=\mathrm{Rs}. \left(10\times 6\right)=\mathrm{Rs}. 60 \\ \therefore \mathrm{Earnings} \mathrm{of} 15 \mathrm{boys} \mathrm{in} 6 \mathrm{days}=\mathrm{Rs}.\left(60\times 15\right) \\ =\mathrm{Rs}. 900 \end{gathered}$$

Question 21:

If 25 men earn Rs 1000 in 10 days, how much will 15 men earn in 15 days?

Answer 21:

$$\begin{gathered} \mathrm{Earnings} \mathrm{of} 25 \mathrm{men} \mathrm{in} 10 \mathrm{days}=\mathrm{Rs}.1000 \\ \therefore \mathrm{Earnings} \mathrm{of} 25 \mathrm{men} \mathrm{in} 1 \mathrm{day}=\mathrm{Rs}.\frac{1000}{10}=\mathrm{Rs}.100 \\ \therefore \mathrm{Earning} \mathrm{of} 1 \mathrm{man} \mathrm{in} 1 \mathrm{day}=\mathrm{Rs}.\frac{100}{25}=\mathrm{Rs}.4 \\ \therefore \mathrm{Earnings} \mathrm{of} 1 \mathrm{men} \mathrm{in} 15 \mathrm{days}=\mathrm{Rs}.\left(15\times 4\right)=\mathrm{Rs}.60 \\ \therefore \mathrm{Earnings} \mathrm{of} 15 \mathrm{men} \mathrm{in} 15 \mathrm{days}=\mathrm{Rs}.\left(60\times 15\right) \\ =\mathrm{Rs}. 900 \end{gathered}$$

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Question 22:

Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days?

Answer 22:

$$\begin{gathered} \mathrm{If} \mathrm{Ashu} \mathrm{works} \mathrm{for} 8 \mathrm{hours} \mathrm{daily}, \mathrm{he} \mathrm{can} \mathrm{finish} \mathrm{the} \mathrm{work} \mathrm{in} 18 \mathrm{days}. \\ \mathrm{If} \mathrm{he} \mathrm{works} \mathrm{for} 1 \mathrm{hour} \mathrm{daily}, \mathrm{he} \mathrm{can} \mathrm{finish} \mathrm{the} \mathrm{work} \mathrm{in} \left(18\times 8\right) \mathrm{days} \mathrm{or} 144 \mathrm{days}. \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{hours} \mathrm{he} \mathrm{should} \mathrm{work} \mathrm{daily} \mathrm{to} \mathrm{finish} \mathrm{the} \mathrm{work} \mathrm{in} 12 \mathrm{days}=\frac{144}{12}=12 \end{gathered}$$

Question 23:

If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour?

Answer 23:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{garlands} \mathrm{made} \mathrm{by} 9 \mathrm{girls} \mathrm{in} 1 \mathrm{hour} =\frac{135}{3}=45 \\ \mathrm{Number} \mathrm{of} \mathrm{garlands} \mathrm{made} \mathrm{by} 1 \mathrm{girl} \mathrm{in} 1 \mathrm{hour} =\frac{45}{9}=5 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{girls} \mathrm{required} \mathrm{to} \mathrm{make} 270 \mathrm{garlands} \mathrm{in} 1 \mathrm{hour} =\frac{270}{5}=54. \end{gathered}$$

Question 24:

A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together?

Answer 24:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{the} \mathrm{first} \mathrm{tap} \mathrm{to} \mathrm{fill} \mathrm{the} \mathrm{cistern}=8 \mathrm{hours} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{the} \mathrm{second} \mathrm{tap} \mathrm{to} \mathrm{fill} \mathrm{the} \mathrm{cistern}=4 \mathrm{hours} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{the} \mathrm{first} \mathrm{tap} \mathrm{in} 1 \mathrm{hour}=\frac{1}{8} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{the} \mathrm{second} \mathrm{tap} \mathrm{in} 1 \mathrm{hour}=\frac{1}{4} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{both} \mathrm{the} \mathrm{taps} \mathrm{in} 1 \mathrm{hour}=\frac{1}{8}+\frac{1}{4} \\ =\frac{1+2}{8}=\frac{3}{8} \\ \mathrm{Thus}, \mathrm{both} \mathrm{the} \mathrm{taps} \mathrm{together} \mathrm{will} \mathrm{fill} \mathrm{the} \mathrm{cistern} \mathrm{in} \frac{8}{3} \mathrm{hours} \mathrm{or} 2\frac{2}{3} \mathrm{hours}. \end{gathered}$$

Question 25:

Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and they B is turned off. How much time will A take to fill the remaining tank?

Answer 25:

$$\begin{gathered} \mathrm{Pipe} \mathrm{A} \mathrm{can} \mathrm{fill} \mathrm{the} \mathrm{tank} \mathrm{in} 10 \mathrm{hours}, \mathrm{and} \mathrm{pipe} \mathrm{B} \mathrm{can} \mathrm{fill} \mathrm{the} \mathrm{tank} \mathrm{in} 15 \mathrm{hours}. \\ \therefore \mathrm{In} 1 \mathrm{hour}, \mathrm{A} \mathrm{can} \mathrm{fill} \frac{1}{10}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{tank}. \\ \mathrm{In} 1 \mathrm{hour}, \mathrm{B} \mathrm{can} \mathrm{fill} \frac{1}{15}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{tank}. \\ \therefore \mathrm{In} 1 \mathrm{hour}, \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{can} \mathrm{fill} \left(\frac{1}{10}+\frac{1}{15}\right) \\ =\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{tank} \\ \therefore \mathrm{In} 4 \mathrm{hours}, \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{can} \mathrm{fill} \left(\frac{1}{6}\times 4\right)=\frac{2}{3}\text{rd} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{tank} \\ \mathrm{Remaining} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{tank}=1-\frac{2}{3}=\frac{1}{3} \\ \mathrm{Now}, \mathrm{A} \mathrm{can} \mathrm{fill} \mathrm{the} \mathrm{tank} \mathrm{in} 10 \mathrm{hours}. \\ \therefore \frac{1}{3}\text{rd} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{tank} \mathrm{can} \mathrm{be} \mathrm{filled} \mathrm{by} \mathrm{A} \mathrm{in} (\frac{1}{3}\times 10) \mathrm{hours} \mathrm{or} \frac{10}{3} \mathrm{hours} \mathrm{or} 3\frac{1}{3}\mathrm{hours}. \end{gathered}$$

Question 26:

A pipe can fill a cistern in 10 hours. Due to a leak in the bottom it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak?

Answer 26:

$$\begin{gathered} \mathrm{When} \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{leakage}, \mathrm{the} \mathrm{pipe} \mathrm{can} \mathrm{fill} \mathrm{the} \mathrm{cistern} \mathrm{in} 10 \mathrm{hours}. \\ \mathrm{Thus}, \mathrm{the} \mathrm{pipe} \mathrm{can} \mathrm{fill} \frac{1}{10}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{in} 1 \mathrm{hour}. \\ \mathrm{When} \mathrm{there} \mathrm{is} \mathrm{leakage}, \mathrm{the} \mathrm{pipe} \mathrm{can} \mathrm{fill} \mathrm{the} \mathrm{cistern} \mathrm{in} 12 \mathrm{hours}. \\ \mathrm{Therefore}, \mathrm{in} \mathrm{case} \mathrm{of} \mathrm{leakage}, \mathrm{the} \mathrm{pipe} \mathrm{can} \mathrm{fill} \frac{1}{12}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{in} 1 \mathrm{hour}. \\ \mathrm{Thus}, \mathrm{in} \mathrm{one} \mathrm{hour}, \mathrm{due} \mathrm{to} \mathrm{leakge}, \left(\frac{1}{10}-\frac{1}{12}\right)\mathrm{th} \mathrm{or} \frac{1}{60}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{is} \mathrm{emptied}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{cistern} \mathrm{will} \mathrm{be} \mathrm{emptied} \mathrm{by} \mathrm{the} \mathrm{leakage} \mathrm{in} 60 \mathrm{hours}. \end{gathered}$$

Question 27:

A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely?

Answer 27:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{tap} \mathrm{A} \mathrm{to} \mathrm{fill} \mathrm{the} \mathrm{cistern} =12 \mathrm{hours} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{tap} \mathrm{B} \mathrm{to} \mathrm{fill} \mathrm{the} \mathrm{cistern} =15 \mathrm{hours} \\ \mathrm{Let} \mathrm{C} \mathrm{be} \mathrm{the} \mathrm{outlet} \mathrm{that} \mathrm{can} \mathrm{empty} \mathrm{the} \mathrm{cistern} \mathrm{in} 10 \mathrm{hours}. \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{tap} \mathrm{C} \mathrm{to} \mathrm{empty} \mathrm{the} \mathrm{cistern} =10 \mathrm{hours} \\ \mathrm{Now}, \\ \mathrm{Tap} \mathrm{A} \mathrm{fills} \frac{1}{12}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{in} 1 \mathrm{hour}. \\ \mathrm{Tap} \mathrm{B} \mathrm{fills} \frac{1}{15}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{in} 1 \mathrm{hour}. \\ \mathrm{Tap} \mathrm{C} \mathrm{empties} \mathrm{out} \frac{1}{10}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{in} 1 \mathrm{hour}. \\ \mathrm{Thus}, \mathrm{in} \mathrm{one} \mathrm{hour}, \left(\frac{1}{12}+\frac{1}{15}-\frac{1}{10}\right)\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{is} \mathrm{filled}. \\ \mathrm{We} \mathrm{have}: \\ \frac{1}{12}+\frac{1}{15}-\frac{1}{10}=\frac{10+8-12}{120}=\frac{6}{120}=\frac{1}{20} \\ \mathrm{Thus}, \mathrm{in} 1 \mathrm{hour}, \frac{1}{20}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{is} \mathrm{filled}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{cistern} \mathrm{will} \mathrm{be} \mathrm{filled} \mathrm{completely} \mathrm{in} 20 \mathrm{hours} \mathrm{if} \mathrm{all} \mathrm{the} \mathrm{three} \text{taps} \mathrm{are} \mathrm{opened} \mathrm{together}. \end{gathered}$$

Question 28:

A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern if both the tap and the pipe are opened together?

Answer 28:

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{the} \mathrm{tap} \mathrm{to} \mathrm{fill} \mathrm{the} \mathrm{cistern}=4 \mathrm{hours} \\ \therefore \mathrm{Tap} \mathrm{fills} \frac{1}{4}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{in} 1 \mathrm{hour}. \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{the} \mathrm{pipe} \mathrm{to} \mathrm{empty} \mathrm{the} \mathrm{cistern}=6 \mathrm{hours} \\ \therefore \mathrm{Pipe} \mathrm{empties} \mathrm{out} \frac{1}{6}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{in} 1 \mathrm{hour}. \\ \mathrm{Thus}, \mathrm{in} 1 \mathrm{hour}, \left(\frac{1}{4}-\frac{1}{6}\right)\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{is} \mathrm{filled}. \\ \mathrm{We} \mathrm{have}: \\ \frac{1}{4}-\frac{1}{6}=\frac{6-4}{24}=\frac{2}{24}=\frac{1}{12} \\ \mathrm{Thus}, \mathrm{in} 1 \mathrm{hour}, \frac{1}{12}\mathrm{th} \mathrm{part} \mathrm{of} \mathrm{the} \mathrm{cistern} \mathrm{is} \mathrm{filled}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{cistern} \mathrm{will} \mathrm{be} \mathrm{filled} \mathrm{in} 12 \mathrm{hours}. \end{gathered}$$