Question 1: $\displaystyle A$ can do a piece of work in $\displaystyle 15$ days while $\displaystyle B$ can do it in $\displaystyle 10$ days. How long will they take together to do it?

$\displaystyle \text{ A's 1 Day } = \frac{1}{15}$

$\displaystyle \text{ B's 1 Day } = \frac{1}{10}$

$\displaystyle \text{ A's + B's 1 Day Work } ( \frac{1}{15} + \frac{1}{10} )= \frac{5}{30} = \frac{1}{6}$

Therefore both can finish the work in 6 Days.

$\displaystyle \\$

Question 2: $\displaystyle A, B$ and $\displaystyle C$ can do a piece of work in $\displaystyle 12$ days, $\displaystyle 15$ days and $\displaystyle 10$ days respectively. In what time will they all together finish it?

$\displaystyle \text{ A's 1 Day } = \frac{1}{12}$

$\displaystyle \text{ B's 1 Day } = \frac{1}{15}$

$\displaystyle \text{ C's 1 Day } \frac{1}{10}$

$\displaystyle \text{ A's + B's + C's 1 Day Work } = ( \frac{1}{12} + \frac{1}{15} + \frac{1}{10} ) = \frac{15}{60} = \frac{1}{4}$

Therefore all three can finish the work in 4 Days.

$\displaystyle \\$

Question 3: $\displaystyle A$ and $\displaystyle B$ together can do a piece of work in $\displaystyle 35$ days, while $\displaystyle A$ alone can do it in $\displaystyle 60$ days. How long would $\displaystyle B$ alone take to do it?

$\displaystyle \text{ A's 1 Day } = \frac{1}{60}$

$\displaystyle \text{ B's 1 Day } = \frac{1}{x}$

$\displaystyle \text{ A's + B's 1 Day Work } = ( \frac{1}{60} + \frac{1}{x} ) = \frac{1}{35}$

Solving for $\displaystyle x = 84$ Days

$\displaystyle \\$

Question 4: $\displaystyle A$ can do a piece of work in $\displaystyle 20$ days while $\displaystyle B$ can do it in $\displaystyle 15$ days. With the help of $\displaystyle C$, they finish the work in $\displaystyle 5$ days. In what time would $\displaystyle C$ alone do it?

$\displaystyle \text{ A's 1 Day } = \frac{1}{20}$

$\displaystyle \text{ B's 1 Day } = \frac{1}{15}$

$\displaystyle \text{ C's 1 Day } = \frac{1}{x}$

$\displaystyle \text{ A's + B's + C's 1 Day Work } = (\frac{1}{20} + \frac{1}{15} + \frac{1}{x} ) = \frac{1}{5}$

Solving for $\displaystyle x = 12$ Days

$\displaystyle \\$

Question 5: $\displaystyle A$ can do a piece of work in $\displaystyle 12$ days and $\displaystyle B$ alone can do it in $\displaystyle 16$ days. They worked together on it for $\displaystyle 3$ days and then $\displaystyle A$ left. How long did $\displaystyle B$ take to finish the remaining work?

$\displaystyle \text{ A's 1 Day } = \frac{1}{12}$

$\displaystyle \text{ B's 1 Day } = \frac{1}{16}$

$\displaystyle \text{ A's + B's 1 Day Work } = ( \frac{1}{12} + \frac{1}{16} ) = \frac{7}{48}$

$\displaystyle \text{The amount of work that is completed in 3 days } = \frac{3\times7}{48} = \frac{7}{16}$

$\displaystyle \text{Amount of work left for B to complete } = 1 - \frac{7}{16} = \frac{9}{16}$

$\displaystyle \text{Therefore the number of days that B will take to finish the work } = \frac {\frac{9}{16}} {\frac{1}{16} } = 9 \text{ days }$

$\displaystyle \\$

Question 6: $\displaystyle A$ can do $\displaystyle \frac{1}{4}$ of a work in $\displaystyle 5$ days, while $\displaystyle B$ can do $\displaystyle \frac{1}{5}$ of the work in $\displaystyle 6$ days. In how many days can both do it together?

If A can do $\displaystyle \frac{1}{4}$ of a work in $\displaystyle 5$ days, then A can do the entire work in $\displaystyle 20$ days.

$\displaystyle \text{Therefore A's 1 Day Work } = \frac{1}{20}$

If B can do $\displaystyle \frac{1}{5}$ of a work in $\displaystyle 6$ days, then B can do the entire work in $\displaystyle 30$ days.

$\displaystyle \text{ Therefore B's 1 Day } = \frac{1}{30}$

$\displaystyle \text{ A's + B's 1 Day Work } = (\frac{1}{20} + \frac{1}{30} ) = \frac{1}{12}$

Therefore both can do the work in $\displaystyle 12$ days.

$\displaystyle \\$

Question 7: $\displaystyle A$ can dig a trench in $\displaystyle 6$ days while $\displaystyle B$ can dig it in $\displaystyle 8$ days. They dug the trench working together and received $\displaystyle 1120$ for it. Find the share of each in it.

$\displaystyle \text{ A's 1 Day } = \frac{1}{6}$

$\displaystyle \text{ B's 1 Day } = \frac{1}{8}$

$\displaystyle \text{Therefore the ratio of work } = \frac {\frac{1}{6}} {\frac{1}{8} } = \frac{8}{6}$

$\displaystyle \text{Therefore A's share } = \frac{8}{14} \times 1120 = 640$

$\displaystyle \text{Therefore B's share }= \frac{6}{14} \times 1120 = 480$

$\displaystyle \\$

Question 8: $\displaystyle A$ can mow a field in $\displaystyle 9$ days; $\displaystyle B$ can mow it in $\displaystyle 12$ days while $\displaystyle C$ can mow it in $\displaystyle 8$ days. They all together mowed the field and received $\displaystyle 1610$ for it. How will the money be shared by them?

$\displaystyle \text{ A's 1 Day } = \frac{1}{9}$

$\displaystyle \text{ B's 1 Day } = \frac{1}{12}$

$\displaystyle \text{ C's 1 Day } = \frac{1}{18}$

$\displaystyle \text{Therefore the ratio of their one day's work } = \frac{1}{9} \colon \frac{1}{12} \colon \frac{1}{18} = 8 \colon 6 \colon 9$

$\displaystyle \text{Hence A's share } = \frac{8}{23} \times 1120 = 560$

$\displaystyle \text{Hence B's share } = \frac{6}{23} \times 1120 = 420$

$\displaystyle \text{Hence C's share } = \frac{9}{23} \times 1120 = 630$

$\displaystyle \\$

Question 9: $\displaystyle A$ and $\displaystyle B$ can do a piece of work in $\displaystyle 30$ days; $\displaystyle B$ and $\displaystyle C$ in $\displaystyle 24$ days; $\displaystyle C$ and $\displaystyle A$ in $\displaystyle 40$ days. How long will it take them to do the work together? In what time can each finish it, working alone?

$\displaystyle \text{ A's + B's 1 Day Work } = \frac{1}{30}$

$\displaystyle \text{(B's + C's) 1 Day work } = \frac{1}{24}$

$\displaystyle \text{(C's + D's) 1 Day work } = \frac{1}{40}$

$\displaystyle \text{Adding the above three } 2\times(A + B + C) \text{ day work }= (\frac{1}{30} + \frac{1}{24} + \frac{1}{40} ) = \frac{1}{10}$

Therefore if they all work together, they will take 20 days to finish the work.

$\displaystyle \\$

Question 10: $\displaystyle A$ can do a piece of work in $\displaystyle 80$ days. He works at it for $\displaystyle 10$ days and then $\displaystyle B$ alone finishes the remaining work in $\displaystyle 42$ days. In how many days could both do it?

$\displaystyle \text{ A's 1 Day } = \frac{1}{80}$

$\displaystyle \text{Work finished by A in 10 days } = \frac{1}{80} \times 10 = \frac{1}{8}$

$\displaystyle \text{B finished the remainder of work } (1 - \frac{1}{8} ) = \frac{7}{8} \text{ in } 42\text{ days }$

$\displaystyle \text{Therefore 1 Days work for B } = \frac{\frac{7}{8}}{42} = \frac{1}{48}$

Hence B can do the work in 48 days

$\displaystyle \text{ A's + B's 1 Day Work } = ( \frac{1}{80} + \frac{1}{48} )= \frac{1}{30}$

Therefore both can do the work in 30 days.

$\displaystyle \\$

Question 11: $\displaystyle A$ and $\displaystyle B$ can together finish a work in $\displaystyle 30$ days. They worked at it for $\displaystyle 20$ days and then $\displaystyle B$ left. The remaining work was done by $\displaystyle A$ alone in $\displaystyle 20$ more days. In how many days can $\displaystyle A$ alone do it?

$\displaystyle \text{ A's + B's 1 Day Work } = \frac{1}{30}$

$\displaystyle \text{Amount of work finished by both in 20 days } = 20 \times \frac{1}{30} = \frac{2}{3}$

$\displaystyle \text{Work left to be finished } = 1 - \frac{2}{3} = \frac{1}{3}$

$\displaystyle \text{Work done by A in 1 Day } = \frac{\frac{1}{3}}{20} = \frac{1}{60}$

Therefore A can do the work alone in 60 days.

$\displaystyle \\$

Question 12: $\displaystyle A$ can do a certain job in $\displaystyle 25$ days which $\displaystyle B$ alone can do in $\displaystyle 20$ days. $\displaystyle A$ started the work and was joined by $\displaystyle B$ after $\displaystyle 10$ days. In how many days was the whole work completed?

$\displaystyle \text{A's 1 Day work } = \frac{1}{25}$

$\displaystyle \text{ B's 1 Day } = \frac{1}{20}$

$\displaystyle \text{ A's + B's 1 Day Work } = ( \frac{1}{25} + \frac{1}{20} ) = \frac{9}{100}$

$\displaystyle \text{ Amount of work finished by A in 10 days } = 10 \times \frac{1}{25}$

$\displaystyle \text{Work left to be finished } = (1- \frac{2}{5} ) = \frac{3}{5}$

$\displaystyle \text{Days taken by both A and B working together } = \frac{\frac{3}{5}}{\frac{9}{100}} = 6 \frac{2}{3}$

$\displaystyle \text{The work got completed in } 10 + 6 \frac{2}{3} = 16 \frac{2}{3}$

$\displaystyle \\$

Question 13: $\displaystyle A$ can do a piece of work in $\displaystyle 14$ days, while $\displaystyle B$ can do in $\displaystyle 21$ days. They begin together. But, $\displaystyle 3$ days before the completion of the work, $\displaystyle A$ leaves off. Find the total number of days taken to complete the work.

$\displaystyle \text{ A's 1 Day work } = \frac{1}{14}$

$\displaystyle \text{ B's 1 Day } = \frac{1}{21}$

$\displaystyle \text{ A's + B's 1 Day Work } = ( \frac{1}{14} + \frac{1}{21} ) = \frac{5}{42}$

$\displaystyle \text{ Amount of work finished by B in 3 days } = (3 \times \frac{1}{21} ) = \frac{1}{7}$

$\displaystyle \text{ Work left to be finished by A and B together } = (1- \frac{1}{7} ) = \frac{6}{7}$

$\displaystyle \text{ Days taken by A + B working together } = \frac{\frac{6}{7}}{\frac{5}{42}} = 7 \frac{1}{5}$

$\displaystyle \text{ The work got completed in } 3+7 \frac{1}{5} = 10 \frac{1}{5}$

$\displaystyle \\$

Question 14: $\displaystyle A$ is thrice as good a workman as $\displaystyle B$ and $\displaystyle B$ is twice as good a workman as $\displaystyle C$. All the three took up a job and received $\displaystyle 1800$ as remuneration. Find the share of each.

$\displaystyle \text{If C takes } x$ days to complete the job

$\displaystyle \text{ Then B will complete the job in } \frac{x}{2}$

$\displaystyle \text{ And A will complete the job in } \frac{x}{3} \text{ days }$

$\displaystyle \text{ Therefore the ratio of 1 day's work of } A \colon B \colon C = \frac{x}{3} \colon \frac{x}{2} \colon \frac{x}{1} = 3 \colon 2 \colon 1$

$\displaystyle \text{ Therefore the share of A } = 1800 \times \frac{3}{6} = 900 \text{ Rs. }$

$\displaystyle \text{ Therefore the share of B } = 1800 \times \frac{2}{6} = 600 \text{ Rs. }$

$\displaystyle \text{ Therefore the share of C } = 1800 \times \frac{1}{6} = 300 \text{ Rs. }$

$\displaystyle \\$

Question 15: $\displaystyle A$ can do a certain job in $\displaystyle 12$ days. $\displaystyle B$ is $\displaystyle 60\%$ more efficient than $\displaystyle A$. Find the number of days taken by $\displaystyle B$ to finish the job.

$\displaystyle \text{Time taken to finish the job } = 12 \text{ }$ days

$\displaystyle \text{ A's 1 Day's work } = \frac{1}{12}$

B is 60% more efficient

$\displaystyle \text{ B's 1 Day's work } = 1.6 \times \frac{1}{12}$

$\displaystyle \text{Therefore the number of days B will take to finish the job } = \frac{1}{\frac{1.6}{12}} = 7 \frac{1}{2}$days

$\displaystyle \\$

Question 16: $\displaystyle A$ is twice as good a workman as $\displaystyle B$ and together they finish a piece of work in $\displaystyle 14$ days. In how many days can $\displaystyle A$ alone do it?

$\displaystyle \text{Let B take } x \text{days to finish the work. }$

$\displaystyle \text{B's 1 day's work } = \frac{1}{x}$

$\displaystyle \text{ The A will take } \frac{x}{2} \text{ days to finish the work. }$

$\displaystyle \text{A's 1 day's work } = \frac{2}{x}$

$\displaystyle \text{ (A+B) one day's work } = ( \frac{1}{x} + \frac{2}{x} ) = \frac{1}{14}$

$\displaystyle \text{Solving for } x=42$

Therefore A will take 21 days to finish the job.

$\displaystyle \\$

Question 17: Two pipes $\displaystyle A$ and $\displaystyle B$ can separately fill a tank in $\displaystyle 36$ minutes and $\displaystyle 45$ minutes respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

$\displaystyle \text{A's 1 minute fill rate } = \frac{1}{36}$

$\displaystyle \text{B's 1 minute fill rate } = \frac{1}{45}$

$\displaystyle \text{A's and B's fill rate together } = ( \frac{1}{36} + \frac{1}{45} ) = \frac{1}{20}$

Therefore if A and B are opened simultaneously, the tank will take 20 minutes to fill up.

$\displaystyle \\$

Question 18: One tap can fill a cistern in $\displaystyle 3$ hours and the waste pipe can empty the full tank in $\displaystyle 5$ hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together?

$\displaystyle \text{Tap's 1 minute fill rate } = \frac{1}{3}$

$\displaystyle \text{ Waste Pipe's 1 minute empty rate } = \frac{1}{5}$

$\displaystyle \text{Therefore the net fill rate } = ( \frac{1}{3} - \frac{1}{5} ) = \frac{2}{15}$

$\displaystyle \text{Therefore if both the tap and the waste pipe are opened simultaneously then it will take } 7\frac{1}{2}$ hours to fill up.

$\displaystyle \\$

Question 19: Two pipes $\displaystyle A$ and $\displaystyle B$ can separately fill a cistern in $\displaystyle 20$ minutes and $\displaystyle 30$ minutes respectively , while a third pipe $\displaystyle C$ can empty the full cistern in $\displaystyle 15$ minutes. If all the pipes are opened together, in what time the empty cistern is filled?

$\displaystyle \text{A's 1 minute fill rate } = \frac{1}{20}$

$\displaystyle \text{ B's 1 minute fill rate } = \frac{1}{30}$

$\displaystyle \text{C's 1 minute empty rate } = \frac{1}{15}$

$\displaystyle \text{ Therefore the net fill rate } = ( \frac{1}{20} + \frac{1}{30} - \frac{1}{15} ) = \frac{1}{60}$

Therefore if all the tap are opened simultaneously then it will take 60 minutes or one hour to fill up.

$\displaystyle \\$

Question 20: A pipe can fill a tank in $\displaystyle 16$ hours. Due to a leak in the bottom, it is filled in $\displaystyle 24$ hours. If the tank is full, how much time will the leak take to empty it?

$\displaystyle \text{ Pipe's fill rate } = \frac{1}{16}$
$\displaystyle \text{Let the leak is at a rate of } = \frac{1}{x}$
$\displaystyle \text{Therefore the net fill rate } = ( \frac{1}{16} - \frac{1}{x} ) = \frac{1}{24}$
$\displaystyle \text{Solving for } x = 48$ hours.