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

A’s 1 Day $=$ $\frac{1}{15}$

B’s 1 Day work $=$ $\frac{1}{10}$

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.

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

A’s 1 Day $=$ $\frac{1}{12}$

B’s 1 Day work $=$ $\frac{1}{15}$

C’s 1 Day work = $\frac{1}{10}$

(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.

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

A’s 1 Day $=$ $\frac{1}{60}$

B’s 1 Day work $=$ $\frac{1}{x}$

(A’s + B’s) 1 Day work $=$ $($ $\frac{1}{60}$ $+$ $\frac{1}{x}$ $) =$ $\frac{1}{35}$

Solving for $x = 84$ Days

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

A’s 1 Day $=$ $\frac{1}{20}$

B’s 1 Day work $=$ $\frac{1}{15}$

C’s 1 Day work $=$ $\frac{1}{x}$

(A’s + B’s + C’s) 1 Day work $=$ $(\frac{1}{20}$ $+$ $\frac{1}{15}$ $+$ $\frac{1}{x}$ $) =$ $\frac{1}{5}$

Solving for $x = 12$ Days

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

A’s 1 Day $=$ $\frac{1}{12}$

B’s 1 Day work $=$ $\frac{1}{16}$

(A’s + B’s) 1 Day work $=$ $($ $\frac{1}{12}$ $+$ $\frac{1}{16}$ $) =$ $\frac{7}{48}$

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

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

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

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

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

Therefore A’s 1 Day Work $=$  $\frac{1}{20}$

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

Therefore B’s 1 Day Work $=$ $\frac{1}{30}$

(A’s + B’s) 1 Day work $=$ $(\frac{1}{20}$ $+$ $\frac{1}{30}$ $) =$ $\frac{1}{12}$

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

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

A’s 1 Day $=$ $\frac{1}{6}$

B’s 1 Day work $=$ $\frac{1}{8}$

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

Therefore A’s share $=$ $\frac{8}{14}$ $\times 1120 = 640$

Therefore B’s share $=$ $\frac{6}{14}$ $\times 1120 = 480$

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

A’s 1 Day $=$ $\frac{1}{9}$

B’s 1 Day work $=$ $\frac{1}{12}$

C’s 1 Day work $=$ $\frac{1}{18}$

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

Hence A’s share $=$ $\frac{8}{23}$ $\times 1120 = 560$

Hence A’s share $=$ $\frac{6}{23}$ $\times 1120 = 420$

Hence A’s share $=$ $\frac{9}{23}$ $\times 1120 = 630$

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

(A’s + B’s) 1 Day work $=$ $\frac{1}{30}$

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

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

Adding the above three $2\times(A + B + C)$ 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.

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

A’s 1 Day $=$ $\frac{1}{80}$

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

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

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

Hence B can do the work in 48 days

(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.

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

(A’s + B’s) 1 Day work $=$ $\frac{1}{30}$

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

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

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.

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

A’s 1 Day work $=$ $\frac{1}{25}$

B’s 1 Day work $=$ $\frac{1}{20}$

(A’s + B’s) 1 Day work $=$ $($ $\frac{1}{25}$ $+$ $\frac{1}{20}$ $) =$ $\frac{9}{100}$

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

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

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

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

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

A’s 1 Day work $=$ $\frac{1}{14}$

B’s 1 Day work $=$ $\frac{1}{21}$

(A’s + B’s) 1 Day work $=$ $($ $\frac{1}{14}$ $+$ $\frac{1}{21}$ $) =$ $\frac{5}{42}$

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

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

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

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

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

If C takes $x$ days to complete the job

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

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

Therefore the ratio of 1 days’ work of $A \colon B \colon C =$ $\frac{x}{3}$ $\colon$ $\frac{x}{2}$ $\colon$ $\frac{x}{1}$ $= 3 \colon 2 \colon 1$

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

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

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

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

Time taken to finish the job $= 12$ days

A’s 1 Day’s work $=$ $\frac{1}{12}$

B is 60% more efficient

B’s 1 Day’s work $=$ $1.6 \times$ $\frac{1}{12}$

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

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

Let B take $x$ days to finish the work.

B’s 1 day’s work $=$ $\frac{1}{x}$

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

A’s 1 day’s work $=$ $\frac{2}{x}$

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

Solving for $x=42$

Therefore A will take 21 days to finish the job.

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

A’s 1 minute fill rate $=$ $\frac{1}{36}$

B’s 1 minute fill rate $=$ $\frac{1}{45}$

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.

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

Tap’s 1 minute fill rate $=$ $\frac{1}{3}$

Waste Pipe’s 1 minute empty rate $=$ $\frac{1}{5}$

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

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

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

A’s 1 minute fill rate $=$ $\frac{1}{20}$

B’s 1 minute fill rate $=$ $\frac{1}{30}$

C’s 1 minute empty rate $=$ $\frac{1}{15}$

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.

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

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