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?

Answer:

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.

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?

Answer:

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.

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?

Answer:

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

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?

Answer:

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

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?

Answer:

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

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?

Answer:

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.

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.

Answer:

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 = $\frac{6}{14}$ $\times 1120 = 480$

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?

Answer:

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 = $\frac{9}{23}$ $\times 1120 = 630$

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?

Answer:

(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.

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?

Answer:

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.

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?

Answer:

(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.

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?

Answer:

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  $10 + 6\frac{2}{3} = 16\frac{2}{3}$

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.

Answer:

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  $3+7\frac{1}{5} = 10\frac{1}{5}$

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.

Answer:

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.

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.

Answer:

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

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?

Answer:

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.

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?

Answer:

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.

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?

Answer:

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.

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?

Answer:

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.

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?

Answer:

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.