Time and Work Tricks

• If \(A\) can complete a piece of work in \(n\) days, then \(\text { A’s } 1 \text { day’s work }=\frac{1}{n} \text {. }\)
• \(
  \text { If } {A’s} \text { ‘ } 1 \text { day’s work }=\frac{1}{n} \text {, then } \mathrm{A} \text { can finish the work in } n \text { days. }
  \)
• If \(A\) and \(B\) can complete a piece of work in \(n_1\) and \(n_2\) days respectively, then if they both start working together, they can finish the same work in \(\frac{n_1 \times n_2}{n_1+n_2}\) days.
• If \(A\) completes \(\frac{1}{n^{\text {th }}}\) of work in \(x\) hours, then to finish the complete work \(A\) will take \(n \times x\) hours.
• If \({A}\) is thrice as good a workman as \({B}\), then :
  Ratio of work done by \({A}\) and \({B}=3: 1\).
  Ratio of times taken by \({A}\) and \({B}\) to finish a work \(=1: 3\).
• If \(x\) number of people can do W1 work, in D1 days, working T1 hours each day and the number of people can do W2 work, in D2 days, working T2 hours each day, then the relation between them will be
  \(
  \frac{\mathrm{M} 1 \times \mathrm{D} 1 \times \mathrm{T} 1}{\mathrm{~W} 1}=\frac{\mathrm{M} 2 \times \mathrm{D} 2 \times \mathrm{T} 2}{\mathrm{~W} 2}
  \)
• \(\mathrm{MDH}\) formula is used to compare the works of the same nature done at two different times using different manpower. The MDH formula can be represented as
  \(
  \frac{M_1 D_1 H_1 E_1}{W_1}=\frac{M_2 D_2 H_2 E_2}{W_2}
  \)
  Where \(\mathrm{M}\) denotes the number of men/women, \(\mathrm{D}\) denotes the number of days, \(\mathrm{H}\) denotes the number of hours in a day, \(\mathrm{E}\) denotes the efficiency of 1 man, and \(\mathrm{W}\) denotes part of work done
• Note: If \(A\) working alone takes ‘ \(d_1\) ‘ days more than \(A\) and \(B\) together, and \(B\) working along takes ‘ \(d_2\) ‘ days more than \(A\) and \(B\) together, then the number of days taken by \(A\) and \(B\) working together will be \(\sqrt{d_1 d_2}\).

Once you are thorough with the simple tricks and their application, the problems related to time and work will become extremely easy to solve. And, thus, we can say that,

• The number of men and the amount of time required to complete one work are inversely proportional to each other.
• Time and work are directly proportional to each other. Thus, in more time, more work can be completed, and vice versa.
• The work can be divided into equal parts, i.e. if a person can complete a piece of work in 5 days, then one day work of that person will be \(\frac{1}{5^{\text {th }}}\) part of that work.

Example 1: Arushi can do a piece of work in 12 days, and Joyita can do the same work in 18 days. How many days will they take to do the work together?

Solution: Arushi can do a piece of work in 12 days.
Therefore,Arushi’s 1-day work \(=\frac{1}{12}\) of the work
Joyita can do a piece of work in 18 days.
Therefore, Joyita’s 1 -day work \(=\frac{1}{18}\) of the work
Thus, Arushi’s and Joyita’s 1-day work \(=\frac{1}{12}+\frac{1}{18}\)
Taking the LCM of denominators, we get
\(
\frac{1}{12}+\frac{1}{18}=\frac{3+2}{36}
\)
\(=\frac{5}{36}\) of the work Hence, to finish the piece of work together, they need \(\frac{1}{\frac{5}{36}}=\frac{36}{5}\) days i.e., \(7 \frac{1}{5}\) days.

Example 2: 6 workers can pack 1500 boxes in 4 days. Working at the same rate, how many boxes can 8 workers pack in 6 days?

Solution:

6 worker takes 4 -days to pack 1500 boxes.
1 worker takes 4 -days to pack \(\frac{1500}{6}\) boxes.
8 workers take 4 -days to pack \(\frac{1500}{6} \times 8=2000\) boxes.
8 workers take 6 -days to pack \(\frac{2000}{4} \times 6=3000\) boxes. Hence, 8 workers can pack 3000 boxes in 6 days.

Example 3: Rachana can do work in 15 days and Swati in 20 days. If they work on it together for 4 days, then what fraction of the work is still left?

Solution: Rachana can do work in 15 days.
Therefore, Rachana’s one day work \(=\frac{1}{15}\)
Swati can do the same work in 20 days.
Therefore, Swati’s one day work \(=\frac{1}{20}\)
Swati’s and Rachana’s 1-day work \(=\left(\frac{1}{15}+\frac{1}{20}\right)=\frac{7}{60}\)
Swati’s and Rachana’s 4-days work \(=\frac{7}{60} \times 4=\frac{7}{15}\)
Therefore, remaining work \(=1-\frac{7}{15}=\frac{15-7}{15}=\frac{8}{15}\). Hence, \(\frac{8}{15}\) part of work is still left.

Example 4: If Roger can do a piece of work in 8 days and Antony can complete the same work in 5 days, in how many days will both of them together complete it? (L.I.C., 2008)

Solution:

Roger’s 1 day’s work \(=\frac{1}{8} ;\) Antony’s 1 day’s work \(=\frac{1}{5}\).
\(\left(\right.\) Roger + Antony)’s 1 day’s work \(=\left(\frac{1}{8}+\frac{1}{5}\right)=\frac{13}{40}\).
\(\therefore\) Both Roger and Antony will complete the work in \(\frac{40}{13}=3 \frac{1}{13}\) days.

Example 5: A and B together can complete a piece of work in 15 days and B alone in 20 days. In how many days can A alone complete the work? (S.S.C., 2010)

Solution:

\((\mathrm{A}+\mathrm{B})\) ‘s 1 day’s work \(=\frac{1}{15} ;\) B’s 1 day’s work \(=\frac{1}{20}\). \(\therefore \quad \mathrm{A}^{\prime}\) s 1 day’s work \(=\left(\frac{1}{15}-\frac{1}{20}\right)=\frac{1}{60}\).
Hence, A alone can complete the work in 60 days.

Example 6: A alone can complete a piece of work of \(₹ 300\) in 6 days; but by engaging an assistant, the work is completed in 4 days. Find the share to be received by the assistant. (Section Officer’s, 2008)

Solution:

Assistant’s 1 day’s work \(=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\).
\(\therefore \quad\) A’s share : Assistant’s share \(=\) Ratio of their 1 day’s work \(=\frac{1}{6}: \frac{1}{12}=2: 1\).
Hence, assistant’s share \(=₹\left(300 \times \frac{1}{3}\right)=₹ 100\).

Example 7: A can do a work in 4 days, \(B\) in 5 days, and \(C\) in 10 days. Find the time taken by \(A, B\) and \(C\) to do the work together. (P.C.S., 2006)

Solution:
\(
\begin{aligned}
& \text { A’s } 1 \text { day’s work }=\frac{1}{4} ; \text { B’s } 1 \text { day’s work }=\frac{1}{5} ; \quad \text { C’s } 1 \text { day’s work }=\frac{1}{10} \\
& (\mathrm{~A}+\mathrm{B}+\mathrm{C}) \text { ‘s } 1 \text { day’s work }=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{10}\right)=\frac{11}{20}
\end{aligned}
\)
\(
\text { Hence, A, B and } C \text { together can do the work in } \frac{20}{11}=1 \frac{9}{11} \text { days. }
\)

Example 8: A and B undertake to do a piece of work for \(₹ 600\). A alone can do it in 6 days while B alone can do it in 8 days. With the help of \(C\), they finish it in 3 days. Find the share of each.

Solution:

\(
\begin{aligned}
& \text { Sol. C’s } 1 \text { day’s work }=\frac{1}{3}-\left(\frac{1}{6}+\frac{1}{8}\right)=\frac{1}{24} \text {. } \\
& \therefore \quad \mathrm{A}: \mathrm{B}: \mathrm{C}=\text { Ratio of their } 1 \text { day’s work } \frac{1}{6}: \frac{1}{8}: \frac{1}{24}=4: 3: 1 \text {. } \\
& \therefore \quad \mathrm{A} \text { ‘s share }=₹\left(600 \times \frac{4}{8}\right)=₹ 300 \text {, B’s share }=₹\left(600 \times \frac{3}{8}\right)=₹ 225 \text {. } \\
& C^{\prime} \text { s share }=₹[600-(300+225)]=₹ 75 . \\
&
\end{aligned}
\)

Example 9: A can do a piece of work in 7 days of 9 hours each and B can do it in 6 days of 7 hours each. How long will they take to do it, working together \(8 \frac{2}{5}\) hours a day?

Solution:A can complete the work in \((7 \times 9)=63\) hours.
B can complete the work in \((6 \times 7)=42\) hours.
\(\therefore \quad\) A’s 1 hour’s work \(=\frac{1}{63}\) and B’s 1 hour’s work \(=\frac{1}{42}\).
\((A+B)\) ‘s 1 hour’s work \(=\left(\frac{1}{63}+\frac{1}{42}\right)=\frac{5}{126}\).
\(\therefore \quad\) Both will finish the work in \(\left(\frac{126}{5}\right) \mathrm{hrs}\).
Number of days of \(8 \frac{2}{5}\) hrs each \(=\left(\frac{126}{5} \times \frac{5}{42}\right)=3\) days.

Example 10: Rahul takes twice as much time as Manick and thrice as much time as Sachin to complete a job. If working together, they can complete the job in 4 days, and find the time taken by each of them separately to complete the work. (I.I.F.T., 2005)

Solution: Suppose Rahul takes \(x\) hours to complete the job.
Then, Manick takes \(\frac{x}{2}\) hours and Sachin takes \(\frac{x}{3}\) hours to do the job.
Rahul’s 1 hour’s work \(=\frac{1}{x} ;\) Manick’s 1 hour’s work \(=\frac{2}{x} ;\) Sachin’s 1 hour’s work \(=\frac{3}{x}\).
\(\therefore \quad 4\left(\frac{1}{x}+\frac{2}{x}+\frac{3}{x}\right)=1 \Rightarrow \frac{6}{x}=\frac{1}{4} \Rightarrow x=24\).
Hence, Rahul takes 24 hours, Manick takes 12 hours and Sachin takes 8 hours to complete the job.

Example 11: \(A\) and \(B\) can do a piece of work in 9 days; \(B\) and \(C\) can do it in 12 days; \(A\) and \(C\) can do it in 18 days. In how many days will \(A, B\) and \(C\) finish it, working together and separately? (S.S.C., 2007)

Solution:

\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{9} ;(\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{12} \\
& (\mathrm{~A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{18}
\end{aligned}
\)
Adding, we get: \(2(A+B+C)^{\prime}\) s 1 day’s work \(=\left(\frac{1}{9}+\frac{1}{12}+\frac{1}{18}\right)=\frac{9}{36}=\frac{1}{4}\).
\(
\therefore \quad(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{8} \text {. }
\)
Thus, \(A, B\) and \(C\) together can finish the work in 8 days.
Now, A’s 1 day’s work \(=[(A+B+C)\) ‘s 1 day’s work \(-(B+C)\) ‘s 1 day’s work \(]\)

\(
=\left(\frac{1}{8}-\frac{1}{12}\right)=\frac{1}{24} \text {. }
\)
\(\therefore \quad\) A alone can finish the work in 24 days.
Similarly, B’s 1 day’s work
\(
=[(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }-(\mathrm{A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }]=\left(\frac{1}{8}-\frac{1}{18}\right)=\frac{5}{72} \text {. }
\)
\(\therefore \quad\) B alone can finish the work in \(\frac{72}{5}=14 \frac{2}{5}\) days.
And, C’s 1 day’s work
\(
=\left[(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }-(\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }\right]=\left(\frac{1}{8}-\frac{1}{9}\right)=\frac{1}{72} \text {. }
\)
\(\therefore \quad C\) alone can finish the work in 72 days.

Example 12: A is twice as good a workman as B and together they finish a piece of work in 18 days. In how many days will A alone finish the work?

Solution:

(A’s 1 day’s work) : (B’s 1 day’s work) \(=2: 1\).
\((A+B)\) ‘s 1 day’s work \(=\frac{1}{18}\). Divide \(\frac{1}{18}\) in the ratio \(2: 1\).
\(
\therefore \quad \text { A’s } 1 \text { day’s work }=\left(\frac{1}{18} \times \frac{2}{3}\right)=\frac{1}{27} \text {. }
\)
Hence, \(\mathrm{A}\) alone can finish the work in 27 days.

Example 13: A can do a certain job in 12 days. \(B\) is \(60 \%\) more efficient than \(A\). How many days does \(B\) alone take to do the same job?

Solution:

The ratio of times taken by A and B \(=160: 100=8: 5\).
Suppose B alone takes \(x\) days to do the job.
Then, \(8: 5: 12: x \Rightarrow 8 x=5 \times 12 \Rightarrow x=7 \frac{1}{2}\) days.

Example 14: 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 much time will A and B, working together, finish the work?

Solution:

Work done by \(\mathrm{A}\) in 10 days \(=\left(\frac{1}{80} \times 10\right)=\frac{1}{8}\). Remaining work \(=\left(1-\frac{1}{8}\right)=\frac{7}{8}\).
Now, \(\frac{7}{8}\) work is done by B in 42 days.
The whole work will be done by B in \(\left(42 \times \frac{8}{7}\right)=48\) days.
\(\therefore \quad\) A’s 1 day’s work \(=\frac{1}{80}\) and B’s 1 day’s work \(=\frac{1}{48}\).
\(\therefore \quad(\mathrm{A}+\mathrm{B})\) ‘s 1 day’s work \(=\left(\frac{1}{80}+\frac{1}{48}\right)=\frac{8}{240}=\frac{1}{30}\).
Hence, both will finish the work in 30 days

Example 15: A can do a piece of work in 10 days and \(B\) in 20 days. They work together but 2 days before the completion of the work, A leaves. In how many days was the work completed? (S.S.C., 2008)

Solution:

\(
\begin{aligned}
& \text { B’s } 2 \text { days’ work }=\left(\frac{1}{20} \times 2\right)=\frac{1}{10} . \text { Remaining work }=\left(1-\frac{1}{10}\right)=\frac{9}{10} \\
& (\mathrm{~A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{10}+\frac{1}{20}\right)=\frac{6}{40}=\frac{3}{20} .
\end{aligned}
\)
Now, \(\frac{3}{20}\) work is done by A and B in 1 day.
\(\therefore \quad \frac{9}{10}\) work is done by \(A\) and \(B\) in \(\left(\frac{20}{3} \times \frac{9}{10}\right)=6\) days. Hence, total time taken \(=(2+6)\) days \(=8\) days.

Example 16: A can complete a work in 10 days, \(B\) in 12 days and \(C\) in 15 days. All of them began the work together, but \(A\) had to leave the work after 2 days of the start and B, 3 days before the completion of the work. How long did the work last? (S.S.C., 2005)

Solution:

A, B and C work together for 2 days. C alone works for 3 days and the remaining work is done by B and C together.
Now, \((\mathrm{A}+\mathrm{B}+\mathrm{C})\) ‘s 2 days’ work \(=2\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}\right)=\left(2 \times \frac{15}{60}\right)=\frac{1}{2}\).
C’s 3 days’ work \(=\left(3 \times \frac{1}{15}\right)=\frac{1}{5}\). Remaining work \(=1-\left(\frac{1}{2}+\frac{1}{5}\right)=\frac{3}{10}\).
But, \((\mathrm{B}+\mathrm{C})\) ‘s 1 day’s work \(=\left(\frac{1}{12}+\frac{1}{15}\right)=\frac{27}{180}=\frac{3}{20}\).
Now, \(\frac{3}{20}\) work is done by \((B+C)\) in 1 day.
\(\therefore \quad \frac{3}{10}\) work is done by \((B+C)\) in \(\left(\frac{20}{3} \times \frac{3}{10}\right)=2\) days.
Hence, total time taken \(=(2+3+2)\) days \(=7\) days.

Example 17: A and B can do a piece of work in 45 and 40 days respectively. They began the work together but A leaves after some days and B finished the remaining work in 23 days. After how many days did A leave? (M.B.A., 2009)

Solution:

B’s 23 days’ work \(=\left(23 \times \frac{1}{40}\right)=\frac{23}{40}\). Remaining work \(=\left(1-\frac{23}{40}\right)=\frac{17}{40}\).
Now, \((\mathrm{A}+\mathrm{B})\) ‘s 1 day’s work \(=\left(\frac{1}{45}+\frac{1}{40}\right)=\frac{17}{360}\).
Thus, \(\frac{17}{360}\) work is done by \((\mathrm{A}+\mathrm{B})\) in 1 day.
\(\therefore \quad \frac{17}{40}\) work is done by \((\mathrm{A}+\mathrm{B})\) in \(\left(\frac{360}{17} \times \frac{17}{40}\right)=9\) days.
Hence, A left after 9 days.

Example 18: A and B working separately can do a piece of work in 9 and 12 days respectively. If they work for a day alternately, A beginning, in how many days, the work will be completed?

Solution:

\((\mathrm{A}+\mathrm{B})\) ‘s 2 days’ work \(=\left(\frac{1}{9}+\frac{1}{12}\right)=\frac{7}{36}\).
Work done in 5 pairs of days \(=\left(5 \times \frac{7}{36}\right)=\frac{35}{36}\).
Remaining work \(=\left(1-\frac{35}{36}\right)=\frac{1}{36}\).
On the 11th day, it is A’s turn. \(\frac{1}{9}\) work is done by him in 1 day.
\(\frac{1}{36}\) work is done by him in \(\left(9 \times \frac{1}{36}\right)=\frac{1}{4}\) day.
\(\therefore \quad\) Total time taken \(=\left(10+\frac{1}{4}\right)\) days \(=10 \frac{1}{4}\) days.

Example 19: A can do a piece of work in 120 days and \(B\) can do it in 150 days. They work together for 20 days. Then, \(B\) leaves, and \(A\) alone continues the work. 12 days after that \(C\) joins \(A\) and the work is completed in 48 days more. In how many days can \(C\) do it if he works alone?

Solution:

\(
\begin{aligned}
& {[(\mathrm{A}+\mathrm{B}) \text { ‘s } 20 \text { days’ work }]+(\mathrm{A} \text { ‘s } 12 \text { days’ work })=20\left(\frac{1}{120}+\frac{1}{150}\right)+\left(12 \times \frac{1}{120}\right)=\frac{2}{5} .} \\
& \text { Remaining work }=\left(1-\frac{2}{5}\right)=\frac{3}{5}=(\mathrm{A}+\mathrm{C})^{\prime} \text { ‘s } 48 \text { days’ work. } \\
& \therefore \quad(\mathrm{A}+\mathrm{C}) \text { ‘s } 1 \text { day’s work }=\left(\frac{3}{5} \times \frac{1}{48}\right)=\frac{1}{80} . \text { C’s } 1 \text { day’s work }=\left(\frac{1}{80}-\frac{1}{120}\right)=\frac{1}{240} .
\end{aligned}
\)
Hence, \(C\) alone can finish the work in 240 days.

Example 20: \(A\) and \(B\) can do a piece of work in 12 days. \(B\) and \(C\) together can do it in 15 days. If \(A\) is twice as good a workman as \(C\), find in what time B alone can do it.

Solution:

A’s 1 day’s work = C’s 2 days’ work.
\(
\therefore \quad(A+B)^{\prime} \text { s } 1 \text { day’s work }=\left(B^{\prime} \text { s } 1 \text { day’s work }\right)+(C^{\prime} \text { s } 2 \text { days’ work })
\)
\(
\Rightarrow \quad \text { (B’s } 1 \text { day’s work })+\left(C^{\prime} \text { s } 2 \text { days’ work }\right)=\frac{1}{12} \dots(i)
\)
\(
\text { But (B’s } 1 \text { day’s work })+(\text { C’s } 1 \text { day’s work })=\frac{1}{15} \dots(ii)
\)
Subtracting (ii) from (i), we get C’s 1 day’s work \(=\left(\frac{1}{12}-\frac{1}{15}\right)=\frac{1}{60}\). \(\therefore \quad\) B’s 1 day’s work \(=\left(\frac{1}{15}-\frac{1}{60}\right)=\frac{3}{60}=\frac{1}{20}\).
Hence, B alone can finish the work in 20 days.

Example 21: 20 men working 10 hours per day can finish \(2 / 3^{\text {rd }}\) of the work in 12 days, while 15 women working 9 hours per day can finish \(2 / 5^{\text {th }}\) of the work in \(X\) days. If one woman is twice as efficient as one man, then what is the value of \(X\)?

Solution:

Here,
\(
M_1=20, D_1=12, H_1=10, W_1=\frac{2}{3}, M_2=15, D_2=X, H_2=9, W_2=\frac{2}{5}
\)
According to the question, one woman is twice as efficient as one man
So, the ratio of efficiency of woman and man is \(2: 1\)
\(
\Rightarrow \frac{E_2}{E_1}=\frac{2}{1}
\)
Now, using MDH formula we get
\(
\begin{aligned}
& \frac{M_1 D_1 H_1 E_1}{W_1}=\frac{M_2 D_2 H_2 E_2}{W_2} \\
& \Rightarrow \frac{20 \times 12 \times 10}{\frac{2}{3}}=\frac{15 \times X \times 9}{\frac{2}{5}} \times \frac{E_2}{E_1} \\
& \Rightarrow \frac{2400 \times 3}{2}=\frac{135 \times 5 \times X}{2} \times 2 \\
& \Rightarrow X=5 \frac{1}{3} \text { days }
\end{aligned}
\)