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\(\mathrm{A}\) and \(\mathrm{B}\) can do a job together in 7 days. \(\mathrm{A}\) is \(1 \frac{3}{4}\) times as efficient as B. The same job can be done by A alone in: (S.S.C., 2003)
(A’s 1 day’s work) : (B’s 1 day’s work \()=\frac{7}{4}: 1=7: 4\).
Let \(A^{\prime}\) s and B’s 1 day’s work be \(7 x\) and \(4 x\) respectively.
Then, \(7 x+4 x=\frac{1}{7} \Rightarrow 11 x=\frac{1}{7} \Rightarrow x=\frac{1}{77}\).
\(\therefore \quad\) A’s 1 day’s work \(=\left(\frac{1}{77} \times 7\right)=\frac{1}{11}\).
Hence, A alone can do the job in 11 days.
Kamal can do a work in 15 days. Bimal is \(50 \%\) more efficient than Kamal. The number of days, Bimal will take to do the same piece of work, is (C.P.O, 2006)
Ratio of times taken by Kamal and Bimal \(=150: 100[latex] [latex]=3: 2\)
Suppose Bimal takes \(x\) days to do the work.
\(3: 2:: 15: x \Rightarrow x=\left(\frac{2 \times 15}{3}\right) \Rightarrow x=10\) days.
A does \(20 \%\) less work than \(B\). If A can complete a piece of work in \(7 \frac{1}{2}\) hours, then \(B\) can do it in (S.S.C. 2006)
Ratio of times taken by \(\mathrm{A}\) and \(\mathrm{B}=100: 80=5: 4\). Suppose B takes \(x\) hours to do the work.
\(5: 4 \because \frac{15}{2}: x \Rightarrow x=\left(\frac{4 \times 15}{2 \times 5}\right) \Rightarrow x=6\) hours.
A is \(30 \%\) more efficient than B. How much time will they, working together, take to complete a job which \(\mathrm{A}\) alone could have done in 23 days?
Ratio of times taken by \(\mathrm{A}\) and \(\mathrm{B}=100: 130=10: 13\). Suppose B takes \(x\) days to do the work.
Then, \(10: 13:: 23: x \Rightarrow x=\left(\frac{23 \times 13}{10}\right) \Rightarrow x=\frac{299}{10}\).
A’s 1 day’s work \(=\frac{1}{23}\); B’s 1 day’s work \(=\frac{10}{299}\).
\((A+B)\) ‘s 1 day’s work \(=\left(\frac{1}{23}+\frac{10}{299}\right)=\frac{23}{299}=\frac{1}{13}\).
\(\therefore \quad\) A and B together can complete the job in 13 days.
A can do a piece of work in 10 days working 8 hours per day. If B is two-thirds as efficient as A, then in how many days can B alone do the same piece of work, working 5 hours per day?
Time taken by A alone to do the work \(=(10 \times 8)\) hrs \(=80 \mathrm{hrs}\).
Since B is two-thirds as efficient as A, so time taken by \(\mathrm{B}\) to do the work
\(
\begin{aligned}
& \quad=\left(80 \times \frac{3}{2}\right) \mathrm{hrs}=120 \mathrm{hrs} . \\
& \therefore \quad \text { Required time }=\left(\frac{120}{5}\right) \text { days }=24 \text { days. }
\end{aligned}
\)
A does half as much work as B in one-sixth of the time. If together they take 10 days to complete a work, how much time shall B alone take to do it? (S.S.C., 2005)
Suppose B takes \(x\) days to do the work.
\(\therefore \quad\) A takes \(\left(2 \times \frac{1}{6} x\right)=\frac{1}{3} x\) days to do it.
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{10} . \\
& \therefore \quad \frac{1}{x}+\frac{3}{x}=\frac{1}{10} \Rightarrow \frac{4}{x}=\frac{1}{10} \Rightarrow x=40 .
\end{aligned}
\)
\(\mathrm{A}\) is \(50 \%\) as efficient as \(\mathrm{B}\). C does half of the work done by \(\mathrm{A}\) and \(\mathrm{B}\) together. If \(\mathrm{C}\) alone does the work in 40 days, then \(A, B\) and \(C\) together can do the work in:
\(
\begin{aligned}
& \text { (A’s } 1 \text { day’s work) : (B’s } 1 \text { day’s work) } \\
& =150: 100=3: 2
\end{aligned}
\)
Let A’s and B’s 1 day’s work be \(3 x\) and \(2 x\) respectively.
Then, C’s 1 day’s work \(=\left(\frac{3 x+2 x}{2}\right)=\frac{5 x}{2}\).
\(
\therefore \quad \frac{5 x}{2}=\frac{1}{40} \text { or } x=\left(\frac{1}{40} \times \frac{2}{5}\right)=\frac{1}{100} \text {. }
\)
A’s 1 day’s work \(=\frac{3}{100} ;\) B’s 1 day’s work \(=\frac{1}{50} ;\)
C’s 1 day’s work \(=\frac{1}{40}\).
\((\mathrm{A}+\mathrm{B}+\mathrm{C})\) ‘s 1 day’s work \(=\left(\frac{3}{100}+\frac{1}{50}+\frac{1}{40}\right)=\frac{15}{200}=\frac{3}{40}\).
So, A, B and C together can do the work in \(\frac{40}{3}=13 \frac{1}{3}\) days.
Two workers A and B working together completed a job in 5 days. If \(\mathrm{A}\) worked twice as efficiently as he actually did and B worked \(\frac{1}{3}\) as efficiently as he actually did, the work would have been completed in 3 days. \(\mathrm{A}\) alone could complete the work in:
None of theseLet \(\mathrm{A}^{\prime}\) s 1 day’s work \(=x\) and B’s 1 day’s work \(=y\).
Then, \(x+y=\frac{1}{5}\) and \(2 x+\frac{1}{3} y=\frac{1}{3}\).
Solving, we get: \(x=\frac{4}{25}\) and \(y=\frac{1}{25}\).
\(\therefore \quad\) A’s 1 day’s work \(=\frac{4}{25}\).
So, A alone could complete the work in \(\frac{25}{4}=6 \frac{1}{4}\) days.
A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is:
\(
\begin{aligned}
& \text { A’s } 1 \text { day’s work }=\frac{1}{15} ; \text { B’s } 1 \text { day’s work }=\frac{1}{20} . \\
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{15}+\frac{1}{20}\right)=\frac{7}{60} . \\
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 4 \text { days’ work }=\left(\frac{7}{60} \times 4\right)=\frac{7}{15} . \\
& \therefore \quad \text { Remaining work }=\left(1-\frac{7}{15}\right)=\frac{8}{15} .
\end{aligned}
\)
A can do a piece of work in 15 days, which B can do in 10 days. B worked at it for 8 days. A can finish the remaining work in (R.R.B., 2004)
\(\begin{aligned}
& \text { B’s } \text { days’ work }=\left(\frac{1}{10} \times 8\right)=\frac{4}{5} \\
& \text { Remaining work }=\left(1-\frac{4}{5}\right)=\frac{1}{5} .
\end{aligned}
\)
Now, \(\frac{1}{15}\) work is done by \(\mathrm{A}\) in 1 day.
\(\therefore \quad \frac{1}{5}\) work is done by \(\mathrm{A}\) in \(\left(15 \times \frac{1}{5}\right)=3\) days.
A and B can complete a work in 18 days and 15 days respectively. They started doing the work together but after 3 days A had to leave and B alone completed the remaining work. The whole work was completed in
\(
\begin{aligned}
& (A+B)^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{18}+\frac{1}{15}\right)=\frac{11}{90} . \\
& (A+B)^{\prime} \text { s } 3 \text { days’ work }=\left(\frac{11}{90} \times 3\right)=\frac{11}{30} . \\
& \text { Remaining work }=\left(1-\frac{11}{30}\right)=\frac{19}{30} .
\end{aligned}
\)
Now, \(\frac{1}{15}\) work is done by B in 1 day.
\(\therefore \quad \frac{19}{30}\) work will be done by B in \(\left(15 \times \frac{19}{30}\right)=\frac{19}{2}\) days \(=9 \frac{1}{2}\) days.
Hence, total time taken \(=\left(3+9 \frac{1}{2}\right)\) days \(=12 \frac{1}{2}\) days.
A, B and C can separately do a work in 12, 15 and 20 days respectively. They started to work together but C left after 2 days. The remaining work will be finished in (R.R.B., 2006)
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{12}+\frac{1}{15}+\frac{1}{20}\right)=\frac{12}{60}=\frac{1}{5} \\
& (\mathrm{~A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 2 \text { days’ work }=\left(\frac{1}{5} \times 2\right)=\frac{2}{5} \\
& \text { Remaining work }=\left(1-\frac{2}{5}\right)=\frac{3}{5} . \\
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{12}+\frac{1}{15}\right)=\frac{9}{60}=\frac{3}{20}
\end{aligned}
\)
Now, \(\frac{3}{20}\) work is done by A and B in 1 day.
\(\therefore \quad \frac{3}{5}\) work will be done by A and \(B\) in \(\left(\frac{20}{3} \times \frac{3}{5}\right)=4\) days.
A can complete a piece of work in 18 days, B in 20 days and C in 30 days. B and C together start the work and are forced to leave after 2 days. The time taken by A alone to complete the remaining work is (S.S.C., 2010)
\(
\begin{aligned}
& (B+C)^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{20}+\frac{1}{30}\right)=\frac{5}{60}=\frac{1}{12} . \\
& (B+C)^{\prime} \text { s } 2 \text { day’s work }=\left(\frac{1}{12} \times 2\right)=\frac{1}{6} . \\
& \text { Remaining work }=\left(1-\frac{1}{6}\right)=\frac{5}{6} .
\end{aligned}
\)
\(\frac{1}{18}\) work is done by \(\mathrm{A}\) in 1 day.
\(\frac{5}{6}\) work will be done by A in \(\left(18 \times \frac{5}{6}\right)=15\) days.
A can complete a piece of work in 10 days, B in 15 days and C in 20 days. A and C worked together for 2 days and then A was replaced by B. In how many days, altogether, was the work completed? (C.P.O., 2007)
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{10}+\frac{1}{20}\right)=\frac{3}{20} . \\
& (\mathrm{A}+\mathrm{C})^{\prime} \text { s } 2 \text { days’ work }=\left(\frac{3}{20} \times 2\right)=\frac{3}{10} . \\
& \text { Remaining work }=\left(1-\frac{3}{10}\right)=\frac{7}{10} . \\
& (\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{15}+\frac{1}{20}\right)=\frac{7}{60} .
\end{aligned}
\)
\(\frac{7}{60}\) work is done by B and \(C\) in 1 day.
\(\frac{7}{10}\) work will be done by B and C in \(\left(\frac{60}{7} \times \frac{7}{10}\right)=6\) days.
Hence, total time taken \(=(2+6)\) days \(=8\) days.
A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 a.m. while machine P is closed at 11 a.m. and the remaining two machines complete the work. Approximately at what time will the work be finished?
\(
\begin{aligned}
& (P+Q+R)^{\prime} \text { s } 1 \text { hour’s work }=\left(\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\right)=\frac{37}{120} \\
& \text { Work done by } P, Q \text { and } R \text { in } 2 \text { hours }=\left(\frac{37}{120} \times 2\right)=\frac{37}{60} \\
& \text { Remaining work }=\left(1-\frac{37}{60}\right)=\frac{23}{60} \\
& (Q+R)^{\prime} \text { s } 1 \text { hour’s work }=\left(\frac{1}{10}+\frac{1}{12}\right)=\frac{11}{60}
\end{aligned}
\)
Now, \(\frac{11}{60}\) work is done by \(Q\) and \(R\) in 1 hour.
So, \(\frac{23}{60}\) nork will be done by \(Q\) and in \(\left(\frac{60}{11} \times \frac{23}{60}\right)=\frac{23}{11}\) hours \(\approx 2 \text { hours. }\)
So, the work will be finished approximately 2 hours after 11 a.m., i.e., around 1 p.m.
A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
\(
\begin{aligned}
& 2(\mathrm{~A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{30}+\frac{1}{24}+\frac{1}{20}\right)=\frac{15}{120}=\frac{1}{8} . \\
& \Rightarrow \quad(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work}=\frac{1}{16} .
\end{aligned}
\)
Work done by A, B and \(C\) in 10 days \(=\frac{10}{16}=\frac{5}{8}\).
Remaining work \(=\left(1-\frac{5}{8}\right)=\frac{3}{8}\).
A’s 1 day’s work \(=\left(\frac{1}{16}-\frac{1}{24}\right)=\frac{1}{48}\).
Now, \(\frac{1}{48}\) work is done by \(\mathrm{A}\) in 1 day.
So, \(\frac{3}{8}\) work will be done by A in \(\left(48 \times \frac{3}{8}\right)=18\) days.
X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?
Work done by \(X\) in 4 days \(=\left(\frac{1}{20} \times 4\right)=\frac{1}{5}\).
Remaining work \(=\left(1-\frac{1}{5}\right)=\frac{4}{5}\).
\((X+Y)\) ‘s 1 day’s work \(=\left(\frac{1}{20}+\frac{1}{12}\right)=\frac{8}{60}=\frac{2}{15}\).
Now, \(\frac{2}{15}\) work is done by \(X\) and \(Y\) in 1 day.
So, \(\frac{4}{5}\) work will be done by \(X\) and \(Y\) in \(\left(\frac{15}{2} \times \frac{4}{5}\right)=6\) days.
Hence, total time taken \(=(6+4)\) days \(=10\) days.
A completes \(\frac{7}{10}\) of a work in 15 days. Then he completes the remaining work with the help of B in 4 days. The time required for \(A\) and \(B\) together to complete the entire work is (S.S.C., 2005)
\(
\begin{aligned}
& (A+B) \text { ‘s } 4 \text { days’ work }=\left(1-\frac{7}{10}\right)=\frac{3}{10} . \\
& (A+B) \text { ‘s } 1 \text { day’s work }=\left(\frac{3}{10} \times \frac{1}{4}\right)=\frac{3}{40} .
\end{aligned}
\)
Hence, A and B together take \(\frac{40}{3}=13 \frac{1}{3}\) days to complete the entire work.
A man and a boy can do a piece of work in 24 days. If the man works alone for the last 6 days, it is completed in 26 days. How long would the boy take to do it alone? (S.S.C., 2005)
\(
\begin{aligned}
& (M+B)^{\prime} \text { s } 1 \text { days’ work }=\frac{1}{24} . \\
& (M+B)^{\prime} \text { s } 20 \text { days’ work }+M^{\prime} \text { s } 6 \text { days’ work }=1 \\
& \Rightarrow \quad \text { M’s } 6 \text { days’ work }=\left(1-\frac{1}{24} \times 20\right)=\frac{4}{24}=\frac{1}{6} \\
& \Rightarrow \quad \text { M’s } 1 \text { day’s work }=\frac{1}{6} \times \frac{1}{6}=\frac{1}{36} . \\
& \therefore \quad \text { B’s } 1 \text { day’s work }=\frac{1}{24}-\frac{1}{36}=\frac{1}{72} .
\end{aligned}
\)
Hence, the boy alone can do the work in 72 days.
A and B can together finish a work in 30 days. They worked together for 20 days and then B left. After another 20 days, A finished the remaining work. In how many days A alone can finish the job? (S.S.C., 2003)
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 20 \text { days’ } \text { work }=\left(\frac{1}{30} \times 20\right)=\frac{2}{3} . \\
& \text { Remaining work }=\left(1-\frac{2}{3}\right)=\frac{1}{3} .
\end{aligned}
\)
Now, \(\frac{1}{3}\) work is done by \(\mathrm{A}\) in 20 days.
Whole work will be done by \(A\) in \((20 \times 3)=60\) days.
X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?
Work done by \(X\) in 8 days \(=\left(\frac{1}{40} \times 8\right)=\frac{1}{5}\).
Remaining work \(=\left(1-\frac{1}{5}\right)=\frac{4}{5}\).
Now, \(\frac{4}{5}\) work is done by \(\mathrm{Y}\) in 16 days.
Whole work will be done by \(Y\) in \(\left(16 \times \frac{5}{4}\right)=20\) days.
\(\therefore \quad\) X’s 1 day’s work \(=\frac{1}{40}, Y^{\prime}\) s 1 day’s work \(=\frac{1}{20}\).
\((X+Y)\) ‘s 1 day’s work \(=\left(\frac{1}{40}+\frac{1}{20}\right)=\frac{3}{40}\).
Hence, \(X\) and \(Y\) will together complete the work in \(\frac{40}{3}=13 \frac{1}{3}\) days.
A, B and C together can complete a piece of work in 10 days. All three started working at it together and after 4 days A left. Then B and C together completed the work in 10 more days. A alone could complete the work in:
Work done by A, B and \(C\) in 4 days \(=\left(\frac{1}{10} \times 4\right)=\frac{2}{5}\). Remaining work \(=\left(1-\frac{2}{5}\right)=\frac{3}{5}\).
Now, \(\frac{3}{5}\) work is done by \(B\) and \(C\) in 10 days.
Whole work will be done by \(B\) and \(C\) in \(\left(10 \times \frac{5}{3}\right)=\frac{50}{3}\) days. \((\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime}\) s 1 day’s work \(=\frac{1}{10}\),
\((\mathrm{B}+\mathrm{C})^{\prime}\) s 1 day’s work \(=\frac{3}{50}\).
A’s 1 day’s work \(=\left(\frac{1}{10}-\frac{3}{50}\right)=\frac{2}{50}=\frac{1}{25}\).
\(\therefore \quad\) A alone could complete the work in 25 days.
A does \(\frac{4}{5}\) of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?
Whole work is done by A in \(\left(20 \times \frac{5}{4}\right)=25\) days.
Now, \(\left(1-\frac{4}{5}\right)\) i.e., \(\frac{1}{5}\) work is done by \(\mathrm{A}\) and \(\mathrm{B}\) in 3 days.
Whole work will be done by \(A\) and \(B\) in \((3 \times 5)=15\) days.
A’s 1 day’s work \(=\frac{1}{25},(\mathrm{~A}+\mathrm{B})\) ‘s 1 day’s work \(=\frac{1}{15}\).
\(\therefore \quad\) B’s 1 day’s work \(=\left(\frac{1}{15}-\frac{1}{25}\right)=\frac{4}{150}=\frac{2}{75}\).
So, B alone would do the work in \(\frac{75}{2}=37 \frac{1}{2}\) days.
A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?
Let \(\mathrm{A}^{\prime}\) s 1 day’s work \(=x\) and \(\mathrm{B}^{\prime}\) s 1 day’s work \(=y\).
Then, \(x+y=\frac{1}{30}\) and \(16 x+44 y=1\).
Solving these two equations, we get : \(x=\frac{1}{60}\) and \(y=\frac{1}{60}\).
\(
\therefore \quad \text { B’s } 1 \text { day’s work }=\frac{1}{60} \text {. }
\)
Hence, B alone shall finish the whole work in 60 days.
A and B together can do a piece of work in 12 days, which B and C together can do in 16 days. After A has been working at it for 5 days and B for 7 days, C finishes it in 13 days. In how many days C alone will do the work?
\(
\begin{aligned}
& \mathrm{A}^{\prime} \text { s } 5 \text { days’ work }+\mathrm{B}^{\prime} \text { s } 7 \text { days’ work }+\mathrm{C}^{\prime} \text { s } 13 \text { days’ work } \\
& =1 \\
& \Rightarrow(\mathrm{A}+\mathrm{B})^{\prime} \text { s } 5 \text { days’ work }+(\mathrm{B}+\mathrm{C})^{\prime} \text { s } 2 \text { days’ work }+ \\
& C^{\prime} \text { s } 11 \text { days’ work }=1 \\
& \Rightarrow \quad \frac{5}{12}+\frac{2}{16}+C^{\prime} \text { s } 11 \text { days’ } \text { work }=1 \\
& \Rightarrow C^{\prime} \text { s } 11 \text { days’ work }=1-\left(\frac{5}{12}+\frac{2}{16}\right)=\frac{11}{24} \\
& \Rightarrow \text { C’s } 1 \text { day’s work }=\left(\frac{11}{24} \times \frac{1}{11}\right)=\frac{1}{24} \text {. } \\
&
\end{aligned}
\)
\(\therefore\) C alone can finish the work in 24 days.
A and B can do a piece of work in 28 and 35 days respectively. They began to work together but A leaves after some time and B completed the remaining work in 17 days. After how many days did A leave?
\(
(\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{28}+\frac{1}{35}\right)=\frac{9}{140}
\)
Work done by B in 17 days \(=\left(\frac{1}{35} \times 17\right)=\frac{17}{35}\).
Remaining work \(=\left(1-\frac{17}{35}\right)=\frac{18}{35}\)
Now, \(\frac{9}{140}\) work was done by \((A+B)\) in 1 day.
So, \(\frac{18}{35}\) work was done by \((A+B)\) in \(\left(\frac{140}{9} \times \frac{18}{35}\right)=8\) days.
\(\therefore \quad\) A left after 8 days.
A can build up a wall in 8 days while B can break it in 3 days. A has worked for 4 days and then B joined to work with A for another 2 days only. In how many days will A alone build up the remaining part of the wall? (M.A.T., 2006)
Part of wall built by \(\mathrm{A}\) in 1 day \(=\frac{1}{8}\).
Part of wall broken by B in 1 day \(=\frac{1}{3}\).
Part of wall built by \(\mathrm{A}\) in 4 days \(=\left(\frac{1}{8} \times 4\right)=\frac{1}{2}\).
Part of wall broken by \((A+B)\) in 2 days \(=2\left(\frac{1}{3}-\frac{1}{8}\right)=\frac{5}{12}\).
Part of wall built in 6 days \(=\left(\frac{1}{2}-\frac{5}{12}\right)=\frac{1}{12}\).
Remaining part to be built \(=\left(1-\frac{1}{12}\right)=\frac{11}{12}\).
Now, \(\frac{1}{8}\) wall is built by \(\mathrm{A}\) in 1 day.
\(\therefore \quad \frac{11}{12}\) wall will be built by \(\mathrm{A}\) in \(\left(8 \times \frac{11}{12}\right)\) \(=\frac{22}{3}\) days \(=7 \frac{1}{3}\) days.
Anuj and Manoj can together paint their house in 30 days. After working for 20 days, Anuj has to go out and Manoj finishes the remaining work in the next 30 days. If Manoj had gone away after 20 days instead of Anuj, then Anuj would have completed the remaining work in (M.B.A., 2006)
\(
\begin{aligned}
& (\text { Anuj }+ \text { Manoj)’s } 20 \text { days’ work }=\left(\frac{1}{30} \times 20\right)=\frac{2}{3} \\
& \text { Remaining work }=\left(1-\frac{2}{3}\right)=\frac{1}{3} . \\
& \text { Manoj’s } 30 \text { days’ work }=\frac{1}{3} . \\
& \therefore \quad \text { Manoj’s } 1 \text { days’ work }=\frac{1}{90} . \\
& \text { Anuj’s } 1 \text { days’ work }=\left(\frac{1}{30}-\frac{1}{90}\right)=\frac{2}{90}=\frac{1}{45} .
\end{aligned}
\)
If Manoj had gone away after 20 days, then the remaining \(\frac{1}{3}\) work would have been done by Anuj.
\(\frac{1}{45}\) work is done by Anuj in 1 day.
\(\frac{1}{3}\) work would be done by Anuj in \(\left(45 \times \frac{1}{3}\right)=15\) days.
A started a work and left after working for 2 days. Then B was called and he finished the work in 9 days. Had A left the work after working for 3 days, B would have finished the remaining work in 6 days. In how many days can each of them, working alone, finish the whole work? (N.M.A.T., 2005)
Suppose A takes \(x\) days to finish the work alone and B takes \(y\) days to finish the work alone.
Then, \(\frac{2}{x}+\frac{9}{y}=1 \dots(i)\)
And, \(\frac{3}{x}+\frac{6}{y}=1 \Leftrightarrow \frac{1}{x}+\frac{2}{y}=\frac{1}{3} \Leftrightarrow \frac{2}{x}+\frac{4}{y}=\frac{2}{3} \dots(ii)\)
Subtracting (ii) from (i), we get: \(\frac{5}{y}=\frac{1}{3}\) or \(y=15\).
Putting \(y=15\) in (i), we get: \(\frac{2}{x}=\frac{2}{5}\) or \(x=5\).
Hence, A alone takes 5 days while B alone takes 15 days to finish the work.
Working together, Asha and Sudha can complete an assigned task in 20 days. However, if Asha worked alone and completed half the work and then Sudha takes over the task and completes the second half of the task, the task will be completed in 45 days. How long will Asha take to complete the task if she worked alone? Assume that Sudha is more efficient than Asha. (M.A.T., 2010)
Suppose Asha takes \(x\) days to complete the task alone while Sudha takes \(y\) days to complete it alone.
Since Sudha is more efficient than Asha, we have \(x>y\).
Asha’s 1 day’s work \(=\frac{1}{x} ;\) Sudha’s 1 day’s work \(=\frac{1}{y}\).
(Asha + Sudha)’s 1 day’s work \(=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{x y .} \dots(i)\)
If Asha and Sudha each does half of the work alone, time taken \(=\left(\frac{x}{2}+\frac{y}{2}\right)\) days \(=\left(\frac{x+y}{2}\right)\) days.
\(
\therefore \quad \frac{x+y}{2}=45 \Rightarrow x+y=90 \dots(ii)
\)
From (i) and (ii), we have: \(\frac{x y}{20}=90\) or \(x y=1800\).
Now, \(x y=1800\) and \(x+y=90 \Rightarrow x=60, y=30\).
\(
[\because x>y]
\)
Hence, Asha alone will take 60 days to complete the task.
A can do a piece of work in 14 days which B can do in 21 days. They begin together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is (G.B.O., 2007)
\(
\begin{aligned}
& \text { B’s } 3 \text { days’ work }=\left(\frac{1}{21} \times 3\right)=\frac{1}{7} . \\
& \text { Remaining work }=\left(1-\frac{1}{7}\right)=\frac{6}{7} .
\end{aligned}
\)
\(
(A+B)^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{14}+\frac{1}{21}\right)=\frac{5}{42}
\)
Now, \(\frac{5}{42}\) work is done by \(A\) and \(B\) in 1 day.
\(\therefore \quad \frac{6}{7}\) work is done by A and B in \(\left(\frac{42}{5} \times \frac{6}{7}\right)=\frac{36}{5}\) days.
Hence, total time taken \(=\left(3+\frac{36}{5}\right)\) days \(=10 \frac{1}{5}\) days.
A, B and C can complete a work separately in 24, 36 and 48 days respectively. They started together but C left after 4 days of start and A left 3 days before the completion of the work. In how many days will the work be completed?
\(
(A+B+C)^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{24}+\frac{1}{36}+\frac{1}{48}\right)=\frac{13}{144} \text {. }
\)
Work done by \((A+B+C)\) in 4 days \(=\left(\frac{13}{144} \times 4\right)=\frac{13}{36}\).
Work done by B in 3 days \(=\left(\frac{1}{36} \times 3\right)=\frac{1}{12}\).
Remaining work \(=\left[1-\left(\frac{13}{36}+\frac{1}{12}\right)\right]=\frac{5}{9}\).
\((A+B)\) ‘s 1 day’s work \(=\left(\frac{1}{24}+\frac{1}{36}\right)=\frac{5}{72}\).
Now, \(\frac{5}{72}\) work is done by \(A\) and \(B\) in 1 day.
So, \(\frac{5}{9}\) work is done by A and B in \(\left(\frac{72}{5} \times \frac{5}{9}\right)=8\) days.
Hence, total time taken \(=(4+3+8)\) days \(=15\) days.
A, B and C can complete a work in 10, 12 and 15 days respectively. They started the work together. But A left the work 5 days before its completion. B also left the work 2 days after A left. In how many days was the work completed? (C.P.O., 2007)
\(
\begin{aligned}
& C^{\prime} \text { s } 3 \text { days’ work }=\left(\frac{1}{15} \times 3\right)=\frac{1}{5} . \\
& (B+C)^{\prime} \text { s } 2 \text { days’ work }=\left[\left(\frac{1}{12}+\frac{1}{15}\right) \times 2\right]=\left(\frac{3}{20} \times 2\right)=\frac{3}{10} \text {. } \\
& \text { Remaining work }=\left[1-\left(\frac{1}{5}+\frac{3}{10}\right)\right]=\left(1-\frac{1}{2}\right)=\frac{1}{2} \text {. } \\
& (\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}\right)=\frac{15}{60}=\frac{1}{4} \text {. } \\
&
\end{aligned}
\)
\(\frac{1}{4}\) work is done by \(A, B\) and \(C\) in 1 day.
\(\therefore \quad \frac{1}{2}\) work is done by \(A, B\) and \(C\) in \(\left(4 \times \frac{1}{2}\right)=2\) days.
The total number of days \(=(3+2+2)=7\).
\(A, B\) and \(C\) are employed to do a piece of work for ₹ 529. A and B together are supposed to do \(\frac{19}{23}\) of the work and B and C together \(\frac{8}{23}\) of the work. What amount should \(A\) be paid?
Work done by \(A=\left(1-\frac{8}{23}\right)=\frac{15}{23}\).
\(\therefore \quad A:(B+C)=\frac{15}{23}: \frac{8}{23}=15: 8\).
So, \(A^{\prime}\) s share \(=₹\left(\frac{15}{23} \times 529\right)=₹ 345\).
Kim can do a work in 3 days while David can do the same work in 2 days. Both of them finish the work together and get \(₹ 150\). What is the share of \(\mathrm{Kim}\)?
Kim’s wages : David’s wages
\(
\begin{aligned}
& =\text { Kim’s } 1 \text { day’s work: David’s } 1 \text { day’s work } \\
& =\frac{1}{3}: \frac{1}{2}=2: 3 .
\end{aligned}
\)
\(
\therefore \quad \text { Kim’s share }=₹\left(\frac{2}{5} \times 150\right)=₹ 60 \text {. }
\)
If \(A\) can do \(\frac{1}{4}\) of a work in 3 days and \(B\) can do \(\frac{1}{6}\) of the same work in 4 days, how much will A get if both work together and are paid \(₹ 180\) in all?
Whole work is done by \(\mathrm{A}\) in \((3 \times 4)=12\) days.
Whole work is done by B in \((4 \times 6)=24\) days.
A’s wages : B’s wages
\(
\begin{aligned}
& \quad=\text { A’s } 1 \text { day’s work : B’s } 1 \text { day’s work }=\frac{1}{12}: \frac{1}{24}=2: 1 . \\
& \therefore \quad \text { A’s share }=₹\left(\frac{2}{3} \times 180\right)=₹ 120 .
\end{aligned}
\)
A man and a boy received \(₹ 800\) as wages for 5 days for the work they did together. The man’s efficiency in the work was three times that of the boy. What are the daily wages of the boy? (S.S.C., 2005)
Ratio of 1 day’s work of man and boy \(=3: 1\).
Total wages of the boy \(=₹\left(800 \times \frac{1}{4}\right)=₹ 200\).
\(\therefore \quad\) Daily wages of the boy \(=₹\left(\frac{200}{5}\right)=₹ 40\).
Two men undertake to do a piece of work for ₹ 1400 . The first man alone can do this work in 7 days while the second man alone can do this work in 8 days. If they working together complete this work in 3 days with the help of a boy, how should the money be divided? (M.A.T., 2007)
Boy’s 1 day’s work \(=\frac{1}{3}-\left(\frac{1}{7}+\frac{1}{8}\right)=\left(\frac{1}{3}-\frac{15}{56}\right)=\frac{11}{168}\).
\(\therefore \quad\) Ratio of wages of the first man, second man and boy \(=\frac{1}{7}: \frac{1}{8}: \frac{11}{168}=24: 21: 11\).
First man’s share \(=₹\left(\frac{24}{56} \times 1400\right)=₹ 600\);
Second man’s share \(=₹\left(\frac{21}{56} \times 1400\right)=₹ 525\);
Boy’s share \(=₹[1400-(600+525)]=₹ 275\).
A sum of money is sufficient to pay A’s wages for 21 days and B’s wages for 28 days. The same money is sufficient to pay the wages of both for : (ICET, 2005; G.B.O., 2007)
Let total money be \(₹ x\). A’s 1 day’s wages \(=₹ \frac{x}{21}\), B’s 1 day’s wages \(=₹ \frac{x}{28}\).
\(\therefore \quad(\mathrm{A}+\mathrm{B})^{\prime}\) ‘s 1 day’s wages \(=₹\left(\frac{x}{21}+\frac{x}{28}\right)=₹ \frac{x}{12}\).
\(\therefore \quad\) Money is sufficient to pay the wages of both for 12 days.
A can do a piece of work in 10 days; B in 15 days. They work for 5 days. The rest of the work was finished by \(C\) in 2 days. If they get \(₹ 1500\) for the whole work, the daily wages of \(B\) and \(C\) are (M.A.T., 2005)
Part of the work done by \(\mathrm{A}=\left(\frac{1}{10} \times 5\right)=\frac{1}{2}\).
Part of the work done by \(B=\left(\frac{1}{15} \times 5\right)=\frac{1}{3}\).
Part of the work done by \(C=1-\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{1}{6}\).
So, (A’s share \()\) : (B’s share \():\left(C^{\prime}\right.\) s share \()=\frac{1}{2}: \frac{1}{3}: \frac{1}{6}=3: 2: 1\).
\(\therefore \quad\) A’s share \(=₹\left(\frac{3}{6} \times 1500\right)=₹ 750\),
B’s share \(=₹\left(\frac{2}{6} \times 1500\right)=₹ 500\),
\(C^{\prime}\) s share \(=₹\left(\frac{1}{6} \times 1500\right)=₹ 250\).
A’s daily wages \(=₹\left(\frac{750}{5}\right)=₹ 150\);
\(
\begin{aligned}
& \text { B’s daily wages }=₹\left(\frac{500}{5}\right)=₹ 100 ; \\
& \text { C’s daily wages }=₹\left(\frac{250}{2}\right)=₹ 125 .
\end{aligned}
\)
\(\therefore \quad\) Daily wages of \(B\) and \(C=₹(100+125)=₹ 225\).
The daily wages of a worker are \(₹ 100\). Five workers can do a work in 10 days. If you pay \(₹ 20\) more daily, they agree to do \(25 \%\) more work daily. If the proposal is accepted, then the total amount that could be saved is
5 workers’ 1 day’s work \(=\frac{1}{10}\).
5 workers’ 1 day’s work on increasing wages
\(
=125 \% \text { of } \frac{1}{10}=\left(\frac{125}{100} \times \frac{1}{10}\right)=\frac{1}{8} \text {. }
\)
So, now the work is done in 8 days.
Original wage bill \(=₹(100 \times 5 \times 10)=₹ 5000\).
New wage bill \(=₹(120 \times 5 \times 8)=₹ 4800\).
\(\therefore \quad\) Amount saved \(=₹(5000-4800)=₹ 200\).
A and B together can complete a work in 12 days. A alone can complete it in 20 days. If B does the work only for half a day daily, then in how many days A and B together will complete the work?
B’s 1 day’s work \(=\left(\frac{1}{12}-\frac{1}{20}\right)=\frac{2}{60}=\frac{1}{30}\).
Now, \((A+B)\) ‘s 1 day’s work \(=\left(\frac{1}{20}+\frac{1}{60}\right)=\frac{4}{60}=\frac{1}{15}\).
\([\because\) B works for half day only]
So, A and B together will complete the work in 15 days.
\(A, B\) and \(C\) completed a work costing \(₹ 1800\). A worked for 6 days, \(B\) for 4 days and \(C\) for 9 days. If their daily wages are in the ratio of \(5: 6: 4\), how much amount will be received by \(A\)? (S.S.C., 2007)
Let the daily wages of \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) be \(₹ 5 x, ₹ 6 x\) and \(₹ 4 x\) respectively.
Then, ratio of their amounts \(=(5 x \times 6):(6 x \times 4):(4 x \times\)
\(9)=30: 24: 36=5: 4: 6\)
\(\therefore \quad\) A’s amount \(=₹\left(1800 \times \frac{5}{15}\right)=₹ 600\).
\(A\) and \(B\) can complete a piece of work in 12 and 18 days respectively. A begins to do the work and they work alternatively one at a time for one day each. The whole work will be completed in (S.S.C., 2007)
\(
(\mathrm{A}+\mathrm{B})^{\prime} \text { s } 2 \text { days ‘ work }=\left(\frac{1}{12}+\frac{1}{18}\right)=\frac{5}{36}
\)
Work done in 7 pairs of days \(=\left(\frac{5}{36} \times 7\right)=\frac{35}{36}\).
Remaining work \(=\left(1-\frac{35}{36}\right)=\frac{1}{36}\).
On 15th day, it is A’s turn.
\(\frac{1}{12}\) work is done by \(\mathrm{A}\) in 1 day.
\(\frac{1}{36}\) work is done by \(\mathrm{A}\) in \(\left(12 \times \frac{1}{36}\right)=\frac{1}{3}\) day.
\(\therefore\) Total time taken \(=14 \frac{1}{3}\) days.
A, B and C can do a piece of work in 11 days, 20 days and 55 days respectively, working alone. How soon can the work be done if A is assisted by B and C on alternate days?
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{11}+\frac{1}{20}\right)=\frac{31}{220} \\
& (\mathrm{~A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{11}+\frac{1}{55}\right)=\frac{6}{55} \\
& \text { Work done in } 2 \text { days }=\left(\frac{31}{220}+\frac{6}{55}\right)=\frac{55}{220}=\frac{1}{4} .
\end{aligned}
\)
Now, \(\frac{1}{4}\) work is done in 2 days.
\(\therefore \quad\) Whole work will be done in \((2 \times 4)=8\) days.
A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
\(
\begin{aligned}
& \text { A’s } 2 \text { days’ work }=\left(\frac{1}{20} \times 2\right)=\frac{1}{10} . \\
& (\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{60}\right)=\frac{6}{60}=\frac{1}{10} . \\
& \text { Work done in } 3 \text { days }=\left(\frac{1}{10}+\frac{1}{10}\right)=\frac{1}{5} .
\end{aligned}
\)
Now, \(\frac{1}{5}\) work is done in 3 days.
\(\therefore \quad\) Whole work will be done in \((3 \times 5)=15\) days.
A can do a piece of work in 90 days, \(B\) in 40 days and \(C\) in 12 days. They work for a day each in turn i.e., first day \(A\) does it alone, \(B\) does it the second day and \(C\) the third day. After that \(A\) does it for another day, and so on. After finishing the work they get \(₹ 240\). If the wages are divided in proportion to the work done by them, find what each will get. (M.A.T., 2006)
\(
\begin{aligned}
& (A+B+C)^{\prime} \text { s } 3 \text { days’ work }=\frac{1}{90}+\frac{1}{40}+\frac{1}{12}=\frac{43}{360} . \\
& (A+B+C)^{\prime} \text { s } 24 \text { days’ } \text { work }=\frac{43}{360} \times 8=\frac{344}{360} . \\
& \text { Remaining work }=\left(1-\frac{344}{360}\right)=\frac{16}{360}=\frac{4}{90} .
\end{aligned}
\)
On 25th day, it is A’s turn.
A’s 1 day’s work \(=\frac{1}{90}\).
Remaining work \(=\left(\frac{4}{90}-\frac{1}{90}\right)=\frac{3}{90}=\frac{1}{30}\).
On 26th day, it is B’s turn.
B’s 1 day’s work \(=\frac{1}{40}\).
Remaining work \(=\left(\frac{1}{30}-\frac{1}{40}\right)=\frac{1}{120}\).
On 27th day, it is C’s turn.
\(\frac{1}{12}\) work is done by \(C\) in 1 day.
\(\frac{1}{120}\) work is done by \(C\) in \(\left(12 \times \frac{1}{120}\right)=\frac{1}{10}\) day.
Hence, the whole work is completed in \(26 \frac{1}{10}\) days out of which A worked for 9 days, B worked for 9 days and \(C\) worked for \(8 \frac{1}{10}\) days.
Ratio of wages of \(\mathrm{A}, \mathrm{B}\) and \(C=\) Ratio of work done by A, B and C
\(
\begin{aligned}
& =\left(\frac{1}{90} \times 9\right):\left(\frac{1}{40} \times 9\right):\left(\frac{1}{12} \times 8 \frac{1}{10}\right)=\frac{1}{10}: \frac{9}{40}: \frac{27}{40} \\
& =4: 9: 27 \text {. } \\
& \text { A’s share }=₹\left(\frac{4}{40} \times 240\right)=₹ 24 \text {; } \\
& \text { B’s share }=₹\left(\frac{9}{40} \times 240\right)=₹ 54 \text {; } \\
& C^{\prime} \text { s share }=₹\left(\frac{27}{40} \times 240\right)=₹ 162 . \\
&
\end{aligned}
\)
\(A\) and \(B\) can finish a work, working on alternate days, in 19 days, when A works on the first day. However, they can finish the work, working on alternate days, in \(19 \frac{5}{6}\) days, when B works on the first day. How many days does \(\mathrm{A}\) alone take to finish the work?
Let the time taken by A alone and B alone to complete the work be \(x\) days and \(y\) days respectively.
When A works on the first day: In this case, A works for 10 days while B works for 9 days.
\(
\therefore \quad \frac{10}{x}+\frac{9}{y}=1 \dots(i)
\)
When B works on the first day:
In this case, \(A\) works for \(9 \frac{5}{6}\) days \(\left(=\frac{59}{6}\right.\) days \()\) while \(B\) works for 10 days.
\(
\therefore \quad \frac{59}{6 x}+\frac{10}{y}=1 \dots(ii)
\)
Multiplying (i) by 10 and (ii) by 9 and then subtracting, we get: \(x=\frac{23}{2}=11 \frac{1}{2}\).
Hence, A alone takes \(11 \frac{1}{2}\) days to complete the work.
A and B can separately do a piece of work in 20 and 15 days respectively. They worked together for 6 days, after which B was replaced by C. If the work was finished in next 4 days, then the number of days in which C alone could do the work will be
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 6 \text { days’ work }=6\left(\frac{1}{20}+\frac{1}{15}\right)=\frac{7}{10} ; \\
& (\mathrm{A}+\mathrm{C})^{\prime} \text { s } 4 \text { days’ work }=\left(1-\frac{7}{10}\right)=\frac{3}{10} ; \\
& (\mathrm{A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{3}{40} . \text { A’s } 1 \text { day’s work }=\frac{1}{20} . \\
& \therefore \quad \text { C’s } 1 \text { day’s work }=\left(\frac{3}{40}-\frac{1}{20}\right)=\frac{1}{40} .
\end{aligned}
\)
Hence, \(\mathrm{C}\) alone can finish the work in 40 days.
A, B and C can do a piece of work in 36, 54 and 72 days respectively. They started the work but A left 8 days before the completion of the work while B left 12 days before the completion. The number of days for which C worked is
Suppose the work was finished in \(x\) days.
Then, \(\mathrm{A}^{\prime}\) s \((x-8)\) days’ work \(+\mathrm{B}^{\prime}\) s \((x-12)\) days’ work + \(C^{\prime}\) s \(x\) days \(^{\prime}\) work \(=1\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{(x-8)}{36}+\frac{(x-12)}{54}+\frac{x}{72}=1 \\
& \Leftrightarrow \quad 6(x-8)+4(x-12)+3 x=216 . \\
& \therefore \quad 13 x=312 \text { or } x=24 .
\end{aligned}
\)
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