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45 men can complete a work in 16 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work?
\((45 \times 16)\) men can complete the work in 1 day.
\(
\begin{aligned}
& \therefore 1 \text { man’s } 1 \text { day’s work }=\frac{1}{720} .45 \text { men’s } 1 \text { days’ work }=\frac{45}{720}=\frac{1}{16} \\
& 45 \text { men’s } 6 \text { days’ work }=\left(\frac{1}{16} \times 6\right)=\frac{3}{8} . \text { Remaining work }=\left(1-\frac{3}{8}\right)=\frac{5}{8} . \\
& 75 \text { men’s } 1 \text { day’s work }=\frac{75}{720}=\frac{5}{48} .
\end{aligned}
\)
Now, \(\frac{5}{48}\) work is done by them in 1 day.
\(\therefore \quad \frac{5}{8}\) work is done by them in \(\left(\frac{48}{5} \times \frac{5}{8}\right)=6\) days.
10 persons begin to work together on a job but after some days 4 persons leave. As a result, the job which could have been completed in 40 days is completed in 50 days. How many days after the commencement of the work did the a persons leave? (S.S.C., 2004)
10 persons can complete the work in 40 days.
\(\therefore 1\) person’s 1 day’s work \(=\frac{1}{40 \times 10}=\frac{1}{400}\).
Suppose 4 persons left after \(x\) days.
Then, \(\frac{1}{400} \times 10 \times x+\frac{1}{400} \times 6 \times(50-x)=1 \Rightarrow \frac{1}{40} x+\frac{3}{200}(50-x)=1\) \(\Rightarrow \frac{1}{40} x+\frac{3}{4}-\frac{3}{200} x=1 \Rightarrow \frac{1}{100} x=\frac{1}{4} \Rightarrow x=25\).
Hence, 4 persons left 25 days after the commencement of the work.
9 children can complete a piece of work in 360 days. 18 men can complete the same piece of work in 72 days
and 12 women can complete it in 162 days. In how many days can 4 men, 12 women, and 10 children together
complete the piece of work? (Bank. P.O., 2006)
\(
\begin{aligned}
& 1 \text { man’s } 1 \text { day’s work }=\frac{1}{72 \times 18}=\frac{1}{1296} \text {. } \\
& 1 \text { woman’s } 1 \text { day’s work }=\frac{1}{162 \times 12}=\frac{1}{1944} . \\
& 1 \text { child’s } 1 \text { day’s work }=\frac{1}{360 \times 9}=\frac{1}{3240} . \\
& \begin{aligned}
(4 \text { men }+12 \text { women }+10 \text { children)’s } 1 \text { day’s work } \\
\qquad=\left(4 \times \frac{1}{1296}+12 \times \frac{1}{1944}+10 \times \frac{1}{3240}\right)=\left(\frac{1}{324}+\frac{1}{162}+\frac{1}{324}\right)=\frac{4}{324}=\frac{1}{81} .
\end{aligned}
\end{aligned}
\)
Hence, 4 men, 12 women and 10 children can complete the work in 81 days.
2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?
Let 1 man’s 1 day’s work \(=x\) and 1 boy’s 1 day’s work \(=y\).
Then, \(2 x+3 y=\frac{1}{10}\) and \(3 x+2 y=\frac{1}{8}\).
Solving, we get : \(x=\frac{7}{200}\) and \(y=\frac{1}{100}\).
\(
\therefore \quad\left(2 \text { men }+1 \text { boy)’s } 1 \text { day’s work }=\left(2 \times \frac{7}{200}+1 \times \frac{1}{100}\right)=\frac{16}{200}=\frac{2}{25}\right. \text {. }
\)
So, 2 men and 1 boy together can finish the work in \(\frac{25}{2}=12 \frac{1}{2}\) days.
3 men and 4 women can earn \(₹ 3780\) in 7 days. 11 men and 13 women can earn \(₹ 15040\) in 8 days. In what time will 7 men and 9 women earn \(₹ 12400\)?
Let 1 man’s 1 day’s earning be \(₹ x\) and 1 woman’s 1 day’s earning be \(₹ y\).
Then, \(3 x+4 y=\frac{3780}{7}=540 \dots(i)\)
And, \(11 x+13 y=\frac{15040}{8}=1880 \dots(ii)\)
Solving (i) and (ii), we get \(x=100, y=60\).
\((7\) men +9 women)’s 1 day’s earning \(=₹(7 \times 100+9 \times 60)=₹ 1240\).
Hence, required time \(=\left(\frac{12400}{1240}\right)\) days \(=10\) days.
Ayesha can complete a piece of work in 16 days. Amita can complete the same piece of work in 8 days. If both of them work together in how many days can they complete the same piece of work? (Bank P.O., 2010)
Ayesha’s 1 day’s work \(=\frac{1}{16}\). Amita’s 1 day’s work \(=\frac{1}{8}\). (Ayesha + Amita)’s 1 day’s work \(=\left(\frac{1}{16}+\frac{1}{8}\right)=\frac{3}{16}\).
\(\therefore \quad\) Both together can complete the work in \(\frac{16}{3}=5 \frac{1}{3}\) days.
A can complete a certain work in 4 minutes, \(B\) in 5 minutes, \(C\) in 6 minutes, \(D\) in 10 minutes, and \(E\) in 12 minutes. The average number of units of work completed by them per minute will be (P.C.S., 2009)
Required average
\(
=\frac{\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{10}+\frac{1}{12}\right)}{5}=\left(\frac{48}{60} \times \frac{1}{5}\right)=\frac{4}{25}=0.16 .
\)
A can finish a work in 18 days and B can do the same work in half the time taken by \(A\). Then, working together, what part of the same work they can finish in a day? (S.S.C., 2002)
\(
\text { A’s } 1 \text { day’s work }=\frac{1}{18} \text { and B’s } 1 \text { day’s work }=\frac{1}{9}
\)
\(
\therefore \quad(A+B)^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{18}+\frac{1}{9}\right)=\frac{1}{6} \text {. }
\)
A tyre has two punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat ?
1 minute’s work of both the punctures \(=\left(\frac{1}{9}+\frac{1}{6}\right)=\frac{5}{18}\). So, both the punctures will make the tyre flat in \(\frac{18}{5}=3 \frac{3}{5} \mathrm{~min}\)
A can knit a pair of socks in 3 days. B can knit the same pair in 9 days. If they are knitting together, then in how many days will they knit two pairs of socks? (R.R.B., 2004)
Number of pairs knit by A and B together in 1 day
\(
=\left(\frac{1}{3}+\frac{1}{9}\right)=\frac{4}{9} \text {. }
\)
\(\therefore \quad\) Required number of days \(=\left(2 \div \frac{4}{9}\right)=\left(2 \times \frac{9}{4}\right)=\frac{9}{2}=4 \frac{1}{2}\)
A can complete a work in 6 days while B can complete the same work in 12 days. If they work together and complete it, the portion of the work done by \(A\) is
\(
(A+B)^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{6}+\frac{1}{12}\right)=\frac{3}{12}=\frac{1}{4} \text {. }
\)
\(\therefore[latex] Both [latex]\mathrm{A}\) and B together can complete the work in 4 days.
Part of the work done by \(\mathrm{A}=\left(\frac{1}{6} \times 4\right)=\frac{2}{3}\).
A can do a piece of work in 8 days and \(B\) can do the same piece of work in 12 days. \(A\) and \(B\) together complete the same piece of work and get \(₹ 200\) as the combined wages. B’s share of the wages will be
\(
\begin{aligned}
& \text { A’s share : B’s share }=\text { Ratio of their } 1 \text { day’s work } \\
& \qquad=\frac{1}{8}: \frac{1}{12}=3: 2 . \\
& \therefore \quad \text { B’s share }=₹\left(200 \times \frac{2}{5}\right)=₹ 80 .
\end{aligned}
\)
George takes 8 hours to copy a 50-page manuscript while Sonia can copy the same manuscript in 6 hours. How many hours would it take them to copy a 100-page manuscript, if they work together? (M.A.T., 2005)
Number of pages typed by George in 1 hour \(=\frac{50}{8}=\frac{25}{4}\). Number of pages typed by Sonia in 1 hour \(=\frac{50}{6}=\frac{25}{3}\). Number of pages typed by George and Sonia together in
\(
\begin{aligned}
& \begin{array}{l}
1 \text { hour }=\left(\frac{25}{4}+\frac{25}{3}\right)=\left(\frac{75+100}{12}\right)=\frac{175}{12} . \\
\therefore \quad \text { Required time }=\left(100 \div \frac{175}{12}\right) \mathrm{hrs}=\left(\frac{100 \times 12}{175}\right) \mathrm{hrs} \\
=\frac{48}{7} \mathrm{hrs}=6 \frac{6}{7} \mathrm{hrs} .
\end{array}
\end{aligned}
\)
A and B together complete a piece of work in \(T\) days. If \(A\) alone completes the work in \(T+3\) days and \(B\) alone completes the piece of work in \(T+12\) days, what is \(T\)? (S.B.I.P.O., 2008)
\(
\begin{aligned}
& \text { A’s } 1 \text { day’s work }=\frac{1}{T+3} \\
& \text { B’s } 1 \text { day’s work }=\frac{1}{T+12} \text {. } \\
& (\mathrm{A}+\mathrm{B}) \text { ‘s } 1 \text { day’s work }=\frac{1}{T} \text {. } \\
& \therefore \quad \frac{1}{T+3}+\frac{1}{T+12}=\frac{1}{T} \Rightarrow \frac{2 T+15}{(T+3)(T+12)}=\frac{1}{T} \\
& \Rightarrow 2 T^2+15 T=T^2+15 T+36 \\
& \Rightarrow T^2=36 \Rightarrow T=6 . \\
&
\end{aligned}
\)
Reena, Aastha, and Shloka can independently complete a piece of work in 6 hours, 4 hours, and 12 hours respectively. If they work together, how much time will they take to complete that piece of work? (Bank P.O., 2004)
Reena’s 1 hour’s work \(=\frac{1}{6}\)
Aastha’s 1 hour’s work \(=\frac{1}{4}\);
Shloka’s 1 hour’s work \(=\frac{1}{12}\).
\(
\begin{aligned}
& \text { (Reena + Aastha }+ \text { Shloka)’s } 1 \text { hour’s work } \\
& =\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{6}{12}=\frac{1}{2}
\end{aligned}
\)
Hence, Reena, Aastha and Shloka together take 2 hours to complete the work.
A man can do a job in 15 days. His father takes 20 days and his son finishes it in 25 days. How long will they take to complete the job if they all work together?
1 day’s work of the three persons \(=\left(\frac{1}{15}+\frac{1}{20}+\frac{1}{25}\right)=\frac{47}{300}\).
So, all the three together will complete the work in \(\frac{300}{47} \simeq 6.4\) days
Amit and Sumit can plough a field in 4 days. Sumit alone can plough the field in 6 days. In how many days will Amit alone plough the field? (R.R.B., 2006)
Amit’s 1 day’s work \(=\left(\frac{1}{4}-\frac{1}{6}\right)=\frac{1}{12}\)
\(\therefore \quad\) Amit alone can plough the field in 12 days.
Two spinning machines \(A\) and \(B\) can together produce \(3,00,000\) metres of cloth in 10 hours. If machine B alone can produce the same amount of cloth in 15 hours, then how much cloth can machine A produce alone in 10 hours? (M.A.T., 2005)
Length of cloth produced by A and B in 10 hrs
\(
=3,00,000 \mathrm{~m}
\)
Length of cloth produced by B in 10 hrs \(=\left(\frac{300000}{15} \times 10\right) \mathrm{m}\) \(=200000 \mathrm{~m}\)
\(\therefore \quad\) Length of cloth produced by A in 10 hrs \(=(300000-200000) \mathrm{m}=100000 \mathrm{~m}\)
\(X, Y\) and \(Z\) complete a work in 6 days. \(X\) or \(Y\) alone can do the same work in 16 days. In how many days \(\mathrm{Z}\) alone can finish the same work?(R.R.B., 2006)
\(
\begin{aligned}
& (X+Y)^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{16}+\frac{1}{16}\right)=\frac{2}{16}=\frac{1}{8} . \\
& \text { Z’s } 1 \text { day’s work }=(X+Y+Z)^{\prime} \text { s } 1 \text { day’s work }-(X+ \\
& Y \text { ‘s } 1 \text { day’s work }=\frac{1}{6}-\frac{1}{8}=\frac{1}{24} .
\end{aligned}
\)
\(\therefore \quad \mathrm{Z}\) alone can finish the work in 24 days.
A can lay railway track between two given stations in 16 days and \(B\) can do the same job in 12 days. With the help of \(C\), they did the job in 4 days only. Then, \(C\) alone can do the job in: (S.S.C., 2003)
\(
(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{4}
\)
A’s 1 day’s work \(=\frac{1}{16}\),
B’s 1 day’s work \(=\frac{1}{12}\).
\(\therefore \quad\) C’s 1 day’s work \(=\frac{1}{4}-\left(\frac{1}{16}+\frac{1}{12}\right)=\left(\frac{1}{4}-\frac{7}{48}\right)=\frac{5}{48}\).
So, \(C\) alone can do the work in \(\frac{48}{5}=9 \frac{3}{5}\) days.
A can complete \(\frac{1}{3}\) of a work in 5 days and B, \(\frac{2}{5}\) of the work in 10 days. In how many days both A and B together can complete the work? (S.S.C., 2010; P.C.S., 2009)
Whole work will be done by \(A\) in \((5 \times 3)=15\) days.
Whole work will be done by B in \(\left(10 \times \frac{5}{2}\right)=25\) days.
A’s 1 day’s work \(=\frac{1}{15} ;\) B’s 1 day’s work \(=\frac{1}{25}\).
\((\mathrm{A}+\mathrm{B})\) ‘s 1 day’s work \(=\left(\frac{1}{15}+\frac{1}{25}\right)=\frac{16}{150}=\frac{8}{75}\).
\(\therefore \quad \mathrm{A}\) and \(\mathrm{B}\) together can complete the work in \(\frac{75}{8}=9 \frac{3}{8}\) days.
\(X\) can do \(\frac{1}{4}\) of a work in 10 days, \(Y\) can do \(40 \%\) of the work in 40 days and \(Z\) can do \(\frac{1}{3}\) of the work in 13 days. Who will complete the work first?
Whole work will be done by \(X\) in \((10 \times 4)=40\) days.
Whole work will be done by \(Y\) in \(\left(40 \times \frac{100}{40}\right)=100\) days.
Whole work will be done by \(\mathrm{Z}\) in \((13 \times 3)=39\) days.
\(\therefore \quad \mathrm{Z}\) will complete the work first.
A man and a boy together can do a certain amount of digging in 40 days. Their speeds in digging are in the ratio of 8: 5. How many days will the boy take to complete the work if engaged alone? (R.R.B., 2005)
Ratio of digging speeds of man and boy \(=8: 5\).
Ratio of times taken by man and boy \(=5: 8\).
Suppose the man takes \(5 x\) days while the boy takes \(8 x\) days to complete the work alone.
Then, \(\frac{1}{5 x}+\frac{1}{8 x}=\frac{1}{40} \Rightarrow \frac{13}{40 x}=\frac{1}{40} \Rightarrow x=13\).
Hence, time taken by the boy to complete the work alone \(=(8 \times 13)\) days \(=104\) days
A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in: (S.S.C., 2002)
Suppose A, B and C take \(x, \frac{x}{2}\) and \(\frac{x}{3}\) days respectively to finish the work.
Then, \(\left(\frac{1}{x}+\frac{2}{x}+\frac{3}{x}\right)=\frac{1}{2} \Rightarrow \frac{6}{x}=\frac{1}{2} \Rightarrow x=12\).
So, B takes 6 days to finish the work.
Work done by A in one day is half of the work done by B in one day. Work done by B is half of the work done by C in one day. If C alone can complete the work in 7 days, in how many days can A, B, and C together complete the work? (S.B.I.P.O., 2008)
\(
\begin{aligned}
& \text { C’s } 1 \text { day’s work }=\frac{1}{7} ; \text { B’s } 1 \text { day’s work }=\left(\frac{1}{2} \times \frac{1}{7}\right)=\frac{1}{14} \\
& \text { A’s } 1 \text { day’s work }=\left(\frac{1}{2} \times \frac{1}{14}\right)=\frac{1}{28} \\
& \therefore \quad(A+B+C)^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{28}+\frac{1}{14}+\frac{1}{7}\right)=\frac{7}{28}=\frac{1}{4}
\end{aligned}
\)
Hence, A, B and C together can complete the work in 4 days.
Rosa can eat 32 rosogollas in one hour. Her sister Lila needs three hours to eat the same number. How much time will they take to eat 32 rosogollas together? (R.R.B., 2005)
Number of rosogollas eaten by Rosa in 1 minute \(=\frac{32}{60}\).
Number of rosogollas eaten by Lila in 1 minute \(=\frac{32}{180}\).
Number of rosogollas eaten by Rosa and Lila together in 1 minute \(=\left(\frac{32}{60}+\frac{32}{180}\right)=\frac{128}{180}\).
\(\therefore\) Required time \(=\left(32 \div \frac{128}{180}\right)=\left(\frac{32 \times 180}{128}\right) \mathrm{min}=45 \mathrm{~min}\).
A conveyor belt delivers baggage at the rate of 3 tons in 5 minutes and a second conveyor belt delivers baggage at the rate of 1 ton in 2 minutes. How much time will it take to get 33 tons of baggage delivered using both the conveyor belts together? (P.C.S., 2006)
Baggage delivered by first belt in \(1 \mathrm{~min}=\left(\frac{3}{5}\right)\) tons.
Baggage delivered by second belt in \(1 \mathrm{~min}=\left(\frac{1}{2}\right)\) ton.
Baggage delivered by both belts in \(1 \mathrm{~min}\)
\(
\begin{aligned}
& =\left(\frac{3}{5}+\frac{1}{2}\right) \text { ton }=\frac{11}{10} \text { tons. } \\
\therefore \quad \text { Required time } & =\left(33 \div \frac{11}{10}\right) \mathrm{min}=\left(\frac{33 \times 10}{11}\right) \mathrm{min} \\
& =30 \mathrm{~min} .
\end{aligned}
\)
A manufacturer builds a machine that will address 500 envelopes in 8 minutes. He wishes to build another machine so that when both are operating together they will address 500 envelopes in 2 minutes. The equation used to find how many minutes x it would require the second machine to address 500 envelopes alone, is (M.B.A., 2011)
Number of envelopes addressed by first machine in 1 \(\min =\frac{500}{8}\)
Number of envelopes addressed by second machine in 1 \(\min =\frac{500}{x}\)
Number of envelopes addressed by both machines in 1 \(\min =\frac{500}{2}\)
\(
\therefore \quad \frac{500}{8}+\frac{500}{x}=\frac{500}{2} \Rightarrow \frac{1}{8}+\frac{1}{x}=\frac{1}{2} .
\)
Computer A takes 3 minutes to process an input while computer B takes 5 minutes. If computers A, B and C can process an average of 14 inputs in one hour, how many minutes does computer C alone take to process one input? (M.C.A., 2009)
Number of units processed by computer \(\mathrm{A}\) in \(1 \mathrm{~min}\) \(=\frac{1}{3}\)
Number of units processed by computer B in \(1 \mathrm{~min}=\frac{1}{5}\).
Number of units processed by A, B and C in 1 min \(=\frac{14 \times 3}{60}=\frac{7}{10}\).
\(\therefore \quad\) Number of units processed by computer \(C\) in \(1 \mathrm{~min}\) \(=\frac{7}{10}-\left(\frac{1}{3}+\frac{1}{5}\right)=\frac{7}{10}-\frac{8}{15}=\frac{5}{30}=\frac{1}{6}\).
Hence, computer \(C\) takes 6 minutes to process one input alone.
Bob and David are two typists. One afternoon, they were each given 40 pages for typing. They divided the work equally but David finished 20 minutes before Bob who took 2 hours for the same. The next afternoon, they were again given 77 pages to type. However, this time they decided to divide the work such that they finished typing simultaneously. How many pages did Bob have to type?
Time taken by Bob to type 20 pages \(=2 \mathrm{hrs}=120 \mathrm{~min}\). Time taken by David to type 20 pages \(=1 \mathrm{hr} 40 \mathrm{~min}=\) \(100 \mathrm{~min}\).
Time taken by Bob to type 1 page \(=\left(\frac{120}{20}\right) \mathrm{min}=6 \mathrm{~min}\). Time taken by David to type 1 page \(=\left(\frac{100}{20}\right) \mathrm{min}=5 \mathrm{~min}\).
Bob’s 1 minute’s work \(=\frac{1}{6}\)
David’s 1 minute’s work \(=\frac{1}{5}\).
\(\therefore \quad\) Ratio of division of work \(=\frac{1}{6}: \frac{1}{5}=5: 6\).
Hence, number of pages to be typed by Bob
\(
=\left(77 \times \frac{5}{11}\right)=35 .
\)
\(P, Q\), and \(R\) are three typists who working simultaneously can type 216 pages in 4 hours. In one hour, \(R\) can type as many pages more than \(Q\) as \(Q\) can type more than \(P\). During a period of five hours, \(R\) can type as many pages as \(P\) can during seven hours. How many pages does each of them type per hour? (M.A.T., 2005)
Let the number of pages typed in one hour by \(P, Q\) and \(\mathrm{R}\) be \(x, y\) and \(z\) respectively.
Then, \(x+y+z=\frac{216}{4}\)
\(
\begin{aligned}
\Rightarrow & x+y+z=54 z \\
& z-y=y-x \\
\Rightarrow & 2 y=x+z \\
& 5 z=7 x \\
\Rightarrow & x=\frac{5}{7} z
\end{aligned}
\)
Solving (i), (ii) and (iii), we get \(x=15, y=18, z=21\).
Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages? (SCMHRD, 2002)
Number of pages typed by Ronald in 1 hour \(=\frac{32}{6}=\frac{16}{3}\).
Number of pages typed by Elan in 1 hour \(=\frac{40}{5}=8\).
Number of pages typed by both in 1 hour \(=\left(\frac{16}{3}+8\right)=\frac{40}{3}\).
\(\therefore \quad\) Time taken by both to type 110 pages
\(
=\left(110 \times \frac{3}{40}\right) \mathrm{hrs}=8 \frac{1}{4} \mathrm{hrs}=8 \mathrm{hrs} 15 \mathrm{~min} .
\)
Cloth Makers Inc. has \(p\) spindles, each of which can produce \(q\) metres of cloth on an average in \(r\) minutes. If the spindles are made to run with no interruption, then how many hours will it take for 20,000 metres of cloth to be produced?
Length of cloth produced in 1 hour
\(
=\left(\frac{p q}{r} \times 60\right) \mathrm{m}=\left(\frac{60 p q}{r}\right) \mathrm{m} .
\)
\(
\therefore \quad \text { Required time }=\left(20000 \div \frac{60 p q}{r}\right) \mathrm{hrs}=\left(\frac{20000 r}{60 p q}\right) \mathrm{hrs}
\)
Two workers \(A\) and \(B\) are engaged to do a work. A working alone takes 8 hours more to complete the job than if both worked together. If \(B\) worked alone, he would need \(4 \frac{1}{2}\) hours more to complete the job than they both working together. What time would they take to do the work together? (M.A.T., 2010)
Let \(\mathrm{A}\) and \(\mathrm{B}\) together take \(x\) hours to complete the work. Then, A alone takes \((x+8)\) hrs and B alone takes \(\left(x+\frac{9}{2}\right)\) hrs to complete the work.
\(
\begin{aligned}
& \text { Then, } \frac{1}{(x+8)}+\frac{1}{\left(x+\frac{9}{2}\right)}=\frac{1}{x} \\
& \Rightarrow \quad \frac{1}{(x+8)}+\frac{2}{(2 x+9)}=\frac{1}{x} \\
& \Rightarrow \quad x(4 x+25)=(x+8)(2 x+9) \\
& \Rightarrow \quad 2 x^2=72 \Rightarrow x^2=36 \Rightarrow x=6 .
\end{aligned}
\)
Three friends Anne, Bob and Chris work together to do a certain job. The time it takes them to do the work together is 6 hours less than Anne would have take alone, 1 hour less than Bob would have taken alone and half the time Chris would have taken working alone. How long did it take them to complete the job, working together? (M.B.A., 2010)
Let the time taken by the three friends together to do the work be \(x\) hours.
Then, time taken by Anne alone \(=(x+6) \mathrm{hrs}\); time taken by Bob alone \(=(x+1)\) hrs;
time taken by Chris alone \(=2 x\) hrs.
\(
\begin{aligned}
& \therefore \quad \frac{1}{x+6}+\frac{1}{x+1}+\frac{1}{2 x}=\frac{1}{x} \\
& \Rightarrow \quad \frac{2 x(x+1)+2 x(x+6)+(x+1)(x+6)}{2 x(x+6)(x+1)}=\frac{1}{x} \\
& \Rightarrow \quad 5 x^2+21 x+6=2\left(x^2+7 x+6\right)
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow 3 x^2+7 x-6=0 \Rightarrow(x+3)(3 x-2)=0 \Rightarrow & x=\frac{2}{3} \\
& {[\because x \neq-3] }
\end{aligned}
\)
\(
\therefore \quad \text { Required time }=\frac{2}{3} \mathrm{hrs}=\left(\frac{2}{3} \times 60\right) \mathrm{min}=40 \mathrm{~min} .
\)
To do a piece of work, B takes 3 times as long as A and C together and C twice as long as A and B together. If A, B and C together can complete the work in 10 days, how long would A take alone to complete it?
\(
\begin{aligned}
& 3 \text { (B’s } 1 \text { day’s work })=(A+C)^{\prime} \text { s } 1 \text { day’s work } \\
& \Rightarrow \quad 4\left(B^{\prime} \text { s } 1 \text { day’s work }\right)=(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work } \\
& \Rightarrow \quad \text { B’s } 1 \text { day’s work }=\left(\frac{1}{4} \times \frac{1}{10}\right)=\frac{1}{40} . \\
& \quad 2\left(C^{\prime} \text { s } 1 \text { day’s work }\right)=(\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work } \\
& \Rightarrow \quad 3\left(C^{\prime} \text { s } 1 \text { day’s work }\right)=(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work } \\
& \Rightarrow \quad \text { C’s } 1 \text { day’s work }=\left(\frac{1}{3} \times \frac{1}{10}\right)=\frac{1}{30} . \\
& \therefore \quad \text { A’s } 1 \text { day’s work }=(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }-(B \\
& +C)^{\prime} \text { s } 1 \text { day’s work } \\
& \quad=\frac{1}{10}-\left(\frac{1}{40}+\frac{1}{30}\right)=\frac{1}{10}-\frac{7}{120}=\frac{5}{120}=\frac{1}{24} .
\end{aligned}
\)
Hence, A alone would take 24 days to complete the work.
P can complete a work in 12 days working 8 hours a day. \(Q\) can complete the same work in 8 days working 10 hours a day. If both \(P\) and \(Q\) work together, working 8 hours a day, in how many days can they complete the work?
P can complete the work in \((12 \times 8)\) hrs \(=96\) hrs. \(Q\) can complete the work in \((8 \times 10) \mathrm{hrs}=80 \mathrm{hrs}\).
\(\therefore \quad[latex] P’s 1 hour’s work [latex]=\frac{1}{96}\)
and \(Q\) ‘s 1 hour’s work \(=\frac{1}{80}\).
\((P+Q)\) ‘s 1 hour’s work \(=\left(\frac{1}{96}+\frac{1}{80}\right)=\frac{11}{480}\).
So, both \(P\) and \(Q\) will finish the work in \(\left(\frac{480}{11}\right)\) hrs.
\(\therefore \quad\) Number of days of 8 hours each
\(
=\left(\frac{480}{11} \times \frac{1}{8}\right)=\frac{60}{11} \text { days }=5 \frac{5}{11} \text { days. }
\)
A and B together can complete a work in 12 days, \(B\) and \(C\) together can complete the same work in 8 days and \(A\) and \(C\) together can complete it in 16 days. In total, how many days do \(A, B\), and \(C\) together take to complete the same work? (Bank P.O., 2009)
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{12} \\
& (\mathrm{~B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{8} \\
& (\mathrm{~A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{16}
\end{aligned}
\)
Adding, we get: \(2(\mathrm{~A}+\mathrm{B}+\mathrm{C})^{\prime}\) s 1 day’s work
\(
=\left(\frac{1}{12}+\frac{1}{8}+\frac{1}{16}\right)=\frac{13}{48} \text {. }
\)
\(\therefore \quad(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime}\) s 1 day’s work \(=\frac{13}{96}\).
So, A, B, and C together can complete the work in \(\frac{96}{13}=7 \frac{5}{13}\) days.
A can do a piece of work in 4 hours, B and C together in 3 hours, and A and C together in 2 hours. How long will B alone take to do it? (S.S.C., 2005)
\(
\begin{aligned}
& \text { A’s } 1 \text { hour’s work }=\frac{1}{4} ; \\
& (\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { hour’s work }=\frac{1}{3} ; \\
& (\mathrm{A}+\mathrm{C})^{\prime} \text { s } 1 \text { hour’s work }=\frac{1}{2} \text {. } \\
& (\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { hour’s work }=\frac{1}{4}+\frac{1}{3}=\frac{7}{12} . \\
& \therefore \quad \text { B’s } 1 \text { hour’s work }=(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { hour’s work }-(\mathrm{A} \\
& +\mathrm{C})^{\prime} \text { s } 1 \text { hour’s work }=\frac{7}{12}-\frac{1}{2}=\frac{1}{12} .
\end{aligned}
\)
So, B alone can complete the work in 12 hours.
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in (S.S.C., 2006)
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{6} ; \\
& (A+B) ‘ s 1 \text { day’s work }=\frac{1}{8} \\
& (\mathrm{~B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{12} \text {. } \\
& \therefore \quad(\mathrm{A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work } \\
& =\left(2 \times \frac{1}{6}\right)-\left(\frac{1}{8}+\frac{1}{12}\right)=\left(\frac{1}{3}-\frac{5}{24}\right)=\frac{3}{24}=\frac{1}{8} . \\
&
\end{aligned}
\)
So, A and \(C\) together will do the work in 8 days.
A and B together can do a job in 2 days; B and \(C\) can do it in 4 days; \(A\) and \(C\) in \(2 \frac{2}{5}\) days. The number of days required for \(A\) to do the job alone is (M.B.A., 2011)
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{2} \\
& (\mathrm{~B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{4} ; \\
& (\mathrm{A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{5}{12} .
\end{aligned}
\)
Adding, we get: \(2(A+B+C)^{\prime}\) s 1 day’s work
\(
\begin{aligned}
& =\left(\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\right)=\frac{14}{12}=\frac{7}{6} \\
\Rightarrow \quad & (\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{7}{12} .
\end{aligned}
\)
So, A’s 1 day’s work \(=\left(\frac{7}{12}-\frac{1}{4}\right)=\frac{4}{12}=\frac{1}{3}\).
\(
\therefore \quad \text { A alone can do the work in } 3 \text { days. }
\)
\(A\) and \(B\) can do a piece of work in 5 days; \(B\) and \(C\) can do it in 7 days; \(A\) and \(C\) can do it in 4 days. Who among these will take the least time if put to do it alone?
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{5} \\
& (\mathrm{~B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{7} \\
& (\mathrm{~A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{4} .
\end{aligned}
\)
Adding, we get: \(2(A+B+C)\) ‘s 1 day’s work
\(
\begin{aligned}
& \quad=\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{4}\right)=\frac{83}{140} . \\
& (\mathrm{A}+\mathrm{B}+\mathrm{C}) \text { ‘s } 1 \text { day’s work }=\frac{83}{280} . \\
& \text { A’s } 1 \text { day’s work }=\left(\frac{83}{280}-\frac{1}{7}\right)=\frac{43}{280} ; \\
& \text { B’s } 1 \text { day’s work }=\left(\frac{83}{280}-\frac{1}{4}\right)=\frac{13}{280} ; \\
& \text { C’s } 1 \text { day’s work }=\left(\frac{83}{280}-\frac{1}{5}\right)=\frac{27}{280} .
\end{aligned}
\)
Thus time taken by A, B, C is \(\frac{280}{43}\) days, \(\frac{280}{13}\) days, \(\frac{280}{27}\) days respectively.
Clearly, the time taken by \(\mathrm{A}\) is least.
A and B can do a piece of work in 12 days, B and C in 8 days and C and A in 6 days. How long would B take to do the same work alone? (S.S.C., 2007)
\(
\begin{aligned}
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{12} \\
& (\mathrm{~B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{8} \\
& (\mathrm{~A}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{6} \\
& {[(\mathrm{~A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }+(\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }]-(\mathrm{A}} \\
& +\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{12}+\frac{1}{8}-\frac{1}{6}=\frac{1}{24} . \\
& \Rightarrow \quad 2\left(\mathrm{~B}^{\prime} \text { s } 1 \text { day’s work }\right)=\frac{1}{24} \\
& \Rightarrow \quad \text { B’s } 1 \text { day’s work }=\frac{1}{48} .
\end{aligned}
\)
Hence, \(B\) alone can do the work in 48 days.
A can build a wall in the same time in which B and C together can do it. If A and B together could do it in 25 days and C alone in 35 days, in what time could B alone do it? (N.M.A.T., 2008)
\(
(\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{25} ; \text { C’s } 1 \text { day’s work }=\frac{1}{35} .
\)
\(
(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime} \text { s } 1 \text { day’s work }=\left(\frac{1}{25}+\frac{1}{35}\right)=\frac{12}{175} \dots(i)
\)
\(
\text { Also, A’s } 1 \text { day’s work }=(B+C)^{\prime} \text { s } 1 \text { day’s work. } \dots(ii)
\)
From (i) and (ii), we get: \(2 \times\left(\right.\) A’s \( 1\) day’s work \()=\frac{12}{175}\)
\(
\begin{aligned}
& \Rightarrow \quad \text { A’s } 1 \text { day’s work }=\frac{6}{175} . \\
& \therefore \quad \text { B’s } 1 \text { day’s work }=\left(\frac{1}{25}-\frac{6}{175}\right)=\frac{1}{175} .
\end{aligned}
\)
Madhu takes twice as much time as Uma to complete a work and Rahul does it in the same time as Madhu and Uma together. If all three working together can finish the work in 6 days, then the time taken by Madhu to finish the work is (M.A.T., 2010)
Suppose Uma takes \(x\) days to complete a work. Then, Madhu takes \(2 x\) days to complete the work.
Uma’s 1 day’s work \(=\frac{1}{x}\);
\(
\begin{aligned}
& \text { Madhu’s } 1 \text { day’s work }=\frac{1}{2 x} . \\
& \begin{aligned}
& \text { Rahul’s } 1 \text { day’s work }=(\text { Madhu }+ \text { Uma)’s } 1 \text { day’s work } \\
&= \frac{1}{2 x}+\frac{1}{x}=\frac{3}{2 x} . \\
& \text { (Rahul }+ \text { Madhu }+ \text { Uma)’s } 1 \text { day’s work } \\
&=\frac{3}{2 x}+\frac{1}{2 x}+\frac{1}{x}=\frac{6}{2 x}=\frac{3}{x} . \\
& \therefore \quad \frac{3}{x}=\frac{1}{6} \Rightarrow x=18 .
\end{aligned}
\end{aligned}
\)
Hence, Madhu takes \((2 \times 18)=36\) days to complete the work.
A takes 5 days more than B to do a certain job and 9 days more than C; A and B together can do the job in the same time as C. How many days A would take to do it?
Suppose A takes \(x\) days to do the job alone.
Then, B takes \((x-5)\) days and C takes \((x-9)\) days.
\((\mathrm{A}+\mathrm{B})\) ‘s 1 day’s work \(=\) C’s 1 day’s work
\(
\begin{aligned}
\Rightarrow & \frac{1}{x}+\left(\frac{1}{x-5}\right)=\frac{1}{x-9} \Rightarrow \frac{(x-5)+x}{x(x-5)}=\frac{1}{(x-9)} \\
\Rightarrow & (2 x-5)(x-9)=x(x-5) \\
\Rightarrow & 2 x^2-23 x+45=x^2-5 x \\
\Rightarrow & x^2-18 x+45=0 \Rightarrow(x-3)(x-15)=0 \Rightarrow x=15 . \\
& {[\because x \neq 3] }
\end{aligned}
\)
Hence, A alone would take 15 days to do the job.
A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work is
Ratio of rates of working of \(\mathrm{A}\) and \(\mathrm{B}=2: 1\).
So, ratio of times taken \(=1: 2\).
\(
\begin{aligned}
& \therefore \quad \text { A’s } 1 \text { day’s work }=\frac{1}{6} ; \text { B’s } 1 \text { day’s work }=\frac{1}{12} . \\
& (\mathrm{A}+\mathrm{B}) \text { ‘s } 1 \text { day’s work }=\left(\frac{1}{6}+\frac{1}{12}\right)=\frac{3}{12}=\frac{1}{4} .
\end{aligned}
\)
So, A and B together can finish the work in 4 days.
A is twice as good a workman as B. If they work together, they can complete a job in 18 days. If A alone does the job, in how many days he will complete the job? (Hotel Management, 2010)
\(
\begin{aligned}
& \left(A^{\prime} \text { s } 1 \text { day’s work }\right):(B \text { ‘s } 1 \text { day’s work })=2: 1 . \\
& (\mathrm{A}+\mathrm{B})^{\prime} \text { s } 1 \text { day’s work }=\frac{1}{18} \\
& \therefore \quad \text { A’s } 1 \text { day’s work }=\left(\frac{1}{18} \times \frac{2}{3}\right)=\frac{1}{27} .
\end{aligned}
\)
Hence, \(A\) alone can finish the work in 27 days.
David and Michael together can finish a job in 4 days 19 hrs 12 min. If David works at two-thirds Michael’s speed, how long does it take Michael alone to finish the same job?
Total time taken by David and Michael together = 4 days \(19 \mathrm{hrs} 12 \mathrm{~min}=4\) days \(19 \frac{1}{5} \mathrm{hrs}\) \(=4\) days \(+\left(\frac{96}{5} \times \frac{1}{24}\right)\) days \(=4 \frac{4}{5}\) days \(=\frac{24}{5}\) days.
\(\left(\right.\) David + Michael)’s 1 day’s work \(=\frac{5}{24}\).
(David’s 1 day’s work) : (Michael’s 1 day’s work) \(=\frac{2}{3}: 1=2: 3\).
\(\therefore \quad\) Michael’s 1 day’s work \(=\left(\frac{5}{24} \times \frac{3}{5}\right)=\frac{1}{8}\).
Hence, Michael alone can finish the job in 8 days.
A is thrice as good a workman as B and so takes 60 days less than B for doing a job. The time in which they can do the job together is (A.A.O. Exam, 2010; M.A.T., 2010)
60 daysRatio of times taken by \(A\) and \(B=1: 3\).
If difference of time is 2 days, \(B\) takes 3 days.
If difference of time is 60 days, B takes \(\left(\frac{3}{2} \times 60\right)=90\) days.
So, A takes 30 days to do the work.
A’s 1 day’s work \(=\frac{1}{30}\); B’s 1 day’s work \(=\frac{1}{90}\).
\((A+B)\) ‘s 1 day’s work \(=\left(\frac{1}{30}+\frac{1}{90}\right)=\frac{4}{90}=\frac{2}{45}\).
\(\therefore \quad\) A and B together can do the work in \(\frac{45}{2}=22 \frac{1}{2}\) days.