Question 15 Marks
A can do a piece of work in $10$ days ; $B$ in $18$ days; and $A, B$, and $C$ together in $4$ days. In what time would $C$ alone do it?
Answer$A$ can do a piece of work in $=10$ days
$B$ can do a piece of work in $=18$ days
$(A+B+C)$ can do a piece of work in $=4$ days
$\therefore A\ 's\ 1-$ day work $=\frac{1}{10}$
$B\ 's \ 1-$ day work $=\frac{1}{18}$
$(A+B)\ 's \ 1-$ day work $=\frac{1}{10}+\frac{1}{18}=\frac{9+5}{90}$
$=\frac{14}{90}=\frac{7}{45}$
$(A+B+C) \ 's\ 1-$ day work $=\frac{1}{4}$
$\therefore C\ 's\ 1-$ day work $=\frac{1}{4}-\frac{7}{45}$
$=\frac{45-28}{180}=\frac{17}{180}$
$\therefore C$ can do the piece of work in $=\frac{180}{17}$ days
$=10 \frac{10}{17}$ days
View full question & answer→Question 25 Marks
A can finish a piece of work in $15$ days and $B$ can do it in $10$ days. They worked together for $2$ days and then $B$ goes away. In how many days will $A$ finish the remaining work?
Answer$A$ can finish a piece of work in $=15$ days
$B$ can finish a piece of work in $=10$ days
$\therefore A\ 's\ 1$ day work $=\frac{1}{15}$
$B\ 's\ 1-$day work $=\frac{1}{10}$
$(A+B)\ 's\ 1-$ day work$=\frac{1}{15}+\frac{1}{10}$
$=\frac{2+3}{30}=\frac{5}{30}=\frac{1}{6}$
$(A+B)\ 's \ 2$ days work $=\frac{1}{6} \times 2=\frac{1}{3}$
Remaining work which will be done by $A$ alone
$=1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$
$\therefore A$ can finish $1$ work in $=15$ days
$A$ can finish $\frac{2}{3}$ work in $=15 \times \frac{2}{3}$ days
$=\frac{30}{3}$ days $=10$ days
View full question & answer→Question 35 Marks
$A$ can do a piece of work in $20$ days and $B$ in $15$ days. They worked together on it for $6$ days and then $A$ left. How long will $B$ take to finish the remaining work?
AnswerGiven,
$A$ can do a piece of work in $=20$ days
$B$ can do a piece of work in $=15$ days
$\therefore A\ 's \ 1-$ day work $=\frac{1}{20}$
$B\ 's\ 1-$ day work $=\frac{1}{15}$
$(A+B)^{\prime} s\ 1 -$ day work $=\frac{1}{20}+\frac{1}{15}$
$=\frac{3+4}{60}=\frac{7}{60}$
$(A+B)\ 's\ 6$ days work $=\frac{7}{60} \times 6=\frac{7}{10}$
Remaining work $=1-\frac{7}{10}$
$=\frac{10-7}{10}=\frac{3}{10}$
$B$ can do $1$ work in $=15$ days
$B$ can do $\frac{3}{10}$ work in $=15 \times \frac{3}{10}$ days
$=\frac{45}{10}$ days $=\frac{9}{2}$ days $=4 \frac{1}{2}$ days
View full question & answer→Question 45 Marks
$A$ can do a piece of work in $24$ days, $A$ and $B$ can do it in $16$ days and $A, B,$ and $C$ in $10 \frac{2}{3}$ days. In how many days can $A$ and $C$ do it working together?
Answer$A$ can do a piece of work in $=24$ days
$(A+B)$ can do a piece of work in $=16$ days
$(A+B+C)$ can do a piece of work in
$=10 \frac{2}{3}$ days $=\frac{32}{3}$ days
$\therefore A\ 's\ 1-$day work $=\frac{1}{24}$
$(A+B) \ 's\ 1$ day work $=\frac{1}{16}$
$(A+B+C)\ 's\ 1$ day work $=\frac{3}{32}$
$C\ 's\ 1$ day work $=\frac{3}{32}-\frac{1}{16}$
$=\frac{3-2}{32}=\frac{1}{32}$
$(A+C)\ 's\ 1-$day work $=\frac{1}{24}+\frac{1}{32}$
$=\frac{4+3}{96}=\frac{7}{96}$
$\therefore(A+C)$ can do the work in $=\frac{96}{7}$ days
$=13 \frac{5}{7}$ days
View full question & answer→Question 55 Marks
Two pipes $P$ and $Q$ would fill an empty cistern in $24$ minutes and $32$ minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern maybe just filled in $16$ minutes.
Answer$P\ '\ s$ one minute's work $=\frac{1}{24}$
$Q\ '\ s$ one minute's work $=\frac{1}{32}$
Let the first pipe must turn off $x$ minutes
The cistern filled in $16$ minutes
Then $P\ 's\ x$ minutes work $+\ Q\ '\ s\ 16$ minutes work $= 1$
$\Rightarrow \frac{1}{24} \times \mathrm{x}+\frac{1}{32} \times 16=1$
$\frac{\mathrm{x}}{24}+\frac{1}{2}=1$
$\Rightarrow \frac{\mathrm{x}}{24}=1-\frac{1}{2}=\frac{1}{2}$
$\Rightarrow \frac{\mathrm{x}}{24}=\frac{1}{2}$
$\Rightarrow \mathrm{x}=\frac{24}{2}=12$
$\therefore$ After $12$ minute pipe $P$ would turned off
View full question & answer→Question 65 Marks
$A$ can do a piece of work in $10$ days, $B$ in $12$ days, and $C$ in $15 $ days. All begin together but $A$ leaves the work after $2$ days and $B$ leaves $3$ days before the work is finished. How long did the work last?
Answer$A\ 's$ one day's work $=\frac{1}{10}$
$B\ 's$ one day's work $=\frac{1}{12}$
$C\ 's$ one day's work $=\frac{1}{15}$
$A, B$ and $C\ 's$ one day's work $=\frac{1}{10}+\frac{1}{12}+\frac{1}{15}$
$=\frac{6+5+4}{60}=\frac{15}{60}=\frac{1}{4}$
Let the work completed in $\mathrm{x}$ days
$\therefore A\ 's\ 2$ days work $+ B\ 's (x-3)$ days work $+ C\ 's\ x$ days work $=1$
$\Rightarrow 2 \times \frac{1}{10}+(\mathrm{x}-3) \times \frac{1}{12}+\mathrm{x} \times \frac{1}{15}=1$
$\Rightarrow \frac{1}{5}+\frac{\mathrm{x}-3}{12}+\frac{\mathrm{x}}{15}=1$
$\frac{12+5 \mathrm{x}-15+4 \mathrm{x}=60}{60} \ldots( \text{L.C.M}$ of $5, 12, 15=60)$
$\Rightarrow 9 x=60-12+15$
$\Rightarrow 9 x=63$
$\Rightarrow \mathrm{x}=\frac{63}{9}=7$
$\therefore$ Work will last for $7$ days
View full question & answer→Question 75 Marks
$A$ and $B$ can do a piece of work in $40$ days; $B$ and $C$ in $30$ days ; and $C$ and $A$ in $24$ days. $(i)$ How long will it take them to do the work together ?$(ii)$ In what time can each finish it working alone?
Answer$A$ and $B$ can do a piece of work in $=40$ days $B$ and $C$ can do a piece of work in $=30$ days $C$ and $A$ can do a piece of work in $=24$ days $(A+B)^{\prime} s\ 1-$day work $=\frac{1}{40}$
$(B+C)\ 's\ 1-$day work $=\frac{1}{30}$
$(\mathrm{C}+\mathrm{A})^{\prime} s \ 1-$day work $=\frac{1}{24}$
$(i) [(A+B)+(B+C)+(C+A)] \ 's\ 1$ day work $=\frac{1}{4}+\frac{1}{30}+\frac{1}{24}$
$=\frac{3+4+5}{120}$
$=\frac{12}{120}=\frac{1}{10}$
i.e. $(A+B+B+C+C+A)^{\prime} s\ 1-$day work $=\frac{1}{10}$
i.e. $2(A+B+C) \ 's\ 1-$day work $=\frac{1}{10}$
$\therefore(A+B+C) \ 's\ 1-$day work $=\frac{1}{10} \times \frac{1}{2}$
$=\frac{1}{20}$
$\therefore(A+B+C)$ can do the work in $=20$ days
$(ii) A\ 's\ 1-$day work $=\frac{1}{20}-\frac{1}{30}$
$=\frac{3-2}{60}=\frac{1}{60}$
$\therefore\ A $ can do the work in $60$ days
$B\ 's\ 1-$day work $=\frac{1}{20}-\frac{1}{24}$
$=\frac{6-5}{120}=\frac{1}{120}$
$\therefore B$ can do the work in $120$ days
now, $C\ 's\ 1-$day work $=\frac{1}{20}-\frac{1}{40}$
$=\frac{2-1}{40}=\frac{1}{40}$
$C$ can do the work in $40$ days
Hence $A$ can do the work in $=60$ days
$B$ can do the work in $=120$ days
$C$ can do the work in $=40$ days
View full question & answer→Question 85 Marks
$A$ and $B$ complete a piece of work in $24$ days. $B$ and $C$ do the same work in $36$ days; and $A, B$, and $C$ together finish it in $18$ days. In how many days will : $(i)\ A$ alone, $(ii)\ C$ alone,$\ (iii)\ A$ and $C$ together, complete the work?
Answer$A$ and $B$ complete a piece of work in $=24$ days
$\mathrm{B}$ and $\mathrm{C}$ complete a piece of work in $=36$ days
$(A+B+C)$ complete a piece of work in $=18$ days
$(A+B)\ 's\ 1-$ day work $=\frac{1}{24}$
$(B+C) \ 's\ 1-$day work $=\frac{1}{36}$
$(A+B+C) \ 's\ 1-$day work $=\frac{1}{18}$
$(i) A\ 's\ 1-$day work $=\frac{1}{18}-\frac{1}{36}$
$=\frac{2-1}{36}=\frac{1}{36}$
$\therefore$ A will complete the work in $=36$ days
$(ii) C\ 's\ 1-$day work $=\frac{1}{18}-\frac{1}{24}$
$=\frac{4-3}{72}=\frac{1}{72}$
$\therefore C$ will complete the work in $=72$ days
$(iii) (A+C)\ 's\ 1-$day work $=\frac{1}{36}+\frac{1}{72}$
$=\frac{2+1}{72}=\frac{3}{72}$
$=\frac{1}{24}$
$\therefore(A+C)$ will complete the work in $=24$ days
View full question & answer→Question 95 Marks
$A$ and $B$ can do a work in $8$ days ; $B$ and $C$ in $12$ days, and $A$ and $C$ in $16$ days. In what time could they do it, all working together?
Answer$A$ and $B$ can do a work in $=8$ days
$\mathrm{B}$ and $\mathrm{C}$ can do a work in $= 12$ days
$A$ and $C$ can do a work in $=16$ days
$(A+B)\ 's\ 1-$day work $=\frac{1}{8}$
$(B+C)\ 's\ 1-$day work $=\frac{1}{12}$
$(A+C) \ 's\ 1-$day work $=\frac{1}{16}$
$\therefore[(A+B)+(B+C)+(A+C)] \ 's\ 1$ day work
$=\frac{1}{8}+\frac{1}{12}+\frac{1}{16}$
i.e. $[A+B+B+C+A+C] \ 's\ 1$ day work
$=\frac{1}{8}+\frac{1}{12}+\frac{1}{16}$
$ =\frac{6+4+3}{48}=\frac{13}{48}$
i.e. $2(A+B+C) \ 's\ 1 -$day work $=\frac{13}{48}$
i.e. $(A+B+C) \ 's\ 1-$day work $=\frac{13}{48} \times \frac{1}{2}=\frac{13}{96}$
$\therefore(A+B+C)$ can do the work in $=\frac{96}{13}$ days
$=7 \frac{5}{13}$ days
View full question & answer→Question 105 Marks
A contractor undertakes to dig a canal, $6$ kilometers long, in $35$ days and employed $90$ men. He finds that after $20$ days only $2\ km$ of the canal has been completed. How many more men must be employed to finish the work on time?
AnswerLength of canal $=6 \mathrm{~km}$
In $20$ days canal made $=2 \mathrm{~km}$
Remaining length of canal $=6-2=4 \mathrm{~km}$
Remaining time $=35-20=15$ days
In $20$ days $2 \mathrm{~km}$ canal is made by $=90$ men
In $1$ day $2 \mathrm{~km}$ canal is made by $=90 \times 20$ men
In $15$ days $2 \mathrm{~km}$ canal is made by $=\frac{90 \times 20}{15}$ men
In $15$ days $1 \mathrm{~km}$ canal is made by $=\frac{90 \times 20}{15 \times 2}$ men
In $15$ days $4 \mathrm{~km}$ canal is made by $=\frac{90 \times 20 \times 4}{15 \times 2}$ men $=6$ $\times 10 \times 4$ men $=240$ men
Number of more men to be employed to finish the work in time $=240-90=150$ men
View full question & answer→Question 115 Marks
A contractor undertakes to build a wall $1000\ m$ long in $50$ days. He employs $56$ men, but at the end of $27$ days, he finds that only $448\ m$ of the wall is built. How many extra men must the contractor employ so that the wall is completed in time?
AnswerNumber of men employed in the beginning $=56$
Length of wall $=1000 \mathrm{~m}$ No. of days $=50$
In the time of $27$ days, only $448 \mathrm{~m}$ of the wall was completed
Remaining period $=50-27=23$ days
and length of wall to be completed $=1000-448=552$
Now in $27$ days, $448 \mathrm{~m}$ long wall was completed by $=1000 \mathrm{~m}$
in $1$ day, $448 \mathrm{~m}$ long was completed by $=56 \times 27$
in $1$ day, $1 \mathrm{~m}$ long wall will be completed by
$=\frac{56 \times 27}{448} \mathrm{~m}$
and in $23$ days $552 \mathrm{~m}$ long wall be completed by
$=\frac{56 \times 27 \times 552}{448 \times 25} \text { men }$
$ =\frac{1 \times 27 \times 24}{8}=81 \text { men }$
$\therefore$ No. of extra men required
$=81-56=25 \text { men }$
View full question & answer→Question 125 Marks
If $12$ men and $16$ boys can do a piece of work in $5$ days and, $13$ men and $24$ boys can do it in $4$ days, how long will $7$ men and $10$ boys take to do it?
Answer$12$ men $+16$ boys can do a piece of work in $=5$ days
$60$ men $+80$ boys can do a piece of work in $=1$ day
and $13$ men $+24$ boys can do the same work in $=4$ days
$52$ men $+96$ boys can do the same work in $=1$ day
From $(i)$ and $(ii)$
$60$ men $+80$ boys $=52$ men $+96$ boys
$\Rightarrow 60$ men $-52$ men $=96$ boys $-80$ boys
$\Rightarrow 8$ men $=16$ boys
$1$ man $=\frac{16}{8}=$ boys
Now, in first case,
$12$ men $+16$ boys $=12 \times 2+16=24+16=40$ boys
In the second case,
$7$ men $+10$ boys $=7 \times 2+10=14+10=24$ boys
Now $40$ boys can do a piece of work in $=5$ days
$1$ boy can do the same work in $=5 \times 40$ days
and $24$ boys will do the same work in $=\frac{5 \times 40}{24}$ days
$=\frac{25}{3}$
$=8 \frac{1}{3}$ days
View full question & answer→Question 135 Marks
A particular work can be completed by $6$ men and $6$ women in $24$ days; whereas the same work can be completed by $8$ men and $12$ women in $15$ days. Find : $(i)$ according to the amount of work done, one man is equivalent to how many women. $(ii)$ the time is taken by $4$ men and $6$ women to complete the same work.
Answer$6$ men and $6$ women can finish the work in $24$ days
$\therefore 144$ men and $144$ women can finish it in $1$ day
$8$ men and $12$ women can finish the work in $15$ days
$\therefore 120$ men and $180$ women can finish it in $1$ day
$(i)\ 144$ men $+ 144$ women $= 120$ men $+ 180$ women
$\Rightarrow 144$ men $– 120$ men $= 180$ women $– 144$ women
$\Rightarrow 24$ men $= 36$ women
$1$ man $=\frac{36}{24}$
$=\frac{3}{2}$ women
$1$ man $=1 \frac{1}{2}$ women
$(ii)\ $ In the first case, $6$ men and $6$ women are equivalent to
$=6 \times \frac{3}{2}+6$
$= 9+6$
$= 15$ women
In the second case, $4$ men and $6$ women are equivalent to
$4 \times \frac{3}{2}+6$
$= 6 + 6$
$= 12$ women
Now, $15$ women can do a piece of work in $= 24$ days
$1$ women will do it in $= 24 \times 15$ days
$12$ women will do it in =$\frac{24 \times 15}{12}$
$= 30$ days
$4$ men and $6$ women will do the same work in $30$ days.
View full question & answer→