Question 13 Marks
$A$ and $B$, working together can finish a piece of work in $6$ days, while $A$ alone can do it in $9$ days. How much time will $B$ alone take to finish it?
Answer$A$ and $B$ both's one days work $=\frac{1}{6}$
As alone's one days work $=\frac{1}{9}$
$\therefore$ $B's$ alone one days work $=\frac{1}{6}-\frac{1}{9}$
$=\frac{3-2}{18}=\frac{1}{18}$
$\therefore$ $B$ can finish the work in $18$ days.
View full question & answer→Question 23 Marks
Pipe $A$ can fill an empty tank in $5$ hours while pipe $B$ can empty the full tank in $6$ hours. If both are opened at the same time in the empty tank, how much time will they take to fill it up completely?
AnswerPipe $A's$ one hour's work for filling the tank $=\frac{1}{5}$
Pipe $B's$ one hour's work for emptying $=\frac{1}{6}$
Both pipes are opened, then,
Both's one hour's work $=\frac{1}{5}-\frac{1}{6}$
$=\frac{6-5}{30}=\frac{1}{30}$
$\therefore$ The empty tank will be filled in $30$ hours.
View full question & answer→Question 33 Marks
Ravi can do a piece of work in $15$ hours while Raman can do it in $12$ hours. How long will both take to do it, working together?
AnswerRavi's one hours $=\frac{1}{15}$
Both's one day's work $=\frac{1}{12}$
Both can finish the work in, $\frac{20}{3}=6\frac{2}{3}\ \text{hours}$ Or $6$ hours, $40$ minutes.
View full question & answer→Question 43 Marks
A pipe can fill a cistern in $9$ hours. Due to a leak in its bottom, the cistern fills up in $10$ hours. If the cistern is full, in how much time will it be emptied by the leak?
AnswerTime taken by the pipe to fill the cistern $= 9$ hours.
Part of the cistern filled in one hour $=\frac{1}{9}$
Suppose the leak empties the full cistern in $x$ hours.
Part of the cistern emptied in one hour $=-\frac{1}{\text{x}}$ (negative sign implies a leak)
Time taken by the cistern to fill completely due to the leak $= 10$ hours.
Part of the cistern filled in one hour due to the leak $=\frac{1}{10}$
$\therefore\frac{1}{10}=\frac{1}{9}-\frac{1}{\text{x}}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{9}-\frac{1}{10}=\frac{1}{90}$ $\text{x}=90\ \text{hours}.$
Therefore, the leak will empty a full cistern in $90$ hours.
View full question & answer→Question 53 Marks
A pipe can fill a cistern in $9$ hours. Due to a leak in its bottom, the cistern fills up in $10$ hours. If the cistern is full, in how much time will it be emptied by the leak?
AnswerThe inlet pipe's $1$ hour's work $=\frac{1}{9}$
The leak and inlet's $1$ hours work $=\frac{1}{10}$
Leak's $1$ hour work $=\frac{1}{9}-\frac{1}{10}$
$=\frac{10-9}{90}$ $=\frac{1}{90}$
The leak can empty the cistern in $= 90$ hours.
View full question & answer→Question 63 Marks
Pipe $A$ can fill a cistern in $6$ hours and pipe $B$ can fill it in $8$ hours. Both the pipes are opened and after two hours, pipe $A$ is closed. How much time will $B$ take to fill the remaining part of the tank?
AnswerInlet pipe $A's$ one hour's work $=\frac{1}{6}$
Inlet pipe $B's$ one hour's work $=\frac{1}{8}$
Both,inlets one hour's work $=\frac{1}{6}+\frac{1}{8}$ $=\frac{4+3}{24}=\frac{7}{24}$
Both inlets $2$ hour's work $=\frac{7}{24}\times2=\frac{7}{12}$
Remaining work $=1-\frac{7}{12}=\frac{12-7}{12}=\frac{5}{12}$
Now pipe $B$ can fill the $\frac{5}{12}$ tank in $=\frac{5}{12}\times8=\frac{10}{3}=3\frac{1}{3}$ hours.
View full question & answer→Question 73 Marks
$A, B$ and $C$ can do a piece of work in $10$ days, $12$ days and $15$ days respectively. How long will they to finish it they work together?
Answer$A's$ one days work $=\frac{1}{10}$
$B's$ one days work $=\frac{1}{12}$
$\therefore$ $A, B$ and $C's$ one days work $=\frac{1}{10}+\frac{1}{12}$
$+\frac{1}{15}$
$=\frac{6+5+4}{60}=\frac{15}{60}=\frac{1}{4}$
$\therefore$ $A, B$ and $C$ can do the same work in $4$ days.
View full question & answer→Question 83 Marks
A cistern has two inlets $A$ and $B$ which can fill it in $12$ minutes and $15$ minutes respectively. An outlet $C$ can empty the full cistern in $10$ minutes. If all the three pipes are opened together in the empty tank, how much time will they take to fill the tank completely?
AnswerInlet $A's$ $1$ minutes work $=\frac{1}{12}$
Inlet $B's$ $1$ minutes work $=\frac{1}{15}$
Outlet $C's$ $1$ minutes work $=\frac{1}{10}$
$\therefore$ Inlet $A$ and $B$ and outlet $C's$ $1$ minutes work,
$=\frac{1}{12}+\frac{1}{15}-\frac{1}{10}$
$=\frac{5+4-6}{60}(\text{LCM}\ \text{of}\ \ 12,15,10=60)$
$=\frac{3}{60}=\frac{1}{20}$
$\therefore$ The empty tank can be filled completely in $= 20$ minutes.
View full question & answer→Question 93 Marks
$A$ can do a piece of work in $24$ hours while $B$ alone can do it in $16$ hours. If $A, B$ and $C$ working together can finish it in $8$ hours, in how many hours can $C$ alone finish the work?
Answer$A's$ $1$ hour work $=\frac{1}{24}$
$B's$ $1$ hours work $=\frac{1}{16}$
$A, B$ and $C's$ $1$ hours work $=\frac{1}{8}$
$\therefore$ $C's$ $1$ hour's work $=\frac{1}{8}-\Big(\frac{1}{16}+\frac{1}{24}\Big)$
$=\frac{6-(2+3)}{48}=\frac{6-5}{48}=\frac{1}{48}$
$\therefore$ $C$ can finish the work in $= 48$ hours.
View full question & answer→Question 103 Marks
$A$ and $B$ can do a piece of work in $15$ days; $B$ and $C$ in $12$ days; $C$ and $A$ in $20$ days. How many days will be taken by $A, B$ and $C$ working together to finish the work?
Answer$(A + B)$ can do a work in $15$ days.
$\therefore$ $(A + B)'s$ $1$ day work $=\frac {1}{15}$ $(B + C)$ can do a work in $12$ days.
$\therefore$ $(B + C)'s$ $1$ day work $=\frac{1}{12}$ $(C + A)$ can do a work in $20$ days.
$\therefore$ $(C + A)'s$ $1$ day work $=\frac{1}{20}$ $2(A+ B + C)'s$
$1$ day work $=\frac{1}{15}+\frac{1}{12}+\frac{1}{20}=\frac{4+5+3}{60}=\frac{12}{60}=\frac{1}{5}$
$(A + B + C)'s$ $1$ day work $=\frac{1}{10}$
$A, B$ and $C$ working together require $10$ days to complete the work.
View full question & answer→Question 113 Marks
Three taps $A, B$ and $C$ can fill an overhead tank in $6$ hours, $8$ hours and $12$ hours respectively. How long would the three taps take to fill the empty tank, if all of them are opened together?
AnswerTap $A's$ one hour's work $=\frac{1}{6}$
Tap $B's$ one hour's work $=\frac{1}{8}$
Tap $C's$ one hour's work $=\frac{1}{12}$
$A, B$ and $C's$ together one hour's work, $=\frac{1}{6}+\frac{1}{8}+\frac{1}{12}$
$=\frac{4+3+2}{24}(\text{LCM}\ \text{of}\ \ 6,8,12=24)$
$=\frac{9}{24}=\frac{3}{8}$
$\therefore$ These three tap will fill the empty tank in, $=\frac{8}{3}$ hour, $=2\frac{2}{3}$ hours = $2$ hours $40$ minutes.
View full question & answer→Question 123 Marks
Tap $A$ can fill a cistern in 8 hours and tap $B$ can empty it in $12$ hours. How long will it take to fill the cistern if both of them are opened together?
AnswerTap $A$ can fill a cistern in $8$ hours.
Part of cistern filled by Tap $A$ in $1$ hour $=\frac{1}{8}$
Tap $B$ empties the cistern in $12$ hours.
Part of cistern emptied by Tap $B$ in $1$ hour $=\frac{1}{12}$ (negative sign shows that tap $B$ drains the tank)
Part of cistern filled in one hour when both taps are opened together $=\frac{1}{8}-\frac{1}{12}=\frac{3-2}{24}=\frac{1}{24}$
Therefore, it will take $24$ hours to fill the cistern.
View full question & answer→Question 133 Marks
$2$ men or $3$ women can do a piece of work in $16$ days. In how many days can $4$ men and $6$ women do the same work?
AnswerWork of $2$ men = Work of $3$ women,
$\Rightarrow $ Work of $1$ man $=\frac{3}{2}$ women.
Three women can do a piece of work in $16$ days.
As $4$ men and $6$ women $=\Big(4\times\frac{3}{2}\Big)$ women + $6$ women
$= 6$ women + $6$ women $= 12$ women, Also, $3$ women can do the work in $16$ days.
So, work done by $3$ women in one day $=\frac{1}{16}$
$\therefore$ Work done by $1$ woman in one day $=\frac{1}{48}$
$\Rightarrow $ Work done by $12$ women in one day $=\frac{1}{4}$
Thus, $4$ men and $6$ women will complete the work in $4$ days.
View full question & answer→Question 143 Marks
$A$ can do a piece of work in $10$ days while $B$ alone can do it in $15$ days. In how many days can both finish the same work?
Answer$A$ can do a piece of work in $10$ days.
$A's$ $1$ day work $=\frac{1}{10}$ $B$ can do a piece of work in $15$ days.
$B's$ $1$ day work $=\frac{1}{15}$ $(A + B)'s$
$1$ day work $=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{300}=\frac{5}{30}=\frac{1}{6}$
$A$ and $B$ working together can complete the work in $6$ days.
View full question & answer→Question 153 Marks
Two motor mechanics, Raju and Siraj working together can overhaul a scoote in $6$ hours. Raju alone can do the job in $15$ hours. In how many hours, can Siraj alone do it?
AnswerRaju and Siraj's $1$ hours work $=\frac{1}{6}$
Raju's alone $1$ hours work $=\frac{1}{15}$
$\therefore$ Siraj's alone $1$ hours work $=\frac{1}{6}-\frac{1}{15}$
$=\frac{5-2}{30}=\frac{3}{30}=\frac{1}{10}$
$\therefore$ Siraj's can do the same work in $10$ hours.
View full question & answer→Question 163 Marks
$A, B$ and $C$ working together can finish a piece of work in $8$ hours. $A$ alone can do it $20$ hours and $B$ alone can do it in $24$ hours. In how many hours will $C$ alone do the same work?
Answer$A, B$ and $C's 1$ hours work $=\frac{1}{8}$
$A's\ 1$ hours work $=\frac{1}{20}$
$B's\ 1$ hours work $=\frac{1}{24}$
$\therefore$ $C's$ $1$ hours work $=\frac{1}{8}-\Big(\frac{1}{20}+\frac{1}{24}\Big)$
$=\frac{15-(6+5)}{120}=\frac{15-11}{120}$
$=\frac{4}{120}=\frac{1}{30}$
$\therefore$ $C$ alone finish the work in $= 30$ hours.
View full question & answer→Question 173 Marks
Pipes $A$ and $B$ can fill an empty tank in $10$ hours and $15$ hours respectively. If both are opened together in the empty tank, how much time will they take to fill it completely?
Answer$A's$ one hours work $=\frac{1}{10}$
$B's$ one hours work $=\frac{1}{15}$
$\therefore$ $A$ and $B's$ one hour's work $=\frac{1}{10}+\frac{1}{15}$
$=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}$
$\therefore$ Pipes $A$ and $B$ can fill the empty tank in $6$ hours.
View full question & answer→Question 183 Marks
Rajan can do a piece of work-in $24$ days while Amit can do it in $30$ days. In how many days can they complete it if they work together.
AnswerRajan's one days work $=\frac{1}{24}$
Amit's one days work $=\frac{1}{30}$
Both's one day's work $=\frac{1}{24}+\frac{1}{30}$
$=\frac{5+4}{120}=\frac{9}{120}[\text{LCM}\ \text{of}\ 24,30=120]$
$=\frac{3}{40}$
$\therefore$ Both can do the work in $=\frac{40}{3}$ days
$=13\frac{1}{3}$ days.
View full question & answer→