Question 15 Marks
Aasheesh can paint his doll in $20$ minutes and his sister Chinki can do so in $25$ minutes. They paint the doll together for five minutes. At this juncture they have a quarrel and Chinki withdraws from painting. In how many minutes will Aasheesh finish the painting of the remaining doll?
AnswerAasheesh can paint a doll in $20$ minutes, and Chinki can do the same in $25$ minutes.
$\therefore$ Work done by Aasheesh in $1$ minute $=\frac{1}{20}$ Work done by Chinki in $1$ minute $=\frac{1}{25}$
$\therefore$ Work done by them together $=\frac{1}{20}+\frac{1}{25}$
$=\frac{5+4}{100}=\frac{9}{100}$
$\therefore$ Work done by them in $5$ minute $=5\times\frac{9}{100}=\frac{9}{20}$
Remaining work $=1-\frac{9}{20}=\frac{11}{20}$ It is given that the remaining work is done by Aasheesh.
The work done by Aasheesh in $20$ minutes.
$\therefore\frac{11}{20}\text{th}$ work will be done by Aasheesh in $\Big(20\times\frac{11}{20}\Big)$minutes or $11$ minutes.
Thus, the remaining work is done by Aasheesh in $11$ minutes.
View full question & answer→Question 25 Marks
$A$ and $B$ can do a piece of work in $18$ days; $B$ and $C$ in $24$ days and $A$ and $C$ in $36$ days. In what time can they do it, all working together?
AnswerTime taken by $(A + B)$ to do the work $= 18$ days
Time taken by $(B + C)$ to do the work $= 24$ days
Time taken by $(A + C)$ to do the work $= 36$ days
Now, Work done by (A + B) $=\frac{1}{18}$
Work done by $(B + C)$ $=\frac{1}{24}$
Work done by (A + C) $=\frac{1}{36}$
$\therefore$ Work done together $= (A + B)+ (B + C) + (A + C)$
$=\frac{1}{18}+\frac{1}{24}+\frac{1}{36}$
$=\frac{4+3+2}{72}=\frac{9}{72}$
$=\frac{1}{8}$
$\therefore$ Work done together $= 2(A + B + C)$ $=\frac{1}{8}$
$\therefore$ Work done by $(A + B + C)$ $=\frac{1}{16}$
Thus, together they can finish the work in $16$ days.
View full question & answer→Question 35 Marks
$A$ and $B$ can do a piece of work in $6$ days and $4$ days respectively. $A$ started the work; worked at it for $2$0 days and then was joined by $B$. Find the total time taken to complete the work.
Answer$A$ can do a work in $6$ days, and $B$ can do the same work in $4$ days.
$\therefore$ Work done by $A$ in $2$ days $=\frac{2}{6}=\frac{1}{3}$
Remaining work $=1-\frac{1}{3}=\frac{2}{3}$
$\therefore$ Work done by $(A + B)$ in $1$ day $=\Big(\frac{1}{6}+\frac{1}{4}\Big)$
$=\frac{2+3}{12}=\frac{5}{12}$
$\because \frac{5}{12}\text{th}$ work is done by $A$ and $B$ in $1$ day
$\therefore\frac{2}{3}\text{rd}$ work will be done by $A$ and $B$ in $\Big(\frac{12}{5}\times\frac{2}{3}\Big)$ days or $\frac{8}{5}$ days
$\therefore$ Total time taken $=\Big(\frac{8}{5}+2\Big)$ days $=\frac{18}{5}$ days $=3\frac{3}{5}$ days.
View full question & answer→Question 45 Marks
$A$ and $B$ can do a piece of work in $12$ days; $B$ and $C$ in $15$ days; $C$ and $A$ in $20$ days. How much time will $A$ alone take to finish the work?
AnswerTime taken by $(A + B)$ to do the work $= 12$ days
Time taken by $(B + C)$ to do the work $= 15$ days
Time taken by $(A + C)$ to do the work $= 20$ days
Now, Work done by $(A + B)$ $=\frac{1}{12}$
Work done by $(B + C)$ $=\frac{1}{15}$
Work done by $(A + C)$ $=\frac{1}{20}$
$\therefore$ Work done together $= (A + B) + (B + C) + (A + C)$
$=\frac{1}{12}+\frac{1}{15}+\frac{1}{20}$
$=\frac{5+4+3}{60}=\frac{12}{60}$
$=\frac{1}{5}$
$\therefore$ Work done together $= 2(A + B + C)$ $=\frac{1}{5}$
$\therefore$ Work done by $(A + B + C)$ $=\frac{1}{10}$
$\therefore$ Work done by $A$ alone $= (A + B + C) - (B + C)$ $\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}$
Thus, $A$ alone can do the work in $30$ days.
View full question & answer→Question 55 Marks
$A, B$ and $C$ can reap a field in $15\frac{3}{4}$ days; $B, C$ and $D$ in $14$ days; $C, D$ and $A$ in $18$ days; $D, A$ and $B$ in $21$ days. In what time can $A, B, C$ and $D$ together reap it?
AnswerTime taken by $(A + B + C)$ to do the work $=15\frac{3}{4}$ days $=\frac{63}{4}$ days
Time taken by $(B + C + D)$ to do the work $= 14$ days
Time taken by $(C + D + A)$ to do the work $= 18$ days
Time taken by $(D + A + B)$ to do the work $= 21$ days Now,
Work done by $(A + B + C)$ $=\frac{4}{63}$
Work done by $(B + C + D)$ $=\frac{1}{14}$
Work done by $(C + D + A)$ $=\frac{1}{18}$
Work done by $(D + A + B)$ $=\frac{1}{21}$
$\therefore$ Work done by working together $= (A + B + C)+ (B + C + D) + (C + A + D) + (D + A + B)$
$=\frac{4}{63}+\frac{1}{14}+\frac{1}{18}+\frac{1}{21}$
$=\frac{4}{63}+\Big(\frac{9+7+6}{126}\Big)=\frac{4}{63}+\frac{22}{126}$
$=\frac{4}{63}+\frac{11}{63}=\frac{15}{63}$
$\therefore$ Work done by working together $= 3(A + B + C + D)$ $=\frac{15}{63}$
$\therefore$ Work done by $(A + B + C + D)$ $=\frac{15}{63\times3}=\frac{5}{63}$
Thus, together they can do the work in $\frac{63}{5}$ days or $12\frac{3}{5}$ days
View full question & answer→Question 65 Marks
Two taps $A$ and $B$ can fill an overhead tank in $10$ hours and $15$ hours respectively. Both the taps are opened for $4$ hours and they $B$ is turned off. How much time will $A$ take to fill the remaining tank?
AnswerPipe $A$ can fill the tank in $10$ hours, and pipe $B$ can fill the tank in $15$ hours.
$\therefore$ In $1$ hour, $A$ can fill $=\frac{1}{10}\text{th}$ part of the tank.
In $1$ hour, $B$ can fill $=\frac{1}{15}\text{th}$ part of the tank.
$\therefore$ In $1$ hour, $A$ and $B$ can fill $\Big(\frac{1}{10}+\frac{1}{15}\Big)$
$=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}$th part of the tank
$\therefore$ In $4$ hours, $A$ and $B$ can fill $\Big(\frac{1}{6}\times4\Big)=\frac{2}{3}\text{rd}$ part of the tank Remaining part of the tank $=1-\frac{2}{3}=\frac{1}{3}$ Now, $A$ can fill the tank in $10$ hours.
$\therefore \frac{1}{3}\text{rd}$ part of the tank can be filled by $A$ in $\Big(\frac{1}{3}\times10\Big)$ hours or $\frac{10}{3}$ hours or $3\frac{1}{3}$ hours.
View full question & answer→Question 75 Marks
A cistern has two inlets $A$ and $B$ which can fill it in $12$ hours and $15$ hours respectively. An outlet can empty the full cistern in $10$ hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely?
AnswerTime taken by tap $A$ to fill the cistern $= 12$ hours
Time taken by tap $B$ to fill the cistern $= 15$ hours Let $C$ be the outlet that can empty the cistern in $10$ hours.
Time taken by tap $C$ to empty the cistern $= 10$ hours Now,
Tap $A$ fills $\frac{1}{12}\text{th}$ part of the cistern in $1$ hour.
Tap $B$ fills $\frac{1}{15}\text{th}$ part of the cistern in $1$ hour.
Tap $C$ empties out $\frac{1}{10}\text{th}$ part of the cistern in $1$ hour.
Thus, in one hour, $\Big(\frac{1}{12}+\frac{1}{15}-\frac{1}{10}\Big)\text{th}$ part of the cistern is filled.
We have: $\frac{1}{12}+\frac{1}{15}-\frac{1}{10}=\frac{10+8-12}{120}=\frac{6}{120}=\frac{1}{20}$
Thus, in $1$ hour, $\frac{1}{20}\text{th}$ part of the cistern is filled.
Hence, the cistern will be filled completely in $20$ hours if all the three taps are opened together.
View full question & answer→Question 85 Marks
$A$ and $B$ can do a piece of work in $20$ days and $B$ in $15$ days. They work together for $2$ days and then $A$ goes away. In how many days will $B$ finish the remaining work?
AnswerIt is given that $A$ can finish the work in $20$ days and $B$ can finish the same work in $15$ days
$\therefore$ Work done by $A$ in $1$ day $=\frac{1}{20}$ Work done by $B$ in $1$ day $=\frac{1}{15}$
$\therefore$ Work done by $(A + B)$ in $1$ day $=\frac{1}{20}+\frac{1}{15}$
$=\frac{3+4}{60}=\frac{7}{60}$
$\therefore$ Work done by $(A + B)$ in $2$ days $=\frac{14}{60}=\frac{7}{30}$
Remaining work $=1-\frac{7}{30}=\frac{23}{30}$ It is given that the remaining work is done by $B$.
$\because$ Complete work is done by $B$ in $15$ days.
$\therefore\frac{23}{30}$ of the work will be done by $B$ in $\Big(15\times\frac{23}{30}\Big)$ days or
$\frac{23}{2}$ days or $11\frac{1}{2}$ days.
Thus, the remaining work is done by $B$ in $11\frac{1}{2}$ days.
View full question & answer→Question 95 Marks
Two taps $A$ and $B$ can fill an overhead tank in $10$ hours and $15$ hours respectively. Both the taps are opened for $4$ hours and they $B$ is turned off. How much time will $A$ take to fill the remaining tank?
AnswerPipe $A$ can fill the tank in $10$ hours, and pipe $B$ can fill the tank in $15$ hours.
$\therefore$ In $1$ hour, $A$ can fill $=\frac{1}{10}\text{th}$ part of the tank.
In $1$ hour, $B$ can fill $=\frac{1}{15}\text{th}$ part of the tank.
$\therefore$ In $1$ hour, $A$ and $B$ can fill $\Big(\frac{1}{10}+\frac{1}{15}\Big)$
$=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}$th part of the tank
$\therefore$ In $4$ hours, $A$ and $B$ can fill $\Big(\frac{1}{6}\times4\Big)=\frac{2}{3}\text{rd}$
part of the tank Remaining part of the tank $=1-\frac{2}{3}=\frac{1}{3}$ Now,
$A$ can fill the tank in $10$ hours.
$\therefore \frac{1}{3}\text{rd}$ part of the tank can be filled by $A$ in $\Big(\frac{1}{3}\times10\Big)$ hours or $\frac{10}{3}$ hours or $3\frac{1}{3}$ hours.
View full question & answer→Question 105 Marks
$A$ can do a piece of work in $40$ days and $B$ in $45$ days. They work together for $10$ days and then $B$ goes away. In how many days will $A$ finish the remaining work?
AnswerIt is given that A can finish the work in $40$ days and $B$ can finish the same work in $45$ days
$\therefore$ Work done by $A$ in $1$ day $=\frac{1}{40}$ Work done by $B$ in $1$ day $=\frac{1}{45}$
$\therefore$ Work done by $(A + B)$ in $1$ day $=\frac{1}{40}+\frac{1}{45}$
$=\frac{9+8}{360}=\frac{17}{360}$
$\therefore$ Work done by $(A + B)$ in $10$ days $=10\times\frac{17}{360}=\frac{17}{36}$
Remaining work $=1-\frac{17}{36}=\frac{19}{36}$ It is given that the remaining work is done by $B$.
$\because$ Complete work is done by $B$ in $45$ days.
$\therefore\frac{19}{36}$ of the work will be done by $B$ in $\Big(45\times\frac{19}{36}\Big)$ days or $23\frac{3}{4}$ days.
Thus, the remaining work is done by $B$ in $23\frac{3}{4}$ days.
View full question & answer→Question 115 Marks
Anil can do a piece of work in $5$ days and Ankur in $4$ days. How long will they take to do the same work, if they work together?
AnswerTime taken by Anil to do the work $= 5$ days
Time taken by Ankur to do the work $= 4$ days
$\therefore$ Work done by Anil in $1$ day $=\frac{1}{5}$
Work done by Ankur in $1$ day $=\frac{1}{4}$
$\therefore$ Work done by Anil and Ankur in one day $=\frac{1}{5}+\frac{1}{4}$
$=\frac{4+5}{20}=\frac{9}{20}$
Thus, Anil and Ankur can do the work in $\frac{20}{9}$ days i. e. $2\frac{2}{9}$ days.
View full question & answer→Question 125 Marks
A cistern can be filled by a tap in $4$ hours and emptied by an outlet pipe in $6$ hours. How long will it take to fill the cistern if both the tap and the pipe are opened together?
AnswerTime taken by the tap to fill the cistern $= 4$ hours
$\therefore$ Tap fills $\frac{1}{4}\text{th}$ part of the cistern in $1$ hour.
Time taken by the pipe to empty the cistern $= 6$ hours
$\therefore$ Pipe empties out $\frac{1}{6}\text{th}$ part of the cistern in $1$ hour.
Thus, in $1$ hour,$\Big(\frac{1}{4}-\frac{1}{6}\Big)\text{th}$part of the cistern is filled.
We have: $\frac{1}{4}-\frac{1}{6}=\frac{6-4}{24}=\frac{2}{24}=\frac{1}{12}$
Thus, in $1$ hour, $\frac{1}{12}\text{th}$ part of the cistern is filled.
Hence, the cistern will be filled in $12$ hours.
View full question & answer→Question 135 Marks
Sita can finish typing a $100$ page document in $9$ hours. Mita in $6$ hours and Rita in $12$ hours. How long will they take to type a $100$ page document if they work together?
AnswerTime taken by Sita to do the work $= 9$ hours
Time taken by Mita to do the work $= 6$ hours
Time taken by Rita to do the work $= 12$ hours
Now, Work done by Sita $=\frac{1}{9}$
Work done by Mita $=\frac{1}{6}$
Work done by Rita $=\frac{1}{12}$
$\therefore$ Work done by them together$=\frac{1}{9}+\frac{1}{6}+\frac{1}{12}$
$=\frac{4+6+3}{36}=\frac{13}{36}$
Thus, together they can do the work in $\frac{36}{13}$ hours.
View full question & answer→Question 145 Marks
$A$ and $B$ can polish the floors of a building in $10$ days. $A$ alone can do $\frac{1}{4}\text{th}$ of it in $12$ days. In how many days can $B$ alone polish the floor?
AnswerIt is given that $A$ and $B$ can polish the floors of the building in $10$ days.
$\therefore$ Work done by $(A + B)$ in $1$ days $=\frac{1}{10}$
Now, $A$ alone can do $\frac{1}{4}\text{th}$ of the work in $12$ days.
$\therefore$ Time taken by $A$ alone to do the complete work $= (4 \times 12) = 48$ days
$\Rightarrow$ Work done by $A$ in $1$ day $=\frac{1}{48}$
Now, work done by $B$ in $1$ day = Work done by $(A + B)$ in $1$ day − Work done by $A$
$=\frac{1}{10}-\frac{1}{48}$
$=\frac{24-5}{240}=\frac{19}{240}$
Thus, $B$ alone can polish the floor in $\frac{240}{19}$ days or $12\frac{12}{19}$ days.
View full question & answer→Question 155 Marks
$A$ and $B$ can finish a work in $20$ days. $A$ alone can do $\frac{1}{5}\text{th}$ of the work in $12$ days. In how many days can $B$ alone do it?
AnswerIt is given that $A$ and $B$ can finish the work in $20$ days.
$\therefore$ Work done by $(A + B)$ in $1$ day $=\frac{1}{20}$
Now, $A$ alone can do $\frac{1}{5}\text{th}$ of the work in $12$ days.
$\therefore$ Time taken by $A$ alone to complete the work $= (5 × 12) = 60$ days
$\Rightarrow$ Work done by $A$ in $1$ day $=\frac{1}{60}$
Now, work done by $B$ in $1$ day = Work done by $(A + B)$ in $1$ day work - Work done by $A$ in $1$ day
$=\frac{1}{20}-\frac{1}{60}$
$=\frac{3-1}{60}=\frac{2}{60}$
Thus, $B$ alone can polish the floor in $\frac{60}{2}$ days or $30$ days.
View full question & answer→Question 165 Marks
$A, B$ and $C$ working together can do a piece of work in $8$ hours. $A$ alone can do it in $20$ hours and $B$ alone can do it in $24$ hours. In how many hours will $C$ alone do the same work?
AnswerTime taken by $A$ to do the work $=20$ hours
Time taken by $B$ to do the work $= 24$ hours
Time taken by $(A+B+C)$ to do the work $= 8$ hours
Now, Work done by $A$ $=\frac{1}{20}$
Work done by $B$ $=\frac{1}{24}$
Work done by $(A + B + C)$ $=\frac{1}{8}$
$\therefore$ Work done by $C$ $=\frac{1}{8}-\Big(\frac{1}{20}+\frac{1}{24}\Big)$
$=\frac{1}{8}-\Big(\frac{6}{120}+\frac{5}{120}\Big)=\frac{1}{8}-\Big(\frac{11}{120}\Big)$
$=\frac{15-11}{120}=\frac{4}{120}=\frac{1}{30}$
Thus, $C$ can do the work in $30$ hours.
View full question & answer→