Question 14 Marks
$A$ and $B$ working together can mow a field in $56$ days and with the help of $C$, they could have mowed it in $42$ days. How long would $C$ take by himself?
Answer$(A+B)$ can mow a field in $=56$ days
$(A+B+C)$ can mow a field in $=42$ days
$\therefore(A+B)$ 's $1$ day work $=\frac{1}{56}$
$(A+B+C)$ 's $1$ day work $=\frac{1}{42}$
$C's \ 1$ day work $=\frac{1}{42}-\frac{1}{56}$
$=\frac{4-3}{168}$
| $2$ |
$42,56$ |
| $7$ |
$21,28$ |
|
$3,4$ |
$=\frac{1}{168}$
$\text{L.C.M}. =2 \times 7 \times 3 \times 4=168$
$\therefore C$ can do the work in $=168$ days View full question & answer→Question 24 Marks
$A, B$, and $C$ can do a piece of work in $6$ days, $12$ days, and $24$ days respectively. In what time will they altogether do it?
Answer$A$ can do a piece of work in $=6$ days
$B$ can do a piece of work in $=12$ days
$C$ can do a piece of work in $=24$ days
$\therefore A\ 's \ 1-$day work $=\frac{1}{6}$
$\therefore B\ 's\ 1-$day work $=\frac{1}{12}$
$\therefore C\ 's\ 1-$day work $=\frac{1}{24}$
$(A+B+C)\ 's\ 1-$day work $=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}$
$=\frac{4+2+1}{24}=\frac{7}{24}$
$\therefore(A+B+C)$ can do the work in $=\frac{24}{7}$ days
$=3 \frac{3}{7} $ days
View full question & answer→Question 34 Marks
$A$ and $B$ together can do a piece of work in $6 \frac{2}{3}$ days, but $B$ alone can do it in $10$ days. How long will A take to do it alone?
Answer$A$ and $B$ can do the work in $=6 \frac{2}{3}$ days $=\frac{20}{3}$ days
$B$ can do the work in $=10$ days
$\therefore(A+B) 's\ 1-$day work $=\frac{3}{20}$
$B's 1-$day work $=\frac{1}{10}$
$A's 1-$day work $=\frac{3}{20}-\frac{1}{10}$
$=\frac{3-2}{20}=\frac{1}{20}$
$\therefore\ A$ can do the work in $=20$
View full question & answer→Question 44 Marks
A can$-$do $\frac{1}{4}$ of work in $5$ days and B can do $\frac{1}{3}$ of the same work in $10$ days. Find the number of days in which both working together will complete the work.
AnswerA can$-$do $\frac{1}{4}$ of work in $= 5$ days
A can$-$do $1$ work in $=5 \times 4$ days
$=20$ days
B can do $\frac{1}{3}$ of work in $=10$ days
B can do $1$ work in $=10 \times 3$ days
$=30$ days
$A's 1-$day work $=\frac{1}{20}$
$B's 1-$day work $=\frac{1}{30}$
$(A+B) 's \ 1$ day work $=\frac{1}{20}+\frac{1}{30}$
$=\frac{3+2}{60}$
$=\frac{5}{60}$
$=\frac{1}{12}$
$\therefore \mathrm{A}$ and $\mathrm{B}$ working together will complete the work in $=12$ days
View full question & answer→Question 54 Marks
$90$ men can complete a work in $24$ days working $8$ hours a day. How many men are required to complete the same work in $18$ days working $7 \frac{1}{2}$ hours a day?
AnswerIn $24$ days, working $8$ hours a day, work can be completed by $=90$ men
In $1$ day, working $8$ hours a day, work can be completed by $=90 \times 24$ men
In $18$ days, working $8$ hours a day, work can be completed by $=\frac{90 \times 24}{18}$ men
In $18$ days, working $1$ hour a day, work can be completed by $=\frac{90 \times 24}{18} \times 8$ men
In $18$ days, working $7 \frac{1}{2}$ i.e. $\frac{15}{2}$ hours a day, work can be completed by $=\frac{90 \times 24 \times 8}{18} \times \frac{2}{15}$ men
$=\frac{90 \times 24 \times 8 \times 2}{18 \times 15}=\frac{1 \times 24 \times 8 \times 2}{3}=128 \text { men }$
View full question & answer→Question 64 Marks
A family of $5$ persons can be maintained for $20$ days with $Rs.2,480$. Find, how long $Rs.6944$ maintains a family of $8$ persons?
AnswerA family of $5$ persons can be maintained with $Rs. 2480$ for $=20$ days
A family of $5$ persons can be maintained with $Rs. 1$ for $=\frac{20}{2480}$ days
A family of $5$ persons can be maintained with $Rs. 6944=\frac{20}{2480} \times 6944=\frac{1}{124} \times 6944$
$=56$ days
A family of $1$ person can be maintained
$=56 \times 5$ days
A family of $8$ persons can be maintained
$=\frac{56 \times 5}{8}$ days
$=7 \times 5$ days
$=35$ days
View full question & answer→Question 74 Marks
If $10$ horses consume $18$ bushels in $36$ days. How long will $24$ bushels last for $30$ horses?
Answer$10$ horses consume $18$ bushels in $=36$ days
$1$ horse consumes $18$ bushels in $=36 \times 10$ days $=360$ days
$30$ horses consume $18$ bushels in $=\frac{360}{30}$ days
$=12$ days
$30$ horse consume $1$ bushel in $=\frac{12}{18}$ days
$30$ horse consume $24$ bushels in
$=\frac{12}{18} \times 24$ days
$ =\frac{2}{3} \times 24$ days
$ =16$ days
View full question & answer→Question 84 Marks
Six taps can fill an empty cistern in $8$ hours. How much more time will be taken, if two taps go out of order? Assume, all the taps supply water at the same rate.
AnswerTotal no. of taps $=6$
Out of order taps $=2$
Taps in working condition $=6-2=4$
$6$ taps can fill an empty cistern in $=8$ hours
$1$ tap can fill an empty cistern in $=6 \times 8$ hours
$4$ taps can fill an empty cistern in $=\frac{48}{4}=12$ hours
More time taken when $2$ taps are out of order $=12-8=4$ hour
View full question & answer→Question 94 Marks
A group of labourers promises to do a piece of work in $10$ days, but five of them become absent. If the remaining labourers complete the work in $12$ days, find their original number in the group.
AnswerTotal period $=10$ days
But work completed in $= 12$ days
No. of men was absent $=5$
Let the number of men in the beginning $=x$
Now $\mathrm{x}$ men can do piece work in $=10$ days
$1$ man will do it in $=10 x \times$ days
and $(x-5)$ will do it in $=\frac{10 \times x}{x-5}$ days
$\therefore \frac{10 x}{x-5}=12$
$\Rightarrow 10 \mathrm{x}=12 \mathrm{x}-60$
$\Rightarrow 12 x-10 x=60 $
$\Rightarrow 2 x=60$
$\Rightarrow \mathrm{x}=\frac{60}{2}=30$
$\therefore$ No. of men in the beginning $=30 \mathrm{C}$
View full question & answer→Question 104 Marks
$4$ men or $6$ women earn $Rs. 360$ in one day. Find, how much will : $(i)$ a man earn in one day? $(ii)$ a woman earn in one day?$(iii) \ 6$ men and $4$ women earn in one day?
Answer$4$ men earn $Rs. 360$ in one day
$(i)\ 1$ man can earn $= Rs. \frac{360}{4}= Rs. 90$ in one day
$(ii) \ 6$ women earn $Rs. 360$ in one day
$1$ woman can earn $Rs. \frac{360}{6}= Rs. 60$ in one day
$(iii) \therefore 6$ men and $4$ women earn in one day
$=6 \times 90+4 \times 60=540+240=\text { Rs. } 780$
View full question & answer→Question 114 Marks
If $\frac{2}{9}$ of property costs $Rs. 2,52,000$; find the cost of $\frac{4}{7}$ of it.
Answer$\frac{2}{9}$ of property costs $=Rs. 252000$
$1$ of a property costs $= Rs. 252000 \times \frac{9}{2}$
$= Rs.\frac{2268000}{2}$
$= Rs. 1134000$
$\frac{4}{7}$ of a property costs $=1134000 \times \frac{4}{7}$
$=162000 \times 4$
$= Rs. 6,48,000$
View full question & answer→Question 124 Marks
$\frac{3}{5}$ quintal of wheat costs $Rs.210$. Find the cost of : $(i) 1$ quintal of wheat $(ii)\ 0.4$ quintal of wheat
Answer$(i) \frac{3}{5}$ quintal of wheat costs $=Rs. 210$
$1$ quintal of wheat costs $=210 \times \frac{3}{5}=70 \times 5= Rs. 350$
$(ii) 1$ quintal of wheat costs $= Rs. 350$
$0.4 $ quintal of wheat costs $=350 \times 0.4= Rs. 140.0= Rs. 140$
View full question & answer→Question 134 Marks
A garrison has sufficient provisions for $480$ men for $12$ days. If the number of men is reduced by $160$ ; find how long will the provisions last?
AnswerStrength of garrison $=480$ men
Strength reduced by $=160$ men
New strength of garrison $=480-160=320$
Provision for $480$ men lasts for $=12$ days
Provision for $1$ man last for $=480 \times 12$ days
Provision for $320$ men lasts for $=\frac{480 \times 12}{320}$ days
$=\frac{3 \times 12}{2}=18 \text { days }$
View full question & answer→Question 144 Marks
A fort had provisions for $450$ soldiers for $40$ days. After $10$ days, $90$ more soldiers come to the fort. Find in how many days will the remaining provisions last at the same rate?
AnswerAfter $10$ days:
For $450$ soldiers, provision is sufficient for $(40-10)$ days $=30$ days
For $1$ soldier, provision is sufficient for $30 \times 450$ days
For $540$ soldiers, the provision is sufficient for
$=\frac{30 \times 450}{540}=\frac{30 \times 50}{60}=\frac{50}{2}=25 \text { days }$
View full question & answer→Question 154 Marks
In order to complete a work in $28$ days, $60$ men are required. How many men will be required if the same work is to be completed in $40$ days?
AnswerLet $\mathrm{x}$ be number of men required $60$ men can do the work in $=28$ days
$1$ man can do the work in $=28 \times 60$ days
$\mathrm{x}$ man can do the work in $=\frac{28 \times 60}{\mathrm{x}}$ day
But $\frac{28 \times 60}{\mathrm{x}}=40$ days $......($Given$)$
$\therefore x=\frac{28 \times 60}{40}=7 \times 6=42$
Required number of men $=42$
View full question & answer→Question 164 Marks
If $3$ men or $6$ boys can finish a work in $20$ days, how long will $4$ men and $12$ boys take to finish the same work?
Answer$3$ men $=6$ boys
$4$ men $=\frac{6}{3} \times 4=8$
Total boys in second case :
$=$ men $+12$ boys $=8+12=20$ boys
$6$ boys can do a piece of work in $20$ days Then let $20$ boys will do the same work in $x$ days
$\therefore 6: 20:: 20: x$
$($More boys,less days$)$
$\Rightarrow 6: 20:: x: 20 \ldots...($By inverse proportion$)$
$x=\frac{20 \times 6}{20}$
$ =6$ days
$\therefore$ They will do the work in $6$ days
View full question & answer→Question 174 Marks
If $12$ men or $18$ women can complete a piece of work in $7$ days, in how many days can $4$ men and $8$ women complete the same work?
Answer$12$ men $=18$ women
$\therefore 4 \mathrm{~m}=\frac{18}{12} \times 4$
$=6$ women
Total women in second case $=4$ men $+8$ women
$6+8=14$ women
$18$ women can do a piece of work in $7$ days
Let $14$ women will do the same work in $x$ days
$\therefore 18: 14:: 7: x ($Less women, more days$)$
$\Rightarrow 18: 14:: x: 7$
$x=\frac{18 \times 7}{14}$
$($Using inverse proportion$)$
$=9$
$\therefore$ They will complete the work in $9$ days
View full question & answer→Question 184 Marks
For $100\ km$, a taxi charges $₹ 1,800.$ How much will it charge for a journey of $120\ km$?
AnswerLet a charge of a car is $₹ x$ in $120\ km$
| Distance in$(km)$ |
$1800$ |
$x$ |
| Taxi charges $(₹)$ |
$100$ |
$120$ |
Since it is the case of direct variation
$\Rightarrow \frac{x_1}{y_1}=\frac{x_2}{y_2}$
$\Rightarrow \frac{1800}{100} \times \frac{x}{120}$
$\Rightarrow 100 x=1800 \times 120$
$ x=\frac{1800 \times 120}{100}$
$=\text{₹} 2160$ View full question & answer→Question 194 Marks
A truck consumes $28$ litres of diesel for moving through a distance of $448\ km$. How much distance will it cover in $64$ litres of diesel?
AnswerLet the truck cover $x$ in $64$ litres of diesel.
| Diesel $($in litres$)$ |
$28$ |
$64$ |
| Distance $($in km$)$ |
$448$ |
$x$ |
It is the case of direct variation
$\Rightarrow \frac{\mathrm{x}_1}{\mathrm{y}_1}=\frac{\mathrm{x}_2}{\mathrm{y}_2}$
$ \Rightarrow \frac{28}{448}=\frac{64}{\mathrm{x}}$
i.e., $28 x=64 \times 448$
$\mathrm{x}=\frac{64 \times 448}{28}$
$ =1024 \mathrm~km$ View full question & answer→Question 204 Marks
If $x$ and $y$ vary directly, find the values of $x, y$, and $z$:
| $x$ |
$3$ |
$x$ |
$y$ |
$10$ |
| $y$ |
$36$ |
$60$ |
$96$ |
$z$ |
Answer$\mathrm{x}$ and $\mathrm{y}$ are in direct variation
$\therefore \frac{3}{36}=\frac{x}{60}=\frac{y}{96}=\frac{10}{z}$
$ \Rightarrow \frac{3}{36}=\frac{x}{60}, \frac{3}{36}=\frac{y}{96}, \frac{3}{36}=\frac{10}{z}$
$ x=\frac{3}{36} \times 60, y=\frac{3}{36} \times 96$
$ z=10 \times \frac{36}{3}$
$ \Rightarrow x=5, y=8, z=120$
| $x$ |
$3$ |
$5$ |
$8$ |
$10$ |
| $y$ |
$36$ |
$60$ |
$96$ |
$120$ |
View full question & answer→Question 214 Marks
A train is moving with a uniform speed of $120\ km$ per hour.$(i)$ How far will it travel in $36$ minutes?$(ii)$ In how much time will it cover $210\ km$?
Answer$(i) $ Speed of train in $60$ minutes $=120 \mathrm{~km}$
i.e. distance covered in $60$ minutes $=\frac{120}{60}$
Distance covered in $36$ minutes $=\frac{120 \times 36}{60}$
$=2 \times 36$
$=72 \mathrm{~km}$
$(ii)$ If distance covered is $120 \mathrm{~km}$ then the time is taken $=60$ minutes
If distance covered is $1 \mathrm{~km}$ then the time is taken $=\frac{60}{120}$
If the distance covered is $210 \mathrm{~km}$ then the time is taken $=\frac{60}{120} \times 210=105$ minutes $=1$ hours $45$ minutes
View full question & answer→Question 224 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
$=10 \frac{10}{17} \text { days }$
View full question & answer→Question 234 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 244 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 254 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 264 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 274 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 284 Marks
$A$ and $B$ can do a piece of work in 4$0$ 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 294 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 304 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 314 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 $2\ \text{km}$ 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 324 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
$=\frac{56 \times 27 \times 552}{448 \times 25}$ men
$ =\frac{1 \times 27 \times 24}{8}=81$ men
$\therefore$ No. of extra men required
$=81-56=25$ men
View full question & answer→Question 334 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
$=\frac{25}{3}$
$=8 \frac{1}{3} \text { days }$
View full question & answer→Question 344 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
$1 \operatorname{man}=\frac{36}{24}$
$=\frac{3}{2}$ women
$1 \operatorname{man}=1 \frac{1}{2}$ 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.
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