Question 13 Marks
$A$ can do a work in $15$ days and $B$ in $20$ days. If they together work on it for $4$ days; what fraction of the work will be left?
Answer$A's 1-$day work $=\frac{1}{15}$
$B's 1-$day work $=\frac{1}{20}$
$(A+B) 's \ 1-$day work $=\frac{1}{15}+\frac{1}{20}$
$=\frac{4+3}{60}=\frac{7}{60}$
$(A+B) 's \ 4$ days work $=\frac{7}{60} \times 4=\frac{7}{15}$
Remaining work =1-$\frac{7}{15}=\frac{15-7}{15}$
$=\frac{8}{15}$
View full question & answer→Question 23 Marks
$A$ can do a piece of work in $10$ days and $B$ in $15$ days. How long will they take together to finish it?
Answer$A's 1-$ day work $=\frac{1}{10}$
$B's 1-$day work $=\frac{1}{15}$
$(A+B) 's 1-$ day work $=\frac{1}{10}+\frac{1}{15}$
$=\frac{3+2}{30}=\frac{5}{30}$
$=\frac{1}{6}$
$\therefore$ Both $A$ and $B$ together can finish the work in $=6$ days
View full question & answer→Question 33 Marks
One tap can fill a cistern in $3$ hours and the waste pipe can empty the full cistern in $5$ hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together?
AnswerOne tap can fill a cistern in $=3$ hours
A waste pipe can empty a cistern in $=5$ hours
$\therefore$ One trap $1-$hour work $=\frac{1}{3}$
Waste pipe $1-$hour work $=\frac{1}{5}$
$\therefore$ One tap and waste pipe together $1-$hour work
$=\frac{1}{3}-\frac{1}{5}=\frac{5-3}{15}$
$ =\frac{2}{15}$
$\therefore$ The empty cistern will be full in $=\frac{15}{2}$ hours
$=7 \frac{1}{2}$ hours
View full question & answer→Question 43 Marks
If $25$ horses consume $18$ quintal in $36$ days, how long will $28$ quintal last for $30$ horses?
Answer$25$ horses consume $18$ quintal in $=36$ days
$1$ horse consume $1$ quintal in $=\frac{36 \times 25}{18 \times 30}$ days
$30$ horses consume $28$ quintal in $=\frac{36 \times 25 \times 28}{18 \times 30}$ days
$=\frac{2 \times 5 \times 28}{6}=\frac{140}{3}=46 \frac{2}{3} \text { days }$
View full question & answer→Question 53 Marks
Twelve typists, all working with the same speed, type a certain number of pages in $18$ days working $8$ hours a day. Find, how many hours per day must sixteen typists work in order to type the same number of pages in $9$ days?
Answer$12$ typists can type in $18$ days with the number of working hours in day $=8$ hours
$1$ typist can type in $18$ days $=8 \times 12$ hour
$1$ typist can type in $9$ days $=2(8 \times 12)$ hour
$16$ typist can type in a day $=\frac{2(8 \times 12)}{16}=12$ hours
View full question & answer→Question 63 Marks
Fifteen men can build a wall in $60$ days. How many more men are required to build another wall of the same size in $45$ days?
AnswerIn $60$ days a wall can be built by $=15$ men
In $1$ day a wall can be built by $=15 \times 60$ men
In $45$ days a wall can be built by $=\frac{15 \times 60}{45}=\frac{900}{45}=20$ men
No. of more men required to build the wall in $45$ days $=20-15=5$ men
View full question & answer→Question 73 Marks
Eight oranges can be bought for $Rs. 10.40$. How many more can be bought for $Rs.16.90$?
AnswerNumber of oranges bought for $Rs. 10.40$ is $8$
Number of oranges can be bought for $Rs. 1$ is $\frac{8}{10.40}$
Number of oranges bought for $Rs. 16.90$
$=\frac{8}{10.40} \times 16.90=\frac{8 \times 1690}{1040}=\frac{13520}{1040}=13$
$13$ oranges can be bought for $Rs. 16.90$
$\therefore$ Number of more oranges which can be bought is $13-8=5$
View full question & answer→Question 83 Marks
Ten men, working for $6$ days of $10$ hours each, finish $\frac{5}{21}$ of a piece of work. How many men working at the same rate and for the same number of hours each day, will be required to complete the remaining work in $8$ days?
AnswerWork does one $=\frac{5}{21}$
Remaining work $=1-\frac{5}{21}=\frac{16}{21}$
$\frac{5}{21}$ of a work can be done in $6$ days working $10$ hours a day by $=10 \mathrm{~m}$
$1$ work can be done in $6$ days working $10$ hours a day by $=\frac{10 \times 21}{5}$
$1$ work can be done in $1$ day working $10$ hours a day by $=\frac{10 \times 21 \times 6}{5}$ men
$\frac{16}{21}$ work can be done in $8$ days working $10$ hours a day by $=\frac{10 \times 21 \times 6 \times 16}{5 \times 21 \times 8}=24$ men
View full question & answer→Question 93 Marks
If $70$ men dig $15,000 sq. m$ of a field in $5$ days, how many men will dig $22,500 \ sq. m$ field in $25$ days?
Answer$70$ men dig $15,000\ sq$. $\mathrm{m}$ of a field in $5$ days
$1 $ man can dig $15,000 sq. \mathrm{m}$ of a field in $=\frac{5 \times 70}{15,000}$
Men required to dig $22,500 sq. m$. a field in $25$ days
$=\frac{5 \times 70 \times 22500}{15,000 \times 25}=\frac{70 \times 900}{3000}=\frac{63}{3}=21$
Hence required to dig $22,500\ sq. \mathrm{m}$ field in $25$ days $=21$
View full question & answer→Question 103 Marks
If $15$ men can complete a piece of work in $30$ days, in how many days will $18$ men complete it?
AnswerSince $15$ men can complete a piece of work in $= 30$ days
$\therefore 1$ man can do the work in $=30 \times 15$ days
$\therefore 18$ men can do the work in $=\frac{30 \times 15}{18}=\frac{5 \times 15}{3}$
$=5 \times 5=25 \text { days }$
Hence $18$ men can do the work in $25$ days.
View full question & answer→Question 113 Marks
Cost of $24$ identical articles is $Rs. 108$, Find the cost of $40$ similar articles.
AnswerCost of $24$ identical articles $= Rs. 108$
$\therefore$ Cost of $1$ identical articles $= Rs.\frac{108}{24}$
$\therefore$ Cost of $40$ identical articles $= Rs.\left(\frac{108}{24} \times 40\right)$
$= Rs. \frac{108}{3} \times 5$
$= Rs.\ 36 \times 5$
$= Rs. 180$
View full question & answer→Question 123 Marks
$50$ labourers can dig a pond in $16$ days. How many labourers will be required to dig another pond, double in size in $20$ days?
AnswerIn $16$ days for diging pond labour reqd.
In $1$ day, labour reqd. $=50 \times 16$
In $20$ days, labour reqd.$=\frac{50 \times 16}{20}$
In $20$ days, with double work, labour reqd
.$=\frac{50 \times 16 \times 2}{20}$
$ =5 \times 8 \times 2$
$ =80$
Hence labourers required $=80$
View full question & answer→Question 133 Marks
A car takes $6$ hours to reach a destination by travelling at a speed of $50\ km$ per hour. How long will it take when the car travels at a speed of $75\ km$ per hour?
AnswerTime: Distance:: Time: Distance
$6: 50 \mathrm{~km}:: \mathrm{x}: 75$
$\therefore$ By inverse proportion
$6 \times 50=x \times 75$
$\Rightarrow \mathrm{x}=\frac{50 \times 6}{75}$
$=4$ hours
View full question & answer→Question 143 Marks
If $56$ workers can build a wall in $180$ hours, how many workers will be required to do the same work in $70$ hours?
AnswerWorkers: Hours:: Workers: Hours
$56: 180:: x: 70$
$\therefore$ By inverse proportion
$56 \times 180=\mathrm{x} \times 70$
$ \Rightarrow \mathrm{x}=\frac{180 \times 56}{70}$
$ =144$ workers
View full question & answer→Question 153 Marks
$72$ men do a piece of work in $25$ days. In how many days will $30$ men do the same work?
AnswerMen: Days:: Men : Days
$72: 25 \text { :. } 30: x$
$\therefore$ By inverse proportion
$72 \times 25=30 \times x$
$ \Rightarrow \mathrm{x}=\frac{25 \times 72}{30}$
$ =60$ days
View full question & answer→Question 163 Marks
In a fort $150$ men had provisions for $45$ days. After $10$ days, $25$ men left the fort. How long would the food last at the same rate?
AnswerAfter $10$ days :
For $150$ men, provision will last $(45 - 10)$
days $=35$ days
$\Rightarrow$ For $1$ man,
the provisions will last
$=150 \times 35$ days
And for $(150-25)=125$ men, the provisions
will last for $=\frac{150 \times 35}{125}$
$=42$ days
View full question & answer→Question 173 Marks
$12$ pipes, all of the same size, fill a tank in $42$ minutes. How long will it take to fill the same tank, if $21$ pipes of the same size are used?
AnswerPipes: Time:: Pipes: Time
$12: 2 x:: 21: 42$
$\therefore$ By inverse proportion
$12 \times 42=21 \times x$
$\Rightarrow \mathrm{x}=\frac{12 \times 42}{21}$
$=24$ minutes
View full question & answer→Question 183 Marks
$36$ men can do a piece of work in $7$ days. How many men will do the same work in $42$ days?
AnswerMen: Days:: Men: Days
$36: 7 \text { :. } x: 42$
$\therefore$ By inverse proportional
$36 \times 7=x \times 42$
$\Rightarrow \mathrm{x}=\frac{36 \times 7}{42}$
$=6$ men
View full question & answer→Question 193 Marks
If $x$ and $y$ vary inversely, find the values of $l, m$, and $n$ :
| $x$ |
$24$ |
$32$ |
$m$ |
$16$ |
| $y$ |
$l$ |
$12$ |
$8$ |
$n$ |
Answer$\because x$ and $y$ are inversely proportional
$\therefore x y$ is equal
Now,
$x y=32 \times 12=384$
$24 \times I=384 \Rightarrow I=\frac{384}{24}=16$
$m \times 8=384 \Rightarrow m=\frac{384}{8}=48$
$16 \times n=384 \Rightarrow n=\frac{384}{16}=24$
View full question & answer→Question 203 Marks
If $x$ and y vary inversely, find the values of $l, m$, and $n$ :
| $x$ |
$4$ |
$8$ |
$2$ |
$32$ |
| $y$ |
$4$ |
$l$ |
$m$ |
$n$ |
Answer$\because x$ and $\mathrm{y}$ are inversely proportional
$\therefore \mathrm{xy}$ is equal
Now,
$x y=4 \times 4=16$
$8 \times I=16$
$\Rightarrow I=\frac{16}{8}=2$
$2 \times m=16$
$\Rightarrow m=\frac{16}{2}=8$
$32 \times n=16$
$\Rightarrow n=\frac{16}{32}=0.5$
View full question & answer→Question 213 Marks
Check whether $x$ and $y$ vary inversely or not.
| $x$ |
$10$ |
$30$ |
$60$ |
$10$ |
| $y$ |
$90$ |
$30$ |
$20$ |
$90$ |
Answer$xy = 10 \times 90 = 900$
$xy= 30 \times 30 = 900$
$xy= 60 \times 20= 1200$
$xy= 10 \times 90 = 900$
$y$ are not equal.
$x$ and $y$ are not inversely proportional.
View full question & answer→Question 223 Marks
Check whether $x$ and $y$ vary inversely or not.
| $x$ |
$30$ |
$120$ |
$60$ |
$24$ |
| $y$ |
$60$ |
$30$ |
$30$ |
$75$ |
Answer$xy = 30 \times 60 = 1800$
$xy= 120 \times 30 = 3600$
$xy = 60 \times 30= 1800$
$xy = 24 \times 75 = 1800$
$xy$ are not equal.
$x$ and $y$ are not inversely proportional.
View full question & answer→Question 233 Marks
Check whether $x$ and $y$ vary inversely or not.
| $x$ |
$4$ |
$3$ |
$12$ |
$1$ |
| $y$ |
$6$ |
$8$ |
$2$ |
$24$ |
Answer$xy = 4 \times 6 = 24$
$xy = 3 \times 8 = 24$
$xy = 12 \times 2 = 24$
$xy = 1 \times 24 = 24$
$xy$ are equal.
$x$ and $y$ are inversely proportional
View full question & answer→Question 243 Marks
If $27$ identical articles cost $₹ 1,890$, how many articles can be bought for $₹ 1,750$?
AnswerLet $x$ number of articles be purchased in $₹1750$
| Cost $(₹)$ |
$1890$ |
$1750$ |
| No. of articles |
$27$ |
$x$ |
Since it is a case of direct variation
$\Rightarrow \frac{1890}{27}=\frac{1750}{\mathrm{x}}$
$ \Rightarrow \mathrm{x}=\frac{1750 \times 27}{1890}$
$ =25$ articles View full question & answer→Question 253 Marks
In which of the following table, $x$ and $y$ vary directly:
| $x$ |
$27$ |
$45$ |
$54$ |
$75$ |
| $y$ |
$81$ |
$180$ |
$216$ |
$225$ |
Answer$\frac{\mathrm{x}_1}{\mathrm{x}_2}=\frac{27}{81}=\frac{1}{3}$
$ \frac{\mathrm{x}_2}{\mathrm{y}_2}=\frac{45}{150}=\frac{15}{50}=\frac{3}{10}$
$ \frac{\mathrm{x}_3}{\mathrm{y}_3}=\frac{54}{216}=\frac{18}{72}=\frac{1}{4}$
$ \frac{\mathrm{x}_4}{\mathrm{y}_4}=\frac{75}{225}=\frac{25}{45}=\frac{5}{9}$
$ \Rightarrow \frac{\mathrm{x}_1}{\mathrm{y}_1} \neq \frac{\mathrm{x}_2}{\mathrm{y}_2} \neq \frac{\mathrm{x}_3}{\mathrm{y}_3} \neq \frac{\mathrm{x}_4}{\mathrm{y}_4}$
$x$ and $y$ are not in direct variation.
View full question & answer→Question 263 Marks
In which of the following table, $x$ and $y$ vary directly:
| $x$ |
$16$ |
$30$ |
$40$ |
$56$ |
| $y$ |
$32$ |
$60$ |
$80$ |
$84$ |
Answer$\frac{x_1}{y_1}=\frac{16}{32}=\frac{1}{2}$
$ \frac{x_2}{y_2}=\frac{30}{60}=\frac{1}{2}$
$ \frac{x_3}{y_3}=\frac{40}{80}=\frac{1}{2}$
$ \frac{x_4}{y_4}=\frac{56}{84}=\frac{2}{3}$
$ \Rightarrow \frac{x_1}{y_1}=\frac{x_2}{y_2}=\frac{x_3}{y_3} \neq \frac{x_4}{y_4}$
$x$ and $y$ are not in direct variation.
View full question & answer→Question 273 Marks
In which of the following table, $x$ and $y$ vary directly:
| $x$ |
$3$ |
$5$ |
$8$ |
$11$ |
| $y$ |
$4.5$ |
$7.5$ |
$12$ |
$16.5$ |
Answer$\frac{\mathrm{x}_1}{\mathrm{x}_2}=\frac{3}{4.5}=\frac{1}{1.5}$
$ \frac{\mathrm{x}_2}{\mathrm{y}_2}=\frac{5}{7.5}=\frac{1}{1.5}$
$ \frac{\mathrm{x}_3}{\mathrm{y}_3}=\frac{8}{12}=\frac{1}{1.5}$
$ \frac{\mathrm{x}_4}{\mathrm{y}_4}=\frac{11}{16.5}=\frac{1}{1.5}$
$ \Rightarrow \frac{\mathrm{x}_1}{\mathrm{y}_1}=\frac{\mathrm{x}_2}{\mathrm{y}_2}=\frac{\mathrm{x}_3}{\mathrm{y}_3}=\frac{\mathrm{x}_4}{\mathrm{y}_4}$
Yes, $x$ and $y$ vary directly.
View full question & answer→