Excercise: 9.1
Question 1
(i) $\frac{5}{15}=\frac{1}{3}, \frac{8}{24}=\frac{1}{3}, \frac{12}{36}=\frac{1}{3}, \frac{15}{60}=\frac{1}{4}, \frac{18}{72}=\frac{1}{4}, \frac{20}{100}=\frac{1}{5}$
Here $\frac{x}{y}=\frac{5}{15}=\frac{8}{24}=\frac{12}{36}=\frac{1}{3} \neq \frac{15}{60}=\frac{18}{72}=\frac{1}{4} \neq \frac{20}{100}=\frac{1}{5}$
ஃ $x$ and $y$ are not proportional.
(ii) $\frac{3}{9}=\frac{1}{3}, \frac{5}{15}=\frac{1}{3}, \frac{7}{21}=\frac{1}{3}, \frac{9}{27}=\frac{1}{3}, \frac{10}{30}=\frac{1}{3} .$
Here $\frac{x}{y}=\frac{3}{9}=\frac{5}{15}=\frac{7}{21}=\frac{9}{27}-\frac{10}{30}=\frac{1}{3}=$ Constant
ஃ $x$ and $y$ are not proportional
Question 2
(i) $\frac{x}{y}=\frac{3}{45}=\frac{5}{y_{2}}=\frac{x_{3}}{90}=\frac{x_{4}}{120}=\frac{x_{5}}{y_{5}}=\frac{1}{15}=$ constant
$\begin{array}{l|l|l}\frac{5}{y_{2}}=\frac{1}{15} & \frac{x_{3}}{90}=\frac{1}{15} & \frac{x_{4}}{120}=\frac{1}{15} \\y_{2}=15 \times 5 & x_{3}=90 / 15 & x_{4}=120 / 15 \\y_{2}=75 & x_{3}=6 & x_{4}=8\end{array}$
$\frac{10}{y_{5}}$= $\frac{1}{15}$
$y_{5}=15 \times 10=150$
$\therefore x_{3}=6, x_{4}-8, y_{2} =75$
(ii) $\frac{x}{4}=\frac{4}{7}=\frac{8}{y_{2}}=\frac{x_{3}}{21}=\frac{20}{4_{4}}=\frac{28}{y_{5}}=$ constant
$\begin{array}{c|c|c}\frac{y}{7}=\frac{8}{y_{2}} & \frac{y}{7}=\frac{x_{1}}{21} & \frac{4}{7}=\frac{20}{y_{4}} \\4 y_{2}=7 \times 8 & x_{3}=\frac{4 \times 13}{7} & y_{4}=\frac{20}{4} \\y_{2}=\frac{7 \times 8}{4} & x_{3}=4 \times 3 & y_{4}=5 \times 7 \\y_{2}=14 & x_{3}=12 & y_{4}=35\end{array}$
$\frac{4}{7}=\frac{28}{45} \Rightarrow y_{5}=\frac{7 \times 28}{4} \Rightarrow y_{5}=49$
$\therefore y_{2}=14, x_{3}=12, y_{4}=35, \quad y_{5}=47$
Question 3
Let the cost of 5.8 mts of cloth be Rs x
Cost of 8 mts of cloth cost Rs 250
so it is a case of direct variation
i .e $\frac{8}{250}=\frac{5.8}{x}$
$x=\frac{250 \times 5.8}{8}=\frac{25 \times 58}{8}$
$x=181.25$
∴ Cost of 5.8 mts of cloth Rs 181.25
Question 4
Labourer earns Rs 672 per week i . e 7 days
Labourer in 7 days he earns Rs 672
in 1 day he earn Rs $\frac{672}{7}$
= Rs 96
Now in 18 days he earns Rs 96 $\times$ 18
= Rs 1458
Question 5
Given 175 dollars cost Rs 7,350
how many dollars cost Rs 24,24
Let the no of dollar be x
$\frac{x}{17 5}=\frac{24024}{7350}$\
x=572
$\therefore 572$ dollars cost Rs 24,024
Question 6
Let the number of kilometers travelled be 'x'
car travels 675 km in 4.5 litres of petrol
x km in 26 .4 litres of petrol
i.e $\frac{x}{67 \cdot 5}=\frac{26 \cdot 4}{4 \cdot 5}$
$x=\frac{26.4 \times 67.5}{4.5}$
$x=396 \mathrm{~km}$
∴ Number of kilometers travelled is 396 km
Question 7
Let the number of sheets of cardboard be $x$
thickness of 12 cardboard sheets is $45 \mathrm{~mm}$
Thickness of x cardboard sheets is 90 cm i .e 900mm
i .e $\frac{x}{12}=\frac{900}{45}$
$\begin{aligned} x &=\frac{900 \times 12}{45} \\ x &=240 \end{aligned}$
∴ Number of sheets of cardboard are 240
Question 8
Mast of shop model is 6cm
Mast of achol shop is 9 m = 900cm
length of achol ship is 33m = 3300cm
Let length of model ship is x
$\begin{aligned} \frac{6}{900} &=\frac{x}{3300} \\ x &=\frac{6 \times 33}{x y} \\ x &=22 \mathrm{~cm} \end{aligned}$
model length of ship = 22cm
Question 9
Mass of aluminium rod varies directly with length
M $\times$ L
Mass of $16 \mathrm{~cm}$ long rod has $192 \mathrm{~g}$
mals of $x \mathrm{~cm}$ longrod has $\operatorname{los} \mathrm{g}$
$x=\frac{105 \times 16}{192}$
i.e $\frac{x}{16}=\frac{105}{192}$
$x=8.75 \mathrm{~cm}$
`105g mass has length of rod is 8.75cm
Question 10
Given Map scale $\mathrm{1cm}=20 \mathrm{~km}$
Anita Measures a distance from village A to
Village $B$ is $3.5 \mathrm{~cm}$
Actual distance between them is $3.5 \times 20$
$\begin{array}{l}=35 \times 2 \\ = & 70 \mathrm{~km}\end{array}$
Question 11
(i) Height of water tank is 23m 74cm i.e 2375 cm
length of shadow is 20m i .e 2000cm
if height of tree is 9 m 50 cm i .e 950 cm
length of shadow is 'x' = ？
i.e
$\begin{aligned} \frac{x}{2000} &=\frac{950}{2335} \\ x &=\frac{950 \times 2000}{2375} \\ x &=800 \mathrm{~cm} \mathrm{i .e 8m} \end{aligned}$
Length of Shadow is 8 m
(ii) if length of shadow is 12m = 1200 cm
Height of free = ? Let = 'x'
$\frac{x}{2375}=\frac{1200}{2000}$
$x=\frac{1200 \times 2375}{2000}$
$x=1425 \mathrm{~cm}$ i.e $14 \mathrm{~m}$ 2 cm
Heigth of free is $14 \mathrm{~m} 25 \mathrm{~cm}$ hight
Question 12
Eaining of $5 \mathrm{men}=$ Earing of 7 women
Earning of $1 \mathrm{man}=$ earing of $\frac{7}{5}$ women
Earing of $10 \mathrm{men}=$ earning of $\left(\frac{7}{8} \times 10\right)$ women
$=$ earing of $14 \mathrm{wom} \mathrm{en} .$
Earning of 10 men and 13 women = earning of (14 + 13 ) women
= earning of 27 women
Let 10 men and 13 women i .e 27 women earn ₹ x in a day
Note that more women will earn more per day
hence , 10 men and 13 women will earn Rs 2025 in a day
Excercise: 9.2
Question 1
(i) more speed , lets time taken by train to corer a fixed distance
Speed and time or inversely proportional
(ii) more people at work less time taken to finish work
less people at work , more time taken to finish work
i .e This is a inverse variation
Question 2
(i) $90 \times 10-900 ; 15 \times 60=900 ; 45 \times 20=900 ; 30 \times 30 \cdot 900 ;$
$20 \times 45=900$
i.e $90 \times 10=15 \times 60=20 \times 45=30 \times 30=45 \times 20=900=$ Constant
ie $x y=$ constant
$\therefore x$ and $y$ are in invesse variation
(ii) $75 \times 10=750: 45 \times 30=1350 ; 30 \times 25=750 ; 20 \times 35.750$
$10 \times 65=650$
ie $75 \times 10 \neq 45 \times 30 \neq 30 \times 25=20 \times 35 \neq 10 \times 15 \neq$ Constant
ஃ $x$ and $y$ are not in invesse variation
Question 3
Given volume inversely proportional to pressure
$V \alpha \frac{1}{p}$
i.e $P V=$ constant
Volume of a mass = 630 $c m^{3}=V_{1}$
pressure of mercury $=360 \mathrm{~mm}=p_{1}$
if volume is $720 \mathrm{~cm}^{3}$, pressur $9 p_{2}=9$
$p_{1} v_{1}=p_{2} v_{2}$
p = 315 mm
∴ prssure of mercury = 315mm
Question 4
No of children are 20
each received 4 sweets
∴ Total no of sweets are 20 $\times$ 4 = 80 sweets
Given number of children reduced by 4 i .e 20 - 4 = 16
Now 16 (hildren were present
$\therefore$ if is a case of invesse Variation
$\therefore$ each children get $\frac{80}{16}$ sweets i.e 5 sweets
Question 5
pooja has money to buy 36 oranges at the rate
of $24.5$ per orange i.e
She has money $36 \times 4.5$
=₹ 162
Now the price of orange is increased by 90 paise
i.e New price of orange is $4.50+0.90={Rs } 5.40$
Now she takes only $\frac{162}{5.4}$ oranges
i.e 30 oranges only
Question 6
In 8 days, numer of men requived to construct a
wall =12
In 6 days, how many. were required ？
ie No of men required $=x$.
$x=\frac{8 \times 12}{6}$
$x=16$
$\therefore 16$ men required fo constract a wall in 6 days
Question 7
Total no fof taps = 8
these takes 27 minutes to fill a tank
out of 8 , 2 taps go out of order i .e
the remainig no of taps are 6
Let the 5 taps taken x minutes
Note that lesser the number of pipes, more will be the
time required to fill the tank '
So , it is a case of inverse variation
$\begin{aligned} 8 \times 27 &=x \times 6 \\ x &=\frac{8 \times 27}{6} \end{aligned}$
x = 36 minutes
∴ Time taken to fill the tank by 6 taps is 36 minutes
Question 8
No. of person contact to complete a part of stadiums
in 9 months = 560 persons
Let the no of persons to complete a part of stadium
in 5 months = x
more month , less person are required
i.e It is a case of inverse variation
$9 \times 560=8 \times x$
$x=9 \times 70$
$x=630$
Now , no of persons extra required is 630- 560 = 70 persons
Question 9
A batch of bottles were packed in 30 boxes with 10 bottles
in each box
Now 12 bottles in each box can be filled or packed in
how many boxes , Let it be x
More boxes ,less bottles were packed
i.e It is a case of inverse variation
$30 \times 10=12 \times x$
$n=\frac{30 \times 10}{12}$
$x=25$
∴ 25 boxes with 12 bottles in each box
Question 10
Vandana takes 24 mimutes to reach school wilt a
Speed of 5 km /h
Now, how mach speed is required fo reach school with
in 20 minutes
More time less speed is required
i.e it is case of inverse variation
∴ Let the speed b 'x'
$\begin{aligned} 5 \times 24=& x \times 20 \\ x &=\frac{5 \times 24}{20} \\ & x=6 \mathrm{~km} / \mathrm{h} \end{aligned}$
∴ Speed required = 6 km/h
Question 11
After 15 days , the food for 80 soldiers for (60 - 15)days
i.e 45 days
So the number of soldiers in ford 80 + 20= 100
Let the food last for x days , whe these are 100
soldiers in food
note that more the number of soldiers in ford , the sooner the food exhaust
$\begin{aligned} 80 \times 45=& 100 \times x \\ x=& \frac{80 \times 45}{100} \\ x=36 \end{aligned}$
∴ The food will last for 36 days
Question 12
1200 Soldiers in a fort had enough food for 28 days
after 4 days some soldiere were sent be x
$\therefore$ The food lasted for 32 more days
If is a case of invesse variation
$\begin{aligned} 1200-x &=\frac{1200 \times 24}{32} \\ 1200-x &=900 \\ x &=1200-900=300 \end{aligned}$
$\therefore$ No of Soldier were left = 300. Soldiers
Excercise: 9.3
Question 1
Farmer can reap a field in 10 days
his wife can reap it in 8 days
Farmer one day work i,e reap a field is $\frac{1}{10} .$
His wife one day work is $\frac{1}{8}$
Fasmer and wife can completed if Key work together
i.e $\frac{1}{10}+\frac{1}{8}$
$\quad=\frac{4+5}{40}=\frac{9}{40}$
∴ Both together can complete in $\frac{40}{9}$ days
Question 2
Since A can complete $\frac{1}{5} k$ of work in 2 days
$\therefore$ A's one day work $=\frac{1}{2}$ of $\frac{1}{5}=\frac{1}{2} \times \frac{1}
{5}=\frac{1}{10}$
Since $B$ Can Complete $\frac{2}{3}$ nd of wok in 8 days
$\therefore B's $ is one day wak $=\frac{2}{3}$ of $\frac{1}{8}=\frac{2}{3} \times \frac{1}
{84}=\frac{1}{12}$
One days work of A and B together = $\frac{1}{10}+\frac{1}{12}$
$\frac{6+5}{60}=\frac{11}{60} .$
∴ A and B working together can complete the work in $\frac{60}{11}$ days
Question 3
'A' tap can fill a tank in 20 minutes
A tap one minutes can fill $\frac{1}{20}th $ of tank
'B" Tap can fill a tank in 12 mint
In one minute, Tap $B$ can fill $\frac{1}{12}th$ of Tank
If both taps were opened then
In one minutes, $\operatorname{Tap} A$ and $B$ can fill $=\frac{1}{20}+\frac{1}{12}$
$=\frac{3+5}{60}$
$=\frac{8}{60}$
∴ Both A and B will fill the tank in $\frac{60}{8}$ minutes
Question 4
A can do a work in 6 days
B Can do a work in 8 days
A's one day wark $=\frac{1}{6}$
B's one day work $=\frac{1}{8}$
One days work of A and B Together = $\frac{1}{6}+\frac{1}{8}$
$\frac{4+3}{24}=\frac{7}{24}$
∴ 2 days work of A and B together = $2 \times \frac{7}{24}=\frac{7}{12}$
∴ Remaining work = $1-\frac{7}{12}=\frac{5}{12}$
∴ The no. OF days taken by A to finish the remaining work
$=\frac{\text { work } to \text { be done }}{\text { A's one day work }}=\frac{5 / 12}{1 / 6}=\frac{5}{122} \times6 =\frac{5}{2}$ day
Hence , A will complete the remaining work in $\frac{5}{2}$ days
Question 5
A can do a piece of work in 40 days
A's one day work = $\frac{1}{40}$
He works for 8 days , he complete $8 \times \frac{1}{40}=\frac{1}{5} th$ wok
Remaining work = $1-\frac{1}{5}=\frac{4}{5} .$
B finisher remaining work in 16 days
i.e B finisher $\frac{4}{5} th$ wok in $16 \times \frac{5}{4}$ day
$=20$ days
$\begin{aligned} A \text { and } B \text { Can completed in } &=\frac{1}{20}+\frac{1}{40} \\ &=\frac{2+1}{40}=\frac{3}{4{0}} . \end{aligned}$
∴ It takes $\frac{40}{2}$ days if they do together.
Question 6
A and B seperately do work in 10 and 15 days
As's one day work $=\frac{1}{10}$ days
B's one day work $=\frac{1}{15}$ days
$\begin{aligned} A \text { and } B \text { one's day wak } &=\frac{1}{10}+\frac{1}{15} \\ &=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6} . \end{aligned}$
A completed the remaining work in 5 days
i .e $\frac{1}{2}$ of the work had completed
Remaining half work had completed by A and B together
i.e $\frac{1}{2} \times 6=3$ days
Question 7
1 women or 5 gilis take 17 days to complete a piece
of work
Since 3 women's work = 5 girls work
1 woman's work $=\frac{5}{3}$ girl's work
$: 7$ women work $=\frac{5}{3} \times 7$ i.e $\frac{35}{3}$ gins work
∴ 7women and 11 girls work $=\frac{35}{3}+11$ i .e $\frac{68}{3}$ girls work
Since 5 girls can do work in 17 days
∴ 1 girl can do work in $5 \times 17$ i .r 85 days
$\frac{68}{3}$ girls can do the work in $\frac{85}{68 /3}$ days
$\frac{83 \times 3}{68}$
$=\frac{15}{4}$ days
Hence 7 women and 11 girls working together will complete
the work in $\frac{15}{4}$ days
Question 8
A's one day work $=\frac{1}{10}$
B's one day wok $=\frac{1}{15}$.
So they divide money in the ratio $\frac{1}{10}: \frac{1}{15}$
i.e $\frac{1}{10} \times 30: \frac{1}{15} \times 30$ i.e $3: su m=3+2=5$
A's share = $\frac{3}{5} \times 3500=3 \times 700=2100$
B's share : $\frac{2}{5} \times 3,500=2 \times 700=1400$
Question 9
A's one day wak $=\frac{1}{2}$
B's one day work $r \frac{1}{6}$
C's one day work $=\frac{1}{3}$.
A, B , C one's days work together =$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
$\frac{3+2+1}{6}=\frac{6}{6}=1$
$A, B, C$ completed their work by working togeter in 1 day
So they divide the maney in the ratio $\frac{1}{2}: \frac{1}{6}: \frac{1}{3}$
i.e
$\begin{aligned} & \frac{1}{2} \times 6: \frac{1}{6} \times 6: \frac{1}{3} \times 6 \\=& 3: 1: 2 \end{aligned}$
Sum of these terms = 3+ 1+ 2 + 6
A's share $=\frac{3}{8} \times 960=3 \times 160=480$
B's share $=\frac{1}{6} \times 960=1 \times 160=160$
c's share $=\frac{2}{6} \times 960=2 \times 160=320$
Question 10
A and B , c together do a piece of work in 15 days
one's days work of A,B and C together is $\frac{1}{15}$
B's one day work $=\frac{1}{30}$
C's one day work $=\frac{1}{40}$.
A's one day work =$\frac{1}{15}-\left\{\frac{1}{30}+\frac{1}{40}\right\}$
$\frac{1}{15}-\left\{\frac{4+3}{120}\right\}$
$\frac{1}{15}-\frac{7}{120}$
$=\frac{8-7}{120}$
A's one day work = $\frac{1}{120}$
∴ A alone do the work in 120 days
Question 11
A's one day +B's one day + C's one day =$\frac{5}{24}$
A's one day $+C$ 's one $d a y=\frac{1}{8}$.
B's one day work
$\begin{aligned}=& \frac{5}{24}-\frac{1}{8} \\ &=\frac{5-3}{2 4} \\ &=\frac{2}{24}=\frac{1}{12} \end{aligned}$
∴ B alone can plough field in 12 days
Question 12
A's one day work +B' s one ofay work $=\frac{1}{10}$
B's one day work +C's is one day work $=\frac{1}{15}$
C's one day wodr +A's As one day work $=\frac{1}{12}$
2 (A's one day work + B's one days work +C's one day work)
$=\frac{1}{10}+\frac{1}{15}+\frac{1}{12}$
$=\frac{6+4+5}{60}=\frac{15}{60}-\frac{1}{4}$
A's one day work + B's one day work + C's one day work = $\frac{1}{4} \times \frac{1}{2}$
$=\frac{1}{8}$
$\therefore A, B, C together can complete in 8 days
A's one day work = $\frac{1}{8}-\frac{1}{15}$
$=\frac{15-8}{120}=\frac{7}{120}$
∴ A alone takes $\frac{120}{7}$ days
B's one day work =$\frac{1}{8}-\frac{1}{12}$
= $\frac{3-2}{24}=\frac{1}{24}$
B alone takes 24 days
C's one day work $=\frac{1}{8}-\frac{1}{10}$
$=\frac{5-4}{40}=\frac{1}{40}$
C alone takes 40 days
Question 13
Pipe fill a tank in 12 hour
In one hour , pipe fills $\frac{1}{12}$ of the tank
A waste pipe is left opened and filled in 16 hour
In one hour , it fills $\frac{1}{16}$ of tank
∴ portion of tank emptied by the waste pipe
in one hour = $\frac{1}{12}-\frac{1}{16}$
=$\frac{4-3}{48}$
$=\frac{1}{48}$
∴ Waste pipe takes 48 hours to empty the tank
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