ML AGGARWAL CLASS 8 CHAPTER 9 Direct and Inverse Variation Exercise 9.1

 Excercise: 9.1

Question 1 

Sol :
(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

Sol :
(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

Sol :
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

Sol :
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

Sol :
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

Sol :
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 km

∴  Number of kilometers travelled is 396 km

Question 7

Sol :
Let the number of sheets of cardboard be x

thickness of 12 cardboard sheets is 45 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

Sol :
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

Sol :
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 cm

105g mass has length of rod is 8.75cm

Question 10

Sol :
Given Map scale $\mathrm{1cm}=20 \mathrm{~km}$

Anita Measures a distance from village A to

Village B is 3.5 cm
Actual distance between them is $3.5 \times 20$

$\begin{array}{l}=35 \times 2 \\ = & 70 \mathrm{~km}\end{array}$

Question 11

Sol :
(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

Sol :
Earning of 5 men= Earing of 7 women

Earning of 1 man=$ earing of $\frac{7}{5}$ women

Earing of 10 man= earning of $\left(\frac{7}{8} \times 10\right)$ women

= earing of 14 women

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 

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