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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