S.chand mathematics solution class 8 chapter 10 Direct and inverse variation

EXERCISE 10 A

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Q1 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper

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Question 1

Given that y is directly proportional to x and y = 40 , when x = 200 . Find the value of
(i) y , when x = 15

Sol: (i)

y ∝ x ⇒ $\dfrac{y}{x}=k$ , where k is the constant of variation

Given , $\dfrac{y}{x}=\dfrac{40}{200}=\dfrac{1}{5}$ ⇒ k = $\dfrac{1}{5}$

Also , $y = \dfrac{x}{5}$

$y=\dfrac{15}{5}$ [given: x=15]

= 3

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(ii) x when y = 8

Sol: (ii)

y ∝ x ⇒ $\dfrac{y}{x}=k$ , where k is the constant of variation

Given , $\dfrac{y}{x}=\dfrac{40}{200}=\dfrac{1}{5}$ ⇒ k = $\dfrac{1}{5}$

Also , 5y = x [given: y=8]

x = 40

 

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Question 2

The length (in cm) stretched by a spring is directly proportional to the amount of force (in kg) applied . Given below are some observations about the force applied and the length stretched by a spring . Find the missing values in the table

Force (in kg)		20	25
Length stretched(in cm)	18	15		28

Sol:

As Force and Length vary directly . So $\dfrac{\text{Force}}{\text{Length}}$ is constant

$\dfrac{\text{Force}}{\text{Length}}=\dfrac{20}{15}=\dfrac{4}{3}$

∴The constant of variation $=\dfrac{4}{3}$

Now , $\dfrac{\text{Force}}{18}=\dfrac{4}{3}$⇒$\text{Force}=\dfrac{4\times 18}{3}$ = 24

$\dfrac{25}{\text{Length}}=\dfrac{4}{3}$⇒$\text{Length}=\dfrac{3\times 25}{4}=18\dfrac{3}{4}$

$\dfrac{\text{Force}}{28}=\dfrac{4}{3}$⇒$\text{Force}=\dfrac{4\times 28}{3}=37\dfrac{1}{3}$

 

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Question 3

Priya takes 4 hours in walking a distance of 20 km . What distance would she cover in 7 hours ?

Sol:

Distance	20 km
Time	4 hours	7 hours

As Distance ∝ Time ⇒$\dfrac{\text{Distance}}{\text{Time}}=k$

So $\dfrac{\text{Distance}}{\text{Time}}$ is constant

$\dfrac{\text{Distance}}{\text{Time}}=\dfrac{20}{4}=5$

∴The constant of variation = 5

Now , $\dfrac{\text{Distance}}{7}=5$

Distance = 7×5 = 35

ALTERNATE METHOD

As we know $\text{Speed}=\dfrac{\text{Distance}}{\text{time}}$

$=\dfrac{20}{4}=5$

= 5 km/hr

7 hours = 7×5 km

= 35 km

 

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Question 4

If 15 burners consume 90 cubic metre of gas in 2 hours , how much will 9 burners consume in the same time ?

Sol:

Gas (m3)	90 m3
Burner	15	9

As Gas consume ∝ Burner ⇒$\dfrac{\text{Gas}}{\text{Burner}}=k$

So $\dfrac{\text{Gas}}{\text{Burner}}$ is constant

$\dfrac{\text{Gas}}{\text{Burner}}=\dfrac{90}{15}=6$

∴The constant of variation = 6

Now , $\dfrac{\text{Gas}}{9}=6$

Distance = 6×9 = 54

= 54 cubic metre or 54 m³

 

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Question 5

The railway charges 5600 to carry a certain amount of luggage for 350 km . What should the charge be carry the same amount of luggage for 425 km ?

Sol :

Charges	5600
Distance (Km)	350 km	425 km

As Charges ∝ Distance ⇒$\dfrac{\text{Charges}}{\text{Distance}}=k$

So $\dfrac{\text{Charges}}{\text{Distance}}$ is constant

$\dfrac{\text{Charges}}{\text{Distance}}=\dfrac{5600}{350}=16$

∴The constant of variation = 16

Now , $\dfrac{\text{Charges}}{425}=16$

Distance = 16×425 = 6800

 

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Question 6

89 litres of oil cost 2091.50  . What is the cost of 15 litres ?

Sol :

Oil Cost	2091.50
Litres	89	15

As Cost ∝ Litres ⇒$\dfrac{\text{Cost}}{\text{Litre}}=k$

So $\dfrac{\text{Cost}}{\text{Litre}}$ is constant

$\dfrac{\text{Cost}}{\text{Litre}}=\dfrac{2091.50}{89}=23.50$

∴The constant of variation = 23.50

Now , $\dfrac{\text{Cost}}{15}=23.50$

Cost = 23.50×15 = 352.50

 

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Question 7

68 packets weigh 1 kg 632 g . What will be weight of 70 packets ?

Sol :

Weigh	1 kg 632 g (1632g)
Packets	68	70

As Weigh ∝ Packets ⇒$\dfrac{\text{Weigh}}{\text{Packets}}=k$

So $\dfrac{\text{Weigh}}{\text{Packets}}$ is constant

$\dfrac{\text{Weigh}}{\text{Packets}}=\dfrac{1632}{68}=24$

∴The constant of variation = 24

Now , $\dfrac{\text{Weigh}}{70}=24$

Cost = 24×70 = 1680 g

1680 g = 1 kg 680 g

 

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Question 8

If a man working for 49 hours earns 1715 , how much will he earn working for 27 hours ?

Sol :

Earns	1715
Hours (hours)	49	27

As Earns ∝ Hours ⇒$\dfrac{\text{Earns}}{\text{Hours}}=k$

So $\dfrac{\text{Earns}}{\text{Hours}}$ is constant

$\dfrac{\text{Earns}}{\text{Hours}}=\dfrac{1715}{49}=35$

∴The constant of variation = 35

Now , $\dfrac{\text{Earns}}{27}=35$

Cost = 35×27 = 945

 

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Question 9

The distance travelled by a ball dropped from an airplane is directly proportional to the square of time t . Given that t=2 seconds when d = 24 meters , find the distance the ball drops in 10 seconds .

Sol:

Distance (m)	24 m
Time (sec)	2 sec	10 sec

As Distance ∝ ${Time}^2$ 

⇒$\dfrac{\text{Distance}}{\text{Time}^2}=k$

So $\dfrac{\text{Distance}}{\text{Time}^2}$ is constant

$\dfrac{\text{Distance}}{\text{Time}}=\dfrac{24}{2^2}=\frac{24}{4}=6$

∴The constant of variation = 6

Now , $\dfrac{\text{Distance}}{{10}^2}=\frac{D}{100}=6$

Distance = 60×10 = 600

= 600 metres

 

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