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

EXERCISE 10 A


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


(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

 


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)2025
Length stretched(in cm)181528

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}

 


Question 3

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

Sol:

Distance20 km
Time4 hours7 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

 


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
Burner15  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

 


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 :

Charges5600
Distance (Km)350 km425 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

 


Question 6

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

Sol :

Oil Cost2091.50
Litres8915

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

 


Question 7

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

Sol :

Weigh1 kg 632 g (1632g)
Packets6870

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

 


Question 8

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

Sol :

Earns1715
Hours (hours)4927

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

 


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 sec10 sec

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

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

\dfrac{\text{Distance}}{\text{Time}}=\dfrac{24}{2}=12

∴The constant of variation = 12

Now , \dfrac{\text{Distance}}{10}=12

Distance = 12×10 = 120

= 120 metres

 


 

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