EXERCISE 10 A
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
(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
Q2 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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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
$\dfrac{25}{\text{Length}}=\dfrac{4}{3}$⇒$\text{Length}=\dfrac{3\times 25}{4}=18\dfrac{3}{4}$
$\dfrac{\text{Force}}{28}=\dfrac{4}{3}$⇒$\text{Force}=\dfrac{4\times 28}{3}=37\dfrac{1}{3}$
Q3 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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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
Q4 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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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
Q5 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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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
Q6 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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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
Q7 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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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
Q8 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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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
Q9 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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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
9 one is wrong sir it's answer is 600
ReplyDeleteThanks. "Fixed for you"
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