Chapter 3 Review 3-1 to 3-6 Y Varies Directly With X Write an Equation for the Direct Variation

MSBSHSE Solutions For Course 8 Maths Part 1 Chapter vii – Variation is provided hither to help students ready for their exams at ease. Students who confront difficulty in understanding the concepts during form hours can apace refer to MSBSHSE Solutions. The facilitators at BYJU'South have formulated the solutions based on the students' grasping abilities. This affiliate mainly deals with direct variation, changed variation, Fourth dimension, Work and Speed concepts. Students can solve exercise wise bug to increment their confidence level earlier appearing for the lath exam. To heave interest amongst students in this chapter, Maharashtra State Board Class eight Textbooks Role 1 pdf links are given here for easy access. Students tin can refer and download the PDF of Maharashtra Board Solutions for Class viii Maths Chapter 7 Variation, which is readily bachelor for download from the links provided.

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Access answers to Maharashtra Board Solutions For Class viii Maths Office ane Chapter vii- Variation.

Exercise set vii.1 PAGE NO: 36

1. Write the following statements using the symbol of variation.
(ane) Circumference (c) of a circle is direct proportional to its radius (r).
(two) Consumption of petrol (I) in a car and distance traveled by that machine (D) are in direct variation.

Solution:

(1) Circumference = c and radius = r

So, c ∝ r or c = kr, where chiliad = constant

(2) Consumption of petrol in a car = I

Distance traveled by that motorcar = D

So, I ∝ D or I = kD, where k = constant

2. Complete the post-obit table considering that the cost of apples and their number are in directly variation.

Number of apples (10)	1	iv	…	12	…
Cost of apples (y)	viii	32	56	…	160

Solution:

Number of apples (ten) and the price of apples (y) are in direct variation.

y ∝ x

y = kx … (i) where grand is abiding of variation

At present permit u.s. consider the conditions,

When, 10 = 1, y = viii

Substitute the value of ten = 1 and y = 8 in (i), we become

viii = k × one

k = viii

Substituting the value of k = 8 back in (i), we become

y = kx

∴ y = 8x … (ii)

This the equation of variation

When, y = 56, x =?

Substituting y = 56 in (two), we go

y = 8x

56 = 8x

ten = 568

∴ 10 = 7

When, x = 12, y =?

Substituting x = 12 in (2), we get

y = 8x

y = 8 × 12

∴ y = 96

When, y = 160, x =?

Substituting y = 160 in (ii), we get

y = 8x

160 = 8x

x = 1608

∴ 10 = 20

Number of apples (x)	i	4	7	12	xx
Cost of apples (y)	8	32	56	96	160

3. If m ∝ n and when m = 154, north = 7. Detect the value of thousand, when n = 14.

Solution:

Given:

m ∝ n

g = kn … (i) where k is abiding of variation.

When g = 154, north = 7

Substitute the value of thousand = 154 and n = vii in (i), we get

chiliad = kn

154 = yard × seven

m = 1547

yard = 22

Now, substitute the value of k = 22 back in (i), we get

m = kn

∴ m = 22n … (ii)

This is the equation of variation.

When, n = 14, m =?

Substitute n = 14 in (two), nosotros get

m = 22n

m = 22 × 14

m = 308

∴ The value of m is 308.

4. If north varies directly as m, complete the following table.

m	3	v	vi.five	…	i.25
n	12	twenty	…	28	…

Solution:

Given:

n varies direct as m

And so, n ∝ m

northward = km …(i) where, one thousand is the constant of variation

Now let us consider the conditions,

When k = 3, n = 12

Substitute the value of m = three and n = 12 in (i), we go

n = km

12 = g × 3

one thousand=123

one thousand = 4

Substitute the value of thou = 4 back in (i), we get

north = km

∴ n = 4m … (ii)

This is the equation of variation.

When yard = 6.5, n =?

Substituting, m = 6.five in (ii), we become

n = 4m

n = 4 × six.5

∴ due north = 26

When n = 28, thou =?

Substituting, n = 28 in (2), nosotros get

northward = 4m

28 = 4m

28 = 4m

thou=284

∴ m = 7

When m = 1.25, n =?

Substituting chiliad = 1.25 in (ii), we get

due north = 4m

north = 4 × 1.25

∴ north = 5

m	3	5	vi.5	7	one.25
n	12	20	26	28	5

5. y varies directly as the foursquare root of 10. When 10 = xvi, y = 24. Find the abiding of variation and equation of variation.

Solution:

Given:

y varies directly as square root of ten.

And so, y ∝ √4x

y = grand √x … (i) where, k is the constant of variation.

When x = 16, y = 24.

Substituting, ten = 16 and y = 24 in (i), nosotros get

y = k√10

24 = k√xvi

24 = 4k

k =244

k = half dozen

Substitute the value of one thousand = six back in (i), nosotros get

y = k√x

y = vi√10

This is the equation of variation.

∴ The constant of variation is 6 and the equation of variation is y = 6√x .

6. The total remuneration paid to laborers, employed to harvest soybeans is indirect variation with the number of laborers. If remuneration of 4 laborers is Rs grand, observe the remuneration of 17 laborers.

Solution:

Let full remuneration paid to laborers be = 'm' and

Number of laborers employed to harvest soybean be = 'n'.

Since, the total remuneration paid to laborers, is in direct variation with the number of laborers.

So, m ∝ n

∴ m = kn … (i) where, one thousand = constant of variation

Remuneration of 4 laborers is Rs 1000.

When n = four, m = Rs g

And then, substitute the value of n = 4 and m = 1000 in (i), we get

m = kn

k = thou × 4

k =10004

chiliad = 250

Now, substitute the value of thousand = 250 dorsum in (i), we get

k = kn

∴ m = 250 north … (2)

This is the equation of variation

At present, permit us find the remuneration of 17 laborers.

When n = 17, m =?

Substituting northward = 17 in (2), nosotros get

thousand = 250 north

k = 250 × 17

m = 4250

∴ The remuneration of 17 laborers is Rs 4250.

Practice prepare seven.2 PAGE NO: 38

1. The information near numbers of workers and the number of days to complete work is given in the post-obit tabular array. Complete the tabular array.

Number of workers	xxx	20		10
Days	vi	9	12		36

Solution:

Permit the number of workers be = 'northward'

Number of days required to complete a piece of work be ='d'

Since, number of workers and number of days to complete a work are in inverse proportion.

n ∝ (1/d)

n = k × (1/d) where chiliad, is the abiding of variation.

∴ north × d = yard …(i)

When northward = 30, d = half dozen

Substitute the value of n = 30 and d = 6 in (i), we go

due north × d = chiliad

30 × six = 1000

chiliad = 180

Now, substitute the value of k = 180 back in (i), we go

n × d = m

∴ due north × d = 180 … (ii)

This is the equation of variation

When d = 12, due north = 7

Substituting d = 12 in (2), we get

n × d = 180

north × 12 = 180

northward = 18012

∴ n = 15

When n = x, d =?

Substituting northward = x in (two), we become

n × d = 180

x × d = 180

d = 18010

∴ d = 18

When d = 36, n =?

Substituting d = 36 in (ii), we become

due north × d = 180

n × 36 = 180

n = 18036

∴ n = 5

Number of workers	30	20	fifteen	x	5
Days	six	ix	12	18	36

2. Find constant of variation and write equation of variation for every example given below:

Solution:

It is given that, p ∝ i/q

p = k × 1/q where, thou is the constant of variation.

∴ p × q = k …(i)

When p = 15, q = four

Substitute the value of p = 15 and q = four in (i), we get

p × q = k

15 × 4 = yard

k = lx

Now, substitute the value of k = 60 back in (i), we get

p × q = k

∴ p × q = 60

This is the equation of variation.

∴ The constant of variation is 60 and the equation of variation is pq = 60.

It is given that, z ∝ one/w

z = k × 1/west where, chiliad is the abiding of variation,

∴ z × w = m …(i)

When z = 2.5, west = 24

Substitute the value of z = 2.v and west = 24 in (i), we get

z × w = k

2.v × 24 = k

thou = 60

Now, substitute the value of k = lx back in (i), we become

z × w = k

∴ z × due west = threescore

This is the equation of variation.

∴ The constant of variation is 60 and the equation of variation is zw = lx.

Information technology is given that, s ∝ 1/t²

s = k × (1/t²) where, 1000 is the constant of variation,

∴ due south × t² = m …(i)

When s = 4, t = 5

Substitute the value of s = 4 and t = 5 in (i), we become

due south × t² = k

4 × (five)² = g

k = iv × 25

k = 100

Substitute the value of k = 100 dorsum in (i), we get

s × t² = k

∴ s × t² = 100

This is the equation of variation.

∴ The constant of variation is 100 and the equation of variation is st² = 100.

Information technology is given that, x ∝ 1/√y

x = k × (1/√y) where, thou is the constant of variation,

∴ x × √y = thou …(i)

When x = 15, y = 9

Substitute the value of x = 15 and y = 9 in (i), nosotros get

x × √y = k

fifteen × √9 = k

k = 15 × 3

1000 = 45

Now, substitute the value of 1000 = 45 back in (i), we get

x × √y = thou

∴ ten × √y = 45.

This is the equation of variation.

∴ The constant of variation is k = 45 and the equation of variation is ten√y = 45.

iii. The boxes are to be filled with apples in a heap. If 24 apples are put in a box and then 27 boxes are needed. If 36 apples are filled in a box how many boxes will be needed?

Solution:

Permit the number of apples in each box exist = 'x'

Total number of boxes required be = 'y'

The number of apples in each box are varying inversely with the total number of boxes.

And so, x ∝ 1/y

10 = k × (i/y) where, 1000 is the constant of variation,

∴ x × y = chiliad … (i)

If 24 apples are put in a box then 27 boxes are needed.

When x = 24, y = 27

Substitute the value of x = 24 and y = 27 in (i), we go

x × y = g

24 × 27 = k

k = 648

Now, substitute the value of k = 648 back in (i), we go

x × y = k

∴ x × y = 648 … (ii)

This is the equation of variation.

Now, allow united states detect number of boxes needed when 36 apples are filled in each box.

So, when x = 36, y =?

Substituting x = 36 in (ii), we get

ten × y = 648

36 × y = 648

y = 64836

y = xviii

∴ If 36 apples are filled in a box then 18 boxes are required.

4. Write the post-obit statements using symbol of variation.

(1) The wavelength of sound (50) and its frequency (f) are in inverse variation.

(ii) The intensity (I) of light varies inversely with the square of the distance (d) of a screen from the lamp.

Solution:

(1) Wavelength of audio (l) and frequency (f) are in inverse proportion.

50 ∝ 1/f

(2) Intensity (I) of light varies inversely

With the square of the distance (d)

l ∝ i/d²

5. 10 ∝ 1/√y and when x = twoscore and then y = sixteen. If x = 10, find y.

Solution:

It is given that, x ∝ i/√y

x = k × (ane/√y) where, k is the constant of variation.

∴ 10 × √y = k …(i)

When x = twoscore, y = 16

Substitute the value of x = 40 and y = 16 in (i), we get

10 × √y = k

40 × √sixteen = one thousand

yard = twoscore × 4

thousand = 160

At present, substitute the value of k = 160 back in (i), we get

x × √y = thousand

∴ x × √y = 160 … (2)

This is the equation of variation.

When x = x, y =?

Substitute the value of ten = x in (two), we get

10 × √y = 160

10 × √y = 160

√y = 16010

√y = 16

Square on both the sides, we become

y = 256

∴ Value of y is 256.

6. x varies inversely equally y, when x = 15 then y = 10, if x = 20 then y =?

Solution:

Given:

x ∝ one/√y

ten = k × (1/√y) where, yard is the abiding of variation.

∴ x × y = k …(i)

When x = xv, y = 10

Substitute the value of ten = 15 and y = ten in (i), we get

x × y = k

fifteen × 10 = k

k = 150

Now, substitute the value of m = 150 back in (i), nosotros go

x × y = yard

∴ 10 × y = 150 … (2)

This is the equation of variation.

When 10 = 20, y =?

Substitute the value of x = xx in (2), we get

x × y = 150

xx × y = 150

y = 15020

y = 7.5

∴ Value of y is vii.5

Do ready 7.three Page NO: 40

1. Which of the following statements is of inverse variation?
(1) The number of workers on a chore and time taken by them to complete the job.
(2) The number of pipes of the same size to fill a tank and the fourth dimension taken by them to fill the tank.
(3) Petrol filled in the tank of a vehicle and its toll.
(four) Area of the circle and its radius.

Solution:

(one) Equally the number of workers increases, the time required to complete the task decreases.

Hence, information technology is of inverse variation.

(2) Equally the number of pipes increases, the time required to make full the tank decreases.

Hence, information technology is of inverse variation.

(3) Every bit the quantity of petrol in the tank increases, its cost increases.

Hence, it is of straight variation.

(4) As the area of circumvolve increases, its radius increases.

Hence, it is of straight variation.

ii. If xv workers tin build a wall in 48 hours, how many workers will exist required to do the aforementioned work in 30 hours?

Solution:

Let the number of workers edifice the wall be = 'n'

The time required exist ='t'

Since, the number of workers varies inversely with the time required to build the wall.

n ∝ 1/t

n = k × (1/t) where, grand is the constant of variation

∴ n × t = chiliad …(i)

15 workers tin can build a wall in 48 hours,

when north = fifteen, t = 48

Substitute the value of n = 15 and t = 48 in (i), we get

n × t = k

15 × 48 = k

k = 720

At present, substitute the value of k = 720 back in (i), we get

north × t = m

∴ n × t = 720 … (2)

This is the equation of variation.

Now, let us discover number of workers required to do the same work in 30 hours.

When t = 30, n =?

Substitute the value of t = thirty in (ii), we get

north × t = 720

due north × thirty = 720

northward = 72030

northward = 24

∴ 24 workers are required to build the wall in xxx hours.

3. 120 bags of half liter milk tin can be filled by a machine within 3 minutes find the time to fill such 1800 bags?

Solution:

Permit the number of numberless of half liter milk exist = 'b'

The time required to fill the bags ='t'

Since, the number of bags and fourth dimension required to fill up the bags varies directly.

b ∝ t

∴ b = kt …(i) where 1000 is the constant of variation.

Since, 120 bags can be filled in 3 minutes

When b = 120, t = 3

Substitute the value of b = 120 and t = 3 in (i), nosotros get

b = kt

120 = grand × 3

k = 1203

thousand = 40

Now, substitute the value of k = 40 back in (i), nosotros become

b = kt

∴ b = 40 t … (two)

This is the equation of variation.

Now, permit united states find the fourth dimension required to fill 1800 numberless

When b = 1800, t =?

Substitute the value of b = 1800 in (two), we become

b = 40 t

1800 = forty t

t = 180040

t = 45

∴ 1800 bags of half liter milk can be filled past the machine in 45 minutes.

four. A motorcar with a speed of 60 km/hr takes eight hours to travel some distance. What should exist the increase in the speed if the aforementioned distance is to be covered in 7 ½ hours?

Solution:

Let the speed of car in km/60 minutes be = 'v'

The time required be ='t'

Since, speed of a car varies inversely as the time required to cover a distance.

five ∝ one/t

5 = chiliad × (ane/t) where, thou is the constant of variation.

∴ five × t = k …(i)

Since, a automobile with speed sixty km/60 minutes takes 8 hours to travel some distance.

When v = 60, t = 8

Substitute the value of v = 60 and t = 8 in (i), we go

v × t = thou

60 × 8 = t

k = 480

At present, substitute the value of k = 480 back in (i), nosotros become

v × t = 1000

∴ 5 × t = 480 … (ii)

This is the equation of variation.

Now, let united states discover the speed of motorcar if the same distance is to exist covered in 7 ½ hours.

When t = 7 ½ = 7.v, v =?

Substitute the value of t = seven.5 in (2), we become

v × t = 480

five × 7.v = 480

v = 480/vii.five

v = 64

The speed of vehicle should be 64 km/hour to cover the aforementioned distance in seven.5 hours.

The increase in speed = 64 – 60

= 4km/hr

∴ The increase in speed of the automobile is four km/60 minutes.

Students can refer to MSBSHSE Solutions to have a skillful hold of the concepts covered in this chapter. Affiliate 7 Variation, which includes Direct variation and Changed variations provides useful resource for students as it helps them in scoring full marks in the examination. Many such exercise problems are given for students to do and to secure expert marks in their examination.

Frequently Asked Questions on Maharashtra State Board Solutions for Class 8 Maths Chapter vii Variation

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Yeah, these Maharashtra Board Grade 8 Maths Chapter 7 Variation Solutions can be very helpful. Students are highly recommended to do these solutions after revising the subjects, as they set the basis for the questions that could get asked often in the board exams.

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