c ∝ r

l ∝ d

x ∝ y          ... (Given)

In a direct variation, the ratio of the two variables remains constant

∴ x = ky     ... (k is a constant.)

When x = 1, y = 8  ∴ 1 = k × 8

∴ k = \(\frac{1}{8}\)

∴ x = \(\frac{1}{8}\)y is the equation of variation.

(i) When y = 56,

x = \(\frac{1}{8}\) × 56 = 7

(ii) When x = 12,

12 = \(\frac{1}{8}\)y

∴ y = 12 × 8 = 96

(iii) When y = 160,

∴ x = \(\frac{1}{8}\) × 160 = 20

Answer : The completed table is given below :

It is given that

m ∝ n

∴ m = kn, where k is constant of variation

When m = 154, n = 7.

∴ 154 = k × 7

∴ \(\frac{154}{7}\) = k

∴ k = 22

Therefore, the equation of variation is m = 22n.

When n = 14,

M = 22 × 14 = 308

Thus, the value of m is 308.

It is given that n varies directly as m, i.e., n ∝ m

∴ n = km, where k is the constant of variation

When m = 3, n = 12.

∴ 12 = k × 3

k = \(\frac{12}{3}\)

∴ k = 4

So, the equation of variation is n = 4m.

When m = 6.5,

n = 4 × 6.5

n = 26

When n = 28,

28 = 4 × m

m = \(\frac{28}{4}\)

∴ m = 7

When m = 1.25,

n = 4 × 1.25

n = 5

Answer : The complete table is given below.

It is given that y varies directly as square root of x i.e. y ∝ \(\sqrt{x}\)

∴ y = k\(\sqrt{x}\), where k is constant of variation

When x = 16, y = 24.

∴ 24 = k × \(\sqrt{16}\)

∴ 24 = 4k

∴ k = \(\frac{24}{4}\) = 6

So, the equation of variation is y = 6\(\sqrt{x}\)

Thus, the constant of variation is 6 and the equation of variation is y = 6\(\sqrt{x}\).

Let the total remuneration paid to labourers be ₹ y and the number of labourers employed be x.

It is given that y varies directly as x, i.e., y ∝ x.

∴ y = kx, where k is constant of variation

When x = 4, y = 1000.

∴ 1000 = k × 4

∴ k = \(\frac{1000}{4}\) = 250

So, the equation of variation is y = 250x.

When x = 17,

y = 250 × 17 = ₹ 4,250

Thus, the total remuneration of 17 labourers is ₹ 4,250.

It can be observed from the given table that if the number of workers is increased from 20 to 30, then the number of days to complete the work decreases from 9 days to 6 days.

So, the number of workers and the number of days to complete a work are in inverse variation.

Let the number of workers be y and the number of days to complete a work be x.

Here, y inversely varies as x, i.e., y ∝ \(\frac{1}{x}\)

∴ y = \(\frac{k}{x}\), where k constant of variation

∴ x.y = k

When y = 30, x = 6.

∴ k = 6 × 30 = 180

So, the equation of variation is xy = 180.

When x = 12,

12y = 180

∴ y = \(\frac{180}{12}\)

∴ y = 15

When y = 10,

10x = 180

x = \(\frac{180}{18}\)

x = 18

When x = 36

36y = 180

∴ y = \(\frac{180}{36}\)

∴ y = 5

Answer : The complete table is given below.

p ∝ \(\frac{1}{q}\)

∴ p = \(\frac{k}{q}\), where k is constant of variation

When p = 15, q = 4.

∴ 15 = \(\frac{1}{4}\)

∴ k = 15 × 4 = 60

So, the equation of variation is p = \(\frac{60}{q}\) or pq = 60.

z ∝ \(\frac{1}{w}\)

∴ z = \(\frac{k}{w}\), where k is constant of variation

When z = 2.5, w = 24.

∴ 2.5 = \(\frac{k}{24}\)

∴ k = 2.5 × 24 = 60     ... (The constant of variation)

So, the equation of variation is z = \(\frac{60}{w}\) or zw = 60.

s ∝ \(\frac{1}{t^2}\)

∴ s = \(\frac{k}{t^2}\), where k is constant of variation

s × t² = k            ... (equestion of variation)

When s = 4 then t = 5

∴ k = 4 × 5² = 4 × 25 = 100 ... (The constant of variation)

s × t² = 100        ... (is the equation of variation)

x ∝ \(\frac{1}{\sqrt{y}}\)

∴ x = \(\frac{k}{\sqrt{y}}\), where k is constant of variation

When x = 15 then y = 9.

∴ 15 = \(\frac{k}{\sqrt{9}}\)

∴ k = 15 × \(\sqrt{9}\) = 15 × 3 = 45         ... (The constant of variation)

So, the equation of variation is x = \(\frac{45}{\sqrt{y}}\)  or xy = 45.

If the number of apples in a box is increased, then the number of boxes decreases. So, the number of apples filled in the box and the number of boxes are in inverse proportion.

Let the number of apples filled in a box be x and the number of boxes be y.

Here, x varies inversely as y, i.e., x ∝ \(\frac{1}{y}\).

x = \(\frac{k}{y}\)., where k is constant of variation

x × y = k

When x = 24, y = 27.

∴ k = 24 × 27 = 648

So, the equation of variation is xy = 648.

When x = 36,

36y = 648

y = 648/36 = 18

Thus, the number of boxes needed is 18.

This statement can be written using the symbol of variation as l ∝ \(\frac{1}{f}\).

This statement can be written using the symbol of variation as I ∝ \(\frac{1}{d^2}\).

It given that x ∝ \(\frac{1}{\sqrt{y}}\)

x = \(\frac{k}{\sqrt{y}}\), where k is constant of variation

x\(\sqrt{y}\) = k

When x = 40, y = 16.

∴ k = 40 × \(\sqrt{16}\) = 40 × 4= 160

So, the equation of variation is x = 160

When x = 10,

10\(\sqrt{y}\) = 160

∴ \(\sqrt{y}\) = 16

∴ y = 16² = 256

It is given that x varies inversely as y, i.e., x ∝ \(\frac{1}{y}\).

x = \(\frac{k}{y}\), where k is constant of variation

∴ x × y = k

When x = 15, y = 10.

∴ K = 15 × 10 = 150

So, the equation of variation is xy = 150.

When x = 20,

20y = 150

∴ y = 150/20 = 7.5

Answer is: y = 7.5

As the number of workers increases, the time taken by them to finish a job decreases. This is an example of inverse variation.

As the number of pipes increases, the time taken to fill the tank decreases. This is an example of inverse variation.

As the quantity of petrol increases, its cost also increases. This is an example not of inverse variation.

As the radius of a circle increases, the area of the circle also increases. This is an example not of inverse variation.

Let the number of workers be n and the time to build a wall in hours be t.

The number of workers varies inversely as the time t (in hours).

n ∝ \(\frac{1}{t}\), ∴ n = \(\frac{k}{t}\)          ... (k is a constant.)

∴ nt = k is the equation of variation.

When n = 5, t =48 hours

∴ 15 × 48 = k

∴ k = 720 is the constant of variation.

∴ nt = 720 is the equation of variation.

Substituting t = 30,

n × 30 = 720

∴ n = \(\frac{720}{30}\)

∴ n = 24

Answer is : 24 workers will be required.

The time taken to fill a bag of milk varies directly as the number of bags.

Let the number of bags be n and the time taken to fill it in minutes be t.

Then n ∝ t

∴ n = kt .   .. (k is the constant.)

∴ \(\frac{n}{t}\) = k is the equation of variation.

When n = 120, t = 3 minutes

∴ \(\frac{120}{3}\) = k

∴ k = 40 is the constant of variation.

\(\frac{n}{t}\) = 40 is the equation of variation.

Substituting n = 1800.

\(\frac{1800}{t}\) = 40

\(\frac{1800}{40}\) = t

∴ t = 45

Answer is: 45 minutes to fill 1800 bags.

Let the speed of the car be s and the time taken to travel some distance be t.

There is inverse variation in the speed and the time.

s ∝ \(\frac{k}{t}\), ∴ s = \(\frac{k}{t}\)         ... (k is a constant.)

∴ st = k is the equation of variation.

When s = 60, t = 8

∴ 60 × 8 = k

∴ k = 480 is the constant of variation.

∴ st = 480 is the equation of variation.

Substituting t = \(7\frac{1}{2}\) = 7.5,

s × 7.5 = 480

∴ s = \(\frac{480}{7.5}\)

∴ s = 64

To cover the distance, speed of the car should be 64 km/h.

s increase in the speed = (64 - 60)km/h = 4 km/h.

Answer is : The speed of the car should be increased by 4 km/h.