Mathematics Solutions Solutions for Class 8 Maths Chapter 7 Variation are provided here with simple step-by-step explanations. These solutions for Variation are extremely popular among Class 8 students for Maths Variation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Maths Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Write the following statements using the symbol of variation.
(1) Circumference (c) of a circle is directly proportional to its radius (r).
(2) Consumption of petrol (l) in a car and distance travelled by that car (d) are in direct variation.

(1)
Circumference (c) of a circle is directly proportional to its radius (r). This statement is written as

$c\propto r$

(2)
Consumption of petrol (l) in a car and distance travelled by that car (d) are in direct variation. This statement is written as

$l\propto d$

#### Question 2:

Complete the following table considering that the cost of apples and their number are in direct variation.

 Number of apples (x) 1 4 ... 12 ... Cost of apples (y) 8 32 56 ... 160

It is given that the cost of apples (y) and number of apples (x) are in direct variation.

$\therefore y\propto x⇒y=kx$, where k is constant of variation

When x = 1, y = 8.

∴ 8 = k × 1

⇒ k = 8

So, the equation of variation is

y = 8x

When y = 56,

56 = 8x

⇒ x = 7

When x = 12,

y = 8 × 12 = 96

When y = 160,

160 = 8x

⇒ x = 20

The complete is given below.

 Number of apples (x) 1 4 7 12 20 Cost of apples (y) 8 32 56 96 160

#### Question 3:

If mn and when m = 154, n = 7. Find the value of m, when n = 14

It is given that $m\propto n$.

∴ mkn, where k is constant of variation

When m = 154, n = 7.

∴ 154 = k × 7

⇒ k$\frac{154}{7}$ = 22

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

When n = 14,

m = 22 × 14 = 308

Thus, the value of m is 308.

#### Question 4:

If n varies directly as m, complete the following table.

 m 3 5 6.5 ... 1.25 n 12 20 ... 28 ...

It is given that n varies directly as i.e. $n\propto m$.

∴ nkm, where k is constant of variation

When m = 3, n = 12.

∴ 12 = k × 3

⇒ k = 4

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

When m = 6.5,

n = 4 × 6.5 = 26

When n = 28,

28 = 4n

⇒ n = 7

When m = 1.25,

n = 4 × 1.25 = 5

The complete table is given below.

 m 3 5 6.5 7 1.25 n 12 20 26 28 5

#### Question 5:

y varies directly as square root of x. When x = 16, y = 24. Find the constant of variation and equation of variation.

It is given that y varies directly as square root of i.e. $y\propto \sqrt{x}$.

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

When x = 16, y = 24.

$\therefore 24=k×\sqrt{16}$

⇒ 24 = 4k

⇒ k = 6

So, the equation of variation is $y=6\sqrt{x}$.

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

#### Question 6:

The total remuneration paid to labourers, employed to harvest soyabeen is in direct variation with the number of labourers. If remuneration of 4 labourers is ₹ 1000, find the remuneration of 17 labourers.

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 i.e. $y\propto x$.

∴ y = kx, where k is constant of variation

When x = 4, y = 1000.

∴ 1000 = k × 4

⇒ k = 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.

#### Question 1:

The information about numbers of workers and number of days to complete a work is given in the following table. Complete the table.

 Number of workers 30 20 10 Days 6 9 12 36

It can be observed from the given table that if 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 worker and 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, inversely varies as x i.e. $y\propto \frac{1}{x}$.

$\therefore y=\frac{k}{x}$, where k constant of variation

⇒ x × yk

When y = 30, x = 6.

∴ k = 6 × 30 = 180

So, the equation of variation is xy = 180.

When x = 12,

12y = 180

⇒ y = 15

When y = 10,

10x = 180

⇒ x = 18

When x = 36

36y = 180

⇒ y = 5

The complete table is given below.

 Number of workers 30 20 15 10 5 Days 6 9 12 18 36

#### Question 2:

Find constant of variation and write equation of variation for every example given below.
(1) p$\frac{1}{q}$ ; if p = 15 then q = 4
(2) z$\frac{1}{w}$ ; when z = 2.5 then w = 24
(3) s$\frac{1}{{t}^{2}}$ ; if s = 4 then t = 5
(4) x$\frac{1}{\sqrt{y}}$ ; if x = 15 then y = 9

(1)
$p\propto \frac{1}{q}$

, where k is constant of variation

When p = 15, q = 4.

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

(2)
$z\propto \frac{1}{w}$

, where k is constant of variation

When z = 2.5, w = 24.

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

(3)
$s\propto \frac{1}{{t}^{2}}$

, where k is constant of variation

When s = 4, t = 5.

So, the equation of variation is $s=\frac{100}{{t}^{2}}$ or st2 = 100.

(4)
$x\propto \frac{1}{\sqrt{y}}$

, where k is constant of variation

When x = 15, y = 9.

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

#### Question 3:

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

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. $x\propto \frac{1}{y}$.

∴ $x=\frac{k}{y}$, where k is constant of variation

⇒ x × yk

When x = 24, y = 27.

∴ k = 24 × 27 = 648

So, the equation of variation is xy = 648.

When x = 36,

36y = 648

⇒ y $\frac{648}{36}$ = 18

Thus, the number of boxes needed are 18.

#### Question 4:

Write the following statements using symbol of variation.
(1) The  wavelength of sound (l) and its frequency (f) are in inverse variation.
(2) The intensity (I) of light varies inversely with the square of the distance (d) of a screen from the lamp.

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

This statement can be written using symbol of variation as

$l\propto \frac{1}{f}$

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

This statement can be written using symbol of variation as

$I\propto \frac{1}{{d}^{2}}$

#### Question 5:

x $\frac{1}{\sqrt{y}}$ and when x = 40 then y = 16. If x = 10, find y.

It given that $x\propto \frac{1}{\sqrt{y}}$.

∴ $x=\frac{k}{\sqrt{y}}$, 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 $x\sqrt{y}=160$.

When x = 10,

$10\sqrt{y}=160\phantom{\rule{0ex}{0ex}}⇒\sqrt{y}=16\phantom{\rule{0ex}{0ex}}⇒y={16}^{2}=256$

#### Question 6:

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

It is given that x varies inversely as y i.e. $x\propto \frac{1}{y}$.

∴ $x=\frac{k}{y}$, 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

⇒ $\frac{150}{20}$ = 7.5

#### Question 1:

Which of the following statements are of inverse variation?
(1) Number of workers on a job and time taken by them to complete the job.
(2) Number of pipes of same size to fill a tank and the time taken by them to fill the tank.
(3) Petrol filled in the tank of a vehical and its cost
(4) Area of circle and its radius.

(1) The number of workers on a job and time taken by them to complete the job are in inverse variation as increase in the number of workers on a job decreases the time taken by the workers to complete the job.

(2) The number of pipes of same size to fill a tank and the time taken by them to fill the tank are in inverse variation as increase in the number of pipes of same size to fill a tank decreses the time taken by them to fill the tank.

(3) The petrol filled in the tank of a vehical and its cost are in direct proportion as increase in the amount of petrol filled in the tank of a vehical increases the cost paid for the petrol.

(4) The area of circle and its radius are in direct proportion as increase in the radius of the circle increases the area of the circle.

#### Question 2:

If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours.

The number of hours required to build a wall is inversely proportional to the number of workers employed.

Let the number of hours be h and the number of workers be w.

Here, h varies inversely as w i.e. $h\propto \frac{1}{w}$.

∴ $h=\frac{k}{w}$, where is k is constant of variation

⇒ h × wk

When h = 48, w = 15.

∴ k = 48 × 15 = 720

So, the equation of variation is hw = 720.

When h = 30,

30w = 720

⇒ w$\frac{720}{30}$ = 24

Thus, the number of workers required to do the same work in 30 hours is 24.

#### Question 3:

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

The time required to fill the bags of milk varies directly as the time required to fill the bags.

Let the number of minutes be x and the number of bags be y.

Here, x varies directly as y i.e. $x\propto y$.

∴ xky, where is k is constant of variation

When x = 3, y = 120.

∴ 3 = k × 120

⇒ k$\frac{3}{120}=\frac{1}{40}$

So, the equation of variation is $x=\frac{1}{40}y$.

When y = 1800,

$x=\frac{1}{40}×1800=45$

Thus, the time required to fill 1800 bags with milk is 45 minutes.

#### Question 4:

A car with speed 60 km/hr  takes 8 hours to travel some distance. What should be the increase in the speed if the same distance is to be covered in 7$\frac{1}{2}$ hours?

The speed of the car is inversely proportional to the time taken by the car to travel the distance.

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

Here, s varies inversely as t i.e. $s\propto \frac{1}{t}$.

∴ $s=\frac{k}{t}$, where is k is constant of variation

⇒ s × t = k

When s = 60, t = 8.

∴ k = 60 × 8 = 480

So, the equation of variation is st = 480.

When t$7\frac{1}{2}$ h = $\frac{15}{2}$ h,

$s×\frac{15}{2}=480$

⇒ $s=\frac{480×2}{15}$ = 64 km/h

∴ Increase in the speed = 64 − 60 = 4 km/h

Thus, the increase in the speed of the car is 4 km/h.

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