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(5.6 × 10⁵) × (6.9 × 10⁸) 

= (5.6 × 6.9) (10⁵⁺⁸)

= 38.64 × 10¹³

= 3.864 × 10¹⁴

To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.

i. (1.5 × 10^-6) – (5.8 × 10^-7) 

= (1.5 × 10^-6) – (0.58 × 10^-6)

= (1.5 – 0.58) × 10^-6

= 0.92 × 10^-6

= 9.2 × 10^-7

ii. (9.8 × 10^-3) – (8.8 × 10^-3) 

= (9.8 – 8.8) × 10^-3 

= 1.0 × 10^-3

iii. (6.5 × 10^-8) – (5.0 × 10^-9) 

= (6.5 × 10^-8) – (0.50 × 10^-8)

= (6.5 – 0.50) × 10^-8

= 6.0 × 10^-8

A chemist has to deal with numbers as large as 602,200,000,000,000,000,000,000 for the molecules of 2 g of hydrogen gas or as small as 0.00000000000000000000000166 g. that is, mass of a H atom.

To avoid the writing of so many zeros in mathematical operations, scientific notations i.e. exponential notations are used.

• Multiple readings of the same quantity are noted to minimize the error.
• If the multiple readings of the same quantity match closely, they are said to have high precision.
• High precision implies reproducibility of the readings.
• High precision is a prerequisite for high accuracy.
• Precision is expressed in terms of deviation (i.e. absolute deviation and relative deviation). 

i. 0.0345 = 3.45 × 10^-2

ii. 0.08 = 8 × 10^-2

iii. 653.00 = 6.5300 × 10²

iv. 34.768 = 3.4768 × 10¹

Nearness of the measured value to the true value is called the accuracy of measurement.

Coordination is a function that is inherent and pervasive. Coordination is not a separate function of management. It is the essence of management.                                 

The coordination is needed to perform all the functions of management. They are:
(i) Coordination in planning In planning, coordination is needed

• between overall plan of the organisation and the departmental plans.
• between objectives and available physical and human resources, e.g., coordination between production department and sales department targets.

(ii) Coordination in organising In organising, coordination is required

• between resources of an organisation and activities to be performed.
  • among authority, responsibility and accountability, e.g. if a supervisor is given responsibility to get the work done, he should also be given the authority to divide the work among his subordinates.

(iii) Coordination in staffing In staffing, coordination is needed

• between the skills of the workers and the jobs assigned to them.
• between the efficiency of the workers and the compensation, etc, e.g. a Chartered Accountant should be given work of financial nature.

(iv) Coordination in directing: In directing, coordination is required

• among orders, instructions, guidelines and suggestions.
• between superiors and subordinates, e.g. a manager instructs the subordinates, motivates them and also supervises their work.

(v) Coordination in controlling: In controlling, coordination is required

• between the standard and the actual performance.
• between correction of deviations and achievement of objectives, e.g. If production target for workers is set as 80 units per hour while they are able to produce only 60 units per hour, then there is a need to coordinate the plans with actual performance.

Thus, coordination makes planning more purposeful, organisation well knit and control more effective. It is the key to the success of management.
(vi) Coordination is the essence of management Management function revolves around making, arranging things, moving things in an organisation in relation to overall objective of the organisation. Thus, coordination can be considered as the core function of management which ensures that all the factors in the business work together smoothly.

Coordination is the orderly arrangement of group efforts to provide unity of action in pursuit of common purpose. It involves unifying, integrating and harmonising the activities of different departments and individuals for the achievement of common goal.

Importance of coordination can be understood by following points:

(i) Growth in size As the organisations grow in size, the number of people in the organisation also increase. Different individuals have different objectives. Coordination is needed to integrate diverse individual objectives with organisational objectives.

(ii) Functional differentiation Different departments in the organisation have their own set of objectives, policies, etc. This creates conflicting situations many a times. Coordination seeks to intergrate the efforts and activities of various departments.

(iii) Specialisation In modern organisations, high level of specialised activities take place, which are performed by specialists. Specialists often consider themselves to be supreme and are not open to suggestions and advice. This brings conflicting situations, which can be resolved by coordination.

Coordination is a continuous process, by which a manager integrates the inter-related activities of different departments in order to achieve the common organisational goals.

Features of coordination are as follows:

(i) Integration of group efforts All business activities are interdependent. Therefore, there should be coordination among them. Coordination enables the business to make efficient use of its available resources.

(ii) Unity of action Coordination enables the manager to secure unity of action in the direction of a common purpose.

(iii) Continuous process It is a continuous process and not a one-time task. A manager has to continuously coordinate the activities of different departments in order to meet the targets by using the available resources efficiently.

(iv) All pervasive function It is an all pervasive function, which runs through all managerial functions from planning till controlling. It is not only needed among different departments but also within the departments at all levels.

(v) Deliberate function A manager has to coordinate the efforts of different individuals working in an organisation in a conscious and deliberate manner.

(i) Growth in size To integrate individual goals with organisational goals.
(ii) Functional differentiation To ensure unity of action among interdependent departments like finance, production, marketing, etc.
(iii) Specialisation To eliminate conflict and integrate efforts of various specialists in the organisation.

Coordination is a continuous process, by which a manager integrates the inter-related activities of different departments in order to achieve the common organisational goals.
Features of coordination are as follows:
(i) Integration of group efforts All business activities are interdependent. Therefore, there should be coordination among them. Coordination enables the business to make efficient use of its available resources.
(ii) Unity of action Coordination enables the manager to secure unity of action in the direction of a common purpose.
(iii) Continuous process It is a continuous process and not a one-time task. A manager has to continuously coordinate the activities of different departments in order to meet the targets by using the available resources efficiently.
(iv) All pervasive function It is an all pervasive function, which runs through all managerial functions from planning till controlling. It is not only needed among different departments but also within the departments at all levels.
(v) Deliberate function A manager has to coordinate the efforts of different individuals working in an organisation in a conscious and deliberate manner.

Coordination is a continuous process, by which a manager integrates the inter-related activities of different departments in order to achieve the common organisational goals.
Features of coordination are as follows:
(i) Integration of group efforts All business activities are interdependent. Therefore, there should be coordination among them. Coordination enables the business to make efficient use of its available resources.
(ii) Unity of action Coordination enables the manager to secure unity of action in the direction of a common purpose.
(iii) Continuous process It is a continuous process and not a one-time task. A manager has to continuously coordinate the activities of different departments in order to meet the targets by using the available resources efficiently.
(iv) All pervasive function It is an all pervasive function, which runs through all managerial functions from planning till controlling. It is not only needed among different departments but also within the departments at all levels.
(v) Deliberate function A manager has to coordinate the efforts of different individuals working in an organisation in a conscious and deliberate manner.

(i) Integration of group efforts         

(ii) Unity of action

(iii)Continuous process                   

(iv) All pervasive function

(v) Deliberate function                    

(vi) Responsibility of all managers

Coordination is a continuous process by which a manager integrates the inter-related activities of different departments in order to achieve the common organisational goals.

Answer: intersecting at (a,b)

The equations are x=a and y=b.
The graph of x=a is a line parallel to y-axis and y=b is a line parallel to x-axis.
∴ the angle between them is 90^o , i.e. they intersect.
So, the equations are consistent.
Now solving simultaneously, the solution is (a,b).
So, the point of intersection is (a,b).

यदि (x – 2), f(x) = 4x³ + 3x² – 4x + k का एक गुणनखण्ड है तो x – 2 = 0 या x = 2 रखने पर

∴ f(2) = 0

4(2)³ + 3(2)² – 4(2) + k = 0

32 + 12 – 8 + k = 0

36 + k = 0 ⇒ k = -36

यदि (x + 1), f(x) = 2x² + kx का एक गुणनखण्ड है तो x + 1 = 0

∴ x = 0 – 1 = -1 रखने पर f(-1) = 0

f(-1) = 2(-1)² + k(-1)

0 = 2 – k ⇒ k = 2

यदि (x + 1) बहुपद 2x³– ax² – (2a – 3)x + 2 का एक गुणनखण्ड है तो

x + 1 = 0 या x = -1 रखने पर।

शेषफल = 0

2(-1)³ – a(-1)² – (2a – 3)(-1) + 2 = 0

-2 – a + 2a – 3 + 2 = 0

a – 3 = 0 ⇒ a = 3

यदि (x – 3), k²x² – kx – 2 का एक गुणनखण्ड है तो x – 3 = 0 या x = 3 रखने पर

शेषफल = 0

k². (3)² – k. 3 – 2 = 0

9k² – 3k – 2 = 0

9k² –(6 – 3)k – 2 = 0

9k² – 6k + 3k – 2 = 0

3k(3k – 2) + 1(k – 2) = 0

(3k – 2)(3k + 1) = 0

यदि 3k – 2 = 0 ∴ k = 2/3

यदि 3k + 1 = 0 ∴ k =1/3

(i) f(x) = x³ – 6x² – 19x + 84 तथा g(x) = x – 7

g(x) = x – 7 = 0 या x = 7 का मान f(x) में रखने पर

f(7) = (7)³ – 6(7)² – 19(7) + 84

= 343 – 294 – 133 + 84

=427 – 427 = 0

अतः g(x), f(x) का एक गुणनखण्ड है।

(ii) f(x) = x³ – 3x² + 4x – 4 तथा g(x) = x – 2

g(x) = 0 या x – 2 = 0 या x = 2 रखने पर

f(2) = (2)³ – 3(2)² + 4(2) – 4

= 8 – 12 + 8 – 4

= 16 – 16 = 0

अतः g(x), f(x) का एक गुणनखण्ड है।

(iii) f(x) = 3x⁴ + 17x³ + 9x² – 7x – 10 तथा g(x) = x + 5

g(x) = 0 या x + 5 = 0 या x = -5 रखने पर

f(-5) = 3(-5)⁴ + 17(-5)³ + 9(-5)² – 7(-5) – 10

= 3 × 625 – 17 × 125 + 9 × 25 + 35 – 10

= 1875 – 2125 + 225 + 25

= 2125 – 2125 = 0

अतः g(x), f(x) का एक गुणनखण्ड है।

(iv) f(x) = 2x³ + 4x + 6 तथा g(x) = x +1

g(x) = 0 या x + 1 = 0 या x = -1 रखने पर

f(-1) = 2(-1)³ + 4(-1) + 6

= -2 – 4 + 6 = 0

अतः g(x), f (x) का एक गुणनखण्ड है।

According to the question,

f and g be real functions defined by f(x) = 2x + 1 and g(x) = 4x – 7

(a) For what real numbers x, f (x) = g (x)

To satisfy the condition f(x) = g(x),

Should also satisfy,

2x + 1 = 4x–7

⇒ 7 + 1 = 4x–2x

⇒ 8 = 2x

Or, 2x = 8

⇒ x = 4

Hence, we get,

For x = 4, f (x) = g (x)

(b) For what real numbers x, f (x) < g (x)

To satisfy the condition f(x) < g(x),

Should also satisfy,

2x + 1 < 4x–7

⇒ 7 + 1 < 4x–2x

⇒ 8 < 2x

Or, 2x > 8

⇒ x > 4

Hence, we get,

For x > 4, f (x) > g (x)

Given function is

f\( f(x) =\begin{cases}\frac{1x -a_1} {x-a}, & \quad x\neq a\\1 & \quad x=a\end{cases}\)

\( f(x) =\begin{cases}\frac{-(x -a)} {x-a}=-1, & \quad x< a\\1, &\quad x=a\\\frac {x-a}{x-a}=1, & \quad x>a\\\end{cases}\)

\( f(x) =\begin{cases}-1 & \quad x< a\\+1 &\quad x=a\\1 & \quad x>a\\\end{cases}\)

f(a^-)=-1, f(a⁺⁾⁼¹&f(a)=1

\(\because\) f(a^-) ≠f(a⁺)

\(\therefore\) f is not continuous at x=a

fg = f(x), g(x) = x. |x| = \(\begin{cases}x^2, & \quad x\geq 0\\-x^2, & \quad x<0\end{cases}\)

\(\frac{f}{g}(x)=\frac{f(x)}{g(x)}=\frac{x}{|x|}\\=\begin{cases}1, & \quad x>0\\-1, & \quad x<0\end{cases}\)

Note: \(\frac{f}{g}\) is not defined at x = 0.

(f– g)(x) = f(x) – g(x) = x - |x| =\(\begin{cases}0, & \quad x\geq 0\\2x, & \quad x<0\end{cases}\)

Class 8 Maths MCQ Questions Direct and Indirect Proportions with Answers are prepared by the mathematical experts of Sarthaks eConnect supported the newest examination pattern.  We have provided Direct and Inverse Proportions Class 8 MCQ Questions with Answers to assist students understands the concept alright. These Questions are very helpful to attain high marks in board exams.

Students can ask these Maths formulas for Maths Class 8 Direct and Indirect Proportions for better exam preparation and score excellent marks. You will get Important MCQ Questions Class 8 supported NCERT Textbook for class 8 here. we have covered Important Questions on Direct and Inverse Proportion for Class 8 Maths subject all topics. Click below link and start Practice all Important MCQ Questions of Direct and Indirect Proportion Class 8 are given.

Practice MCQ Question for Class 8 Maths chapter-wise

1. 10 meters of cloth cost Rs 1000. What will 4 meters cost?

(a) Rs 400
(b) Rs 800
(c) Rs 200
(d) Rs 100

2. 20 trucks can hold 150 metric tonnes. How much will 12 trucks hold?

(a) 80 metric tonnes
(b) 90 metric tonnes
(c) 60 metric tonnes
(d) 40 metric tonnes

3. If x = 20 and y = 40, then x and y are:

(a) Directly proportional
(b)  Inversely proportional
(c) Neither directly nor inversely proportional
(d) Cannot be determined

4.  If x = 15 and y = 1/30, then x and y are:

(a) Directly proportional
(b)  Inversely proportional
(c) Neither directly nor inversely proportional
(d) Cannot be determined

5. If x and y are directly proportional, then which of the following is correct?

(a) x+y=constant
(b) x-y=constant
(c) xy=constant
(d) x/y=constant

6. If x∝y and x₁=5,y₁=210 and x₂ = 2, then find y₂?

(a) 200
(b) 84
(c) 99
(d) 70

7. If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 2500 grams?

(a) 750
(b) 800
(c) 850
(d) 950

8. If 400 Kg of tar costs 8000₹, then find the cost of 160 kg of tar?

(a) 1200₹ 
(b) 2400₹ 
(c) 3200₹ 
(d) 4200₹

9. If x and y are inversely proportional, then which one is true?

(a) x₁/y₁ = x₂/y₂
(b) x₁ /x₂ = y₁/y₂
(c) x₁ /x₂ = y₂/y₁
(d) x₁ .x₂ = y₁.y₂

10. 6 pipes are required to fill a tank in 1 hour 20 minutes. If we use 5 such types of pipes, how much time it will take to fill the tank?

(a) 120 minutes
(b) 96 minutes
(c) 80 minutes
(d) 85 minutes

11. If 12 workers can build a wall in 50 hours, how many workers will be required to do the same work in 40 hours?

(a) 10
(b) 13
(c) 14
(d) 15

12. A man walks 20 km in 5 hours. How much time it will take for him to walk 32 km?

(a) 3 hours
(b) 4 hours
(c) 6 hours
(d) 8 hours

13. If 300 Kg of coal cost 6000₹, then find the cost of 120 kg of coal?

(a) 1200₹
(b) 2400₹
(c) 3200₹
(d) 4200₹

14. Manvi types 200 words in 30 minutes. How many words she will type in 12 minutes?

(a) 100
(b) 80
(c) 60
(d) 50

15. If it takes 40 days for 120 men to complete a work, how long will it take for 80 men to complete the same work?

(a) 50 days
(b) 60 days
(c) 80 days
(d) 100 days

16. If 16 labours can construct a road in 40 hours, how many labours will be required to construct the same road in 20 hours?

(a) 20
(b) 16
(c) 30
(d) 32

17. A car takes 18 hours to ride 720 kilometers. Time taken by the car to travel 360 kilometers is:

(a) 10 hours
(b) 11 hours
(c) 9 hours
(d) 16 hourstt

18. If x = ky and when y = 4, x = 8 then k =

(a) 1
(b) 2
(c) 3
(d) 4

19. The scale of a map is given as 1: 300. Two cities are 4 km apart on the map. The actual distance between them is:

(a) 1000 km
(b) 1100 km
(c) 1200 km
(d) 1300 km

20. A garrison of 500 men had provision for 27 days. After 3 days a reinforcement of 300 men arrived. For how many more days will the remaining food last now?

(a) 15
(b) 16
(c) 12
(d) 13

21. When one quantity is increased, the other quantity is also increased. This proportion is called _______

(a) Kally proportion
(b) Direct proportion
(c) Inverse proportion
(d) None of these

22. A van covers 432 km with 36 liters of diesel. How much distance would it cover with 25 litres of diesel?

(a) 200 km
(b) 300 km
(c) 100 km
(d) 350 km

23. A journey by bus takes 45 minutes at 40 km/hour. How fast must a car go to undertake the same journey in 25 minutes?

(a) 36 km/h
(b) 48 km/h
(c) 72 km/h
(d) None of these

24. If the cost of 48 bags of paddy is Rs. 16800, then cost of 36 bags of paddy is

(a) Rs. 12000
(b) Rs. 16000
(c) Rs. 12600
(d) None of these

25. Priya types 160 words in 40 minutes. How many words she will type in 12 minutes?

(a) 100
(b) 48
(c) 70
(d) 50

Answer:

 1. Answer:  (a) Rs 400

Explanation: \(\frac{10}{1000}=\frac{4}{x}\)

x = 400

2. Answer:  (b) 90 metric tonnes

Explanation: \(\frac{20}{150}=\frac{12}{x}\)

x = 90

3. Answer:  (a) Directly proportional

Explanation: x = 20 and y = 40

Clearly, 40 = 2 x 20

y=2x

y∝x, where 2 is the proportionality constant.

4. Answer: (b)  Inversely proportional

Explanation: x = 15, y = 1/30

1/30 = 1/(2 × 15)

y = 1/2x

Hence, y ∝ 1/x, where 1/2 is the proportionality constant.

5. Answer:  (d) x/y=constant

Explanation: x∝y

x=ky

k=x/y, where k is a constant.

6. Answer: (b) 84

Explanation: x∝y

x=ky

x/y=k

x₁/y₁ = x₂/y₂

5/210 = 2/y₂

y₂ = 84

7. Answer: (a) 750

Explanation: For 12 sheets, the weight of the paper is 40 grams

Let a number of sheets for 2500 is x.

Using direct proportion concept:

12/40 = x/2500

x=(12×2500)/40

x=750

8. Answer: (c) 3200₹

Explanation: total weight =400kg

cost of 1 kg = 8000/400

= 20rupees

160kg = 20×160

= 32000rupees

9. Answer: (c) x₁ /x₂ = y₂/y₁

Explanation: ∝1/y

x=k(1/y)

xy=k

x₁y1=x₂y₂

x₁/x₂=y₂/y₁

10. Answer:  (b) 96 minutes

Explanation: For 6 pipes, it takes 1 hour 20 minutes

1 hour 20 minutes = 60+20 = 80 minutes

For 5 pipes, let the time taken be x.

This is inverse proportion case:

80 × 6 = x × 5

x = 480/5 

= 96

11. Answer:  (d) 15

Explanation: 12 × 50 = x × 40

x = (12×50)/40 

= 15

12. Answer:  (d) 8

Explanation: A man walks 20 km in → 5 hours

That means it will take more time to walk 32 km. This is the case of direct proportion. 

20/5 = 32/x

x = 32/4

= 8

13. Answer: (b) 2400₹

Explanation: total weight =300kg

cost of 1 kg = 6000/300

=20rupees

120kg = 20×120

= 2400rupees

14. Answer: (b) 80

Explanation: 30/200 = 12/x

x = 12 x 200/30

x = 80

15. Answer: (b) 60 days

Explanation: 40 × 120 = 80 × x 

⇒ x = 60

16. Answer: (d) 32

Explanation: 16 × 40 = x × 20

x = (16 × 40)/20

 = 32

17. Answer: (c) 9 hours

Explanation: 720/18 = 360/x

x = 9 hours

18. Answer: (b) 2

Explanation: 8 = 4k 

⇒ k = 2

19. Answer: (c) 1200 km

Explanation: 1 : 300 = x : y

1/300 = 4/y

y = 4 x 300 = 1200

Hence, the distance between the two cities is 1200 km.

20. Answer: (a) 15

Explanation: Let's remaining food will last for x days. 500 men had provisions for (27−3)=24 days.

 (500+300)  men had provisions for x  days. More men, less days

∴ 800:500:: 24:x

800 × x = 500 × 24

\(x=\frac{500\times24}{800}\)

= 15

21. Answer: (b) Direct proportion

Explanation: When two quantities X and Y increase together or decrease together, they are said to be directly proportional or they are in direct proportion with each other. It is also known as a direct variation.

22. Answer: (b) 300 km

Explanation:In 36 litre of diesel a van cover 432 km 

In 1 litre of diesel a van cover 432/36 =12 km

In 25 litre of diesel a van cover 12×25=300 km

23. Answer: (c) 72 km/h

Explanation: Distance covered by bus in 45 minutes with speed of 40 km/h = \(40\times\frac{45}{60}\) [∵speed= 
Distance/Time] = 30 km.

the speed that can be to covers the same journey in 25 minutes = \(\frac{30}{\frac{25}{60}}=\frac{30\times60}{25} km/h\) = 72 km/h

24. Answer:  (c) Rs. 12600

Explanation: Cost of 48 bags = Rs.16800

Cost of 1 bag = Rs. 16800/48 = Rs 3500

Cost of 36 bags = Rs.3500 ×36 = Rs.12600

25. Answer: (b) 48

Explanation:40/160 = 12/x

x = 12 x 160/40

x = 48

Click here Practice MCQ Question for Direct and Indirect Proportions Class 8

(i) We have, h = {(4,6),(3,9), (-11,6), (3,11)}.

Since pre-image 3 has two images 9 and 11, it is not a function.

(ii) We have, f = {(x, x) | x is a real number}

Since every element in the domain has unique image, it is a function.

(iii) We have, g= {(n, 1/n) | nis a positive integer}

For n, it is a positive integer and 1/n is unique and distinct. Therefore,every element in the domain has unique image. So, it is a function.

(iii) We have, s = {(n, n² ) | n is a positive integer}

Since the square of any positive integer is unique, every element in the domain has unique image. Hence, ibis a function.

(iv) We have, t = {(x, 3)| x is a real number}.

Since every element in the domain has the image 3, it is a constant function.

Answer : (c) 5220

The numbers between 107 and 253 divisible by 5 are  

110, 115, 120, .......... , 245, 250. 

This is an A.P with first term (a) = 110 and common difference (d) = 5. 

Let the last term be the nth term.

∴ T_n = a + (n – 1)d 

⇒ 110 + (n – 1) × 5 = 250 

⇒ 5n = 250 – 105 = 145 

⇒ n = 29.

∴ Required sum = \(\frac{n}{2}\) (a + T_n) = \(\frac{29}{2}\)(110 + 250)

= \(\frac{29}{2}\) × 360 = 5220

We have,

\(\sqrt{2{\frac{1}{4}}}=\sqrt{\frac{9}{4}}\)

= \(\frac{\sqrt{9}}{\sqrt{4}}\)

= \(\frac{\sqrt{3\times3}}{\sqrt{2\times2}}\)

= \(\frac{3}{2}\)

= \(1\frac{1}{2}\)

Therefore, option (C) is correct

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Practice MCQ Question for Class 8 Maths chapter-wise

1. a⁰ is equal to

(a) 0
(b) 1
(c) -1
(d) a

2. 10^-1 is equal to

(a) 10
(b) -1
(c) 1/10
(d) –1/10

3. Multiplicative inverse of (2/−3) is

(a) 2/3
(b) -3/2
(c) 3/2
(d) –1/10

4. 5³ × 5^-1 is equal to

(a) 5
(b) 5³
(c) 5^-1
(d) 5²

5. (- 2)^-5 x (-2)⁶ is equal to

(a) 2
(b) -2
(c) -5
(d) 6

6. 3² × 3^-4 × 3⁵ is equal to

(a) 3
(b) 3²
(c) 3³
(d) 3⁵

7. (2⁰ + 4^-1) × 2² is equal to

(a) 2
(b) 3
(c) 4
(d) 5

8. (2^-1 + 3^-1 + 5^-1)⁰ is equal to

(a) 2
(b) 3
(c) 5
(d) 1

9. (-2)^m+1 × (-2)⁴ = (-2)⁶ ⇒ m =

(a) 0
(b) 1
(c) -1
(d) none of these

10. (-1)⁵⁰ is equal to

(a) -1
(b) 1
(c) 50
(d) – 50

11. 2² x 2³ x 2⁴ is equal to:

(a) 2²⁴
(b) 2^-5
(c) 2⁹
(d) 2^-9

12. 3^-2 x 3^-5 is equal to:

(a) 3^-7
(b) 3^-3
(c) 3^-10
(d) 3⁷

13. 5⁴/5² is equal to:

(a) 5⁶
(b) 5^-6
(c) 5^-2
(d) 5²

14. The value of (3⁴)³ is:

(a) 3
(b) 3¹²
(c) 3⁷
(d) None of the above

15. 3² x 4² is equal to:

(a) 121
(b) 49
(c) 144
(d) 156

16. 5⁷/6⁷ will give the value:

(a) (5/6)⁷
(b) (5/6)⁰
(c) (5/6)^-7
(d) (6/5)^-7

17. 100⁰+20⁰+5⁰ is equal to

(a) 125
(b) 25
(c) 1/125
(d) 3

18. If (–3)^m+1 × (-3)⁵ = (-3)⁷, then the value of m is:

(a) 5
(b) 7
(c) 1
(d) 3

19. Very small numbers can be expressed in standard form using __________ exponents.

(a) equal
(b) negative
(c) positive
(d) none of these

20. The multiplicative inverse of 7^-2 is:

(a) 7²
(b) 7
(c) 1/7²
(d) 1/7

21. (1/3)² is equal to:

(a) 9
(b) -9
(c) -1/9
(d) 1/9

22. Reciprocal of 1 is

(a) 0
(b) 1
(c) -1
(d) none of these

23. 384467000 is equal to

(a) 3.84467 × 10⁸
(b) 3.84467 × 10³
(c) 3.84467 × 10⁷
(d) 3.84467 × 10⁶

24. When we have to add numbers in standard form, we convert them into numbers with the ________ exponents.

(a) equal
(b) different
(c) not equal
(d) None of these

25. Find the value of the expression a² for a = -10.

(a) 100
(b) 1
(c) 10
(d) None of these

Answer:

 1. Answer: (b) 1

Explanation: The value of any non - zero number raised to the power 0 is 1.

2.  Answer: (c) 1/10

Explanation: 10^-1 = 1/10¹ 

= 1/10

3.  Answer: (b) -3/2

Explanation: Multiplicative inverse of a/b is b/a So,

The multiplicative inverse of (−2/3) is (−3/2)

4.  Answer: (d) 5²

Explanation:5³ × 5^-1

= 5^3-1

= 5²

5.  Answer: (b) -2

Explanation:(- 2)^-5 x (-2)⁶

 = (-2)^-5+6 

= (-2)¹ 

= -2

6.  Answer: (c) 3³

Explanation: 3² × 3^-4 × 3⁵ 

= 3^{2 - 4 + 5} 

= 3³

7. Answer: (d) 5

Explanation: (2⁰ + 4^-1) × 2²

\((1+\frac{1}{4})\times4\)

\(\frac{5}{4}\times4\)

 = 5

8.  Answer: (d) 1

Explanation: (2^-1 + 3^-1+ 5^-1)⁰ = 1 [∵ a⁰ = 1]

9.  Answer: (b) 1

Explanation: (-2)^m+1 × (-2)⁴ = (-2)⁶

⇒ (-2)^m+1+4 = (-2)⁶

⇒ m + 5 = 6

 ⇒ m = 1

10.  Answer: (b) 1

Explanation: (-1) ^{even natural number} = 1

11.  Answer: (c) 2⁹

Explanation:laws of exponents:

a^m x aⁿ = a^m+n

2² x 2³ x 2⁴

= 2²⁺³⁺⁴ 

= 2⁹

12.  Answer: (a) 3^-7

Explanation: 3^-2 x 3^-5

= (1/3²) x (1/3⁵)

= (1/3²⁺⁵)

= (1/3⁷)

= 3^-7

13.  Answer :(d) 5²

Explanation: Exponent law:

a^m/aⁿ = a^m-n

= 5⁴/5² 

= 5^4-2 

= 5²

14.  Answer: (b) 3¹²

Explanation:By law of exponent:

(a^m)ⁿ = a^mn

(3⁴)³ = 3^4×3 

= 3¹²

15.  Answer: (c) 144

Explanation:exponent law;

a^m x b^m = (ab)^m

= 3² x 4² 

= (3×4)²

 = 12² 

= 144

16. Answer: (a) (5/6)⁷

Explanation:Exponent law:

a^m/b^m = (a/b)^m

= 5⁷/6⁷ 

= (5/6)⁷

17. Answer: (d) 3

Explanation: exponent law we know:

a⁰ = 1

= 100⁰ + 20⁰ + 5⁰ 

= 1 + 1 + 1 

= 3

18.  Answer: (c) 1

Explanation: (-3)^m+1 × (-3)⁵ = (-3)⁷

(-3)^m+1+5 = (-3)⁷

(-3)^m+6 = (-3)⁷

Since, base are equal on both the sides, hence if we compare the powers,

m + 6 = 7

m = 7 – 6

= 1

19.  Answer: (b) negative

Explanation:A very small number can be expressed in standard form by using 'negative' exponents.

20.  Answer: (a) 7²

Explanation:The multiplicative inverse of any value is the one which when multiplied by the original value gives a value equal to 1.

7^-2 = 1/7²

Hence, 7² x 1/7² 

= 1

21.  Answer: (d) 1/9

Explanation: (1/3)² = (1/3 x 1/3) 

= 1/9

22.  Answer: (b) 1

Explanation: multiplicative inverse of a number is a number which when multiplied with the original number produces 1. Therefore, the reciprocal of 1 is 1

23.  Answer: (a) 3.84467 × 10⁸

Explanation: 384467000 = 3.84467 × 10⁸

24.  Answer: (a) equal

Explanation: To add the numbers given in the standard form, we first convert them into numbers with "equal" exponents.

25. Answer: (a) 100

Explanation: Put the value of ’a’ in given equation

a² = (−10)²

= 100 

Therefore, the value of a² for a =−10 is 100.

Click here Practice MCQ Question for Exponents and Powers Class 8

We know that,

Any number which is ending with 2, 3, 7 and 8 is not a perfect square

Therefore,

1222 is not a perfect square as it is ending with digit 2

Hence, option (D) is correct

Thickness of each book = 20 mm

Hence, thickness of 5 books = (5 × 20) mm = 100 mm Thickness of each paper sheet = 0.016 mm

Hence, thickness of 5 paper sheets = (5 × 0.016) mm = 0.080 mm

Total thickness of the stack = Thickness of 5 books + Thickness of 5 paper sheets

= (100 + 0.080) mm

= 100.08 mm

= 1.0008 × 10² mm

We know that, 

As per the properties of square, 

All the numbers that end with digits 2, 3, 7 or 8 are not a perfect square 

Hence, 

Considering the property, we get 

The number 1843 is not a perfect square 

As the last digit of the number is 3. 

Therefore, 

Option (A) is the correct option.

Let a and b be two numbers 

a × b = 1296 

a = 16b 

= 16 b × b 

= 1296 

b² = 81 

b = 9 

Therefore, 

a = 144 and b = 9

(i) 1/1000000 - 1 x 10^-6

(ii) 0.000, 000, 000, 000, 000, 000, 16 = 1.6 × 10⁻¹⁹

(iii) 0.0000005 = 5 × 10⁻⁷

(iv) 0.00001275 = 1.275 × 10⁻⁵

(v) 0.07 = 7 × 10⁻²

Let total residents be a 

Therefore, each paid Rs. a 

Total collection = a (a) = a² 

given, Total Collection = 202500 

Hence,

a = \(\sqrt{202500}\) = √(2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 × 5)a = 2 × 3 × 3 × 5 × 5a = 450

Therefore,

Total residents = 450

Let there were a members 

Therefore, each attributed a paise 

Therefore, 

a (a), i.e. total cost collected = 9216 paise 

a² = 9216 

a = \(\sqrt9216\) 

= 2 × 2 × 2 × 12 

= 96 

Therefore, there were 96 members and each contributed 96 paise

Let, a be number of school students 

Therefore, each student contributed a paise 

Total money obtained = a²paise

= 230400 paise

a = \(\sqrt{230400}\)

= \(\sqrt{2304}\) x \(\sqrt{100}\)

= 10 \(\sqrt{2304}\)

a = 10 x 2 x 2 x 12

a = 480

Therefore, there were 480 students

(i) 23805 

Resolving 23805 into prime factors, we get 

23805 = 3 × 3 × 23 × 23 × 5 

Obtained factors can be paired into equal factors except for 5 

To pair it equally multiply with 5 

23805 × 5 = 3 × 3 × 5 × 5 × 23 × 23 

Again, 

23805 × 5 = (3× 5 × 23) × (3 × 5 × 23) 

= 345 × 345 

= (345)² 

Therefore, product is the square of 345.

(ii) 12150 

Resolving 12150 into prime factors, we get

12150 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 × 2 

Obtained factors can be paired into equal factors except for 2 

To pair it equally multiply with 2 

12150 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3

 Again, 

12150 × 2 = (5 × 3 × 2 × 2 × 2) × (5 × 3 × 2 × 2 × 2)

= 120 × 120 

= (120)² 

Therefore, product is the square of 120.

(iii) 7688 

Resolving 7688 into prime factors, we get 

7688 = 2 × 2 × 31 × 31 × 2 

Obtained factors can be paired into equal factors except for 2 

To pair it equally multiply with 2 

7688 × 2 = 2 × 2 × 2 × 2 × 31 × 31 

Again, 

7688 × 2 = (2× 2 × 31) × (2 × 2 × 31) 

= 124 × 124 

= (124)² 

Therefore, product is the square of 124.

180 = 2² × 3² × 5 

= (2 × 2) × (3 × 3) × 5 

To make the unpaired 5 into paired, multiply the number with 5 Therefore, 

180 × 5 = 2² × 3² × 5²

Hence, square root of number = \(\sqrt{180}\) x \(\sqrt{5}\) = 2 x 3 x 5

= 30

The Perfect square from 1 to 500 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484.

(i) 243 = 3 x 3 x 3 x 3

Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required. 

In that case, 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 is a perfect cube. Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required. Then, we obtain 

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 is a perfect cube.  Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii) 72 = 2 × 2 × 2 × 3 × 3

Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required. Then, we obtain 

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 is a perfect cube. Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv) 675 = 3 × 3 × 3 × 5 × 5

Here, two 5s are left which are not in a triplet. To make 675 a cube, one more 5 is required. Then, we obtain  

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375 is a perfect cube. Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v) 100 = 2 × 2 × 5 × 5

Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5. Then, we obtain 

100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10

.

200, 201, 204, 205, 206, 209.

(i) We know that, 

Any number which ends with an odd number of zeros is not a perfect square 

Also, the given number 360 is ending with the digit 0 

Therefore, 

The given number is not a perfect square 

(ii) We know that, 

Any number which ends with an odd number of zeros is not a perfect square 

Also, the given number 6400 is ending with the digit 0 

Therefore, 

The given number is not a perfect square 

(iii) We know that, 

Any number which ends with an odd number of zeros is not a perfect square 

Also, the given number 2500000 is ending with the digit 0 

Therefore, 

The given number is not a perfect square

Hence, 7, 8, 3, 2 as ending numbers respectively. 

As mentioned above ending with 2, 3, 7, 8 are not perfect square. 

So, these given numbers are not perfect squares.

Numbers ending with 2, 3, 7 or 8 are not perfect squares. So, 

(i) 1547 

(ii) 45743 

(iii) 8948 

(iv) 333333 are not perfect squares

The square of numbers may end with any one of the digits 0, 1, 5, 6, or 9. Also, a perfect square has even number of zeroes at the end of it.

(i) 1057 has its unit place digit as 7. Therefore, it cannot be a perfect square.

(ii) 23453 has its unit place digit as 3. Therefore, it cannot be a perfect square.

(iii) 7928 has its unit place digit as 8. Therefore, it cannot be a perfect square.

(iv) 222222 has its unit place digit as 2. Therefore, it cannot be a perfect square.

(v) 64000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

(vi) 89722 has its unit place digit as 2. Therefore, it cannot be a perfect square.

(vii) 222000 has three zeroes at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square

(viii) 505050 has one zero at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square

(i) 32² = (30 + 2)²

= 30 (30 + 2) + 2 (30 + 2)

= 302 + 30 × 2 + 2 × 30 + 2²

= 900 + 60 + 60 + 4

= 1024

(ii) The number 35 has 5 in its unit’s place. Therefore,

35² = (3) (3 + 1) hundreds + 25

= (3 × 4) hundreds + 25

= 1200 + 25 = 1225

(iii) 86² = (80 + 6)²

= 80 (80 + 6) + 6 (80 + 6)

= 802 + 80 × 6 + 6 × 80 + 6²

= 6400 + 480 + 480 + 36

= 7396

(iv) 93² = (90 + 3)²

= 90 (90 + 3) + 3 (90 + 3)

= 902 + 90 × 3 + 3 × 90 + 3²

= 8100 + 270 + 270 + 9

= 8649

(v) 71² = (70 + 1)²

= 70 (70 + 1) + 1 (70 + 1)

= 702 + 70 × 1 + 1 × 70 + 1²

= 4900 + 70 + 70 + 1

= 5041

(vi) 46² = (40 + 6)²

= 40 (40 + 6) + 6 (40 + 6)

= 402 + 40 × 6 + 6 × 40 + 6²

= 1600 + 240 + 240 + 36

= 2116

Given \(\overrightarrow{CF}|| \overrightarrow{BD}, \overrightarrow{BE}\) is transversal. 

∠CAE = 135°

∠BAF = ∠CAE = 135° (vertically opposite angles)

∴ ∠BAF = 135° ∠BAF + ∠ABD = 180°

(co-interior angles are supplementary)

(i) transversal line

(ii) corresponding 

(iii) alternate interior 

(iv) co-interior