[For Session 2025-26] Class 8 Math Ganita Prakash Chapter 2 Power Play

Updated on July 30, 2025
by Tiwari Academy

NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 2 Power Play are prepared for Session 2025–26. Grade 8th Math chapter 2 explains powers, exponents and their real-life use. Students learn to simplify large numbers using exponents. The solutions include step-by-step explanations for each question. They follow the latest syllabus and help in clear understanding. These answers are useful for school exams and practice. Concepts are explained in easy language. Diagrams and examples make learning fun and simple.
Class 8 Maths Ganita Prakash Chapter 2 MCQs
Class 8 Maths NCERT Solutions

Class 8 Maths Ganita Prakash Chapter 2 Solutions

Page 22

Figure it Out

1. Express the following in exponential form:
(i) 6 × 6 × 6 × 6
(ii) y × y
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c × c × d
See Solutions(i) 6 × 6 × 6 × 6 = 6⁴
(ii) y × y = y²
(iii) b × b × b × b = b⁴
(iv) 5 × 5 × 7 × 7 × 7 = 5² × 7³
(v) 2 × 2 × a × a = 2² × a²
(vi) a × a × a × c × c × c × c × d = a³ × c⁴ × d

2. Express each of the following as a product of powers of their prime factors in exponential form.
(i) 648 (ii) 405 (iii) 540 (iv) 3600
See Solutions(i) 648
648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2³ × 3⁴
(ii) 405
405 = 3 × 3 × 3 × 3 × 5 = 3⁴ × 5
(iii) 540
540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5
(iv) 3600
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴ × 3² × 5²

3. Write the numerical value of each of the following:
(i) 2 × 10³ (ii) 7² × 2³ (iii) 3 × 4⁴
(iv) (– 3)² × (– 5)² (v) 3² × 10⁴ (vi) (– 2)⁵ × (– 10)⁶
See Solutions(i) 2 × 10³ = 2 × 1000 = 2000
(ii) 7² × 2³ = 49 × 8 = 392
(iii) 3 × 4⁴ = 3 × 256 = 768
(iv) (–3)² × (–5)² = 9 × 25 = 225
(v) 3² × 10⁴ = 9 × 10,000 = 90,000
(vi) (–2)⁵ × (–10)⁶ = (–32) × 1,000,000 = –32,000,000

Ganita Prakash Class 8 Maths Chapter 2 Question Answers

Page 44

Figure it Out

1. Find out the units digit in the value of 2²²⁴ ÷ 4³²? [Hint: 4 = 2²]
See SolutionsExpressing everything with base 2:
4 = 2², so: 4² = (2²)³² = 2⁶⁴
So the expression becomes:
2²²⁴ ÷ 2⁶⁴
= 2²²⁴⁻⁶⁴
= 2¹⁶⁰
Now, we have to find out unit digit of 2¹⁶⁰:
The unit digit of powers of 2 follows a cycle of 4:
2¹ = 2 → unit digit = 2
2² = 4 → unit digit = 4
2³ = 8 → unit digit = 8
2⁴ = 16 → unit digit = 6
Then it repeats: 2, 4, 8, 6, …
So, divide the exponent by 4: 160 ÷ 4 = 40
If divisible exactly, the 4th number in the cycle is the unit digit, i.e., 6.

2. There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?
See SolutionsEach container has 5 bottles, and 1 container is added every day.
So, after 40 days, the total number of containers = 40
Therefore, total number of bottles = 40 × 5 = 200

3. Write the given number as the product of two or more powers in three different ways. The powers can be any integers.
(i) 64³
See Solutions(i) 64³
We know:
64 = 2⁶
So,
64³ = (2⁶)³ = 2¹⁸
64³ = 8⁶, since 8 = 2³
64³ = 4⁹, since 4 = 2²

(ii) 192⁸
See Solutions(ii) 192⁸
After factorisation, we get 192⁸ = 2⁶ × 3
So,
192⁸ = (2⁶ × 3)⁸ = 2⁴⁸ × 3⁸
192⁸ = (64 × 3)⁸ = 64⁸ × 3⁸
192⁸ = (2³ × 2³ × 3)⁸ = 2⁴⁸ × 3⁸

(iii) 32⁻⁵
See Solutions(iii) 32⁻⁵
We know that: 32 = 2⁵
So,
32⁻⁵ = (2⁵)⁻⁵ = 2⁻²⁵
32⁻⁵ = (2³ × 2²)⁻⁵, since 32 = 2⁵
32⁻⁵ = (4³ × 2⁻¹)⁻⁵

Ganita Prakash Class 8 Maths Chapter 2 Solutions

4. Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True’ or ‘Never True’. Explain your reasoning.
(i) Cube numbers a re also square numbers.
See SolutionsOnly Sometimes True
Reason:
A cube number is of the form 𝑛³ and a square number is of the form 𝑚².
A number that is both a perfect square and a perfect cube must have 6 power, i.e., 𝑥⁶
Example: 64 = 4³ = 8² is both.
But 8 = 2³ is not a square.
So, only some cube numbers are also square numbers.

(ii) Fourth powers are also square numbers.
See SolutionsAlways True
Reason:
Any number raised to the power 4 is also a square because:
𝑛⁴ = (𝑛²)²
Hence, all fourth powers are perfect squares.

(iii) The fifth power of a number is divisible by the cube of that number.
See SolutionsAlways True
Reason:
Any number 𝑎⁵ is clearly divisible by 𝑎³, because:
𝑎⁵ = 𝑎³⋅𝑎²
So, the cube 𝑎³ is a factor of 𝑎⁵ for all 𝑎 ≠ 0.

(iv) The product of two cube numbers is a cube number.
See SolutionsAlways True
Reason:
Let two cube numbers be 𝑎³ and 𝑏³.
Then, 𝑎³⋅𝑏³ = (𝑎𝑏)³
So, the product is again a perfect cube.

(v) 𝑞⁴⁶ is both a 4th power and a 6th power (where 𝑞 is a prime number).
See SolutionsNever True
Reason:
A number is both a 4th and a 6th power if its exponent is a multiple of LCM(4, 6) = 12.
Since, 46 = 2 × 23 (not divisible by 4 or 6)
But the question says 𝑞⁴⁶ is both a 4th and 6th power. That’s impossible unless 46 is divisible by both 4 and 6.
So the correct answer is: Never True

5. Simplify and write these in the exponential form.
(i) 10⁻² × 10⁻⁵
(ii) 5⁷ ÷ 5⁴
(iii) 9⁻⁷ ÷ 9⁴
(iv) (13⁻²)⁻³
(v) m⁵n¹²(mn)⁹
See Solutions(i) 10⁻² × 10⁻⁵ = 10⁻²⁻⁵ = 10⁻⁷
(ii) 5⁷ ÷ 5⁴ = 5⁷⁻⁴ = 5³
(iii) 9⁻⁷ ÷ 9⁴ = 9⁻⁷⁻⁴ = 9⁻¹¹
(iv) (13⁻²)⁻³ = 13⁶
(v) m⁵n¹²(mn)⁹ = m⁵n¹².m⁹.n⁹ = m⁵⁺⁹n¹²⁺⁹ = m¹⁴n²¹

Class 8 Maths Chapter 2 Exercises

6. If 12² = 144 what is
(i) (1.2)²
See SolutionsGiven that 12² = 144
(i) (1.2)²
= (12/10)² = 12²/10²
= 144/100 = 1.44

(ii) (0.12)²
See Solutions(ii) (0.12)²
= (12/100)² = 12²/100²
= 144/10000 = 0.0144

(iii) (0.012)²
See Solutions(iii) (0.012)²
= (12/1000)² = 12²/1000000²
= 144/1000000 = 0.000144

(iv) 120²
See Solutions(iv) 120²
= (12 × 10)² = 12² × 10²
= 144 × 100 = 14400

7. Circle the numbers that are the same —
• 2⁴ × 3⁶
• 6⁴ × 3²
• 6¹⁰
• 18² × 6²
• 6²⁴
See SolutionsSimplifying all the numbers:
• 2⁴ × 3⁶
• 6⁴ × 3² = (2 × 3)⁴ × 3² = 2⁴ × 3⁴ × 3² = 2⁴ × 3⁶
• 6¹⁰ = (2 × 3)¹⁰ = 2¹⁰ × 3¹⁰
• 18² × 6² = (2×3²)² × (2×3)² = 2²×3⁴ × 2²×3² = 2⁴ × 3⁶
• 6²⁴ = (2 × 3)²⁴ = 2²⁴ × 3²⁴
After simplification, we can see that (2⁴ × 3⁶), (6⁴ × 3²) and (18² × 6²) are same.

8. Identify the greater number in each of the following —
(i) 4³ or 3⁴
See Solutions(i) 4³ or 3⁴
4³ = 64
3⁴ = 81
So, 3⁴ is greater.

(ii) 2⁸ or 8²
See Solutions(ii) 2⁸ or 8²
2⁸ = 256
8² = 64
So, 2⁸ is greater.

(iii) 100² or 2¹⁰⁰
See Solutions(iii) 100² or 2¹⁰⁰
100² = 10000
2¹⁰⁰ = (2¹⁰)¹⁰ = (1024)¹⁰ [Since, 2¹⁰ = 1024]
So, 2¹⁰⁰ is a huge number, thus it is greater.

Class 8 Maths Chapter 2 All Questions

9. A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of?
See SolutionsA dairy plans to produce 8.5 billion = 8.5 × 10⁹ packets.
Each packet needs a unique numeric code using digits 0–9.
The number of unique codes that can be made with n digits = (10)^𝑛
We must find the smallest n such that: 10^𝑛 ≥ 8.5 × 10⁹
If we put 𝑛 = 9, we get:
10⁹ = 1,000,000,000 less than 8.5 billion.
If we put 𝑛 = 10, we get:
10¹⁰ = 10,000,000,000 more than 8.5 billion.
So, The code must have at least 10 digits, since 10¹⁰ is the smallest power of 10 greater than 8.5 billion.

10. 64 is a square number (8²) and a cube number (4³). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?
See SolutionsThe numbers that are both squares and cubes:
• 1 = 1² = 1³
• 64 = 8² = 4³
• 729 = 27² = 9³
• 4096 = 64² = 16³
• 15625 = 125² = 25³
The way to describe such numbers in general:
All numbers that are sixth powers (like 1⁶, 2⁶, 3⁶, …) are both perfect squares and perfect cubes.
So the general form is:
Numbers of the form 𝑛 = 𝑘⁶ are both squares and cubes.

11. A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW and 003AZ. How many such codes are possible?
See SolutionsIt is given that each passcode is 5 characters long.
Each character can be alphanumeric, i.e., A–Z (26 letters) and 0–9 (10 digits)
So, total choices per position = 26 + 10 = 36
Now, each of the 5 positions in the code can be filled in 36 ways.
So, total number of such codes = 36 × 36 × 36 × 36 × 36 = 36⁵
Thus, 36⁵ = 60,466,176
Therefore, 60,466,176 codes are possible.

12. The worldwide population of sheep (2024) is about 10⁹ and that of goats is also about the same. What is the total population of sheep and goats?
(ii) 20⁹ (ii) 10¹¹ (iii) 10¹⁰
(iv) 10¹⁸ (v) 2 × 10⁹ (vi) 10⁹ + 10⁹
See SolutionsGiven that:
Population of sheep = 10⁹
Population of goat = 10⁹
So, the total population of sheep and goat:
= 10⁹ + 10⁹
= 2 × 10⁹
Hence, the obtions (v) and (vi) are correct.

13. Calculate and write the answer in scientific notation:
(i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.
See Solutions(i) World population ≈ 8 × 10⁹ (approx.)
Each person has 30 clothes
So, total clothes:
30 × 8 × 10⁹
= 240 × 10⁹
= 2.4 × 10¹¹

(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
See Solutions(ii) Given that 100 million bee colonies, 50,000 bees per colony.
So, the number of bee colonies
= 100 million
= 1 × 10⁸
Each colony has bees = 50,000
= 5 × 10⁴
Therefore, the total bees:
(Number of bee colonies)×(Number of bees in each colony)
= (1 × 10⁸) × (5 × 10⁴)
= 5 × 10⁸ × 10⁴
= 5 × 10¹² bees

(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.
See Solutions(iii) World population ≈ 8 × 10⁹ (approx.)
The human body has bacterial cells = 38 trillion
38 trillion = 3.8 × 10¹³
So, total Bacteria = (8 × 10⁹) × (3.8 × 10¹³)
= 30.4 × 10⁹ × 10¹³
= 30.4 × 10²²
= 3.04 × 10²³

(iv) Total time spent eating in a lifetime in seconds.
See Solutions(iv) Average eating time per day = 1.5 hours
= 1.5 × 60 × 60 = 5400 seconds
Average lifespan = 70 years
Number of days in 70 years = 70 × 365 = 25,550 days
Total eating time in lifespan:
= (5400) × 25,550
= 137,970,000 seconds
= 1.3797 × 10⁸ seconds

14. What was the date 1 arab/1 billion seconds ago?
See SolutionsIn the Indian number system:
1 arab = 1,00,00,00,000 = 10⁹ = 1 billion seconds
Now, convert 10⁹ seconds into years.
We know that:
1 minute = 60 seconds
1 hour = 60 minutes = 3600 seconds
1 day = 24 hours = 86,400 seconds
1 year ≈ 365.25 days (to include leap years)
So, seconds in a year
= 365.25 × 24 × 60 × 60
= 31,557,600
Therefore, the number of years in 10⁹ seconds
= 10⁹/31,557,600
= 1,00,00,00,000/31,557,600
= 31.7 years
Let’s assume today is July 29, 2025.
Go back 31.7 years
⇒ approximately 31 years and 8.5 months
So, subtract 31 years
⇒ July 29, 1994
Go back approx. 8.5 months
⇒ Around mid-November 1993
That means 1 arab seconds ago, it was around November 1993.

Is Class 8 Maths Ganita Prakash Chapter 2 difficult?

Class 8 Maths Ganita Prakash Chapter 2 Power Play is not difficult if you understand the basic rules of powers and exponents. At first, terms like exponents, bases and scientific notation may seem new, but once you practice them using examples, it becomes easy. The chapter builds on your previous knowledge of multiplication and division. It teaches you to deal with very large or very small numbers in a smart and simple way. The key is to go slow in the beginning and practice each law of exponents one by one. Using visual aids, class notes and solved examples in the NCERT textbook makes it easier. Once you get the hang of it, the chapter becomes fun and useful in many real-life situations like time, population and measurement.

How can I score full marks in Class 8 Maths Chapter 2?

To get maximum marks in Class 8 Maths Ganita Prakash Chapter 2 Power Play, you need to focus on understanding the rules of exponents and how they are applied. Don’t just memorize the laws—try to understand how and why they work. Practice problems from each type: multiplying powers, dividing powers, negative exponents and converting to scientific notation. Attempt all NCERT exercises and revise your mistakes. Use solved examples to guide you and test yourself with extra problems. Learn to write steps clearly, as Maths marks are awarded for each step shown. Also, revise word problems like those involving population, time or large values. In the exam, read questions carefully and manage your time. With consistent practice and a calm mind, scoring full marks is totally possible.

How can I complete 8th Ganita Prakash Chapter 2 in a day?

If you want to finish Class 8 Maths Ganita Prakash Chapter 2 Power Play in one day, you need a clear and focused plan. Start with watching a quick revision video or reading the summary in your textbook to understand the main concepts. Then divide your time into blocks—spend one hour on laws of exponents, one hour on negative powers and one hour on scientific notation. Solve a few questions from each section, focusing more on understanding than quantity. Once confident, try to do the NCERT examples and exercises without looking at the solutions. If stuck, refer to the step-by-step NCERT solutions. Don’t skip the real-life application questions—they are often asked in tests. By evening, revise everything quickly and test yourself with a short mock quiz. One day is enough if you stay focused and avoid distractions.

What is the best way to remember Class 8 Maths Ganita Prakash Chapter 2?

The best way to remember exponent rules from Class 8 Maths Ganita Prakash Chapter 2 Power Play is to create short memory tricks or flashcards. Each rule can be written with a small example and a name—for example, product rule, quotient rule, or zero exponent rule. Use colorful charts or sticky notes and stick them on your study wall. Practice a few questions daily using only those rules to build strong memory. You can also teach these rules to a friend or try explaining them aloud—it helps you remember better. Try grouping similar rules together and solving mixed problems to test your memory. Creating patterns and relating powers to real-life examples like mobile storage or population also makes them easy to recall. With regular revision and simple tricks, these rules will stay in your memory for long.

Why is Class 8 Maths Chapter 2 important for future learning?

Class 8 Maths Ganita Prakash Chapter 2 is important because it introduces you to powers and exponents, which are used in higher classes and even in science subjects. Whether it’s calculating large distances in space, estimating population or understanding data in scientific research, exponent knowledge is very useful. It forms the base for topics like algebra, square roots, cube roots and scientific notation used in physics and chemistry. The chapter also helps you improve your mental math and logical reasoning. Once you understand how powers work, you’ll find many future concepts easier to learn. It’s a small chapter, but it plays a big role in strengthening your number sense and making you confident in handling very big or very small numbers in daily life as well as competitive exams.

Content Reviewed: July 30, 2025

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Saikat Chakravarty

Providing help in science for class 6 to 10. Adviser in Tiwari Academy for the science related subjects subject as well as videos contents. Ample teaching experience in schools. I am not only working for Tiwari Academy but also provide new ideas for the website and apps.