NCERT Solutions for Class 8 Mathematics Ganita Prakash I Chapter 2

NCERT solutions for Class 8 Maths Chapter 2 Power Play – complete answers & explanations

Class 8 Maths from the Ganit Prakash Part 1 textbook brings some of the most exciting and thought-provoking topics in the curriculum, and Chapter 2 – Power Play – is one of the highlights. This chapter introduces students to the world of exponents and powers, helping them understand how numbers can grow at incredible rates when multiplied by themselves repeatedly. From paper-folding activities to stories about diamonds, magical ponds, and scientific notation used to describe astronomical distances, the chapter makes abstract mathematical ideas come alive through real-world contexts. Understanding exponential notation is a foundational skill, not just for Class 8, but for higher mathematics and science as well. Students who build a strong grip on powers and their rules here will find topics like standard form, large number calculations, and algebraic expressions much easier in later classes. This blog covers clear and reliable NCERT solutions for every question, activity, and figure it out section in this chapter. Download the worksheet and practice alongside solutions for better clarity. If your child needs additional support to go deeper into the concepts, book a free trial now to get expert guidance.

NCERT Solutions for Class 8 Mathematics Ganita Prakash I Chapter 2.png

What this NCERT chapter covers?

1. The chapter opens with an engaging activity called An Impossible Venture, where students explore what happens to the thickness of a sheet of paper when it is folded repeatedly. This introduces the concept of exponential growth in a hands-on way.
2. Students learn about exponential notation and how any number can be expressed as a base raised to a power, for example, 32400 = 2^4 x 5^2 x 3^4.
3. The chapter covers prime factorisation in exponential form, helping students break down large numbers into their basic prime factors and represent them using powers.
4. Key laws of exponents are taught step by step, including the product rule (n^a x n^b = n^(a+b)), the quotient rule (n^a / n^b = n^(a-b)), and the power of a power rule ((n^a)^b = n^(ab)).
5. Students explore the concept of zero exponent, understanding why any non-zero number raised to the power 0 equals 1.
6. Negative exponents are introduced, teaching students that n^(-a) = 1/n^a, with several practice problems to reinforce the idea.
7. The chapter includes rich story-based sections like The Stones that Shine, Magical Pond, and How Many Combinations, which make learning laws of exponents contextual and enjoyable.
8. Scientific notation is introduced to help students express very large and very small numbers in a compact and standard form, such as 3 x 10^20 or 5.976 x 10^24.
9. The concept of linear growth vs. exponential growth is explored through real-life examples, helping students understand the difference in scale between the two.
10. The chapter also includes interesting sections on getting a sense of large numbers, age expressed in days and seconds, and a fun puzzle activity called Tremendous in Ten, making this one of the most concept-rich chapters of Class 8 Maths.

How to use these NCERT solutions?

1. Begin by reading each question or activity carefully in your worksheet before looking at the answer. Attempting questions on your own first builds understanding and confidence.
2. After attempting, refer to the solution for that specific question. Do not jump ahead – the solutions are arranged in the exact same order as the worksheet, section by section.
3. For questions that involve calculation steps, follow each step in the solution carefully. Try to reproduce the working in your own notebook to build fluency.
4. For activity-based sections like An Impossible Venture and Puzzle Time, read the explanation given and understand the process described, as these are often discussed in class and asked in assessments.
5. Parents can use this blog to cross-check their child's work at home. Each answer is aligned with NCERT answers and follows the correct approach to answering, making it easy to verify.
6. Teachers can use these solutions as a reference when explaining concepts in class or during revision sessions.
7. For open-ended or estimation-based questions, note that assumptions may vary. Focus on the method and logic used rather than expecting a single fixed answer.
8. Revisit sections where your answers did not match. Try to understand where the approach went wrong before moving on.
9. Use the solutions alongside your textbook. The combination of reading the chapter explanation and practising with solutions gives the best results.
10. After completing each section, try solving a similar problem on your own without looking at the answer, to test your understanding.

Important tips & tricks for students

1. Always write the full working when solving problems involving laws of exponents. Simply writing the answer without steps may cost marks in exams.
2. A very common mistake is to add bases instead of adding exponents when multiplying powers with the same base. Remember: 2^3 x 2^4 = 2^7, not 4^7.
3. When applying the power of a power rule, multiply the exponents – do not add them. For example, (2^3)^4 = 2^12, not 2^7.
4. For negative exponents, always remember the rule: n^(-a) = 1/n^a. Do not confuse a negative exponent with a negative number.
5. When expressing numbers in standard form (scientific notation), make sure the first factor is always between 1 and 10. For example, 59,853 = 5.9853 x 10^4, not 59.853 x 10^3.
6. In story-based questions like The Stones that Shine and Magical Pond, read carefully and identify the repeated multiplication pattern. Expressing the repeated factor as a power is the key step.
7. For activity questions and estimation questions, always state your assumptions clearly. The method of solving matters more than getting the exact value.
8. Practice the laws of exponents together as a group: product rule, quotient rule, power rule, zero exponent, and negative exponent. These five rules cover most of the chapter.
9. When comparing large numbers written in scientific notation, first compare the powers of 10. A higher power means a larger number, regardless of the coefficient in front.
10. For Figure it Out sections, read the full problem before writing anything. These questions often require two or more laws of exponents applied together.

NCERT solutions – complete answer key

An impossible venture! (Activity)
How many times can you fold it over and over?
In practice, a sheet of paper can be folded only about 7 times, regardless of its size, because each fold doubles the thickness, making it too thick and stiff to fold further.

What would the thickness be after 30 folds? Make a guess.
After 30 folds, the thickness would be approximately 10.7 km (about the height at which aeroplanes fly). This is 0.001 cm x 2^30 ≈ 10,737 km.

After 46 folds, thickness = 0.001 x 2^46 ≈ 7,03,687,441 km (more than 7,00,000 km). This confirms the paper can reach the Moon (≈ 3,84,400 km away) and beyond!

Notice the change in thickness after two folds. By how much does it increase?
After 2 folds, the thickness increases by 4 times (= 2 x 2 = 2^2). For example, from Fold 4 (0.016 cm) to Fold 6 (0.064 cm): 0.064 / 0.016 = 4 times.

2.2 Exponential notation and operations
Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is represented by the letter-number v.
The correct answer is (v) 2^10v. Each fold doubles the thickness, so after 10 folds, the thickness = v x 2 x 2 x … (10 times) = v x 2^10 = 2^10v.

Express the number 32400 as a product of its prime factors and represent the prime factors in their exponential form.
32400 = 2 x 2 x 2 x 2 x 5 x 5 x 3 x 3 x 3 x 3
In exponential form: 32400 = 2^4 x 5^2 x 3^4

What is (-1)^5? Is it positive or negative? What about (-1)^56?
(-1)^5 = -1 (negative), because an odd number of negative factors gives a negative product. (-1)^56 = +1 (positive), because an even number of negative factors gives a positive product.

Is (-2)^4 = 16? Verify
(-2)^4 = (-2) x (-2) x (-2) x (-2) = 4 x 4 = 16. Yes, it is 16.

What is 0^2, 0^5? What is 0^n?
0^2 = 0 x 0 = 0; 0^5 = 0 x 0 x 0 x 0 x 0 = 0. 0^n = 0 for any positive integer n (0 multiplied by itself any number of times is always 0).

Figure it out – Exercise 1
Q1. Express the following in exponential form:
(i) 6 x 6 x 6 x 6 = 6^4
(ii) y x y = y^2
(iii) b x b x b x b = b^4
(iv) 5 x 5 x 7 x 7 x 7 = 5^2 x 7^3
(v) 2 x 2 x a x a = 2^2 x a^2
(vi) a x a x a x c x c x c x c x d = a^3 x c^4 x d

Q2. Express each number as a product of prime factors in exponential form:
(i) 648
648 = 2 x 324 = 2 x 2 x 162 = 2 x 2 x 2 x 81 = 2 x 2 x 2 x 3 x 27 = 2 x 2 x 2 x 3 x 3 x 9 = 2 x 2 x 2 x 3 x 3 x 3 x 3
648 = 2^3 x 3^4
(ii) 405
405 = 5 x 81 = 5 x 3 x 27 = 5 x 3 x 3 x 9 = 5 x 3 x 3 x 3 x 3
405 = 3^4 x 5
(iii) 540
540 = 2 x 270 = 2 x 2 x 135 = 2 x 2 x 3 x 45 = 2 x 2 x 3 x 3 x 15 = 2 x 2 x 3 x 3 x 3 x 5
540 = 2^2 x 3^3 x 5
(iv) 3600
3600 = 36 x 100 = 4 x 9 x 4 x 25 = 2^2 x 3^2 x 2^2 x 5^2
3600 = 2^4 x 3^2 x 5^2

Q3. Evaluate the following:
(i) 2 x 10^3 = 2 x 1000 = 2000
(ii) 7^2 x 2^3 = 49 x 8 = 392
(iii) 3 x 4^4 = 3 x 256 = 768
(iv) (-3)^2 x (-5)^2 = 9 x 25 = 225
(v) 3^2 x 10^4 = 9 x 10000 = 90000
(vi) (-2)^5 x (-10)^6 = (-32) x 10,00,000 = -3,20,00,000

The stones that shine
How many rooms were there altogether?
3 daughters x 3 baskets x 3 keys = 3^3 rooms per daughter's keys. Total rooms = 3 x 3 x 3 x 3 = 3^4 = 81 rooms.

How many diamonds were there in total?
Diamonds = 3^7 = 3 x 3 x 3 x 3 x 3 x 3 x 3 = 2187 diamonds. (3 daughters x 3 baskets x 3 keys x 3 rooms x 3 tables x 3 necklaces x 3 diamonds = 3^7)

3^7 can also be written as 3^2 x 3^5. Can you reason out why?
Using the law n^a x n^b = n^(a+b): 3^2 x 3^5 = 3^(2+5) = 3^7.

Write the product p^4 x p^6 in exponential form
p^4 x p^6 = p^(4+6) = p^10

Use n^a x n^b = n^(a+b) to compute: (i) 2^9 (ii) 5^7 (iii) 4^6
(i) 2^9 = 2^4 x 2^5 = 16 x 32 = 512
(ii) 5^7 = 5^3 x 5^4 = 125 x 625 = 78125
(iii) 4^6 = 4^3 x 4^3 = 6^4 x 6^4 = 4096

Write the following expressions as a power of a power in at least two different ways: (i) 8^6 (ii) 7^15 (iii) 9^14 (iv) 5^8
(i) 8^6 = (8^2)^3 = (8^3)^2
(ii) 7^15 = (7^3)^5 = (7^5)^3
(iii) 9^14 = (9^2)^7 = (9^7)^2
(iv) 5^8 = (5^2)^4 = (5^4)^2

Magical pond
On which day was the pond half full?
The pond is fully covered on Day 30. Since the number of lotuses doubles every day, on the previous day (Day 29) the pond was half full.

Write the number of lotuses (in exponential form) when the pond was: (i) fully covered (ii) half covered
Let the initial lotus on Day 0 = 1 = 2^0. (i) Fully covered (Day 30): 2^30 lotuses. (ii) Half covered (Day 29): 2^29 lotuses.

After 4 days in the doubling pond, Damayanti transfers flowers to the tripling pond. How many lotuses after 4 more days?
After 4 days in doubling pond: 1 x 2^4 = 2^4 = 16 lotuses.
After 4 more days in tripling pond: 2^4 x 3^4 = 16 x 81 = 1296 lotuses.

If Damayanti changed the order: tripling pond first, then doubling pond. How many?
1 x 3^4 x 2^4 = 3^4 x 2^4 = 81 x 16 = 1296 lotuses (same result).

Can 3^4 x 2^4 be expressed as m^n?
Yes. Using m^a x n^a = (mn)^a: 3^4 x 2^4 = (3 x 2)^4 = 6^4 = 1296.

Use m^a x n^a = (mn)^a to compute 2^5 x 5^5.
2^5 x 5^5 = (2 x 5)^5 = 10^5 = 1,00,000.

Simplify 10^4 / 5^4 and write in exponential form.
10^4 / 5^4 = (10 / 5)^4 = 2^4 = 16.

How many combinations
Estu has 4 dresses and 3 caps. How many different ways can Estu combine them?
Total combinations = 4 x 3 = 12 combinations.

Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up?
Total combinations = 7 x 2 x 3 = 42 combinations.

Estu and Roxie try every 5-digit password to open a safe. How many passwords did they check?
Each digit has 10 choices (0-9). For 5 digits: 10 x 10 x 10 x 10 x 10 = 10^5 = 1,00,000 passwords.

How many passwords are possible with a 6-slot lock using letters A to Z?
Each slot has 26 choices (A to Z). For 6 slots: 26^6 = 3,08,91,57,76 ≈ 3.09 x 10^8 passwords. This is far more than the 10^5 = 1,00,000 possible for a 5-digit numeric lock. Estu is right — it is much safer.

The other side of powers
What is 2^100 / 2^25 in powers of 2?
2^100 / 2^25 = 2^(100-25) = 2^75.

Why can't n be 0 in n^a / n^b = n^(a-b)?
If n = 0, then n^b = 0, and division by zero is undefined. Therefore n ≠ 0 is a necessary condition.

What is 2^0?
2^0 = 2^(4-4) = 2^4 / 2^4 = 16 / 16 = 1. In general, x^0 = 1 for any x ≠ 0.

Write equivalent forms of the following:
(i) 2^(-4) = 1 / 2^4 = 1/16
(ii) 10^(-5) = 1 / 10^5 = 1/1,00,000
(iii) (-7)^(-2) = 1 / (-7)^2 = 1/49
(iv) (-5)^(-3) = 1 / (-5)^3 = -1/125
(v) 10^(-100) = 1 / 10^100

Simplify and write the answers in exponential form:
(i) 2^(-4) x 2^7 = 2^(-4+7) = 2^3 = 8
(ii) 3^2 x 3^(-5) x 3^6 = 3^(2-5+6) = 3^3 = 27
(iii) p^3 x p^(-10) = p^(3-10) = p^(-7) = 1/p^7
(iv) 2^4 x (-4)^(-2) = 16 x (1/16) = 1
(v) 8^p x 8^q = 8^(p+q)

Power lines
How many times larger than 4^(-2) is 4^2?
4^2 / 4^(-2) = 4^(2-(-2)) = 4^4 = 256 times larger.

Use the power line for 7 to answer the following:
2,401 x 49 = 7^4 x 7^2 = 7^6 = 1,17,649
49^3 = (7^2)^3 = 7^6 = 1,17,649
343 x 2,401 = 7^3 x 7^4 = 7^7 = 8,23,543
16,807 / 49 = 7^5 / 7^2 = 7^3 = 343
7 / 343 = 7^1 / 7^3 = 7^(-2) = 1/49
16,807 / 8,23,543 = 7^5 / 7^7 = 7^(-2) = 1/49
1,17,649 x (1/343) = 7^6 x 7^(-3) = 7^3 = 343
(1/343) x (1/343) = 7^(-3) x 7^(-3) = 7^(-6) = 1/1,17,649

Powers of 10
Write these numbers in expanded form using powers of 10: (i) 172 (ii) 5642 (iii) 6374
(i) 172 = (1 x 10^2) + (7 x 10^1) + (2 x 10^0)
(ii) 5642 = (5 x 10^3) + (6 x 10^2) + (4 x 10^1) + (2 x 10^0)
(iii) 6374 = (6 x 10^3) + (3 x 10^2) + (7 x 10^1) + (4 x 10^0)

Scientific notation
Write the large-number facts in scientific notation:
(i) Distance of Sun from centre of Milky Way: 3,00,00,00,00,00,00,00,00,000 m = 3 x 10^20 m
(ii) Number of stars in our galaxy: 1,00,00,00,00,000 = 1 x 10^11
(iii) Mass of the Earth: 5,97,60,00,00,00,00,00,00,00,00,000 kg ≈ 5.976 x 10^24 kg

The distance between the Sun and Saturn is 1.4335 x 10^12 m, between Saturn and Uranus is 1.439 x 10^12 m, and between the Sun and Earth is 1.496 x 10^11 m. Which is smallest?
The exponent of Sun-Earth distance is 10^11, while both Saturn and Uranus have 10^12. Therefore, the Sun-Earth distance (1.496 x 10^11 m) is the smallest.

Express the following numbers in standard form:
(i) 59,853 = 5.9853 x 10^4
(ii) 65,950 = 6.595 x 10^4
(iii) 34,30,000 = 3.43 x 10^6
(iv) 70,04,00,00,000 = 7.004 x 10^10

2.5 Did you ever wonder?
What is the worth of donated jaggery and wheat?
Assuming Roxie's weight = 45 kg, cost of jaggery = 70/kg: Worth of jaggery = 45 x 70 = 3150. Assuming Estu's weight = 50 kg, cost of wheat = 50/kg: Worth of wheat = 50 x 50 = 2500.
Note: Assumptions may vary. The method of solving is more important than the exact answer.

Roxie wonders: Instead of jaggery, how many 1-rupee coins are needed to equal her weight?
Step 1 – Guess: The number of coins would be in the thousands.
Step 2 – Calculate: Weight of Roxie ≈ 45 kg = 45,000 g. Weight of 1-rupee coin ≈ 3.76 g (standard weight). Number of coins = 45,000 / 3.76 ≈ 11,968 coins ≈ 1.2 x 10^4 coins.

What if 5-rupee coins or 10-rupee notes are used? How much money?
5-rupee coin weighs ≈ 6 g. Number of coins = 45,000 / 6 = 7,500 coins. Value = 7,500 x 5 = 37,500 (≈ 3.75 x 10^4).
10-rupee note weighs ≈ 1 g. Number of notes = 45,000 notes. Value = 45,000 x 10 = 4,50,000 (= 4.5 x 10^5).
Note: Assumptions may vary. Answers are estimates.

Fossil of Kelenken Guillermoi dated to 15 million years ago ≈ ___ seconds.
15 million years = 15 x 10^6 x 365 x 24 x 3600 ≈ 15 x 10^6 x 3.156 x 10^7 ≈ 4.73 x 10^14 seconds.

Plants on land started 470 million years ago ≈ ___ seconds.
470 x 10^6 x 3.156 x 10^7 ≈ 1.48 x 10^16 seconds ≈ 1.5 x 10^16 seconds.

Calculate using scientific notation:
(i) Stars = 2 x 10^23. At 1 per second, time = 2 x 10^23 seconds.
(ii) Total water on Earth ≈ 1.386 x 10^21 litres = 1.386 x 10^24 ml. Glasses = (1.386 x 10^24) / 200 ≈ 6.93 x 10^21. Time = 6.93 x 10^21 x 10 seconds ≈ 6.93 x 10^22 seconds.

Linear growth vs. exponential growth
How many 20 cm steps to reach the Moon (3,84,400 km)?
3,84,400 km = 3,84,400 x 1,00,000 cm = 3.844 x 10^10 cm. Number of steps = (3.844 x 10^10) / 20 = 1.922 x 10^9 = 1,92,20,00,000 steps.

Can you come up with examples of linear growth and exponential growth?
Linear growth examples: Walking a fixed distance per day; saving a fixed amount of money every month; filling a tank at a constant rate.
Exponential growth examples: Population of bacteria doubling every hour; compound interest; spread of a viral disease; number of WhatsApp forwards.

10^5 seconds ≈ 1.16 days and 10^6 seconds ≈ 11.57 days. Think of events of these orders:
(i) 10^5 seconds (≈ 1 day): Time for Earth to rotate once on its axis = 8.64 x 10^4 seconds ≈ 10^5 seconds.
(ii) 10^6 seconds (≈ 11.5 days): Duration of a school examination term; time taken for a long sea voyage.

Getting a sense for large nos.
Global starling population ≈ 1.3 billion = 1.3 x 10^9.
Mosquito population ≈ 11 neel/110 trillion = 1.1 x 10^14.

With 8 x 10^9 humans and 4 x 10^5 African elephants, are there nearly 20,000 people per elephant?
People per elephant = (8 x 10^9) / (4 x 10^5) = 2 x 10^4 = 20,000. Yes!

Calculate using scientific notation:
(i) Ants per human = (2 x 10^16) / (8.2 x 10^9) ≈ 2.44 x 10^6 (about 24 lakh ants per human).
(ii) Number of starling flocks (10,000 birds each): (1.3 x 10^9) / (10^4) = 1.3 x 10^5 flocks.
(iii) Total leaves = trees x leaves per tree = (3 x 10^12) x 10^4 = 3 x 10^16 leaves.
(iv) Sheets of paper to reach Moon: Distance = 3,84,400 km = 3.844 x 10^10 cm. Thickness of one sheet = 0.001 cm = 10^(-3) cm. Sheets needed = (3.844 x 10^10) / 10^(-3) = 3.844 x 10^13 sheets.

A different way to say your age!
Roxie is 4840 days old. How many hours old is she?
4840 days x 24 hours/day = 1,16,160 hours ≈ 1.16 x 10^5 hours.

Estu is 4070 days old. Find his date of birth.
4070 days before 26 March 2026 = approximately 7 November 2014. Note: The current date used for calculation is 26 March 2026 as given in the problem context.

If you have lived for a million seconds, how old would you be?
1,000,000 seconds / 60 / 60 / 24 ≈ 11.57 days. So if you have lived a million seconds, you would be about 11-12 days old.

Roxie and Estu hear someone who did p day tra for about 400 km. How long ago would they have started?
Average walking speed ≈ 4 km/hour, walking ≈ 8 hours/day. Distance per day = 32 km. Days taken = 400 / 32 ≈ 12.5 days ≈ about 12-13 days ago. Note: Answers depend on assumptions of speed and hours walked per day.

How many times can a person circumnavigate the Earth in their lifetime if they walk non-stop? Distance around Earth = 40,000 km.
Average lifespan ≈ 70 years = 70 x 365 x 24 = 6,13,200 hours. At 4 km/hr: Distance covered = 6,13,200 x 4 = 24,52,800 km. Number of circumnavigations = 24,52,800 / 40,000 ≈ 61 times.

A pinch of history
What does the first part of each name (million, billion, trillion…) denote?
The prefix denotes the Latin number for the power of 1000:
mi- (one) → million = 10^6 = 1000^1 x 1000
bi- (two) → billion = 10^9 = 1000^3
tri- (three) → trillion = 10^12
quad- (four) → quadrillion = 10^15
quin- (five) → quintillion = 10^18, and so on.
Each prefix is a Latin number indicating how many thousands are multiplied together.

Figure it out
Q1.
4^32 = (2^2)^32 = 2^64.
2^224 / 2^64 = 2^(224-64) = 2^160.
Powers of 2 cycle in units digit: 2, 4, 8, 6, 2, 4, 8, 6, … (period 4). 160 / 4 = 40 (remainder 0), so units digit of 2^160 = units digit of 2^4 = 6. Units digit = 6.

Q2.
After Day 1: 5 bottles. After Day 2: 5 + 5 = 10. The pattern: each day a new container of 5 is added. After 40 days: 5 x 40 = 200 bottles. Note: This is linear growth — 5 bottles added each day.

Q3.
(i) 64^3
Way 1: 64^3 = (8^2)^3 = 8^6
Way 2: 64^3 = (4^3)^3 = 4^9
Way 3: 64^3 = (2^6)^3 = 2^18
(ii) 192^8
192 = 2^6 x 3, so 192^8 = 2^48 x 3^8
Way 1: (2^48) x (3^8)
Way 2: (2^24)^2 x (3^4)^2
Way 3: (2^6 x 3)^8 = 192^8
32^(-5): 32 = 2^5, so 32^(-5) = (2^5)^(-5) = 2^(-25)
Way 1: 2^(-25)
Way 2: (2^(-5))^5
Way 3: (2^5)^(-5)

Q4.
(i) Cube numbers are also square numbers. Only Sometimes True. E.g., 64 = 4^3 = 8^2 (both square and cube). But 8 = 2^3 is not a square.
(ii) Fourth powers are also square numbers. Always True. n^4 = (n^2)^2, which is always a perfect square.
(iii) The fifth power of a number is divisible by the cube of that number. Always True. n^5 / n^3 = n^2, which is always a whole number for any non-zero n.
(iv) The product of two cube numbers is a cube number. Always True. a^3 x b^3 = (ab)^3, which is always a cube.
(v) q^46 is both a 4th power and a 6th power (q is a prime number). Never True for prime q, since neither 4 nor 6 divides 46 evenly.

Q5.
(i) 10^(-2) x 10^(-5) = 10^(-2+(-5)) = 10^(-7)
(ii) 5^7 / 5^4 = 5^(7-4) = 5^3 = 125
(iii) 9^(-7) / 9^4 = 9^(-7-4) = 9^(-11)
(iv) (13^(-2))^(-3) = 13^((-2)x(-3)) = 13^6
(v) m^5 n^12 (mn)^9 = m^5 n^12 x m^9 n^9 = m^(5+9) n^(12+9) = m^14 n^21

Q6.
(i) (1.2)^2 = (12 / 10)^2 = 144 / 100 = 1.44
(ii) (0.12)^2 = (12 / 100)^2 = 144 / 10000 = 0.0144
(iii) (0.012)^2 = (12 / 1000)^2 = 144 / 10,00,000 = 0.000144
(iv) 120^2 = (12 x 10)^2 = 144 x 100 = 14400

Q7.
2^4 x 3^6, 6^4 x 3^2, and 18^2 x 6^2 are all equal (= 11,664). Note: 2^4 x 3^6 = 2^4 x 3^4 x 3^2 = 6^4 x 9 = 64 x 32. Also 18^2 x 6^2 = (18x6)^2 = 108^2? Let's verify: 18^2 = 324, 6^2 = 36, 324 x 36 = 11664. 6^4 = 1296, 3^2 = 9, 1296 x 9 = 11664.

Q8.
(i) 4^3 or 3^4: 4^3 = 64; 3^4 = 81. 3^4 is greater.
(ii) 2^8 or 8^2: 2^8 = 256; 8^2 = 64. 2^8 is greater.
(iii) 100^2 or 2^100: 100^2 = 10,000 = 10^4; 2^100 ≈ 1.27 x 10^30. 2^100 is much greater.

Q9.
8.5 billion = 8.5 x 10^9 ≈ 10^10. For a 9-digit code: 10^9 = 1,000,000,000 (only 1 billion — not enough). For a 10-digit code: 10^10 = 10,000,000,000 (10 billion — enough). The code should consist of 10 digits.

Q10.
Yes. Numbers that are both perfect squares and perfect cubes are perfect 6th powers. General form: n^6 (i.e., n^(2x3)). Examples: 1^6=1, 2^6=64, 3^6=729, 4^6=4096, ... Such numbers are of the form n^6 for any positive integer n.

Q11.
Total characters = 10 digits + 26 letters = 36 characters. For each of the 5 positions, there are 36 choices. Total codes = 36^5 = 36 x 36 x 36 x 36 x 36 = 6,04,66,176 codes ≈ 6.05 x 10^7.

Q12.
Total = 10^9 + 10^9 = 2 x 10^9. Correct option: (vi) 10^9 + 10^9 = (v) 2 x 10^9. Note: Options (i) 20^9, (ii) 10^11, (iii) 10^10, (iv) 10^18 are incorrect. The correct answer is 2 x 10^9.

Q13.
(i) Clothing: Global population ≈ 8.2 x 10^9. Pieces per person = 30. Total = 30 x 8.2 x 10^9 = 2.46 x 10^11 pieces.
(ii) Honeybees: 100 million colonies = 10^8. 50,000 = 5 x 10^4 per colony. Total = 10^8 x 5 x 10^4 = 5 x 10^12 bees.
(iii) Bacterial cells: Per human = 38 trillion = 3.8 x 10^13. Human population = 8.2 x 10^9. Total = 3.8 x 10^13 x 8.2 x 10^9 ≈ 3.116 x 10^23 bacterial cells.
(iv) Time spent eating: Average 2 hours/day x 365 days x 70 years = 51,100 hours = 1.8396 x 10^8 seconds ≈ 1.84 x 10^8 seconds.

Q14.
10^9 seconds / (3.156 x 10^7 seconds/year) ≈ 31.7 years. 31.7 years before 26 March 2026 ≈ around August 1994. Note: 1 billion seconds = approximately 31 years 8 months.

Puzzle time – Tremendous in Ten!
Activity: Find a partner. In 10 seconds, write a number/expression using only digits 0–9 and arithmetic operations. The larger number wins.

Round 2: Roxie wrote 10^1000 + 10^1000 + 10^1000 + 10^1000 and Estu wrote (10^1000000) x 9000. Which is greater?
Roxie's number: 4 x 10^1000.
Estu's number: 9000 x 10^1000000 = 9 x 10^3 x 10^1000000 = 9 x 10^1000003.
Estu's number is much greater, since 10^1000003 >> 10^1000.

Activity explanation: Play in pairs. Set a 10-second timer. Each player writes a number or expression using only digits 0–9 and the allowed arithmetic operations (as per the round's conditions). Both players reveal their numbers and compare. The one with the larger value wins the round. Explore different rounds with different rules: only addition, only addition and multiplication, with exponents and only addition, or with exponents and any operation.

Why NCERT solutions help students?

Having access to clear and reliable NCERT solutions gives students a significant advantage when preparing for exams. Chapter 2 of Class 8 Maths covers topics like exponential notation, laws of powers, scientific notation, and exponential growth that are directly tested in assessments and form the foundation for higher classes. When students can verify their answers against correct, step-by-step solutions, they identify gaps in understanding much faster. This builds genuine concept clarity rather than surface-level memorisation. Parents feel more confident helping their children at home, and students feel reassured that they are on the right track. Following the correct approach to answering also trains students to present their working neatly and logically, which is what examiners look for. Over time, regular practice with well-structured solutions builds the confidence and exam readiness every Class 8 student needs.

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