NCERT Solutions for Class 8 Maths Chapter 2 Power Play

This worksheet provides complete and accurate NCERT Solutions for Class 8 Maths Chapter 2 Power Play from Ganit Prakash Part 1. This chapter introduces students to the concept of exponents and powers in a deeply engaging and exploratory way. Rather than presenting rules as dry formulas, the chapter builds understanding through real-world contexts such as paper folding, pond growth, password combinations, scientific notation, and large-number estimation. This makes it one of the most thought-provoking chapters in the Class 8 Maths syllabus. The NCERT Solutions provided in this worksheet are carefully worked out, follow the exact structure of the chapter, and help students understand not just the answers but the reasoning behind each step.

Chapter summary: stories, poems & themes

Chapter 2 Power Play does not contain stories or poems in the traditional sense. Instead, it is built around a series of interconnected real-world activities and exploratory contexts that make the learning of exponents meaningful and memorable. The chapter opens with "An Impossible Venture," a paper folding activity that helps students discover exponential growth by observing how the thickness of a sheet of paper doubles with every fold. This leads naturally into the concept of exponential notation. The chapter then progresses through several thematic mini-sections: "The Stones that Shine," which uses a layered riddle involving daughters, baskets, keys, rooms, and diamonds to practise powers of 3; "Magical Pond," which uses doubling lotuses to explore the law of multiplication of powers with the same base; "How Many Combinations," which covers the law of multiplying powers with the same exponent using outfit combinations and pond transfer problems; "The Other Side of Powers," which introduces zero exponents, negative exponents, and the division law; "Power Lines," which uses a number line of powers of 7 to practise all laws together; "Powers of 10," which covers expanded notation and scientific notation; "Getting a Sense for Large Numbers," which uses real-world data on ants, starlings, trees, and honeybees; "A Different Way to Say Your Age," which uses time conversions involving powers of 10; "Linear Growth vs. Exponential Growth," which compares the two types of growth using real-world examples; "A Pinch of History," which explains the Latin origins of number names like million, billion, and trillion; and "Puzzle Time – Tremendous in Ten!," which is a student-generated activity involving writing the largest possible number within ten seconds. The central theme of the chapter is understanding how exponential growth differs from linear growth and how the laws of exponents make working with very large and very small numbers efficient and manageable.

What this NCERT chapter covers?

The chapter covers the following key learning areas as reflected in the worksheet:

- Understanding the concept of repeated multiplication and how it leads to exponential notation
- Writing numbers in exponential form using prime factorisation
- Applying the law of multiplication of powers with the same base: n^a × n^b = n^(a+b)
- Applying the law of power of a power: (n^a)^b = n^(a×b)
- Applying the law of multiplication of powers with the same exponent: m^a × n^a = (mn)^a
- Applying the law of division of powers with the same base: n^a ÷ n^b = n^(a-b), where n ≠ 0
- Understanding zero exponents: x^0 = 1 for any x ≠ 0
- Understanding negative exponents and writing their equivalent forms
- Writing numbers in standard scientific notation
- Using powers of 10 to write numbers in expanded form
- Comparing very large and very small numbers using scientific notation
- Distinguishing between linear growth and exponential growth with real-life examples
- Applying all laws of exponents in multi-step problems involving decimals and algebraic expressions
- Exploring puzzle-based and estimation-based problems that use exponents creatively

How to use these NCERT solutions?

Students should first read each question carefully and try to solve it on their own before looking at the solutions in this worksheet. Attempting questions independently helps build confidence and reveals areas where more practice is needed. Parents can use these solutions to verify their child's work at home and to understand the step-by-step reasoning expected at the Class 8 level. Teachers can use this worksheet as a ready reference to explain solutions in class or to help students who need additional support. All solutions in this worksheet follow the exact order and structure of the NCERT chapter, making it easy to locate any specific question. The worksheet is also highly useful during revision before exams, as it brings together all the key concepts of the chapter in one place with clear, worked-out explanations.

Student tips & learning tricks

- Always write the base and exponent clearly. A common mistake is confusing 2^3 (which equals 8) with 3^2 (which equals 9). The base is the number being multiplied and the exponent tells how many times.
- When multiplying powers with the same base, add the exponents — do not multiply them. For example, 3^2 × 3^5 = 3^7, not 3^10.
- When applying the power of a power rule, multiply the exponents. For example, (2^3)^4 = 2^12, not 2^7.
- Remember that any non-zero number raised to the power zero is always 1. Zero itself raised to any positive power is always 0.
- For negative exponents, remember that a^(-n) = 1/a^n. A negative exponent does not make the result negative — it makes it a fraction.
- In scientific notation, the coefficient must always be between 1 and 10. When comparing numbers in scientific notation, compare the exponents first; the larger exponent gives the larger number.
- For estimation-based problems in this chapter, focus on the method and the approach — the NCERT chapter itself acknowledges that assumptions may vary and that the method is more important than the exact numerical answer.
- When finding the units digit of a large power of 2, use the fact that units digits of powers of 2 follow a repeating cycle of 4: 2, 4, 8, 6, 2, 4, 8, 6, …

Why NCERT solutions are important?

NCERT Solutions for Class 8 Maths are essential because they align perfectly with the curriculum prescribed by the Central Board of Secondary Education and state boards that follow NCERT. Chapter 2 Power Play introduces foundational concepts of exponents and scientific notation that students will continue to use in higher classes, including Class 9 and Class 10, as well as in competitive examinations. Having access to accurate, step-by-step NCERT solutions helps students understand the correct approach to each type of problem, build a strong conceptual base, and avoid common errors. When students practise regularly using these solutions, they develop both the speed and the accuracy needed to perform well in school assessments. For parents, NCERT-aligned solutions provide reassurance that their child is learning exactly what the school curriculum requires, with no gaps.

Complete answer key – NCERT solutions

An Impossible Venture (Activity)

How many times can you fold a sheet of paper 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?
After 30 folds, the thickness would be approximately 10.7 km (about the height at which aeroplanes fly). This is 0.001 cm × 2^30 ≈ 10,737 km.

Fill the table: Thickness after each fold
(Refer to the table in the worksheet. Selected values are: Fold 8 ≈ 262 cm, Fold 9 ≈ 524 cm, Fold 10 ≈ 10.4 m, Fold 17 ≈ 1.34 km, Fold 18 ≈ 2.68 km, Fold 21 ≈ 2097 cm = 20.97 m, Fold 22 ≈ 41.94 m, Fold 23 ≈ 83.89 m, Fold 24 ≈ 167.77 m, Fold 25 ≈ 335.54 m, Fold 26 ≈ 671.08 m = 670 m, Fold 29 ≈ 5.37 km, Fold 30 ≈ 10.73 km, Fold 31 ≈ 21.47 km, Fold 32 ≈ 42.95 km, Fold 33 ≈ 85.9 km, Fold 34 ≈ 171.8 km, Fold 35 ≈ 343.6 km, Fold 36 ≈ 687.19 km, Fold 37 ≈ 1374 km, Fold 38 ≈ 2748 km, Fold 39 ≈ 5497 km, Fold 41 ≈ 21,990 km, Fold 42 ≈ 43,980 km, Fold 43 ≈ 87,960 km, Fold 44 ≈ 1,75,921 km)

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

After 46 folds, can the paper reach the Moon?
After 46 folds, thickness = 0.001 × 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.

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 2^10v. Each fold doubles the thickness, so after 10 folds, the thickness = v × 2 × 2 × … (10 times) = v × 2^10 = 2^10v.

Express the number 32400 as a product of its prime factors in exponential form.
32400 = 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3 × 3 × 3
In exponential form: 32400 = 2^4 × 5^2 × 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) × (–2) × (–2) × (–2) = 4 × 4 = 16. Yes, it is 16.

What is 0^2, 0^5? What is 0^n?
0^2 = 0 × 0 = 0; 0^5 = 0 × 0 × 0 × 0 × 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

1.
(i) 6 × 6 × 6 × 6 = 6^4
(ii) y × y = y^2
(iii) b × b × b × b = b^4
(iv) 5 × 5 × 7 × 7 × 7 = 5^2 × 7^3
(v) 2 × 2 × a × a = 2^2 × a^2
(vi) a × a × a × c × c × c × c × d = a^3 × c^4 × d

2.
(i) 648
648 = 2 × 324 = 2 × 2 × 162 = 2 × 2 × 2 × 81 = 2 × 2 × 2 × 3 × 27 = 2 × 2 × 2 × 3 × 3 × 9 = 2 × 2 × 2 × 3 × 3 × 3 × 3
648 = 2^3 × 3^4

(ii) 405
405 = 5 × 81 = 5 × 3 × 27 = 5 × 3 × 3 × 9 = 5 × 3 × 3 × 3 × 3
405 = 3^4 × 5

(iii) 540
540 = 2 × 270 = 2 × 2 × 135 = 2 × 2 × 3 × 45 = 2 × 2 × 3 × 3 × 15 = 2 × 2 × 3 × 3 × 3 × 5
540 = 2^2 × 3^3 × 5

(iv) 3600
3600 = 36 × 100 = 4 × 9 × 4 × 25 = 2^2 × 3^2 × 2^2 × 5^2
3600 = 2^4 × 3^2 × 5^2

3.
(i) 2 × 10^3 = 2 × 1000 = 2000
(ii) 7^2 × 2^3 = 49 × 8 = 392
(iii) 3 × 4^4 = 3 × 256 = 768
(iv) (–3)^2 × (–5)^2 = 9 × 25 = 225
(v) 3^2 × 10^4 = 9 × 10000 = 90000
(vi) (–2)^5 × (–10)^6 = (–32) × 10,00,000 = –3,20,00,000

The Stones that Shine

How many rooms were there altogether?
3 daughters × 3 baskets × 3 keys = 3^3 rooms per daughter's keys.
Total rooms = 3 × 3 × 3 × 3 = 3^4 = 81 rooms.

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

Magical Pond

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

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

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

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

How Many Combinations

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 × 2^4 = 2^4 = 16 lotuses.
After 4 more days in tripling pond: 2^4 × 3^4 = 16 × 81 = 1296 lotuses.

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

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

Use m^a × n^a = (mn)^a to compute 2^5 × 5^5.
2^5 × 5^5 = (2 × 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.

Estu has 4 dresses and 3 caps. How many different ways can Estu combine them?
Total combinations = 4 × 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 × 2 × 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 × 10 × 10 × 10 × 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 × 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) × 2^7 = 2^(–4+7) = 2^3 = 8
(ii) 3^2 × 3^(–5) × 3^6 = 3^(2–5+6) = 3^3 = 27
(iii) p^3 × p^(–10) = p^(3–10) = p^(–7) = 1/p^7
(iv) 2^4 × (–4)^(–2) = 16 × (1/16) = 1
(v) 8^p × 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 × 49 = 7^4 × 7^2 = 7^6 = 1,17,649
49^3 = (7^2)^3 = 7^6 = 1,17,649
343 × 2,401 = 7^3 × 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 × (1/343) = 7^6 × 7^(–3) = 7^3 = 343
(1/343) × (1/343) = 7^(–3) × 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 × 10^2) + (7 × 10^1) + (2 × 10^0)
(ii) 5642 = (5 × 10^3) + (6 × 10^2) + (4 × 10^1) + (2 × 10^0)
(iii) 6374 = (6 × 10^3) + (3 × 10^2) + (7 × 10^1) + (4 × 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 × 10^20 m
(ii) Number of stars in our galaxy: 1,00,00,00,00,000 = 1 × 10^11
(iii) Mass of the Earth: 5,97,60,00,00,00,00,00,00,00,00,000 kg ≈ 5.976 × 10^24 kg

The distance between the Sun and Saturn is 1.4335 × 10^12 m, the distance between Saturn and Uranus is 1.439 × 10^12 m, and the distance between the Sun and Earth is 1.496 × 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 × 10^11 m) is the smallest.

Express the following numbers in standard form:
(i) 59,853 = 5.9853 × 10^4
(ii) 65,950 = 6.595 × 10^4
(iii) 34,30,000 = 3.43 × 10^6
(iv) 70,04,00,00,000 = 7.004 × 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 × 70 = ₹3150.
Assuming Estu's weight = 50 kg, cost of wheat = ₹50/kg: Worth of wheat = 50 × 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 × 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 × 5 = ₹37,500 (≈ 3.75 × 10^4).
10-rupee note weighs ≈ 1 g. Number of notes = 45,000 notes.
Value = 45,000 × 10 = ₹4,50,000 (= 4.5 × 10^5).
Note: Assumptions may vary. Answers are estimates.

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

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

Calculate using scientific notation:
(i) Stars = 2 × 10^23. At 1 per second, time = 2 × 10^23 seconds.
(ii) Total water on Earth ≈ 1.386 × 10^21 litres = 1.386 × 10^24 ml. Glasses = (1.386 × 10^24) ÷ 200 ≈ 6.93 × 10^21. Time = 6.93 × 10^21 × 10 seconds ≈ 6.93 × 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 × 1,00,000 cm = 3.844 × 10^10 cm.
Number of steps = (3.844 × 10^10) ÷ 20 = 1.922 × 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 × 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 Numbers

Global starling population ≈ 1.3 billion = 1.3 × 10^9
Mosquito population ≈ 11 neel/110 trillion = 1.1 × 10^14

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

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

A Different Way to Say Your Age!

Roxie is 4840 days old. How many hours old is she?
4840 days × 24 hours/day = 1,16,160 hours ≈ 1.16 × 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 a pada yatra 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 × 365 × 24 = 6,13,200 hours. At 4 km/hr: Distance covered = 6,13,200 × 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 × 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 (Final Exercise)

1. 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.

2. 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 × 40 = 200 bottles.
Note: This is linear growth — 5 bottles added each day.

3. (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 × 3, so 192^8 = 2^48 × 3^8
Way 1: (2^48) × (3^8)
Way 2: (2^24)^2 × (3^4)^2
Way 3: (2^6 × 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)

4.
(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 × 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. 46/4 = 11.5 (not an integer) and 46/6 is also not an integer. So q^46 cannot be expressed as a perfect 4th power or 6th power of an integer when q is prime.

5.
(i) 10^(–2) × 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)×(–3)) = 13^6
(v) m^5n^12(mn)^9 = m^5n^12 × m^9n^9 = m^(5+9)n^(12+9) = m^14n^21

6.
(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 × 10)^2 = 144 × 100 = 14400

7. 2^4 × 3^6, 6^4 × 3^2, and 18^2 × 6^2 are all equal (= 11,664).
Note: 2^4 × 3^6 = 2^4 × 3^4 × 3^2 = 6^4 × 9 = 64 × 32. Also 18^2 × 6^2 = (18×6)^2 = 108^2? Verify: 18^2 = 324, 6^2 = 36, 324 × 36 = 11,664. 6^4 = 1296, 3^2 = 9, 1296 × 9 = 11,664.

8.
(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 × 10^30. 2^100 is much greater.

9. 8.5 billion = 8.5 × 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.

10. Yes. Numbers that are both perfect squares and perfect cubes are perfect 6th powers.
General form: n^6 (i.e., n^(2×3)).
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.

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

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

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

14. 10^9 seconds ÷ (3.156 × 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! (Student-generated activity)

This is a student-generated activity. Students find a partner, set a 10-second timer, and write a number or expression using only digits 0–9 and the allowed arithmetic operations. The player with the larger number wins. The activity is played in different rounds with different rules: only addition; only addition and multiplication; with exponents and only addition; or with exponents and any operation.

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

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