What are the laws of exponents covered in Chapter 2 Power Play?

The laws are: n^a times n^b = n^(a+b); n^a divided by n^b = n^(a-b); (n^a)^b = n^(a times b); m^a times n^a = (mn)^a; m^a divided by n^a = (m/n)^a; n^0 = 1 and n^(-a) = 1/n^a, all for non-zero bases.

Class 8 Maths Ganita Prakash Chapter 2 Solutions (NCERT 2026–27) – Power Play

Q: What is Class 8 Maths Ganita Prakash Chapter 2 Power Play about?

Power Play introduces exponents through fun situations such as paper folding, doubling lotuses and combination locks. It builds all the laws of exponents, the zero and negative exponents, powers of 10, expanded form of decimals, scientific notation, and compares linear growth with exponential growth.

Q: How do you write a number in scientific notation?

Write the number as x times 10 to the power y, where the coefficient x is at least 1 but less than 10 and y is an integer. For example, 3430000 = 3.43 times 10^6 and 308100000 = 3.081 times 10^8.

Q: Are Class 8 Maths Ganita Prakash Chapter 2 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 2 Power Play are free and follow the official NCERT textbook for 2026-27.

These Class 8 Maths Ganita Prakash Chapter 2 solutions cover Power Play from the new NCF-2023 textbook (Part I, 2026–27). Every “Figure it Out” exercise, Math Talk and Try This box is solved step by step, with verified answers for the laws of exponents, powers of 10 and scientific notation so you can master the whole chapter quickly.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 2 Exercises: Figure it Out (Set 1 & Set 2) Session: 2026–27

On this page

Chapter overview
Key concepts & definitions
Important formulas (laws of exponents)
Figure it Out — Set 1 (Q1–Q3)
Figure it Out — Set 2 (Q1–Q14)
Math Talk & Try This (in-text answers)
Additional textbook activities
Common mistakes to avoid
Practice MCQs & Assertion–Reason
Quick revision summary
FAQs

Chapter 2 Overview

Chapter 2 of Ganita Prakash, Power Play, uses playful situations — folding a sheet of paper, doubling lotuses in a magical pond, counting combinations of dresses and caps, and unlocking number-pad locks — to build the idea of exponents. You learn that n^a means n multiplied by itself a times, develop all the laws of exponents (product, quotient, power of a power, products and quotients with the same exponent, the zero exponent and negative exponents), and then apply them to powers of 10, the expanded form of decimals, and scientific (standard) notation for very large numbers. The chapter ends by comparing linear growth with exponential growth and getting a feel for huge quantities. Use the contents to jump to the exercises, in-text questions or additional activities.

Key Concepts & Definitions

Exponential form: n^a means n × n × … × n (n multiplied by itself a times). Here n is the base and a is the exponent (power). e.g. 5⁴ = 5 × 5 × 5 × 5 = 625.

Exponential / multiplicative growth: a quantity that keeps doubling (or tripling) grows by a fixed factor each step — far faster than linear growth, which adds a fixed amount each step.

Zero exponent: n⁰ = 1 for any n ≠ 0.

Negative exponent: n^−a = 1/n^a and n^a = 1/n^−a (n ≠ 0).

Powers of 10: place value in our number system is built on powers of 10; decimals use negative powers, e.g. 10⁻¹ = 1/10.

Scientific (standard) form: a positive number written as x × 10^y, where 1 ≤ x < 10 (the coefficient) and y is an integer (the exponent).

Important Formulas (Laws of Exponents)

Product law: n^a × n^b = n^a+b

Quotient law: n^a ÷ n^b = n^a−b (n ≠ 0)

Power of a power: (n^a)^b = (n^b)^a = n^a×b

Same exponent, product: m^a × n^a = (m × n)^a

Same exponent, quotient: m^a ÷ n^a = (m ÷ n)^a (n ≠ 0)

Zero & negative powers: n⁰ = 1 (n ≠ 0); n^−a = 1/n^a (n ≠ 0)

Scientific notation: any positive number = x × 10^y, with 1 ≤ x < 10 and y an integer.

Class 8 Maths Ganita Prakash Chapter 2 Solutions — Figure it Out (Set 1)

This first “Figure it Out” set appears in Section 2.2 (Exponential Notation and Operations).

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

SOLUTION (i) 6 occurs 4 times ⇒ 6⁴. (ii) y occurs 2 times ⇒ y². (iii) b occurs 4 times ⇒ b⁴. (iv) 5 twice, 7 thrice ⇒ 5² × 7³. (v) 2 twice, a twice ⇒ 2² × a². (vi) a thrice, c four times, d once ⇒ 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

SOLUTION (i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2³ × 3⁴. (Check: 8 × 81 = 648.) (ii) 405 = 3 × 3 × 3 × 3 × 5 = 3⁴ × 5. (Check: 81 × 5 = 405.) (iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5. (Check: 4 × 27 × 5 = 540.) (iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴ × 3² × 5². (Check: 16 × 9 × 25 = 3600.)

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

SOLUTION (i) 2 × 1000 = 2000. (ii) 49 × 8 = 392. (iii) 3 × 256 = 768. (iv) 9 × 25 = 225 (both squares are positive). (v) 9 × 10000 = 90000. (vi) (−2)⁵ = −32 and (−10)⁶ = 1000000, so (−32) × 1000000 = −32000000.

Class 8 Maths Ganita Prakash Chapter 2 Solutions — Figure it Out (Set 2)

This is the main end-of-chapter “Figure it Out” exercise (Q1–Q14).

1. Find out the units digit in the value of 2²²⁴ ÷ 4³²? [Hint: 4 = 2²]

SOLUTION 4³² = (2²)³² = 2⁶⁴, so 2²²⁴ ÷ 4³² = 2²²⁴ ÷ 2⁶⁴ = 2²²⁴⁻⁶⁴ = 2¹⁶⁰. Units digits of powers of 2 repeat in the cycle 2, 4, 8, 6 (period 4): 2¹=2, 2²=4, 2³=8, 2⁴=6, … 160 ÷ 4 leaves remainder 0, which matches 2⁴, so the units digit is 6.

The repeating units digits 2, 4, 8, 6 correspond to exponent remainders 1, 2, 3, 0 on division by 4. — After simplifying to 2¹⁶⁰, the exponent has remainder 0 modulo 4, so the units digit is 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?

SOLUTION Each day one container of 5 bottles is added, so after 40 days the number of bottles = 5 × 40 = 200 = 2 × 10² bottles. (This is linear growth — a fixed amount is added each day, not multiplied.)

A table compares adding five per day with doubling from one each day; after five steps the values are 25 and 32. — The bottles question adds one five-bottle container each day. It uses multiplication by 40, not an exponent of 40.

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 (ii) 192⁸ (iii) 32⁻⁵

SOLUTION Some of the ways are: (i) 64³ = (2⁶)³ = 2¹⁸, which can be split as 2¹⁰ × 2⁸, 4⁵ × 4⁴ (= 2¹⁰ × 2⁸), 8³ × 8³ (= 2⁹ × 2⁹ = 2¹⁸). (ii) 192 = 2⁶ × 3, so 192⁸ = 2⁴⁸ × 3⁸ = 2⁴⁸ × 3⁸, 2⁴⁰ × 2⁸ × 3⁸, 2⁴⁰ × 6⁸ (since 2⁸ × 3⁸ = 6⁸). (iii) 32 = 2⁵, so 32⁻⁵ = 2⁻²⁵ = 2⁻¹⁰ × 2⁻¹⁵, 2⁻⁵ × 2⁻²⁰, 4⁻¹² × 2⁻¹ (= 2⁻²⁴ × 2⁻¹ = 2⁻²⁵).

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 are also square numbers. (ii) Fourth powers are also square numbers. (iii) The fifth power of a number is divisible by the cube of that number. (iv) The product of two cube numbers is a cube number. (v) q⁴⁶ is both a 4th power and a 6th power (q is a prime number).

SOLUTION (i) Only Sometimes True. A cube n³ is also a square only when n itself is a perfect square, i.e. when the number is a 6th power, n⁶ = (n²)³ = (n³)². e.g. 64 = 4³ = 8² works, but 8 = 2³ is not a square. (ii) Always True. A fourth power n⁴ = (n²)², which is the square of n². (iii) Always True for positive integers n. n⁵ ÷ n³ = n⁵⁻³ = n², a whole number, so n³ divides n⁵. (iv) Always True. (n₁³) × (n₂³) = (n₁ × n₂)³, again a cube. (v) Never True. q⁴⁶ is a 4th power only if 4 divides 46, and a 6th power only if 6 divides 46. Neither divides 46, so (with prime q) it is neither.

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

SOLUTION (i) 10^−2+(−5) = 10⁻⁷ (= 1/10⁷). (ii) 5⁷⁻⁴ = 5³. (iii) 9⁻⁷⁻⁴ = 9⁻¹¹ (= 1/9¹¹). (iv) (13⁻²)⁻³ = 13^{(−2)×(−3)} = 13⁶. (v) (mn)⁹ = m⁹n⁹, so m⁵n¹² × m⁹n⁹ = m⁵⁺⁹n¹²⁺⁹ = m¹⁴n²¹.

6. If 12² = 144 what is (i) (1.2)² (ii) (0.12)² (iii) (0.012)² (iv) 120²

SOLUTION Using 12² = 144 and shifting the decimal point: (i) (1.2)² = 1.44. (ii) (0.12)² = 0.0144. (iii) (0.012)² = 0.000144. (iv) (120)² = 14400.

7. Circle the numbers that are the same— 2⁴ × 3⁶ 6⁴ × 3² 6¹⁰ 18² × 6² 6²⁴

SOLUTION Write each in prime powers: 2⁴ × 3⁶; 6⁴ × 3² = (2·3)⁴ × 3² = 2⁴ × 3⁶; 18² × 6² = (2·3²)² × (2·3)² = 2⁴ × 3⁶. So 2⁴ × 3⁶, 6⁴ × 3² and 18² × 6² are the same (each equals 16 × 729 = 11664). 6¹⁰ and 6²⁴ are different.

8. Identify the greater number in each of the following— (i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰

SOLUTION (i) 4³ = 64, 3⁴ = 81, so 3⁴ is greater. (ii) 2⁸ = 256, 8² = 64, so 2⁸ is greater. (iii) 100² = 10⁴ = 10000, while 2¹⁰⁰ = (2¹⁰)¹⁰ > 1000¹⁰ = 10³⁰, so 2¹⁰⁰ is far greater.

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?

SOLUTION 8.5 billion = 8,500,000,000 packets. With 10 digits (0–9), allowing repetition and leading zeros, an n-digit code gives 10ⁿ unique codes. We need 10ⁿ ≥ 8,500,000,000. Since 10⁹ = 1,000,000,000 < 8,500,000,000 and 10¹⁰ = 10,000,000,000 > 8,500,000,000, we take n = 10. So the code should have at least 10 digits.

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?

SOLUTION Yes, there are infinitely many. A number that is both a perfect square and a perfect cube must be a perfect 6th power: n⁶ = (n³)² = (n²)³. Examples: 1⁶ = 1, 2⁶ = 64, 3⁶ = 729, 4⁶ = 4096, … So all numbers of the form n⁶ 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?

SOLUTION Each position can be any of 26 uppercase letters A–Z or 10 digits, giving 36 choices. There are 5 positions; repetition and leading zeros are allowed. Total codes = 36 × 36 × 36 × 36 × 36 = 36⁵ = 6,04,66,176 (about 6 crore).

Five slots each have 36 choices, giving 36 to the fifth or 60,466,176 passcodes. — Treat codes as strings of characters: examples such as 003AZ are valid, so leading zeros count.

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? (i) 20⁹ (ii) 10¹¹ (iii) 10¹⁰ (iv) 10¹⁸ (v) 2 × 10⁹ (vi) 10⁹ + 10⁹

SOLUTION Total = 10⁹ + 10⁹ = 2 × 10⁹. So both (v) 2 × 10⁹ and (vi) 10⁹ + 10⁹ are correct (they are equal). Note: you cannot “add the exponents” — 10⁹ + 10⁹ is not 10¹⁸. Exponents add only when powers are multiplied.

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. (ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees. (iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world. (iv) Total time spent eating in a lifetime in seconds.

SOLUTION (Using the chapter’s estimate of world population ≈ 8.2 × 10⁹.) (i) Total clothing = 8.2 × 10⁹ × 30 = 246 × 10⁹ = 2.46 × 10¹¹ pieces. (ii) Colonies = 100 million = 10⁸; bees per colony = 50,000 = 5 × 10⁴. Total bees = 10⁸ × 5 × 10⁴ = 5 × 10¹² honeybees. (iii) Cells per body = 38 trillion = 38 × 10¹²; people = 8.2 × 10⁹. Total = 38 × 8.2 × 10¹²⁺⁹ = 311.6 × 10²¹ = 3.116 × 10²³ bacterial cells. (iv) Assume lifetime ≈ 70 years and ≈ 1 hour (= 3600 s) of eating per day. Total ≈ 3600 × 365 × 70 = 9,19,80,000 = 9.198 × 10⁷ seconds. (Answers vary with the assumptions.)

14. What was the date 1 arab/1 billion seconds ago?

SOLUTION1 billion seconds = 11,574 days, 1 hour, 46 minutes and 40 seconds. This is about 31.7 years, but subtracting rounded years does not give an exact date.Choose a reference date, time and time zone, then subtract that exact duration. For example, 4 October 2026 at 00:00 IST − 1,000,000,000 seconds = 25 January 1995 at 22:13:20 IST. This uses ordinary civil time, ignoring leap seconds.For a different reference date or time, recalculate; the example above is fixed rather than a claim about “today”.

Math Talk & Try This — In-text Answers

These are the “?”, Math Talk and Try This prompts inside the chapter, in order of appearance.

Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is the letter-number v. (i) 10v (ii) 10 + v (iii) 2 × 10 × v (iv) 2¹⁰ (v) 2¹⁰v (vi) 10²v

SOLUTIONThe thickness doubles each fold, so after 10 folds it is v × 2¹⁰ = (v) 2¹⁰v.

Zero through four folds produce 1, 2, 4, 8 and 16 layers; thickness is 0.001 times 2 to the power n centimetres. — In the ideal model, every additional ten folds multiply the thickness by 2¹⁰ = 1024.

What is (−1)⁵? Is it positive or negative? What about (−1)⁵⁶? Is (−2)⁴ = 16? Verify.

SOLUTION (−1)⁵ = −1 (negative) — an odd power keeps the negative sign. (−1)⁵⁶ = +1 (positive) — an even power makes it positive. (−2)⁴ = (−2)(−2)(−2)(−2) = 4 × 4 = 16. Yes, verified (even power ⇒ positive).

Express the number 32400 as a product of its prime factors and represent the prime factors in their exponential form.

SOLUTION32400 = 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3 × 3 × 3 = 2⁴ × 5² × 3⁴.

Math Talk: 3⁷ can also be written as 3² × 3⁵. Can you reason out why? Use the product law to compute (i) 2⁹ (ii) 5⁷ (iii) 4⁶.

SOLUTION 3⁷ = (3 × 3) × (3 × 3 × 3 × 3 × 3) = 3² × 3⁵, since 2 + 5 = 7 (product law n^a × n^b = n^a+b). (i) 2⁹ = 2⁴ × 2⁵ = 16 × 32 = 512. (ii) 5⁷ = 5³ × 5⁴ = 125 × 625 = 78125. (iii) 4⁶ = 4³ × 4³ = 64 × 64 = 4096.

A product of 3 squared and 3 to the fifth has seven factors; 3 squared raised to the fourth has eight factors. — Use the same-base product law for multiplication, and the power-of-a-power law for nested powers.

Write the following expressions as a power of a power in at least two different ways: (i) 8⁶ (ii) 7¹⁵ (iii) 9¹⁴ (iv) 5⁸.

SOLUTION (i) 8⁶ = (8²)³ = (8³)²; also 8⁶ = (2³)⁶ = 2¹⁸. (ii) 7¹⁵ = (7³)⁵ = (7⁵)³. (iii) 9¹⁴ = (9²)⁷ = (9⁷)²; also 9¹⁴ = (3²)¹⁴ = 3²⁸. (iv) 5⁸ = (5²)⁴ = (5⁴)².

Magical Pond: Write the number of lotuses (in exponential form) when the pond was (i) fully covered (ii) half covered.

SOLUTIONUse day 0 for the single initial lotus. After n complete days of doubling, the number is 2ⁿ.Thus full coverage on day 30 corresponds to 2³⁰ lotuses. Half coverage was on day 29, with 2²⁹ lotuses. Working backward by one day halves the count.

Three pond circles show one quarter full on day 28, half full on day 29 and completely full on day 30. — Working backward, divide by 2 each day. Using day 0 for the initial lotus gives 2³⁰ lotuses on day 30.

Use m^a × n^a = (mn)^a to compute 2⁵ × 5⁵. Simplify 10⁴/5⁴ and write it in exponential form.

SOLUTION 2⁵ × 5⁵ = (2 × 5)⁵ = 10⁵ = 100000. 10⁴ ÷ 5⁴ = (10 ÷ 5)⁴ = 2⁴ = 16.

How Many Combinations: Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up? A 6-slot lock with letters A–Z: how many passwords are possible? What is 2¹⁰⁰ ÷ 2²⁵ in powers of 2?

SOLUTION Roxie’s outfits = 7 × 2 × 3 = 42 ways. 6-slot lock with 26 letters each = 26 × 26 × 26 × 26 × 26 × 26 = (26)⁶ passwords. 2¹⁰⁰ ÷ 2²⁵ = 2¹⁰⁰⁻²⁵ = 2⁷⁵.

Seven dress groups each contain six cells for two hats and three pairs of shoes, giving 42 outfits. — In each group, rows represent the two hats and columns represent the three shoe pairs.

Math Talk (2.3): Why can’t n be 0 in n^a ÷ n^b? Can a and b be any integers in the laws? How many times larger than 4⁻² is 4²?

SOLUTIONFor a positive exponent b, 0^b = 0, so n = 0 would put zero in the denominator. Zero and negative powers of 0 are not defined in these laws.The exponent laws extend to all integer exponents when the bases are non-zero. Positive powers of zero still make sense: 0² = 0⁵ = 0, and 0ⁿ = 0 for positive integers n.4² ÷ 4⁻² = 4²⁻⁽⁻²⁾ = 4⁴ = 256. Therefore 4² is 256 times as large as 4⁻².

Lengths halve from 16 through 8, 4, 2, 1, one half and one quarter, labelled 2 to the powers 4 through minus 2. — Each step divides the value by 2 and decreases the exponent by 1: 2⁰ = 1 and 2⁻² = 1/4.

Write equivalent forms of: (i) 2⁻⁴ (ii) 10⁻⁵ (iii) (−7)⁻² (iv) (−5)⁻³ (v) 10⁻¹⁰⁰. Then simplify: (i) 2⁻⁴ × 2⁷ (ii) 3² × 3⁻⁵ × 3⁶ (iii) p³ × p⁻¹⁰ (iv) 2⁴ × (−4)⁻² (v) 8^p × 8^q.

SOLUTION Equivalent forms: (i) 1/2⁴ (ii) 1/10⁵ (iii) 1/(−7)² (iv) 1/(−5)³ (v) 1/10¹⁰⁰. (i) 2⁻⁴⁺⁷ = 2³. (ii) 3^2+(−5)+6 = 3³. (iii) p^3+(−10) = p⁻⁷. (iv) (−4)² = 16 = 2⁴, so 2⁴ × (−4)⁻² = 2⁴ ÷ 16 = 1. (v) 8^p × 8^q = 8^p+q.

Power line for 7: evaluate 2,401 × 49; 49³; 343 × 2,401; 16,807/49; 7/343; 16,807/8,23,543; 1,17,649 × (1/343); (1/343) × (1/343).

SOLUTION Using 7¹=7, 7²=49, 7³=343, 7⁴=2401, 7⁵=16807, 7⁶=117649, 7⁷=823543: 2,401 × 49 = 7⁴ × 7² = 7⁶. 49³ = (7²)³ = 7⁶. 343 × 2,401 = 7³ × 7⁴ = 7⁷. 16,807/49 = 7⁵ ÷ 7² = 7³. 7/343 = 7¹ ÷ 7³ = 7⁻². 16,807/8,23,543 = 7⁵ ÷ 7⁷ = 7⁻². 1,17,649 × (1/343) = 7⁶ × 7⁻³ = 7³. (1/343) × (1/343) = 7⁻³ × 7⁻³ = 7⁻⁶.

Powers from 7 to the minus fourth through 7 to the seventh have values from 1/2401 to 823543. — The labels show consecutive exponents, not equally spaced numerical values. Continue from 7¹ at the end of the first row to 7² on the second.

Powers of 10: Write 172, 5642 and 6374 using powers of 10.

SOLUTION 172 = (1 × 10²) + (7 × 10¹) + (2 × 10⁰). 5642 = (5 × 10³) + (6 × 10²) + (4 × 10¹) + (2 × 10⁰). 6374 = (6 × 10³) + (3 × 10²) + (7 × 10¹) + (4 × 10⁰).

Scientific Notation: Which of the three distances is smallest? Express in standard form: (i) 59,853 (ii) 65,950 (iii) 34,30,000 (iv) 70,04,00,00,000.

SOLUTION Comparing exponents: Sun–Saturn 1.4335 × 10¹², Saturn–Uranus 1.439 × 10¹², Sun–Earth 1.496 × 10¹¹. The Sun–Earth distance is the smallest (exponent 11 < 12). (i) 5.9853 × 10⁴ (ii) 6.595 × 10⁴ (iii) 3.43 × 10⁶ (iv) 7.004 × 10¹⁰.

On the given Sun-to-Saturn number line, Earth is about 10.4 percent of the total distance from the Sun. — Divide 1.496 × 10¹¹ by 1.4335 × 10¹² to get about 0.1044. This is the textbook’s distance comparison, not a diagram of current planetary positions.

Try This (p.38): (i) ants per human (ii) number of starling flocks of 10,000 birds (iii) total leaves if each tree has 10⁴ leaves (iv) sheets of paper to reach the Moon.

SOLUTION Using ants ≈ 2 × 10¹⁶, humans ≈ 8 × 10⁹, starlings ≈ 1.3 × 10⁹, trees ≈ 3 × 10¹²: (i) (2 × 10¹⁶) ÷ (8 × 10⁹) = 0.25 × 10⁷ = 2.5 × 10⁶ ants per human. (ii) (1.3 × 10⁹) ÷ 10⁴ = 1.3 × 10⁵ flocks. (iii) (3 × 10¹²) × 10⁴ = 3 × 10¹⁶ leaves. (iv) Earth–Moon ≈ 3,84,400 km = 3.844 × 10¹⁰ cm; each sheet 0.001 cm. Sheets needed = 3.844 × 10¹⁰ ÷ 10⁻³ = 3.844 × 10¹³ sheets.

Try This (p.42): (i) seconds to count all stars one per second (ii) time to drink all water on Earth at one 200 ml glass per 10 s. If you have lived for a million seconds, how old are you?

SOLUTION Using stars ≈ 2 × 10²³ and water drops ≈ 2 × 10²⁵ (16 drops/ml): (i) One star per second ⇒ 2.0 × 10²³ seconds. (ii) Volume of water = (2 × 10²⁵) ÷ 16 ml = 1.25 × 10²⁴ ml = 6.25 × 10²¹ glasses (200 ml each); at 10 s per glass ⇒ 6.25 × 10²² seconds. A million seconds = 10⁶ ÷ 86400 ≈ 11.6 days (about 12 days) old.

Additional textbook activities and visual solutions

Complete the folding tables (pages 19–21)

Assume the paper starts at 0.001 cm and each fold doubles its thickness. Actual paper cannot be folded indefinitely; this is an ideal mathematical model. Two additional folds multiply thickness by 4, three by 8, and ten by 1024.

Fold	Thickness (approximately)	Fold	Thickness (approximately)
18	0.002621 km	32	42.95 km
19	0.005243 km	33	85.899 km
20	0.010486 km	34	171.799 km
21	0.020972 km	35	343.597 km
22	0.041943 km	36	687.195 km
23	0.083886 km	37	1,374.39 km
24	0.167772 km	38	2,748.779 km
25	0.335544 km	39	5,497.558 km
26	0.671089 km	40	10,995.116 km
27	1.342177 km	41	21,990.233 km
28	2.684355 km	42	43,980.465 km
29	5.368709 km	43	87,960.93 km
30	10.737418 km	44	1,75,921.86 km
31	21.474836 km	45	3,51,843.721 km

Useful conversions: fold 18 = 262.144 cm; fold 19 = 524.288 cm; fold 20 = 10.48576 m; fold 26 ≈ 671.089 m; fold 30 ≈ 10.737 km; fold 45 ≈ 351,843.721 km; fold 46 ≈ 703,687.442 km. The ideal stack exceeds the chapter’s Earth–Moon distance of 384,400 km at fold 46.

The Stones that Shine (pages 23–24)

A seven-stage multiplication chain gives 3 daughters, 9 baskets, 27 keys, 81 rooms, 243 tables, 729 necklaces and 2187 diamonds. — There are 3⁴ rooms and 3⁷ diamonds. Group the factors as 3⁴ × 3³ = 81 × 27 = 2187.

Multiply once for each level in the story: 3 daughters × 3 baskets × 3 keys × 3 rooms × 3 tables × 3 necklaces × 3 diamonds = 3⁷ = 2187 diamonds. There are 3⁴ = 81 rooms, each ultimately providing 3³ = 27 diamonds. Also, p⁴ × p⁶ = p¹⁰ and 2¹⁰ = (2⁵)² = 32 × 32 = 1024.

Transfer between doubling and tripling ponds (page 25)

Four days of doubling give 2⁴ = 16 lotuses. Four more days of tripling give 16 × 3⁴ = 16 × 81 = 1296. Reversing the ponds gives 3⁴ × 2⁴, the same result. The common exponent lets us write the product as (2 × 3)⁴ = 6⁴.

Locks and real-world codes (pages 26–27)

Estu’s 4 dresses and 3 caps make 12 outfits. A five-digit lock, allowing repetition and leading zeros, has 10⁵ = 100,000 codes, from 00000 to 99999. A six-slot lock using A–Z has 26⁶ = 308,915,776 possible codes.

For the open investigation about postal, mobile and vehicle codes, first record the actual rules: which positions allow letters or digits, whether the first position is restricted, and which combinations are reserved. Multiply the allowed choices only when each choice is independent. The number of mathematically possible strings may exceed the number actually assigned. Record the official source and date when researching a real code system.

Expand a decimal with powers of 10 (page 30)

The digits of 561.903 occupy the hundreds, tens, ones, tenths, hundredths and thousandths columns. — The zero in the hundredths place contributes 0 × 10⁻². It keeps the 3 in the thousandths place.

561.903 = 5 × 10² + 6 × 10¹ + 1 × 10⁰ + 9 × 10⁻¹ + 0 × 10⁻² + 3 × 10⁻³.

Large-number notation and the distance diagram (pages 31–32)

The chapter’s three large-number facts become 3 × 10²⁰ m, 1 × 10¹¹ stars and 5.976 × 10²⁴ kg. These are the textbook’s example estimates, not freshly measured figures.

On the Sun–Saturn line, Earth belongs about 10.4% of the distance from the Sun, since (1.496 × 10¹¹)/(1.4335 × 10¹²) ≈ 0.1044. The diagram above uses that ratio. An approximate age such as 70 million years does not justify claiming precision to 15 extra years: the original estimate was rounded.

Thought experiments: state your assumptions (pages 33–35)

These problems accept different reasonable estimates. The following are sample assumptions for calculation, not current prices, standard body weights or universal coin weights.

Question	Assumptions and calculation
Donation value	45 kg of jaggery at ₹70/kg: 45 × 70 = ₹3150. 50 kg of wheat at ₹50/kg: 50 × 50 = ₹2500.
₹1 coins matching 45 kg	Assume the particular coin weighs 4 g. 45,000/4 = 11,250 coins, worth ₹11,250. Measure the coin used if doing this activity.
₹5 coins or ₹10 notes matching 45 kg	Assume 6 g per ₹5 coin: 45,000/6 = 7,500 coins, worth ₹37,500. Assume 1 g per ₹10 note: 45,000 notes, worth ₹450,000.
Notebooks matching an adult’s weight	Assume 60 kg and 0.2 kg per notebook: 60/0.2 = 300 notebooks; one each could benefit 300 students.
Grain matching an adult’s weight	Assume 60 kg of grain and 0.1 kg per serving: 60/0.1 = 600 grain servings. This estimates the grain component, not a complete meal.
400 km walking journey	Assume 5 km/h for 8 hours/day: 40 km/day, so 400/40 = 10 walking days. Rest days would add time.
Non-stop lifetime walk around Earth	For the hypothetical model, assume 70 years at 5 km/h, using 365 days/year: 5 × 24 × 365 × 70 = 3,066,000 km; divide by 40,000 to get about 76.65 circuits. This is not physically feasible.

For a ladder to the Moon with 20 cm step spacing: 384,400 km = 3.844 × 10¹⁰ cm, so the model needs 1.922 × 10⁹ step intervals. Adding the same distance each step is linear growth; doubling paper thickness is exponential growth. Adding ₹10 to savings daily is another linear example; an ideal population doubling every generation is exponential.

Fill the large-number blanks (pages 37–42)

Using the chapter’s estimates: 1.3 billion starlings = 1.3 × 10⁹; 110 trillion mosquitoes = 1.1 × 10¹⁴. Humans per African elephant ≈ (8 × 10⁹)/(4 × 10⁵) = 2 × 10⁴ = 20,000.

Using 365 days per year, 15 million years ≈ 15 × 10⁶ × 31,536,000 = 4.7304 × 10¹⁴ seconds (about 4.7 × 10¹⁴). Similarly, 470 million years ≈ 1.482192 × 10¹⁶ seconds (about 1.5 × 10¹⁶).

Examples of time scales: a one-day school trip lasts 8.64 × 10⁴ seconds, of order 10⁵; a 12-day holiday lasts 1.0368 × 10⁶ seconds, of order 10⁶.

Age in days, hours and seconds (page 39)

Roxie’s 4,840 days equal 4,840 × 24 = 116,160 hours. To find the birthday of someone aged 4,070 days, subtract 4,070 calendar days from the reference date. For example, 4 October 2026 − 4,070 days = 13 August 2015. A million seconds is exactly 11 days, 13 hours, 46 minutes and 40 seconds.

The unexplained age “69,70,710” needs a unit. If it means minutes, it is 6,970,710/(60 × 24 × 365) ≈ 13.26 years, a plausible interpretation for the character. If it means seconds, it is only about 80.68 days. The number alone is insufficient.

Names of large numbers (page 43)

Name	Power of ten
Lakh	10⁵
Million	10⁶
Crore	10⁷
Arab / billion	10⁹
Kharab	10¹¹
Trillion	10¹²
Neel	10¹³
Padma / quadrillion	10¹⁵
Shankh	10¹⁷
Maha shankh	10¹⁹

In the short-scale international sequence, million is 1000², billion is 1000³, trillion is 1000⁴, and so on. For n = 1, 2, 3, … the corresponding “-illion” value is 10³⁽ⁿ⁺¹⁾. A googol is 10¹⁰⁰ and a googolplex is 10^(10¹⁰⁰).

Tremendous in Ten! (page 47)

Round 1: Roxie’s 10¹³ exceeds Estu’s 999999², which is less than (10⁶)² = 10¹².

Round 2: Roxie writes 4 × 10¹⁰⁰⁰. Estu writes 9000 × 10^1000000 = 9 × 10^1000003, which is far greater. Compare the exponent first.

For your own rounds, agree on allowed digits, operations, parentheses and whether exponent towers are permitted. Under a ten-second writing limit there is no single largest possible answer independent of those rules. Compare legal expressions by place value, grouping factors and exponent laws.

Common Mistakes to Avoid

Watch out for these

Confusing aⁿ with a × n — 2¹⁰ = 1024, but 2 × 10 = 20. Repeated multiplication, not multiplication by the exponent.
Adding exponents when adding powers: 10⁹ + 10⁹ = 2 × 10⁹, not 10¹⁸. Exponents add only when powers are multiplied.
Sign of negative bases: an even power is positive, an odd power is negative. e.g. (−1)⁵⁶ = +1, (−1)⁵ = −1.
Negative exponent does not make a number negative: 2⁻⁴ = 1/2⁴ = 1/16, a positive fraction.
Scientific form needs the coefficient between 1 and 10: write 5900 = 5.9 × 10³, not 59 × 10².
For powers of a power, multiply the exponents: (13⁻²)⁻³ = 13⁶, not 13⁻⁵.

Practice MCQs & Assertion–Reason

1. In 5⁴ = 625, the number 4 is called the:

(a) base (b) exponent (c) coefficient (d) product

2. 2³ × 2⁵ equals:

(a) 2⁸ (b) 2¹⁵ (c) 4⁸ (d) 2²

3. The value of 7⁰ is:

(a) 0 (b) 7 (c) 1 (d) undefined

4. (3²)⁴ is equal to:

(a) 3⁶ (b) 3⁸ (c) 9⁸ (d) 3¹⁶

5. 2⁻³ is the same as:

(a) −8 (b) −6 (c) 1/8 (d) 1/6

6. The number 3.081 × 10⁸ in usual form is:

(a) 30810000 (b) 308100000 (c) 3081000000 (d) 30810

7. 10⁹ ÷ 10⁴ equals:

(a) 10¹³ (b) 10³⁶ (c) 10⁵ (d) 1⁵

8. 2⁵ × 5⁵ equals:

(a) 10⁵ (b) 10¹⁰ (c) 10²⁵ (d) 7⁵

9. Which of these is the greatest?

(a) 4³ (b) 3⁴ (c) 2⁵ (d) 5²

10. The standard (scientific) form of 34,30,000 is:

(a) 3.43 × 10⁵ (b) 34.3 × 10⁵ (c) 3.43 × 10⁶ (d) 0.343 × 10⁷

Answer key: 1-(b), 2-(a), 3-(c), 4-(b), 5-(c), 6-(b), 7-(c), 8-(a), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: 2⁰ = 1.

Reason: For any non-zero base x, x^a ÷ x^a = x⁰ = 1.

A-R 2. Assertion: 10⁹ + 10⁹ = 10¹⁸.

Reason: When two powers with the same non-zero base are multiplied, their exponents are added.

A-R 3. Assertion: 5⁻² = 1/25.

Reason: For n ≠ 0, n^−a = 1/n^a.

A-R 4. Assertion: (−2)⁴ = 16.

Reason: (−2)⁴ = [(−2) × (−2)] × [(−2) × (−2)] = 4 × 4 = 16.

A-R 5. Assertion: 5900 written in scientific form is 59 × 10².

Reason: In scientific form x × 10^y, the coefficient x must satisfy 1 ≤ x < 10.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(D). [A-R2: Assertion false (it equals 2×10⁹), Reason true. A-R5: Assertion false (correct is 5.9×10³), Reason true.]

Quick Revision Summary

n^a means n multiplied by itself a times; base = n, exponent = a.
Exponential (multiplicative) growth is far faster than linear (additive) growth.
Product law: n^a × n^b = n^a+b; Quotient law: n^a ÷ n^b = n^a−b (n ≠ 0).
Power of a power: (n^a)^b = (n^b)^a = n^a×b.
Same exponent: m^a × n^a = (mn)^a; m^a ÷ n^a = (m/n)^a.
n⁰ = 1 (n ≠ 0); n^−a = 1/n^a (n ≠ 0).
Scientific form: x × 10^y with 1 ≤ x < 10 and y an integer; e.g. 308100000 = 3.081 × 10⁸.

How to score full marks in this chapter

Quote the law you are using before each step (product, quotient, power of a power, etc.), and never “add” exponents when you are adding numbers. Keep answers in exponential form unless a numerical value is asked, write scientific notation with the coefficient between 1 and 10, and double-check the sign of negative bases (even power → positive, odd power → negative). For estimation questions, state your assumptions clearly — reasonable assumptions earn marks.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 2 Power Play about?

Power Play introduces exponents through fun situations (paper folding, doubling lotuses, combination locks). It builds all the laws of exponents, the zero and negative exponents, powers of 10, expanded form of decimals, and scientific (standard) notation, and compares linear with exponential growth.

What are the laws of exponents covered in this chapter?

n^a × n^b = n^a+b; n^a ÷ n^b = n^a−b; (n^a)^b = n^a×b; m^a × n^a = (mn)^a; m^a ÷ n^a = (m/n)^a; n⁰ = 1 and n^−a = 1/n^a (all for non-zero bases).

How do you write a number in scientific notation?

Write it as x × 10^y, where the coefficient x is at least 1 but less than 10 and y is an integer. For example, 34,30,000 = 3.43 × 10⁶ and 308100000 = 3.081 × 10⁸.

Are these Class 8 Maths Ganita Prakash Chapter 2 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 2 (Power Play) are free and follow the official NCERT textbook for 2026–27.

Textbook reference: NCERT Ganita Prakash, Chapter 2 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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