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

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
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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 na 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. These Class 8 Maths Ganita Prakash Chapter 2 solutions solve every part in order.

Key Concepts & Definitions

Exponential form: na 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/na and na = 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 number written as x × 10y, where 1 ≤ x < 10 (the coefficient) and y is an integer (the exponent).

Important Formulas (Laws of Exponents)

Product law: na × nb = na+b

Quotient law: na ÷ nb = na−b (n ≠ 0)

Power of a power: (na)b = (nb)a = na×b

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

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

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

Scientific notation: any number = x × 10y, 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.

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.)

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. 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), 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 letters + 10 digits = 36 characters, and there are 5 positions. Total codes = 36 × 36 × 36 × 36 × 36 = 365 = 6,04,66,176 (about 6 crore).

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?

SOLUTION 1 billion = 109 seconds. Dividing by 60 × 60 × 24 = 86,400 seconds/day gives 109 ÷ 86400 ≈ 11,574 days ≈ 31.7 years ago. So count back about 31 years 8 months from today’s date to get the exact calendar date. (As of 17 June 2026, that is roughly mid-October 1994.)

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.

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 na × nb = na+b). (i) 29 = 24 × 25 = 16 × 32 = 512. (ii) 57 = 53 × 54 = 125 × 625 = 78125. (iii) 46 = 43 × 43 = 64 × 64 = 4096.

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.

SOLUTIONThe number doubles daily and the pond is full on day 30, so (i) fully covered = 230, (ii) half covered (day 29) = 229.

Use ma × na = (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.

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

SOLUTION If n = 0, division by 0b = 0 is undefined (and 00 is not defined), so we need n ≠ 0. Yes — the laws still hold when a and b are any integers (not just counting numbers). 42 ÷ 4−2 = 42−(−2) = 44, so 42 is 44 = 256 times larger than 4−2.

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) 8p × 8q.

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) 32+(−5)+6 = 33. (iii) p3+(−10) = p−7. (iv) (−4)2 = 16 = 24, so 24 × (−4)−2 = 24 ÷ 16 = 1. (v) 8p × 8q = 8p+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 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.

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.

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) 94 only    (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, xa ÷ xa = x0 = 1.

A-R 2. Assertion: 109 + 109 = 1018.

Reason: When two powers are multiplied, their exponents are added.

A-R 3. Assertion: 5−2 = 1/25.

Reason: For n ≠ 0, n−a = 1/na.

A-R 4. Assertion: (−2)4 = 16.

Reason: An even power of a negative number is positive.

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

Reason: In scientific form x × 10y, 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

  • na means n multiplied by itself a times; base = n, exponent = a.
  • Exponential (multiplicative) growth is far faster than linear (additive) growth.
  • Product law: na × nb = na+b;   Quotient law: na ÷ nb = na−b (n ≠ 0).
  • Power of a power: (na)b = (nb)a = na×b.
  • Same exponent: ma × na = (mn)a;   ma ÷ na = (m/n)a.
  • n0 = 1 (n ≠ 0);   n−a = 1/na (n ≠ 0).
  • Scientific form: x × 10y 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?

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

How do you write a number in scientific notation?

Write it as x × 10y, 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.

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