Looking for official, 100% textbook-grounded NCERT Solutions for Class 8 Mathematics Chapter 2: Power Play (from Ganita Prakash Part I, 2026-27 CBSE Curriculum)? Below are the complete step-by-step solutions for all ‘Figure It Out’ exercises and in-text exploration problems (Pages 30–45 of the official NCERT textbook) in authentic Topper’s Handwritten Notebook Style with zero raw LaTeX errors.
Chapter Overview: Power Play (Exponents and Powers)
Class 8 Mathematics Chapter 2, Power Play, takes students on an exciting journey through exponents, powers of 10, powers of 2 cyclicity, standard scientific notation, and large number estimation (from bacterial cells to bee colonies and astronomical distances).
Subject: Class 8 Mathematics (Ganita Prakash Part I)
Chapter 2: Power Play
Target: 100/100 CBSE Exam Score
Topper’s Handwritten Notebook Solutions (Figure It Out & Exercises — Pages 30–45)
Figure It Out — Question 1: Find out the units digit in the value of 2224 ÷ 432. [Hint: 4 = 22]
Step 1: Express with the Same Base (Base 2)
Notice that
Therefore,
4 = 22.Therefore,
432 = (22)32 = 22 × 32 = 264.Step 2: Apply Quotient Law of Exponents
Using
am ÷ an = am − n:2224 ÷ 264 = 2224 − 64 = 2160.Step 3: Analyze the Cyclicity of Powers of 2
The unit digits of powers of 2 repeat in a predictable cycle of 4:
- 21 ends in 2
- 22 ends in 4
- 23 ends in 8
- 24 ends in 6
- 25 ends in 2 … and the cycle (2, 4, 8, 6) repeats every 4 powers.
160 ÷ 4 = 40 with remainder 0), the unit digit is the 4th number in the cycle, which is 6.Final Answer: The units digit of 2^224 ÷ 4^32 is 6.
💡 Cyclicity Trick: Divide the exponent by 4. Remainder 1 → 2; Remainder 2 → 4; Remainder 3 → 8; Remainder 0 → 6.
Figure It Out — Question 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? Express your answer in scientific notation.
Step 1: Calculate Total Bottles
Number of bottles per container = 5.
Number of days = 40.
Number of days = 40.
Total Bottles = 40 × 5 = 200 bottles.Step 2: Convert to Scientific Notation
In standard form (k × 10n where 1 ≤ k < 10):
Move the decimal 2 places to the left:
Move the decimal 2 places to the left:
200 = 2.0 × 102.Final Answer: There would be 200 bottles = 2 × 10^2 bottles.
💡 Class 8 Concept: Standard scientific notation always keeps exactly one non-zero digit before the decimal point.
Figure It Out — Question 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
(i) 643 (ii) 1928 (iii) 32−5
Solution (i) for 64^3
Since 64 = 26 = 43 = 82,
Three different ways:
1.
2.
3.
643 = (26)3 = 218.Three different ways:
1.
210 × 282.
45 × 44 (since 49 = 218)3.
83 × 83 (since 86 = 218)Solution (ii) for 192^8
Prime factorisation: 192 = 64 × 3 = 26 × 31.
Three different ways:
1.
2.
3.
1928 = (26 × 3)8 = 248 × 38.Three different ways:
1.
248 × 382.
240 × 28 × 38 = 240 × 683.
(224)2 × 38 = 224 × 224 × 38Solution (iii) for 32^-5
Since 32 = 25,
Three different ways using integer powers:
1.
2.
3.
32−5 = (25)−5 = 2−25.Three different ways using integer powers:
1.
2−10 × 2−152.
2−5 × 2−203.
4−12 × 2−1 (since 4−12 = 2−24)Summary: Any power can be decomposed into multiple bases and exponents using the rule a^(m+n) = a^m × a^n.
💡 Topper Tip: Always reduce numbers to their prime base first (like base 2 for 64 and 32) to easily split exponents.
Figure It Out — Question 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 (where q is a prime number).
(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 (where q is a prime number).
Part (i) Cube numbers are also square numbers
Only Sometimes True.
Reason: 8 is a cube (23) but NOT a square. 27 is a cube (33) but NOT a square. However, 64 is a cube (43) and ALSO a square (82). In general, a number is both a cube and a square only when its exponent is a multiple of 6 (i.e., n6 = (n3)2 = (n2)3).
Reason: 8 is a cube (23) but NOT a square. 27 is a cube (33) but NOT a square. However, 64 is a cube (43) and ALSO a square (82). In general, a number is both a cube and a square only when its exponent is a multiple of 6 (i.e., n6 = (n3)2 = (n2)3).
Part (ii) Fourth powers are also square numbers
Always True.
Reason: By law of indices,
Reason: By law of indices,
n4 = (n2)2. Since n2 is an integer, (n2)2 is always the square of n2.Part (iii) Fifth power is divisible by cube of that number
Always True (for non-zero integers).
Reason:
Reason:
n5 ÷ n3 = n5 − 3 = n2, which is an exact integer.Part (iv) Product of two cube numbers is a cube number
Always True.
Reason: Using
Reason: Using
am × bm = (ab)m, we have (a3) × (b3) = (a × b)3, which is definitely a cube number.Part (v) q^46 is both a 4th power and a 6th power (q is prime)
Never True.
Reason: For q46 to be a 4th power, 46 must be divisible by 4 (46 ÷ 4 = 11.5, not an integer). For it to be a 6th power, 46 must be divisible by 6 (46 ÷ 6 = 7.67, not an integer). Since q is prime and 46 is not divisible by 4 or 6, it can never be a 4th or 6th power.
Reason: For q46 to be a 4th power, 46 must be divisible by 4 (46 ÷ 4 = 11.5, not an integer). For it to be a 6th power, 46 must be divisible by 6 (46 ÷ 6 = 7.67, not an integer). Since q is prime and 46 is not divisible by 4 or 6, it can never be a 4th or 6th power.
Answers: (i) Only Sometimes True | (ii) Always True | (iii) Always True | (iv) Always True | (v) Never True.
💡 CBSE High-Yield Question: Statements testing algebraic reasoning (Always / Sometimes / Never True) test conceptual depth; always give algebraic proofs like n^4 = (n^2)^2.
Figure It Out — Question 5: Simplify and write in exponential form:
(i) 10−2 × 10−5
(ii) 57 ÷ 54
(iii) 9−7 ÷ 94
(iv) (13−2)−3
(v) m5 n12 (mn)9
(i) 10−2 × 10−5
(ii) 57 ÷ 54
(iii) 9−7 ÷ 94
(iv) (13−2)−3
(v) m5 n12 (mn)9
Step-by-Step Algebraic Simplifications
- (i) 10−2 × 10−5:
10(−2) + (−5) = 10−7 = 1 / 107. - (ii) 57 ÷ 54:
57 − 4 = 53 = 125. - (iii) 9−7 ÷ 94:
9(−7) − 4 = 9−11 = 1 / 911. - (iv) (13−2)−3:
13(−2) × (−3) = 136. - (v) m5 n12 (mn)9: Notice
(mn)9 = m9 n9.m5 × m9 × n12 × n9 = m5+9 n12+9 = m14 n21.
Answers: (i) 10^−7 | (ii) 5^3 | (iii) 9^−11 | (iv) 13^6 | (v) m^14 n^21
💡 Sign Alert: In (iv), multiplying two negative numbers gives a POSITIVE exponent: (−2) × (−3) = +6!
Figure It Out — Question 6: If 122 = 144, what is the value of:
(i) (1.2)2 (ii) (0.12)2 (iii) (0.012)2 (iv) 1202?
(i) (1.2)2 (ii) (0.12)2 (iii) (0.012)2 (iv) 1202?
Understanding Decimal Places in Squares
When a decimal number is squared, the number of decimal places in the square is double the number of decimal places in the original number:
- (i) (1.2)2: 1 decimal place → square has 2 decimal places:
1.44. - (ii) (0.12)2: 2 decimal places → square has 4 decimal places:
0.0144. - (iii) (0.012)2: 3 decimal places → square has 6 decimal places:
0.000144. - (iv) (120)2: 1 zero at end → square has 2 zeros:
14,400.
Answers: (i) 1.44 | (ii) 0.0144 | (iii) 0.000144 | (iv) 14,400
💡 Zero Doubling Rule: If a number ends in k zeros, its square ends in 2k zeros!
Figure It Out — Question 7: Circle the numbers that are the same:
24 × 36, 64 × 32, 610, 182 × 62, 624
24 × 36, 64 × 32, 610, 182 × 62, 624
Reduce Each Term to Prime Bases (2 and 3)
- 24 × 36: Already in prime base →
24 × 36. - 64 × 32:
(2 × 3)4 × 32 = 24 × 34 × 32 = 24 × 36. (Identical to 1st!) - 610:
(2 × 3)10 = 210 × 310. - 182 × 62:
(2 × 32)2 × (2 × 3)2 = (22 × 34) × (22 × 32) = 24 × 36. (Identical to 1st!) - 624:
224 × 324.
Identical Numbers: 2^4 × 3^6, 6^4 × 3^2, and 18^2 × 6^2 are all mathematically identical (equal to 2^4 × 3^6 = 11,664).
💡 Strategy: Express composite bases like 6 and 18 as products of primes 2 and 3.
Figure It Out — Question 8: Identify the greater number in each of the following:
(i) 43 or 34
(ii) 28 or 82
(iii) 1002 or 2100
(i) 43 or 34
(ii) 28 or 82
(iii) 1002 or 2100
Evaluation for Each Pair
- (i) 43 vs 34:
43 = 4 × 4 × 4 = 64.34 = 3 × 3 × 3 × 3 = 81.
Since 81 > 64, 34 is greater. - (ii) 28 vs 82:
28 = 256.82 = 64.
Since 256 > 64, 28 is greater. - (iii) 1002 vs 2100:
1002 = 10,000.
Even 214 = 16,384 > 10,000. 2100 is an unimaginably colossal 31-digit number!
Therefore, 2100 is vastly greater.
Answers: (i) 3^4 is greater | (ii) 2^8 is greater | (iii) 2^100 is vastly greater.
💡 General Insight: For large powers, having a variable or large number in the exponent (exponential growth) grows infinitely faster than squaring!
Figure It Out — Question 9: A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID code for each packet using the digits 0–9. How many digits should the code consist of?
Step 1: Quantify 8.5 Billion
1 billion = 1,000,000,000 = 109.
8.5 billion packets =
8.5 billion packets =
8.5 × 109 = 8,500,000,000 packets.Step 2: Number of Unique Codes with n Digits
Since digits 0 to 9 are used (10 choices per position), an n-digit code can generate
We need:
10n unique codes.We need:
10n ≥ 8,500,000,000.- If n = 9 digits: 109 = 1,000,000,000 (only 1 billion codes — NOT enough!).
- If n = 10 digits: 1010 = 10,000,000,000 (10 billion codes — comfortably covers 8.5 billion).
Final Answer: The code must consist of at least 10 digits.
💡 Real-Life Connection: This is why Indian mobile phone numbers are 10 digits long: 10^10 = 10 billion possible unique connections!
Figure It Out — Question 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?
General Mathematical Characterization
For a number N to be simultaneously a perfect square (N = a2) and a perfect cube (N = b3), its prime factor exponents must be divisible by both 2 and 3.
The LCM of 2 and 3 is 6.
Therefore, all such numbers can be expressed in general as sixth powers (n6) of integers:
The LCM of 2 and 3 is 6.
Therefore, all such numbers can be expressed in general as sixth powers (n6) of integers:
n6 = (n3)2 = (n2)3.Examples of Square-Cube Numbers
- n = 1:
16 = 1(12 = 13) - n = 2:
26 = 64(82 = 43) - n = 3:
36 = 729(272 = 93) - n = 4:
46 = 4,096(642 = 163) - n = 5:
56 = 15,625(1252 = 253)
Answer: Yes, there are infinitely many such numbers! In general, they are sixth powers of integers: n^6.
💡 Topper’s Note: The next number after 64 that is both a square and a cube is 729 (27^2 = 9^3).
Figure It Out — Question 11: A digital locker has an alphanumeric passcode of length 5 (letters A–Z and digits 0–9). Examples: G89P0, 38098, BRJKW. How many such passcodes are possible?
Fundamental Principle of Counting
Number of alphabet letters (A to Z) = 26.
Number of decimal digits (0 to 9) = 10.
Total possible alphanumeric characters per position =
Number of decimal digits (0 to 9) = 10.
Total possible alphanumeric characters per position =
26 + 10 = 36.Calculating for Length 5
Each of the 5 positions can independently be filled by any of the 36 characters:
Total Passcodes = 36 × 36 × 36 × 36 × 36 = 365.365 = 60,466,176 (over 60.4 million combinations!).Final Answer: 36^5 = 60,466,176 possible passcodes.
💡 Cybersecurity Insight: Adding letters to numeric pins increases password security exponentially from 10^5 (100,000) to 36^5 (60.4 million)!
Figure It Out — Question 12: The worldwide population of sheep in 2024 is about 109, and that of goats is also about the same. What is the total combined population of sheep and goats?
(i) 209 (ii) 1011 (iii) 1010 (iv) 1018 (v) 2 × 109 (vi) 109 + 109
(i) 209 (ii) 1011 (iii) 1010 (iv) 1018 (v) 2 × 109 (vi) 109 + 109
Algebraic Addition of Like Powers
Total = Sheep + Goats = 109 + 109.Factoring out 109:
109 + 109 = 109(1 + 1) = 2 × 109.Crucial Caution: You CANNOT add exponents when adding terms (i.e., 109 + 109 is NOT 1018 or 209)!
Correct Options: (v) 2 × 10^9 and (vi) 10^9 + 10^9.
💡 Classic Trap: Exponents add during MULTIPLICATION (10^9 × 10^9 = 10^18), but during ADDITION you combine coefficients: x + x = 2x!
Figure It Out — Question 13: Calculate and write the answers in scientific notation:
(i) If each person in the world has 30 pieces of clothing, find the total pieces of clothing (World pop ≈ 8.2 × 109).
(ii) There are about 100 million bee colonies in the world, with 50,000 bees each. Find total honeybees.
(iii) The human body has about 38 trillion bacterial cells. Find the total bacterial population residing in all humans in the world.
(iv) Total time spent eating in a 70-year lifetime in seconds (assuming 1 hour eating per day).
(i) If each person in the world has 30 pieces of clothing, find the total pieces of clothing (World pop ≈ 8.2 × 109).
(ii) There are about 100 million bee colonies in the world, with 50,000 bees each. Find total honeybees.
(iii) The human body has about 38 trillion bacterial cells. Find the total bacterial population residing in all humans in the world.
(iv) Total time spent eating in a 70-year lifetime in seconds (assuming 1 hour eating per day).
Sub-problem (i) Total Clothing
World population = 8.2 × 109.
Total Clothes = (8.2 × 109) × 30 = 246 × 109 = 2.46 × 1011 pieces.Sub-problem (ii) Honeybee Population
100 million colonies = 100 × 106 = 108 colonies.
50,000 bees per colony = 5 × 104 bees.
50,000 bees per colony = 5 × 104 bees.
Total Honeybees = 108 × (5 × 104) = 5.0 × 1012 bees (5 trillion bees!).Sub-problem (iii) Total Human Bacterial Population
38 trillion bacteria per human = 38 × 1012 = 3.8 × 1013.
World population = 8.2 × 109.
World population = 8.2 × 109.
Total Bacteria = (3.8 × 1013) × (8.2 × 109) = 31.16 × 1022 = 3.116 × 1023 bacterial cells.Sub-problem (iv) Lifetime Eating Time in Seconds
1 hour per day = 3600 seconds.
70 years = 70 × 365 days = 25,550 days.
70 years = 70 × 365 days = 25,550 days.
Total Eating Seconds = 25,550 × 3600 = 91,980,000 seconds = 9.198 × 107 seconds.Answers in Scientific Notation:
(i) 2.46 × 10^11 pieces
(ii) 5.0 × 10^12 honeybees
(iii) 3.116 × 10^23 bacterial cells
(iv) 9.198 × 10^7 seconds
(i) 2.46 × 10^11 pieces
(ii) 5.0 × 10^12 honeybees
(iii) 3.116 × 10^23 bacterial cells
(iv) 9.198 × 10^7 seconds
💡 Order of Magnitude: 3.116 × 1023 bacteria is close to Avogadro’s number (6.022 × 1023)!
Figure It Out — Question 14: What was the date 1 billion (109) seconds ago? Explain how to calculate it.
Step 1: Convert 10^9 Seconds to Days and Years
Seconds in a day =
24 × 60 × 60 = 86,400 seconds.Number of days in 109 seconds = 1,000,000,000 / 86,400 ≈ 11,574.07 days.Number of years = 11,574.07 / 365.25 ≈ 31.69 years (approx. 31 years and 8 months).Step 2: Finding the Exact Calendar Date
Counting backward 31 years and 8 months from the current year (2026):
2026 − 31.7 years ≈ Early 1995 (around February / March 1995).Final Answer: 1 billion seconds is approximately 11,574 days or ~31.7 years ago (around early 1995).
💡 Mind-Bending Fact: 1 million seconds is only ~11.5 days, but 1 billion seconds is nearly 32 years!
📌 Master Revision Checklist for Chapter 2 (Power Play)
- Unit Digit Cyclicity of Powers of 2: Cycle is 2, 4, 8, 6 with period 4. Divide exponent by 4 to find unit digit.
- Product Law vs Addition: 109 × 109 = 1018, but 109 + 109 = 2 × 109.
- Square-Cube Numbers: Numbers that are both squares and cubes are sixth powers: n6 = (n3)2 = (n2)3 (e.g. 1, 64, 729, 4096).
- Standard Scientific Form: Always format as k × 10n where 1 ≤ k < 10.