Class 8 Maths Chapter 2 Power Play Ganita Prakash NCERT Solutions Looking for the Class 8 Maths Chapter 2 Power Play Ganita Prakash NCERT Solutions? You’re at the right place. This chapter introduces the concept of powers and exponents through simple explanations and practical examples. Our step-by-step NCERT solutions help students understand every question clearly and prepare effectively for school exams.
Textbook Page 20
The following table lists the thickness after each fold. Observe that the thickness doubles after each fold.
(We use the sign ‘≈’ to indicate ‘approximately equal to’.)
After 10 folds, the thickness is just above 1 cm (1.024 cm). After 17 folds, the thickness is about 131 cm (a little more than 4 feet).
Q. Now, what do you think the thickness would be after 30 folds? 45 folds? Make a guess.
Solution:
I think that the thickness after 30 folds would be 10 km, and after 45 folds it would be 20,000 km.
Q. Fill the table below.
Solution:
Q. After 26 folds, the thickness is approximately 670 m. Burj Khalifa in Dubai, the tallest building in the world, is 830 m tall.
Solution:
Q. After 30 folds, the thickness of the paper is about 10.7 km, the typical height at which planes fly. The deepest point discovered in the oceans is the Mariana Trench, with a depth of 11 km.
Solution:
Textbook Page 22
Q. 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.
(i) 10v (ii) 10 + v (iii) 2 × 10 × v
(iv) 210 (v) 210v (vi) 102v
Solution:
Initial thickness of sheet = v
Since the thickness of the sheet doubles after every fold.
Therefore, the thickness of a sheet of paper after 10 folds = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × v = 210v.
Therefore, (v) 210v is the correct answer.
Q. What is (-1)5? Is it positive or negative? What about (-1)56?
Solution:
(-1)odd = −1 → Negative
(-1)even = 1 → Positive
(i) (-1)5
Since 5 is an odd number, (-1)5 = -1, which is negative.
(ii) (-1)56
Since 56 is an even number, (-1)56 = 1, which is positive.
Q. Is (-2)4 = 16? Verify.
Solution:
(-2)4 = (-2) × (-2) × (-2) × (-2)
(-2)4 = (-2) × (-2) × (-2) × (-2)
(-2)4 = 4 × 4
(-2)4 = 16.
So yes, (-2)4 = 16 is correct.
Q. What is 02, 05? What is 0n?
Solution:
02 = 0 × 0 = 0.
05 = 0 × 0 × 0 × 0 × 0 = 0.
Similarly, 0n = 0.
Textbook Page 22 – 23
Figure it Out
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 × 6 × 6 × 6 = 64
(ii) y × y = y2
(iii) b × b × b × b = b4
(iv) 5 × 5 × 7 × 7 × 7 = 52 × 73.
(v) 2 × 2 × a × a = 22 × a2.
(vi) a × a × a × c × c × c × c × d = 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.
(ii) 405 = 3 × 3 × 3 × 3 × 5 = 34 × 5.
(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 22 × 33 × 5.
(iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 × 32 × 52.
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 × 103
= 2 × (10 × 10 × 10)
= 2 × 1000
= 2000.
(ii) 72 × 23
= (7 × 7) × (2 × 2 × 2)
= 49 × 8
= 392.
(iii) 3 × 44
= 3 × (4 × 4 × 4 × 4)
= 3 × (16 × 16)
= 3 × 256
= 768.
(iv) 32 × 104
= {(-3) × (-3)} × {(-5) × (-5)}
= 9 × 25
= 225.
(v) 32 × 104
= (3 × 3) × (10 × 10 × 10 × 10)
= 9 × 10000
= 90000.
(vi) (-2)5 × (-10)6
= {(-2) × (-2) × (-2) × (-2) × (-2)} × {(-10) × (-10) × (-10) × (-10) × (-10) × (-10)}
= {4 × 4 × (-2)} × {100 × 100 × 100}
= (-32) × (1000000)
= -32000000.
Textbook Page 24
Q. 37 can also be written as 32 × 35. Can you reason out why?
Solution:
Yes, 37 can also be written as 32 × 35 because 37 = (3 × 3) × (3 × 3 × 3 × 3 × 3), which becomes 32 × 35.
Q. na × nb = n(a+b), where a and b are counting numbers.
Use this observation to compute the following.
(i) 29
(ii) 57
(iii) 46
Solution:
(i) 29 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = 23 × 23 × 23 = 8 × 8 × 8 = 512.
(ii) 57 = (5 × 5) × (5 × 5) × (5 × 5) × 5 = 52 × 52 × 52 × 5 = 25 × 25 × 25 × 5 = 78,125.
(iii) 46 = (4 × 4) × (4 × 4) × (4 × 4) = 42 × 42 × 42 = 16 × 16 × 16 = 4096.
Q. 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
Two possible ways:
86 = (83)2
86 = (82)3
(ii) 715
Two possible ways:
715 = (73)5
715 = (75)3
(iii) 914
Two possible ways:
914 = (92)7
914 = (97)2
(iv) 58
Two possible ways:
58 = (52)4
58 = (54)2
Textbook Page 25
In the middle of a beautiful, magical pond lies a bright pink lotus. The number of lotuses doubles every day in this pond. After 30 days, the pond is completely covered with lotuses. On which day was the pond half full?
If the pond is completely covered by lotuses on the 30th day, how much of it is covered by lotuses on the 29th day?
Since the number of lotuses doubles every day, the pond should be half covered on the 29th day.
Q. Write the number of lotuses (in exponential form) when the pond was —
(i) fully covered (ii) half covered
Solution
The number of lotuses doubles every day.
On the 30th day, the pond is fully covered.
So, on the 29th day, the pond must be half full.
Number of lotuses:
Day 1 → 1 = 20
Day 2 → 21
Day 3 → 22
Day 4 → 23
…….
Day 29 → 228
Day 30 → 229
(i) The number of lotuses when the pond was fully covered (Day 30) = 229.
(ii) The number of lotuses when the pond was fully covered (Day 29) = 228.
Q. ma × na = (mn)a, where a is a counting number.
Use this observation to compute the value of 25 × 55.
Solution:
25 × 55 = (2 × 5)5 = (10)5.
Q. Simplify 104/54 and write it in exponential form.
Solution:
Textbook Page 26
How Many Combinations
Q. Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up?
Hint: Try drawing a diagram like the one above.
Solution:
Each outfit includes one dress, one hat, and one pair of shoes.
Dresses: D1, D2, D3, D4, D5, D6, D7
Hats: H1, H2
Shoes: S1, S2, S3
Total combination of outfits = 7 × 2 × 3 = 42.
Textbook Page 27
Q. Estu says, “Next time, I will buy a lock that has 6 slots with the letters A to Z. I feel it is safer.”
How many passwords are possible with such a lock?
Solution:
Choices of letters for each slot = 26.
Total possible password combinations of a lock with 6 slots = 26 × 26 × 26 × 26 × 26 × 26 = 266.
Q. What is 2100 ÷ 225 in powers of 2?
Solution:
Textbook Page 29
Q. Consider the following general forms we have identified.
na × nb = na + b
(na)b = (nb)a = na × b
na ÷ nb = na – b
We had required a and b to be counting numbers. Can a and b be any integers? Will the generalised forms still hold true?
Solution:
Yes, the generalised forms are valid for all integers a and b, as long as the base n ≠ 0.
For example:
na × n-b = na + (-b) = na – b
(na)-b = (n-b)a = na × (-b)
n-a ÷ nb = n-a – b
Q. Write equivalent forms of the following.
(i) 2-4
(ii) 10-5
(iii) (-7)–2
(iv) (-5)–3
(v) 10-100
Solution:
Q. Simplify and write the answers in exponential form.
(i) 2-4 × 27
(ii) 32 × 3-5 × 36
(iii) p3 × p-10
(iv) 24 × (-4)–2
(v) 8p × 8q
Solution:
Textbook Page 30
Power Lines
Q. How many times larger than 4-2 is 42 ?
Solution:
Since, 42 ÷ 4-2 = 44.
Therefore, 42 is 256 (44) times larger than 4-2.
Q. Use the power line for 7 to answer the following questions.
Solution:
2401 × 49 = 74 × 72 = 74 + 2 = 76.
493 = (72)3 = 72 × 3 = 76.
343 × 2401 = 73 × 74 = 73 + 4 = 77.
Powers of 10
We have used numbers like 10, 100, 1000, and so on when writing Indian numerals in an expanded form. For example:
47561 = (4 × 10000) + (7 × 1000) + (5 × 100) + (6 × 10) + 1.
This can be written using powers of 10 as
47561 = (4 × 104) + (7 × 103) + (5 × 102) + (6 × 101) + (1 × 100).
Q. Write these numbers in the same way: (i) 172, (ii) 5642, (iii) 6374.
Solution:
(i) 172 = (1 × 100) + (7 × 10) + 2 = (1 × 102) + (7 × 101) + (2 × 100).
(ii) 5642 = (5 × 1000) + (6 × 100) + (4 × 10) + 2 = (5 × 103) + (6 × 102) + (4 × 101) + (2 × 100).
(iii) 6374 = (6 × 1000) + (3 × 100) + (7 × 10) + 4 = (6 × 103) + (3 × 102) + (7 × 101) + (4 × 100).
Textbook Page 32
The distance between the Sun and Saturn is 14,33,50,00,00,000 m = 1.4335 × 1012 m.
The distance between Saturn and Uranus is 14,39,00,00,00,000 m = 1.439 × 1012 m.
The distance between the Sun and Earth is 1,49,60,00,00,000 m = 1.496 × 1011 m.
Q. Can you say which of the three distances is the smallest?
Solution:
Since 1011 < 1012.
Therefore, the distance between the Sun and Earth is the smallest.
Q. The number line below shows the distance between the Sun and Saturn (1.4335 × 1012 m). On the number line below, mark the relative position of the Earth. The distance between the Sun and the Earth is 1.496 × 1011 m.
Solution:
Q. Express the following numbers in standard form.
(i) 59,853
(ii) 65,950
(iii) 34,30,000
(iv) 70,04,00,00,000
Solution:
(i) 59,853 = 5.9853 × 104.
(ii) 65,950 = 6.595 × 104.
(iii) 34,30,000 = 3.43 × 106.
(iv) 70,04,00,00,000 = 7.004 × 1010.
Textbook Page 43
Q. Continuing this, a thousand trillion is a quadrillion (1015). This pattern continues. Observe the names million (106), billion (109), trillion (1012), quadrillion (1015), quintillion (1018), sextillion (1021), septillion (1024), octillion (1027), nonillion (1030), decillion (1033).
What does the first part of each name denote?
Solution:
The first part of each name denotes a Latin or Greek prefix indicating the number of groups of three zeros that follow after the initial 1,000 or 103.
Textbook Page 44 – 45
Figure it Out
1. Find out the units digit in the value of 2224 ÷ 432 ?
Solution:
2224 ÷ 432
= 2224 ÷ (22)32
= 2224 ÷ 22 × 32
= 2224 ÷ 264
= 2224 – 64 = 2160
21 = 2 (units digit 2)
22 = 4 (units digit 4)
23 = 8 (units digit 8)
24 = 16 (units digit 6)
25 = 32 (units digit 2)
26 = 64 (units digit 4)
27 = 128 (units digit 8)
………..
………..
Here, the pattern repeats after every 4 steps.
So, the unit’s digit of 2160 is the same as that of 24, which 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:
Number of containers added every day = 1
Number of containers after 40 days = 40 containers
Number of bottles in a container = 5
Total bottles in 40 containers = 5 × 40 = 200 bottles
Therefore, there would be 200 bottles after 40 days.
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:
(i) 643
64 = 2 × 2 × 2 × 2 × 2 × 2 = 26
643 = (26)3 = 218
Three different ways:
1. 643 = 218 = 29 × 29
2. 643 = 218 = 210 × 28
3. 643 = 218 = 26 × 26 × 26
(ii) 1928
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 26 × 3
1928 = (26 × 3)8 = 26 × 8 × 38 = 248 × 38
Three different ways:
1. 1928 = 248 × 38 = 240 × 28 × 38 = 240 × (2×3)8 = 240 × 68
2. 1928 = 248 × 38 = (224×34)×(224×34)
3. 1928 = 248 × 38 = (26)8 × 38
(iii) 32-5
32 = 2 × 2 × 2 × 2 × 2 = 25
32-5 = (25)-5 = 2-25
Three different ways:
1. 32-5 = 2-25 = 2-15 × 2-10
2. 32-5 = 2-25 = 2−5 × 2−5 × 2−5 × 2−5 × 2−5
3. 32-5 = 2-25 = 2-5 × 2-20
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.
Explanation: 64 = 26 = (23)2 = (22)3 is both a cube and a square.
But 8 = 23 is a cube, not a square.
(ii) Always true.
Eplanation: 34 = (32)2 = 92.
54 = (52)2 = 252.
(iii) Always true.
Explanation: a5 = a3 × a2 and is divisible by a3.
(iv) Always true.
Explanation: 8 = 23, 27 = 33
8 × 27 = 216, which is 63.
(v) Never true.
Explanation: Since 46 is not divisible by 4 or 6. Therefore, there is no prime number q such that q46 is both a perfect fourth power and a perfect sixth power.
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 × 10–5 = 10-2 – 5 = 10-7
(ii) 57 ÷ 54 = 57 – 4 = 53
(iii) 9–7 ÷ 94 = 9-7 + 4 = 9–3
(iv) (13–2)–3 = 13-2 × -3 = 136
(v) m5n12(mn)9 = 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:
7. Circle the numbers that are the same —
24 × 36, 64 × 32, 610, 182 × 62, 624
Solution:
(i) 24 × 36
(ii) 64 × 32 = (2 × 3)4 × 32 = 24 × 34 × 32 = 24 × 36
(iii) 610 = (2 × 3)10 = 210 × 310
(iv) 182 × 62 = (2 × 3 × 3)2 × (2 × 3)2 = 22 × 32 × 32 × 22 × 32 = 24 × 36
(v) 624 = (2 × 3)24 = 224 × 324
Therefore, 24 × 36, 64 × 32, and 182 × 62 are the same.
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 or 34
43 = 64 , 34 = 81
So, 34 is greater.
(ii) 28 or 82
28 = 256, 82 = 64
So, 28 is greater.
(iii) 1002 or 2100
1002 = 10,000
2100 = (210)10 = 102410, which is far greater than 10,000.
So, 2100 is greater than 1002.
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
Therefore, the code should contain at least 10 digits to get a unique ID for each packet.
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 other numbers that are both squares and cubes. For example:
729 = 93 (cube number) = 272 (perfect square)
4096 = 163 (cube number) = 642 (perfect square)
General Rule: The sixth power of any number (i.e., n6) is both a square and a cube. i.e.
16 = 1
26 = 64
36 = 729
46 = 4096
56 = 15,625
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:
Length of passcode = 5
Total choices of alphabets and letters for each slot = 26 + 10 = 36
Total possible codes = 36 × 36 × 36 × 36 × 36 = 365
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:
Population of sheep = 109
Population of goats = 109
Total population of sheep and goats = 109 + 109 or 2 × 109
Therefore, (v) 2 × 109 and (vi) 109 + 109 are the correct answers.
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:
(i) Estimated world population: ≈ 8 billion = 8 × 109
Number of pieces of clothing each person had = 30 pieces
Total number of pieces of clothing = 8×109 × 30 = 2.4 × 1011
(ii) Number of bee colonies in the world = 100 million = 1×108
Number of bees in each colony = 50,000 = 5 × 104
Total number of honeybees = 1 × 108 × 5 × 104 = 5 × 1012
(iii) World population ≈ 8 × 109
Number of bacterial cells in human body = 38 trillion = 3.8 × 1013
Total bacterial population residing in all humans in the world = (3.8 × 1013)× (8 × 109) = 30.4 × 1022 = 3.04 × 1023
(iv) Average person’s lifespan = 80 years
Time spent eating daily = 1.5 hours
Total eating time per day in seconds = 1.5 × 3600 = 5400 seconds
Number of days in 80 years = 80 × 365 = 29,200
Total seconds spent eating in a lifetime in seconds = 5400 × 29,200 = 1.5768 × 108
14. What was the date 1 arab/1 billion seconds ago?
Solution:
1 billion seconds = 1,000,000,000 seconds
1,000,000,000 ÷ (60 × 60 × 24 × 365) = 31.71 years
31.71 years = 31 full years and 0.71 × 365 = 259 days
Today is: 28 July 2025
Subtracting 31 years → 28 July 1994
Going back 259 days from 28 July 1994 → 11 November 1993.
So, 1 billion seconds ago was November 11, 1993.
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