[New Book] Solutions POWER PLAY Ncert Class 8 Chapter 2

Solutions POWER PLAY Ncert Class 8 Chapter 2 – provides clear and step-by-step solutions to the Class 8 Chapter 2 Ganit Prakashan book. It will help you understand the concepts of powers, exponents, and scientific notation.

The aim of this page is not just to give answers, but to explain the logic, formulas, and reasoning used in the chapter so that students can confidently solve similar problems in exams and assignments.

To get the best out of this post, first try solving the questions on your own from the NCERT textbook.

Then compare your answers with the solutions provided here.

Each solution is written in a clean, exam-ready format, making it easy for students to revise concepts, check their work, and quickly prepare for tests.

2.1 Experiencing the Power Play |
Page No. 20-21

Paper Folding Thickness Table

Question. Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is represented by the variable v.

Options:

(i) 10v
(ii) 10 + v
(iii) 2 × 10 × v
(iv) 2¹⁰
(v) 2¹⁰ v
(vi) 10² v

Solution:

Therefore, after n folds, the thickness becomes:

After 10 Folds

(v) 2¹⁰ V Ans

2.2 Exponential Notation and Operations

Question. What is (–1)⁵? Is it positive or negative? What about (–1)⁵⁶?

Solution:

Understanding the Pattern

Let’s observe a small pattern:

So the sign alternates between negative and positive.

Value of (–1)⁵

Since 5 is odd, the result is negative.

Answer: –1 (Negative)

Value of (–1)⁵⁶

Since 56 is an even number, all the negative signs cancel in pairs.

Answer: +1 (Positive)

✔ Rule to Remember:

Question. Is (–2)⁴ = 16? Verify.

Solution:

This means –2 is multiplied by itself 4 times.

Now multiply step by step:

Therefore,

Why the Result is Positive

• Four negative numbers are being multiplied.
• Multiplying negative numbers in pairs yields a positive result.

So:

Verified Answer: Yes, (–2)⁴ = 16.

Figure it Out

Page 22 – 23

1. Express the following in exponential form

Solution :

2. Express each number as a product of powers of its prime factors

Solution:

3. Write the numerical value of each of the following

solution:

Page 24

The Formula Used :
$$(a^m)^n = a^{mn}$$

Question. 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) $5^8$

Solutions:

(i) $8^6$

(ii) $7^{15}$

(iii) $9^{14}$

(iv) $5^8$

First form:$$5^8 = (5^2)^4$$

Second form:$$5^8 = (5^4)^2$$

Page 25

Magical Pond

Q. Simplify and write it in exponential form.
$$\frac{10^4}{5^4}$$

Step 1: Use the exponent rule

Step 2: Apply the rule

Step 3: Simplify inside the bracket

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?

Roxie has:

• 7 dresses
• 2 hats
• 3 pairs of shoes

Each dress can be worn with any hat and any pair of shoes.

Roxie can dress up in 42 different ways.

2.3 The Other Side of Powers

Page 27

Q. What is ${2}^{100}$ ÷ $2^{25}$ in powers of 2?

Formula Used

$$a^n \div a^b = a^{\,n-b}$$

Solution

$$2^{100} \div 2^{25}$$$$= 2^{100-25}$$

Page 29

Q. Write equivalent forms of the following.

$(i) 2^{-4}$
$$2^{-4}=\frac{1}{2^4}=\frac{1}{16}$$

$(ii) 10^{-5}$$$10^{-5}=\frac{1}{10^5}=\frac{1}{100000}$$

$(iii) (-7)^{-2}$$$(-7)^{-2}=\frac{1}{(-7)^2}=\frac{1}{49}$$

$(iv) (-5)^{-3}$$$(-5)^{-3}=\frac{1}{(-5)^3}=\frac{1}{-125}=-\frac{1}{125}$$

$$10^{-100}=\frac{1}{10^{100}}$$

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 × 8ᑫ

Page 30

Q. How many times larger than $4^{-2}$ is $4^2 \$?
$$\frac{4^2}{4^{-2}}$$$$= 4^{\,2-(-2)}$$$$= 4^4$$$$= 256$$

Answer: $4^2$ 256 times larger than $4^{-2}$

Q. Write the numbers using powers of 10

(i) 172$$172 = (1 \times 10^2) + (7 \times 10^1) + (2 \times 10^0)$$

(ii) 5642$$5642 = (5 \times 10^3) + (6 \times 10^2) + (4 \times 10^1) + (2 \times 10^0)$$

(iii) 6374$$6374 = (6 \times 10^3) + (3 \times 10^2) + (7 \times 10^1) + (4 \times 10^0)$$

Scientific Notation

Page 32

Q.Express the following numbers in standard form

(i) 59,853$$59,853 = 5.9853 \times 10^4$$

(ii) 65,950$$65,950 = 6.595 \times 10^4$$

(iii) 34,30,000$$34,30,000 = 3.43 \times 10^6$$

(iv) 70,04,00,00,000$$70,04,00,00,000 = 7.004 \times 10^9$$

Page 33

Roxie’s Question

Q. “Instead of jaggery, if we use ₹1 coins, how many coins are needed to equal my weight?”

Assume Roxie’s weight = 45 kg

Weight of one ₹1 coin ≈ 3.76 g

Convert Roxie’s weight to grams

$$45 \text{ kg} = 45 \times 1000$$$$= 45000 \text{ g}$$

Number of ₹1 coins needed

$$\text{Number of coins} = \frac{\text{Total weight}}{\text{Weight of one coin}}$$
Number of coins=Weight of one coin, Total weight​
$$= \frac{45000}{3.76}$$$$\approx 11968$$

Final Answer

$$\boxed{\text{About } 12,000 \text{ coins}}$$

So, about 12,000 one-rupee coins would be needed to equal Roxie’s weight of 45 kg.

page 34

Q. How many people might benefit from each of these offerings in a
year? Again, guess first before finding out.

Estu’s Idea – Donating Notebooks

Assume Estu’s weight = 50 kg

Cost of notebooks equal to his weight in rupees:
$$\text{Value of notebooks} = 50 \text{ kg} = ₹50$$

Assume one notebook costs ₹10$$\text{Number of notebooks} = \frac{50}{10} = 5$$

So, about 5 students could receive notebooks.

Roxie’s Idea – Annadāna (Donating Meals)

Assume Roxie’s weight = 45 kg

Value of food equal to her weight in rupees:
$$\text{Value of food} = ₹45$$

Assume one meal costs ₹15$$\text{Number of meals} = \frac{45}{15} = 3$$

So, about 3 people could receive a meal.

Linear Growth vs. Exponential Growth

Page 38-39

Q. With a global human population of about $8 × 10^9$ and about $4 × 10^5$
African elephants, can we say that there are nearly 20,000 people for
Every African elephant is?

Human population $= 8 \times 10^9$

African elephants $= 4 \times 10^5$$$\text{People per elephant}=\frac{8\times10^9}{4\times10^5}$$$$=\frac{8}{4}\times10^{9-5}$$$$=2\times10^4$$$$=20{,}000$$$$\boxed{\text{Yes, there are about } 20{,}000 \text{ people for every African elephant.}}$$

Q. Calculate and write the answer using scientific notation:

(i) How many ants are there for every human in the world?

Number of ants ≈ $2 \times 10^{16}$

Human population ≈ $8 \times 10^9$$$\frac{2 \times 10^{16}}{8 \times 10^9}$$$$= \frac{2}{8} \times 10^{16-9}$$$$= 0.25 \times 10^7$$$$= 2.5 \times 10^6$$$$\boxed{2.5 \times 10^6}$$

So, about $2.5×10^6$ ants for every human.

(ii) If a flock of starlings contains $10^4$ birds

Total starlings in the world ≈ $5 \times 10^8$$$\frac{5 \times 10^8}{10^4}$$$$= 5 \times 10^{8-4}$$$$= 5 \times 10^4$$$$\boxed{5 \times 10^4}$$

So, about $5×10^4$ flocks.

(iii) Total number of leaves on all trees

Number of trees ≈ $3 \times 10^{12}$

Leaves per tree ≈ $10^4$104$$(3 \times 10^{12}) \times 10^4$$$$= 3 \times 10^{12+4}$$$$= 3 \times 10^{16}$$$$\boxed{3 \times 10^{16}}$$

Total leaves ≈ $3×10^{16}$

(iv) Sheets of paper needed to reach the Moon

Distance to Moon ≈ $3.84 \times 10^8$ m

Thickness of one sheet ≈ $10^{-4}$m
$$\frac{3.84 \times 10^8}{10^{-4}}$$$$= 3.84 \times 10^{8-(-4)}$$$$= 3.84 \times 10^{12}$$$$\boxed{3.84 \times 10^{12}}$$

So, about $3.84×10^{12}$ sheets of paper are needed to reach the Moon.

Page 39

Q. If you have lived for a million seconds, how old would you be?

1 million seconds:$$1,000,000 \text{ seconds}$$

1 day = 86,400 seconds$$\frac{1,000,000}{86400}$$$$\approx 11.57 \text{ days}$$

So,$$\boxed{\text{1 million seconds } \approx 11.6 \text{ days}}$$

Answer: About 11½ days old.

Page 42

Q. Calculate and write the answer using scientific notation:

(i) If one star is counted every second, how long would it take to
count all the stars in the universe? Answer in terms of the
number of seconds using scientific notation.

(ii) If one could drink a glass of water (200 ml) every 10 seconds,
How long would it take to finish the entire volume of water
on Earth?

(i) Counting all the stars in the universe

Stars in the universe:$$10^{22}$$

If 1 star is counted every second:
$$\text{Time} = 10^{22} \text{ seconds}$$$$\boxed{10^{22} \text{ seconds}}$$

(ii) Drinking all the water on Earth

Total water on Earth:$$1.4 \times 10^{21} \text{ litres}$$

1 glass = 200 ml = 0.2 litres

Number of glasses:$$\frac{1.4 \times 10^{21}}{0.2}$$$$= 7 \times 10^{21}$$

If one glass is drunk every 10 seconds:$$(7 \times 10^{21}) \times 10$$$$= 7 \times 10^{22} \text{ seconds}$$$$\boxed{7 \times 10^{22} \text{ seconds}}$$

Figure it Out: End-of-Chapter Questions

Q1. Find out the units digit in the value of $2^{224} \div 4^{32}$[Hint: 4 = 22]

Solution :

First convert 4 into powers of 2 using the hint.$$4 = 2^2$$

So,$$4^{32} = (2^2)^{32}$$

Using the power rule:$$(2^2)^{32} = 2^{64}$$

Now substitute in the expression:$$\frac{2^{224}}{4^{32}} = \frac{2^{224}}{2^{64}}$$

Using the law of exponents:$$2^{224-64}$$$$= 2^{160}$$

Finding the Units Digit

The units digits of powers of 2 follow a repeating cycle:

The cycle repeats every 4 powers.

So we find:$$160 \div 4$$

Since 160 is divisible by 4, the power corresponds to the 4th position in the cycle.

Units digit = 6

$$2^{224} \div 4^{32} = 2^{160}$$

Units digit = 6 Ans.​

Q2. 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:

Concept Used

Bottles in 1 container = 5

Containers brought in 40 days = 40

$$\text{Total bottles} = 5 \times 40$$$= 200 \ Ans.$

So, 200 bottles would be there after 40 days

Q3. Write the given number as the product of two or more powers in
three different ways. The powers can be any integers.
(i)$64^3$ (ii) $192^8$ (iii) $32^{–5}$

(i) $64^3$

$$64^3 = 2^{18}$$$$= 2^{10} \times 2^{8}$$$$= 2^{6} \times 2^{6} \times 2^{6}$$$$= (2^{9}) \times (2^{9}) \quad \mathbf{Ans.}$$

(ii) $192^8$

$$192 = 2^{6} \times 3$$$$192^8 = 2^{48} \times 3^{8}$$$$= (2^{24}) \times (2^{24}) \times 3^{8}$$$$= 2^{32} \times 2^{16} \times 3^{8} \quad \mathbf{Ans.}$$

(iii) $32^{-5}$

$$32^{-5} = 2^{-25}$$$$= 2^{-10} \times 2^{-15}$$$$= (2^{-5}) \times (2^{-5}) \times (2^{-5}) \times (2^{-5}) \times (2^{-5}) \quad \mathbf{Ans.}$$

Q4. 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) $q^{46}$ is both a 4th power and a 6th power (q is a prime number).

Solution:

(i) Cube numbers are also square numbers.

Example:$$2^3 = 8$$

8 is not a square number.

But,$$1^3 = 1$$

1 is a square number.

Therefore, this statement is Only Sometimes True.

Ans. Only Sometimes True

ii) Fourth powers are also square numbers.

$$a^4 = (a^2)^2$$

So every fourth power can be written as the square of a number.

Therefore, the statement is Always True.

Ans. Always True

(iii) The fifth power of a number is divisible by the cube of that number.

$$\frac{a^5}{a^3} = a^{5-3}$$$$= a^2$$

Since the result is a whole number, $a^5$ is divisible by $a^3$.

Therefore, the statement is Always True.

Ans. Always True

(iv) The product of two cube numbers is a cube number.

Let the numbers be:$$a^3 \text{ and } b^3$$

Their product:$$a^3 \times b^3 = (ab)^3$$

Which is again a cube number.

Therefore, the statement is Always True.

Ans. Always True

(v) $q^{46}$ is both a 4th power and a 6th power (q is a prime number).

For a number to be a 4th power, the exponent must be divisible by 4.
$$46 \div 4 = 11.5$$

Not divisible.

For a number to be a 6th power, the exponent must be divisible by 6.
$$46 \div 6 \neq \text{integer}$$

Therefore, it is neither a 4th power nor a 6th power.

Ans. Never True

Q5. Simplify and write these in the exponential form.

(i) $10^{– 2} × 10^{– 5}$ (ii) $5^7 ÷ 5^4$

(iii) $9^{– 7 }÷ 9^4$ (iv) $(13^{-2})^{– 3}$

(v) $m^5n^{12}(mn)^9$

Solution:

(i) $10^{-2} \times 10^{-5}$

$$10^{-2} \times 10^{-5} = 10^{-2-5}$$$$= 10^{-7} \quad \mathbf{Ans.}$$

(ii) $5^{7} \div 5^{4}$

$$5^{7} \div 5^{4} = 5^{7-4}$$$$= 5^{3} \quad \mathbf{Ans.}$$

(iii) $9^{-7} \div 9^{4}$

$$9^{-7} \div 9^{4} = 9^{-7-4}$$$$= 9^{-11} \quad \mathbf{Ans.}$$

(iv) $(13^{-2})^{-3}$

$$(13^{-2})^{-3} = 13^{-2 \times (-3)}$$$$= 13^{6} \quad \mathbf{Ans.}$$

(v) $m^{5}n^{12}(mn)^{9}$

$$(mn)^9 = m^9 n^9$$$$m^5 n^{12} \times m^9 n^9$$$$= m^{5+9} n^{12+9}$$$$= m^{14} n^{21} \quad \mathbf{Ans.}$$

Q6. If $12^2 = 144$ what is

$(i) (1.2)^2 \ \ \\ (ii) (0.12)^2 \\ (iii) (0.012)^2\\ (iv) 120^2$

Solution :

(i) $(1.2)^2$

$$(1.2)^2 = (12/10)^2$$$$= \frac{12^2}{10^2}$$$$= \frac{144}{100}$$$$= 1.44 \quad \mathbf{Ans.}$$

(ii) $(0.12)^2$

$$(0.12)^2 = (12/100)^2$$$$= \frac{12^2}{100^2}$$$$= \frac{144}{10000}$$$$= 0.0144 \quad \mathbf{Ans.}$$

(iii) $(0.012)^2$

$$(0.012)^2 = (12/1000)^2$$$$= \frac{12^2}{1000^2}$$$$= \frac{144}{1000000}$$$$= 0.000144 \quad \mathbf{Ans.}$$

(iv) $120^2$

$$120^2 = (12 \times 10)^2$$$$= 12^2 \times 10^2$$$$= 144 \times 100$$$$= 14400 \quad \mathbf{Ans.}$$

Q. 7. Circle the numbers that are the same —

$a)2^4 × 3^6\\\\b)6^4 × 3^2\\\\c)6^{10}\\\\d)18^2 × 6^2\\\\e)6^{24}$

$$2^4 \times 3^6$$$$= (2^2)^2 \times (3^3)^2$$$$= (2^2 \times 3^3)^2$$$$= (4 \times 27)^2$$$$= 108^2$$

$$6^4 \times 3^2$$$$= (2 \times 3)^4 \times 3^2$$$$= 2^4 \times 3^4 \times 3^2$$$$= 2^4 \times 3^6$$

So,$$2^4 \times 3^6 = 6^4 \times 3^2$$

$$18^2 \times 6^2$$$$= (3 \times 6)^2 \times 6^2$$$$= 3^2 \times 6^2 \times 6^2$$$$= 3^2 \times 6^4$$$$= 2^4 \times 3^6$$

Numbers that are the same

$$2^4 \times 3^6,\; 6^4 \times 3^2,\; 18^2 \times 6^2$$$$\mathbf{Ans.}$$

Q8. Identify the greater number in each of the following —

$(i) 4^3 or 3^4 \\\\(ii) 2^8 or 8^2 \\\\(iii) 100^2 or 2^{100}$

Solution:

(i) $4^3$ or $3^4$

$$4^3 = 64$$$$3^4 = 81$$$$3^4 > 4^3$$$$\mathbf{Ans.\; 3^4}$$

(ii) $2^8$or $8^2$

$$2^8 = 256$$$$8^2 = 64$$$$2^8 > 8^2$$$$\mathbf{Ans.\; 2^8}$$

(iii) $100^2$ or $2^{100}$

$$100^2 = 10000$$$$2^{100} \text{ is much larger than } 10000$$$$2^{100} > 100^2$$$$\mathbf{Ans.\; 2^{100}}$$

Q9. 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:

Number of packets:$$8.5 \text{ billion} = 8.5 \times 10^9$$

If a code has n digits, the total number of possible codes is:
$$10^n$$

Now find the smallest $n$n such that:
$$10^n \ge 8.5 \times 10^9$$$$10^9 = 1 \times 10^9 \quad (\text{not enough})$$$$10^{10} = 10 \times 10^9$$$$10^{10} = 10,000,000,000$$

This is greater than $8.5×10^9$

$$\mathbf{Ans.\; 10\ digits}$$

Q10. 64 is a square number$(8^2)$ and a cube number$(4^3)$. Are there
other numbers that are both squares and cubes? Is there a way
to describe such numbers in general?

Solution:

If a number is both a square and a cube, then it must satisfy:
$$\text{Number} = a^2 = b^3$$

Such numbers are powers of 2 and 3 that are common multiples.

The LCM of 2 and 3 is 6, so these numbers can be written as:
$$n^6$$

Examples

$$1 = 1^6$$$$64 = 2^6$$$$729 = 3^6$$$$4096 = 4^6$$

General Form

All numbers that are both squares and cubes are sixth powers.
$$\text{Number} = n^6$$$$\mathbf{Ans.\; Numbers\ of\ the\ form\ } n^6$$

Q11. 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 in the code can be:

• 26 letters (A–Z)
• 10 digits (0–9)

Total possible characters:$$26 + 10 = 36$$

For a 5-character passcode, the number of possible codes is:
$$36^5$$$$= 60,466,176$$$$\mathbf{Ans.\; 36^5 \; (= 60,466,176)}$$

Q12. 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?

$(ii) 20^9\\\\ (ii) 10^{11} \\\\(iii) 10^{10 }\\\\(iv) 10^{18} \\\\(v) 2 × 10^9 \\\\(vi) 10^9 + 10^9$

Population of sheep:$$10^9$$

Population of goats:$$10^9$$

Total population:$$10^9 + 10^9$$$$= 2 \times 10^9$$

Correct forms representing the answer:$$2 \times 10^9$$$$10^9 + 10^9$$$$\mathbf{Ans.\; (v)\; 2 \times 10^9 \; \text{and} \; (vi)\; 10^9 + 10^9}$$

Q13. 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) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.

World population ≈ $8 \times 10^9$

Pieces of clothing per person = $30 = 3 \times 10^1$
$$(8 \times 10^9) \times (3 \times 10^1)$$$$= 24 \times 10^{10}$$$$= 2.4 \times 10^{11}$$$$\mathbf{Ans.\; 2.4 \times 10^{11}}$$

(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.

Number of colonies:$$100\ \text{million} = 10^8$$100 million=108

Bees per colony:$$50,000 = 5 \times 10^4$$$$(10^8) \times (5 \times 10^4)$$$$= 5 \times 10^{12}$$$$\mathbf{Ans.\; 5 \times 10^{12}}$$

(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.

Bacteria per human:$$38\ \text{trillion} = 3.8 \times 10^{13}$$

World population:$$8 \times 10^9$$$$(3.8 \times 10^{13}) \times (8 \times 10^9)$$$$= 30.4 \times 10^{22}$$$$= 3.04 \times 10^{23}$$$$\mathbf{Ans.\; 3.04 \times 10^{23}}$$

(iv) Total time spent eating in a lifetime in seconds.

Assume:

3 meals per day × 30 minutes per meal = 90 minutes per day
$$90\ \text{minutes} = 5400\ \text{seconds}$$

Lifetime ≈ 80 years

Days in 80 years:$$80 \times 365 = 29200$$$$29200 \times 5400$$$$= 157680000$$$$= 1.5768 \times 10^8$$$$\mathbf{Ans.\; 1.58 \times 10^8\ seconds}$$

Q14. What was the date 1 arab/1 billion seconds ago?

Solution :

Concept Used

Conversion of seconds → days → years using scientific notation.
$$1\ \text{billion} = 10^9\ \text{seconds}$$

1 day:$$24 \times 60 \times 60 = 86400\ \text{seconds}$$

Convert seconds to days

$$\frac{10^9}{86400}$$$$\approx 11574\ \text{days}$$

Convert days to years

$$\frac{11574}{365} \approx 31.7\ \text{years}$$

So 1 billion seconds ≈ 31.7 years.

Find the date

Current year ≈ 2026$$2026 – 31.7 \approx 1994$$

More precisely,$$1{,}000{,}000{,}000\ \text{seconds before 14 March 2026}$$

≈ 5 July 1994

$$\mathbf{Ans.\; \text{About 5 July 1994}}$$

Also Read | A SQUARE AND A CUBE NCERT Solutions