Power Play Class 8 Solutions Maths Ganita Prakash Chapter 2

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Class 8 Maths Chapter 2 Power Play Solutions

Ganita Prakash Class 8 Chapter 2 Solutions

Class 8 Maths Ganita Prakash Chapter 2 Solutions Power Play

Intext (NCERT Page 22)

Question 1.
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) 2x 10 xv
(iv) 210
(v) 210v
(vi) 102v
Solution:
(v) Given, the initial thickness = v
After 1 fold, the thickness = 2 × v = 2¹v
After 2 folds, the thickness = 2 × (2v) = 2²
∵ After 3 folds, the thickness = 2 × (2²v) = 2³v
……………..
……………..
After 10 folds, the thickness = 210 v

Question 2.
(i) What is (-1)⁵? Is it positive or negative?
What about (-1)⁵⁶?
Solution:
We know that when a negative number is raised to an odd power, the result is negative.
So, (-1)⁵ = -1
And when a negative number is raised to an even power, the result is positive.
So, (-1)⁵⁶ = 1

(ii) What are 0² and 0⁵?
Solution:
We have,
0² = 0 × 0 = 0
and 0⁵ = 0 × 0 × 0 × 0 × 0 = 0

(iii) What is 0ⁿ?
Solution:
We know that for any positive integer n,
0ⁿ = 0

Question 3.
Is (-2)⁴ =16? Verify.
Solution:
On taking LHS, we get
(-2)⁴ = (-2) × (-2) × (-2) × (-2)
= 4 × 4
= 16 = RHS
Since, the result of (-2)⁴ is 16, so the given statement is true.

Figure it Out (NCERT Page 22-23)

Question 1.
Express the following in exponential form
(i) 6 × 6 × 6 × 6
Solution:
We have, 6 × 6 × 6 × 6
Here, 6 is multiplied 4 times.
So, 6 × 6 × 6 × 6 = 6⁴

(ii) y × y
Solution:
We have y × y
Here, y is multiplied by itself 2 times.
So, y × y = y²

(iii) b × b × b × b
Solution:
We have, b × b × b × b
Here, b is multiplied by itself 4 times.
So, b × b × b × b = b⁴

(iv) 5 × 5 × 7 × 7 × 7
Solution:
We have, 5 × 5 × 7 × 7 × 7
Here, 5 is multiplied 2 times and 7 is multiplied 3 times.
So, 5 × 5 × 7 × 7 × 7 = 5² × 7³

(v) 2 × 2 × a × a
Solution:
We have, 2 × 2 × a × a
Here, 2 is multiplied 2 times and a is multiplied 2 times.
So, 2 × 2 × a × a = 2² × a²

(vi) a × a × a × c × c × c × c × d
Solution:
We have, a × a × a × c × c × c × c × d
Here, a is multiplied 3 times, c is multiplied 4 times and d is multiplied 1 time.
So, a × a × a × c × c × c × c × d = a³ × c⁴ × d

Question 2.
Express each of the following as a product of powers of their prime factors in exponential form.
(i) 648
Solution:
We have, 648

Prime factors of 648 are 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2³ × 3⁴
So, the exponential form of 648 = 2³ × 3⁴

(ii) 405
Solution:
We have, 405

Prime factors of405 are 3 × 3 × 3 × 3 × 5 = 3⁴ × 5
So, the exponential form of 405 = 3⁴ × 5

(iii) 540
Solution:
We have, 540

Prime factors of 540 are 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5
So, the exponential form of 540 = 2² × 3³ × 5

(iv) 3600
Solution:
We have, 3600

Prime factors of 3600 are 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴ × 3² × 5²
So, the exponential form of 3600
= 2⁴ × 3² × 5²

Question 3.
Write the numerical value of each of the following.
(i) 2 × 10³
Solution:
We have, 2 × 10³
= 2 × (10 × 10 × 10)
= 2 × 1000
= 2000

(ii) 7² × 2³
Solution:
We have, 7² × 2³
= (7 × 7) × (2 × 2 × 2)
= 49 × 8
= 392

(iii) 3 × 4⁴
Solution:
We have, 3 × 4⁴
= 3 × (4 × 4 × 4 × 4)
= 3 × 256
= 768

(iv) (-3)² × (-5)²
Solution:
We have, (-3)² × (-5)²
= [(- 3) × (- 3)] × [(- 5) × (- 5)]
= (+9) × (+25)
= 225

(v) 3² × 10⁴
Solution:
We have, 3² × 10⁴
= (3 × 3) × (10 × 10 × 10 × 10)
= 9 × 10000
= 90000

(vi) (-2)⁵ × (-10)⁶
Solution:
We have, (-2)⁵ × (-10)⁶
= [(- 2) × (- 2) × (- 2) × (- 2) × (- 2)] × [(- 10) × (- 10) × (- 10) × (- 10) × (- 10) × (- 10)]
= -32 × 1000000
= -32000000

Intext (NCERT Page 24)

Question 1.
Use n^a × n^b = n^a+b observation.
(i) 2⁹
Solution:
We have, 2⁹ = 2⁽⁴⁺⁵⁾
= 2⁴ × 2⁵
= 16 × 32
= 512

(ii) 5⁷
Solution:
We have, 5⁷ = 5⁽⁴⁺³⁾
= 5⁴ × 5³
= 625 × 125
= 78125

(iii) 4⁶
Solution:
We have, 4⁶ =4⁽³⁺³⁾
= 4³ × 4³
= 6⁴ × 6⁴
= 4096

Question 2.
Write the following expressions as a power of a power in atleast two different ways.
(i) 8⁶
Solution:
We have, 8⁶ = (2 × 2 × 2)⁶ =(2³)⁶
Also, 8⁶ = 8^2×3
=(8²)³

(ii) 7¹⁵
Solution:
We have,7¹⁵ = 7^3×5 = (7³)⁵
Also, 7¹⁵ = 7^5×3
= (7⁵)³
=(16807)³

(iii) 9¹⁴
Solution:
We have, 9¹⁴ = (3×3)¹⁴ = (3²)¹⁴
Also, 9¹⁴ = 9^2×7
= (9²)⁷

(iv) 5⁸
Solution:
We have, 5⁸ = 5⁸ = (5²)⁴
Also, 5⁸ = 5^4×2 = (5⁴)²
=(5 x5 x5 x5)²
=(625)²

(NCERT Page 25)

Question 1.
in the middle of a beautiful, magical pond lies a bright pink lotus. The number of lotuses doubles everyday 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 everyday, the pond should be half covered on the 29th day.
Write the number of lotuses (in exponential form) when the pond was
(i) fully covered
(ii) half covered
Solution:
Let n be the initial number of lotus in the pond.
Since, the number of lotus doubles everyday.
(i) The number of lotus on first day = n × 2
The number of lotus on second day = n × 2²
The number of lotus on third day = n × 2³
……………………………..
……………………………..
The number of lotus on 29th day = n × 2²⁹
The number of lotus on 30th day = n × 2³⁰
Since, the pond is fully covered on 30th day.
So, number of lotus when the pond was full covered
= n × 2³⁰

(ii) Since, the number of lotus doubles everyday, therefore the pond should be half covered on the 29th day.
So, number of lotus when the pond was half covered
= n × 2²⁹

Question 2.
Simplify \(\frac{10^4}{5^4}\) and write it in exponential form.
Solution:
We have, \(\frac{10^4}{5^4}=\frac{(2 \times 5)^4}{5^4}\)
= \(\frac{2^4 \times 5^4}{5^4}\)
= 2⁴ [∵ (ab)^m = a^m × b^m]

Intext (NCERT Page 26-27)

Question 1.
Roxie has 7 dresses, 2 hats and 3 pairs of shoes. How many different ways can Roxie dress up?
Solution:
Given, number of dresses = 7,
number of hats = 2 and number of pairs of shoes = 3
So, the total number of different ways can Roxie dress up = 7 × 2 × 3 = 42

Question 2.
Think about how many combinations are possible in different contexts. Some examples are
(i) Pincodes of places in India— The Pincode of Vidisha in Madhya Pradesh is 464001. The Pincode of Zemabawk in Mizoram is 796017.
Solution:
Since, the first digit of a Pincode indicates the region and it cannot be 0.
For Pincode of Vidisha in Madhya Pradesh 464001
For the remaining 5 digits, each can be any number 4, 6,0 and 1. So, we have 4 choices for every number and 3 choices for the first number.
Therefore, the total number of combinations of 6 digit Pincode = 3 × 4 × 4 × 4 × 4 × 4 = 3072
For Pincode of Zemabawk in Mizoram 796017 For the remaining 5 digits, each can be any number 1, 6, 7, 0 and 9.
So, we have 5 choices for every number and 4 choices for the first number.
Therefore, the total number of combinations of 6 digit Pincode = 4 × 5 × 5 × 5 × 5 × 5 = 12500

(ii) Mobile numbers.
Solution:
We know that Indian mobile numbers are typically 10-digit numbers.
And the first digit of mobile number in India usually starts with 6,7,8 or 9. So, there are 4 choices for the first digit.
For the remaining 9 digits, each can be any number from 0 to 9, giving 10 choices for each position.
So, the number of possible combinations
= 4 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 4 × 10⁹
= 4 × 1000000000
= 4000000000

(iii) Vehicle registration numbers.
Try to find out how these numbers or codes are allotted/generated.
Solution:
We know that Indian vehicle registration numbers follow a specific formats XX YY ZZ NNNN
XX represents the State/Union Territor code (2 letters). So, 26 letters choices for each of the two letters.
YY represents the RTO code within the State/UT (2 digits 1 to 9.)
ZZ is a two-letter alphabetical series, which increments as numbers in the NNNN part are exhausted.
NNNN is a numerical part, typically a four-digit number from 0001 to 9999, assigned sequentially.
So, the combination of these elements ensures a unique identification for each registered vehicle in the country.

Question 3.
What is 2¹⁰⁰ ÷ 2²⁵ in powers of 2?
Solution:
We have, 2¹⁰⁰ ÷ 2²⁵ = \(\frac{2^{100}}{2^{25}}\) = 2^100-25 [∵ \(\frac{n^a}{n^b}\) = n^a-b]
= 2⁷⁵

Intext (NCERT Page 29)

Question 1.
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:
We know that the generalised forms of the laws of exponents are
n^a x n^b = n^a+b
(n^a)^b = (n^b)^a = n^a×b
n^a ÷ n^b = n^a-b
Yes, these generalised forms of exponent rules hold true even when a and b are any integers (positive, negative or zero). These rules are fundamental properties of exponents and are derived from the definition of exponents for all integer values.

Question 2.
Write equivalent forms of the following.
(i) 2^-4
Solution:
We know that a^-m = \(\frac{1}{a^m}\)
We have, 2^-4 = \(\frac{1}{2^4}\)

(ii) 10^-5
Solution:
We have, 10^-5 = \(\frac{1}{10^5}\)

(iii) (-7)^-2
Solution:
We have, (-7)^-2 = \(\frac{1}{(-7)^2}=\frac{1}{(-1 \times 7)^2}\)
= \(\frac{1}{(-1)^2 \times 7^2}=\frac{1}{7^2}\)

(iv) (-5)^-3
Solution:
We have, (-5)^-3 = \(\frac{1}{(-5)^3}\)
= \(\frac{1}{(-1 \times 5)^3}=\frac{1}{(-1)^3 \times 5^3}=\frac{1}{-(5)^3}\)

(v) 10^-100
Solution:
We have, 10^-100 = \(\frac{1}{10^{100}}\)

Question 3.
Simplify and write the answers in exponential form.
(i) 2^-4 × 2⁷
Solution:
We have, 2^-4 × 2⁷
= 2^(-4+7)
= 2³ [∵ a^m x aⁿ = a^m+n]

(ii) 3² × 3⁵ × 3⁶
Solution:
We have, 3² × 3⁵ × 3⁶
= 3^(2-5+6)
= 3³
[∵ a^m × aⁿ = a^m+n]

(iii) p³ × p^-10
Solution:
We have, p³ × p^-10
= p^(3-10)
= p^-7 [∵ a^m × aⁿ = a^m+n]

(iv) 2⁴ × (-4)²
Solution:
We have, 2⁴ × (-4)²
= 2⁴ × (—2 × 2)^-2
= 2 × (-1)² × (2²)² [∵ (ab)^m = a^m × b^m)
= 2⁴ × 2^-4 [∵ (a^m)ⁿ = a^mn]
= 2⁰ [∵ a^m × aⁿ = a^m+n]
= 1 [∵ a⁰ = 1]

(v) 8^p × 8^q
Solution:
We have, 8^p × 8^q
= 8^(p+q) [∵a^m × aⁿ = a^m+n]

Intext (NCERT Page 30)

Question 1.
How many times larger than 4^-2 is 4²?
Solution:
We know that a^-n = \(\frac{1}{a^n}\)
∴ 4^-2 = \(\frac{1}{4^2}\)
On dividing 4^-2 by 4², we get
\(\frac{4^2}{4^{-2}}=\frac{4^2}{\frac{1}{4^2}}\)
= 4² × 4²
= 4⁽²⁺²⁾
= 4⁴ [∵ a^m × aⁿ = a^m+n]
4⁴ = 4 × 4 × 4 × 4
= 16 × 16
= 256
So, 4² is 256 times larger than 4^-2.

Question 2.
Use the power line for 7 to answer the following questions

(i) 2,401 × 49
Solution:
We have, 2401 × 49
From the power line, 2401 = 7⁴ and 49 = 7²
∴ 2401 × 49 = 7⁴ × 7² = 7⁽⁴⁺²⁾ = 7⁶ [∵ a^m × aⁿ = a^m+n]
= 117649

(ii) (49³)
Solution:
We have, (49)³
From the power line, 49 = 7²
∴ (49)³ = (7²)³
= 7^2×3
= 7⁶ [∵(a^m)ⁿ = a^mn]
= 117649

(iii) 343 × 2,401
Solution:
We have, 343 × 2401
From the power line, 343 = 73 and 2401 = 74
∴ 343 × 2401 = 7³ × 7⁴ = 7⁽³⁺⁴⁾ = 7⁷ [∵ a^m × aⁿ = a^m+n]
= 823543

(iv) \(\frac{16,807}{49}\)
Solution:
We have, \(\frac{16,807}{49}\)
From the power line, 16807 = 7⁵ and 49 = 7²
∴ \(\frac{16807}{49}=\frac{7^5}{7^2}\)
= 7^(5-2)
= 7³ [∵ \(\frac{a^m}{a^n}\) = a^m-n]
= 343

(v) \(\frac{7}{343}\)
Solution:
We have, \(\frac{7}{343}\)
From the power line, 7 = 7¹ and 343 = 7³
∴ \(\frac{7}{343}=\frac{7}{7^3}\)
= 7^(1-3)
= 7^-2 [∵ \(\frac{a^m}{a^n}\) = a^m-n]
= \(\frac{1}{49}\)

(vi) \(\frac{16,807}{8,23,543}\)
Solution:
We have, \(\frac{16,807}{8,23,543}\)
From the power line, 16807 = 7⁵ and 823543 = 7⁷
∴ \(\frac{16807}{823543}=\frac{7^5}{7^7}\)
= 7^(5-7)
= 7^-2 [∵ \(\frac{a^m}{a^n}\) = a^m-n]
= \(\frac{1}{49}\)

(vii) 1,17,649 × \(\frac{1}{343}\)
Solution:
We have, 1,17,649 × \(\frac{1}{343}\)
From the power line, 1,17,649 = 7⁶ and 343 = 7³
∴ 117649 × \(\frac{1}{343}=\frac{7^6}{7^3}\)
= 7^(6-3)
= 7³ [∵ \(\frac{a^m}{a^n}\) = a^m-n]
= 343

(viii) \(\frac{1}{343} \times \frac{1}{343}\)
Solution:
We have, \(\frac{1}{343} \times \frac{1}{343}\)
From the power line, \(\frac{1}{343}\) = 7^-3
∴ \(\frac{1}{343} \times \frac{1}{343}\) = 7^-3 × 7^-3 [∵ a^m × aⁿ = a^m+n]
= 7^-6
= \(\frac{1}{7^6}\) [∵ a^-m = \(\frac{1}{a^m}\) ]
= \(\frac{1}{117649}\)

Question 4.
Write these numbers in power of 10.
(i) 172
Solution:
We have, 172 = 1 × 100 + 7 × 10 + 2
= 1 × 10² + 7 × 10¹ + 2 × 10⁰

(ii) 5642
Solution:
We have, 5642 = 5 × 1000 + 6 × 100 + 4 × 10 + 2
= 5 × 10³ + 6 × 10² + 4 × 10¹ + 2 × 10⁰

(iii) 6374
Solution:
We have, 6374 = 6 × 1000 + 3 × 100 + 7 × 10 + 4
= 6 × 10³ + 3 × 10² + 7 × 10¹ + 4 × 10⁰

Intext (NCERT Page 32)

Question 1.
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 1.495 × 10¹¹ m.

Solution:
Given, the distance between Sun and Saturn
= 14335 × 10¹² m
and the distance between Sun and Earth
= 1.496 × 10¹¹ m
= 0.1496 × 10¹² m
Here, 0.1496 × 10¹² m (Earth – Sun) is significantly smaller than 14335 × 10¹² m (Saturn – Sun).
Since, the Earth’s distance from the Sun is much smaller than the Saturn’s distance from the Sun, the Earth’s position on the number line will be very close to Sun, much closer than Saturn. The number line shows ‘Sun’ at one end and ‘Saturn’ at the other end. The Earth would be marked very near the ‘Sun’ end of the line.

Question 2.
Express the following numbers in standard form.
(i) 59853
Solution:
We have, 59853
= 59853 × 10⁴

(ii) 65950
Solution:
We have, 65950
= 65950 × 10⁴

(iii) 3430000
Solution:
We have, 3430000
= 3.43 × 10⁶

(iv) 70040000000
Solution:
We have, 70040000000
= 7.004 × 10¹⁰

Intext (NCERT Page 38-39)

Question 1.
Calculate and write the answer using scientific notation.
(i) How many ants are there for every human in the world?
(ii) If a flock of starlings contains 10,000 birds, how many flocks could there be in the world?
(iii) If each tree had about 104 leaves, find the total number of leaves on all the trees in the world.
(iv) If you stacked sheets of paper on top of each other, how many would you need to reach the Moon?
Solution:
(i) We know that the population of ants in the world
= 20 quadrillion
= 2 × 10¹⁶
And the population of human in the world
= 82 billion
= 82 × 10⁹
So, the number of ants for every human
= \(\frac{2 \times 10^{16}}{8.2 \times 10^9}\) = 0243 × 10⁷
= 2.43 × 10⁶

(ii) Given, the number of birds contained by a flock of starlings = 10000 = 104
We know that total global population of starlings = 500 million
= 500000000
= 5 × 10⁸
So, the number of flocks in the world
= \(\frac{\text { Total global population of starlings }}{\text { Number of birds per flock }}\)
= \(\frac{5 \times 10^8}{10^4}\) [∵ \(\frac{a^m}{a^n}\) = a^m-n]
= 5 × 10^(8-4)
= 5 × 10⁴
= 50000

(iii) Given, the number of leaves on a tree = 104
We know that the total number of trees in the world
= 3 × 10¹²
So, the total number of leaves on all the trees in the world
= 3 × 10¹² × 10⁴
= 3 × 10⁽¹²⁺⁴⁾ [∵ a^m × aⁿ = a^m-n]
= 3 × 10¹⁶

(iv) We know that the distance from the Earth to Moon
= 384400000 km
= 3.844 × 10⁸ m
and the thickness of a standard paper = 0.1mm
= 0.0001m
= 1 × 10^-4 m

So, number of sheets = \(\frac{\text { Distance to Moon }}{\text { Thickness of one sheet }}\)
= \(\frac{3.844 \times 10^8 \mathrm{~m}}{1 \times 10^{-4} \mathrm{~m}}\)
= 3.844 × 10¹²
Therefore, we need approximately 3.844 × 10¹² sheets of paper stacked on top of each other to reach the Moon.

Question 2.
A DIFFERENT WAY TO SAY YOUR AGE!
‘How old are you?’ asked Estu.
‘I completed 13 years a few weeks ago!’ said Roxie. ‘How old are you?’ asked Estu again.
‘I’m 4840 days old today!’ said Roxie.
‘How old are you?’ asked Estu again.
‘I’m ……… hours old!’ said Roxie.
Make an estimate before finding this number.
Estu: ‘I am 4070 days old today. Can you find out my date of birth?’
Solution:
Given, the age of Roxie = 4840 days
So, age of Roxie in hours = 4840 × 24 [∵ 1 day = 24 hours]
= 116160

Since, Estu 4070 days old today.
\(\frac{4070}{365}\) days = 11 years 5 days
[∵ 1 year = 365 days assuming non leap year] Let today’s date is 31 July 2025.

This means Estu’s birth year would be 2025 – 11 = 2014
Now, subtract the remaining 5 days from July 31, 2014 which take us to July 27, 2014.

Intext (NCERT Page 42)

Question 1.
Calculate and write the answer using scientific notation.
(i) If one star is counted every second, how long would it take to count all stars in the universe?
Answer in terrrwofTlTe number of seconds using scientific notations.
(ii) If one could drink a glass of water (200 ml) every 10 seconds, how long would it take finish the entire volume of water on Earth?
Solution:
(i) Given, one star is counted every second.
We know that the estimated number of stars in the observable universe is 2 × 10²³.
Thus, time to count = 2 × 10²³ seconds.

(ii) Given, volume of one glass = 200 ml
and time for one glass =10 seconds
Since, estimated number of drops of water on the Earth is 2 × 10²⁵ drops (assuming 16 drops per ml).
Volume of water on Earth = \(\frac{2 \times 10^{25}}{16}\)ml
= 1.25 × 10²⁴ ml
∴ Number of glasses = \(\frac{1}{2}\)
[∵ volume of one glass = 200 ml]
= 0.00625 × 10²⁴ glasses
= 6.25 × 10²¹ glasses
.’. Time to finish = 6.25 × 10²¹ × 10 seconds
[∵ time to finish one glass = 10 seconds]
= 6.25 × 10²² seconds

Figure It Out (NCERT Page 44-45)

Question 1.
Find out the unit digit in the value of 2²²⁴ ÷ 4³²?
[Hint 4 = 2²]
Solution:
We have,
2²²⁴ ÷ 4³²
= 2²²⁴ ÷ (2²)³²
= 2²²⁴ ÷ 2⁶⁴
= 2^(224-64)
= 2¹⁶⁰
We know that 2¹ = 2²
2² = 4
2³ = 8
2⁴ = 16
2⁵ = 32
2⁶ = 64
Here, the cycle of unit digits for powers of 2 is 2, 4, 8, 6 which repeats after every 4 powers.
So, \(\frac{160}{4}\) = 40 4
When the remainder is 0, the unit digit is the last digit in the cycle, which is 6.
So, the unit digit in the value of 2²²⁴ ÷ 4³² is 6.

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?
Solution:
Given, there are 5 bottles in a container initially.
And everyday, a new container is brought in since each container has 5 bottles, this means 5 new bottles are added each day.
So, the total number of bottles after 40 days = 40 × 5 + 5 initial bottles = 200 + 5 = 205

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
Solution:
We know that
(a^m)ⁿ = a^mn and a^m+n = a^m × aⁿ
(i) We have, 64³ = (8²)³ = 8^2×3
= 8² × 8³

Also, (64)³ = (4³)³ = 4^3×3
= 4³ × 4³

(ii) We have, (192)⁸ = (64 × 3)⁸ = (82 × 31)⁸
= 8^2×8 × 3⁸
= 8¹⁶ × 3⁸

Also, (192)⁸ =(64 × 3)⁸
= (4³ × 3¹)⁸
= 4²⁴× 3⁸

Also, (192)⁸ = (64 × 3)⁸
= (26 × 31)⁸
= 2⁴⁸ × 3⁸

(iii) We have, (32)^-5 = (2⁵)^-5
= 2^5×(-5)

Also, (32)^-5 = (2⁵)^-5
= 2^-25
= 2^(-5-10-10)
= 2^-5 × 2^-10 × 2^-10

Also,(32)= (2⁵)^-5 = 2^(10-35)
= 2¹⁰ × 2^-35

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.
Solution:
This statement is only sometimes true.
e.g. 64 = 4³ = 8² is both a cube and a square number. However, 8 = 2³ is a cube but not a square number and 9 = 3² is a square but not a cube.

(ii) Fourth powers are also square numbers.
Solution:
This statement is always true.
A fourth power can be written as x⁴. This can be rewritten as (x²)², which is the square of x².

(iii) The fifth power of a number is divisible by the cube of that number.
Solution:
This statement is always true.
Let the number be n. Its fifth power is n⁵ and it’s cube is n³. Since, n⁵ = n³ × n², n⁵ is always divisible by n³ for any integer.

(iv) The product of two cube numbers is a cube number.
Solution:
This statement is always true.
Let the two cube numbers be a³ and b³. Their product is a³ × b³ = (a x b)³. Since, the product can be expressed as the cube of another number (a × b). It is always a cube number.

(v) q⁴⁶ is both a 4th power and a 6th power (q is a prime number).
Solution:
This statement is never true.
For q⁴⁶ to be a 4th power, 46 must be divisible by 4, which it is not (as 46 + 4 = 115).
For q⁴⁶ to be a 6th power, 46 must be divisible by 6, which is not (as 46 + 6 = 7.67). Therefore, q46 cannot be both a 4th power and a 6th power.

Question 5.
Simplify and write these in the exponential form.
(i) 10² × 10⁵
Solution:
We have, 10² × 10⁵
= 10^(-2-5)
= 10^-7 [∵ a^m × aⁿ = a^(m+n)]

(ii) 5⁷ ÷ 5⁴
Solution:
We have, 5⁷ ÷ 5⁴
= 5^(7-4)
= 5³ [∵ a^m ÷ aⁿ = a^(m-n)]

(iii) 9^-7 ÷ 9⁴
Solution:
We have, 9^-7 ÷ 9⁴
= 9^(-7-4)
= 9^-11 [∵ a^m ÷ aⁿ = a^(m-n)]

(iv) (13^-2)^-3
Solution:
We have, (13^-2)^-3
= 13^-2×(-3)
= 13⁶ [∵ (a^m)ⁿ = a^mn]

(v) m⁵n¹²(mn)⁹
Solution:
We have, m⁵n¹²(mn)⁹
= m⁵n¹²m⁹n⁹ [∵ (ab)^m = a^mb^m]
= m⁽⁵⁺⁹⁾n⁽¹²⁺⁹⁾
= m¹⁴n²¹ [∵ a^m × aⁿ = a^(m+n)]

Question 6.
If 12² =144, what is
(i) (1.2)²
Solution:
We have, (1.2)²
(1.2)² = \(\left(\frac{12}{10}\right)^2=\frac{(12)^2}{(10)^2}=\frac{144}{100}\)
[∵ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)]
= 1.44

(ii) (0.12)²
Solution:
We have, (0.12)²
(0.12)² = \(\left(\frac{12}{100}\right)^2=\frac{(12)^2}{(100)^2}=\frac{144}{10000}\)
[∵ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)]
= 0.0144

(iii) (0.012)²
Solution:
We have, (0.012)²
(0.012)² = \(\left(\frac{12}{1000}\right)^2=\frac{(12)^2}{(1000)^2}=\frac{144}{1000000}\)
[∵ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)]
= 0.000144

(iv) 120²
Solution:
We have, 120²
120² = (12 × 10)² = (12)² × (10)²
[∵ (a × b)² = a² × b²]
= 144 × 100
= 14400

Question 7.
Circle the numbers that are the same
(i) 2⁴ × 3⁶
Solution:
We have, 2⁴ × 3⁶

(ii) 6⁴ × 3²
Solution:
We have, 6⁴ × 3²
= (2 × 3)⁴ × 3²
= 2⁴ × 3⁴ × 3²
[∵ a^m × aⁿ = a^m+n]

(iii) 6¹⁰
Solution:
We have, 6¹⁰ = (2 × 3)¹⁰
= 2¹⁰ × 3¹⁰ [∵ (ab)^m = a^m × b^m]

(iv) 18² × 6²
Solution:
(iv) We have, 18² × 6² = (2 × 3 × 3)² × (2 × 3)²
= (2 × 3²)² × 2² × 3²
= 2² × 3⁴ × 2² × 3²
= 2⁴ × 3⁶ [∵ a^m × aⁿ = a^m+n]

(v) 6²⁴
Solution:
We have, 6²⁴
6²⁴ is a very large number, significantly larger than the others.
From the calculations, the expressions that result in the same value are 2⁴ × 3⁶, 6⁴ × 3² and 18² × 6² all of these equal to 11664.

Question 8.
Identify the greater number in each of the following
(i) 4² or 3⁴
Solution:
We have, 4³ or 3⁴
4³ = 4 × 4 × 4 = 64
3⁴ = 3 × 3 × 3 × 3 = 81
∵ 64 < 81
∴ 4³ < 3⁴

(ii) 2⁸ or 8²
Solution:
We have, 2⁸ or 8²
8² = (2³)² = 2⁶
∴ 2⁸ > 8²

(iii) 100² or 2¹⁰⁰
Solution:
We have, (100)² or 2¹⁰⁰
(100)² = 100 × 100 = 10000
2¹⁰⁰ = (2¹⁰)¹⁰ = (1024)¹⁰
Since, 1024 is already greater than (100)2, raising it to the power of 10 will result in larger number than 10000
∴ (100)² < 2¹⁰⁰

Question 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:
Given, number of packets = 85 billion
= 8.5 × 10⁹
Let n be the smallest number such that
10ⁿ >8.5 × 10⁹
∵ 10⁹ is less than 8.5 × 10⁹ and 1010 is greater than 85 × 10⁹, so the dairy needs atleast 10 digits to create a unique ID for each of the 8.5 billion packets.

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?
Solution:
Yes, there are other numbers besides 64 that are both perfect square and perfect cubes,
e.g. 1⁶ = 1 (which is 1² and 1³)
3⁶ = 729 (which is 27² and 9³)
4⁶ = 4096 (which is 64² and 16³)
……………………………………..
For N to be both a² and b³, the prime factorisation of N must have exponents that are multiples of both 2 and 3. The least common multiple of 2 and 3 is 6.
Therefore, any number that is both a perfect square and a perfect cube must be of the form k⁶ for some integer k.

Question 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:
An alphanumeric character can be a digit (0 – 9, which are 10 options) or a letter (A-Z which are 26 options).
So, there are 10 + 26 = 36 possible characters for each position.
Since, the pass code has a length of 5.
So, the total number of possible codes
= 36 × 36 × 36 × 36 × 36
= 365
= 60466176

Question 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:
(v), (vi) Given, the population of sheep = population of goats = 109
So, the total population of sheep and goats =109 + 109
= 2 × 10⁹

Question 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.
Solution:
Given, the number of clothes having by each person
= 30
We know that the population of the world = 8 × 10⁹
So, the total number of clothes in the world
= 8 × 10⁹ × 30
= 240 × 10⁹
= 2.4 × 10¹¹

(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
Solution:
Given, the number of bees in a colony = 50000
= 5 × 10⁴
and the number of colonies in the world = 100 million = 100 × 1000000= 10⁸
So, the total number of honeybees in the world = 5 × 10⁴ × 10⁸
= 5 × 10^{(4 + 8)}
= 5 × 10¹²
[∵ a^m × aⁿ =a^m+n]

(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.
Solution:
Given, number of bacterial cells in a human body
= 38 trillion = 38 × 10¹² = 3.8 × 10¹³
We know that the population of world = 8 × 10⁹
So, the bacterial population residing in all humans in the world = 3.8 × 10¹³ × 8 × 10⁹
= 30.4 × 10²²
= 3.04 × 10²³ [∵ a^m × aⁿ =a^m+n]

(iv) Total lime spent eating in a lifetime in seconds.
Solution:
Assuming an average lifespan of 70 years.
Assuming an average of 1 hour per day spent eating.
So, the time spent in eating in a day = 1 hour
= 3600 seconds
and the time spent in eating in a year = 3600 × 365
= 1314000
Total time spent eating in a life time . = 1314000 × 70
= 9.1980000 seconds
= 9.198 × 10⁷ seconds.

Question
What was the date 1 arab/1 billion seconds ago?
Solution:
Let today’s date is 31 July 2025.
On converting 1 arab second to minutes, we get
1000000000 ÷ 60 = 1666666667 minutes
[∵ 1 second = \(\frac{1}{60}\) minute]

On converting 1666666667 minutes to hours, we get
1666666667 ÷ 60 = 27777778 hours
[∵ 1 minute = \(\frac{1}{60}\) hour]

On converting 27777778 hours to days, we get
27777778 ÷ 24 = 1157407 days

On converting 1157407 days to years, we get
1157407 ÷ 36525 = 3L68 years
[∵ 365.25 for leap years]
It means 1 arab seconds is approximately 31 years, 7 months.
On subtracting this duration from July 31,2025, we get 2025 – 31= 1994 and also subtracting 7 months, we get the date 31 December 1994.