Class 8 Maths Ganita Prakash Chapter 2 Solutions
Page 22
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
See
Solutions(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
See
Solutions(i) 648
648 = 2 × 2
× 2 × 3 × 3 × 3 × 3 = 2³ × 3⁴
(ii) 405
405 = 3 × 3
× 3 × 3 × 5 = 3⁴ × 5
(iii) 540
540 = 2 × 2
× 3 × 3 × 3 × 5 = 2² × 3³ × 5
(iv) 3600
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
See
Solutions(i) 2 × 10³ = 2 × 1000 = 2000
(ii) 7² × 2³
= 49 × 8 = 392
(iii) 3 × 4⁴
= 3 × 256 = 768
(iv) (–3)² ×
(–5)² = 9 × 25 = 225
(v) 3² × 10⁴
= 9 × 10,000 = 90,000
(vi) (–2)⁵ ×
(–10)⁶ = (–32) × 1,000,000 = –32,000,000
Ganita
Prakash Class 8 Maths Chapter 2 Question Answers
Page 44
Figure it
Out
1. Find out
the units digit in the value of 2²²⁴ ÷ 4³²? [Hint: 4 = 2²]
See
SolutionsExpressing everything with base 2:
4 = 2², so:
4² = (2²)³² = 2⁶⁴
So the
expression becomes:
2²²⁴ ÷ 2⁶⁴
= 2²²⁴⁻⁶⁴
= 2¹⁶⁰
Now, we have
to find out unit digit of 2¹⁶⁰:
The unit
digit of powers of 2 follows a cycle of 4:
2¹ = 2 →
unit digit = 2
2² = 4 →
unit digit = 4
2³ = 8 →
unit digit = 8
2⁴ = 16 →
unit digit = 6
Then it
repeats: 2, 4, 8, 6, …
So, divide
the exponent by 4: 160 ÷ 4 = 40
If divisible
exactly, the 4th number in the cycle is the unit digit, i.e., 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?
See
SolutionsEach container has 5 bottles, and 1 container is added every day.
So, after 40
days, the total number of containers = 40
Therefore,
total number of bottles = 40 × 5 = 200
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
See
Solutions(i) 64³
We know:
64 = 2⁶
So,
64³ = (2⁶)³
= 2¹⁸
64³ = 8⁶,
since 8 = 2³
64³ = 4⁹,
since 4 = 2²
(ii) 192⁸
See
Solutions(ii) 192⁸
After
factorisation, we get 192⁸ = 2⁶ × 3
So,
192⁸ = (2⁶ ×
3)⁸ = 2⁴⁸ × 3⁸
192⁸ = (64 ×
3)⁸ = 64⁸ × 3⁸
192⁸ = (2³ ×
2³ × 3)⁸ = 2⁴⁸ × 3⁸
(iii) 32⁻⁵
See
Solutions(iii) 32⁻⁵
We know
that: 32 = 2⁵
So,
32⁻⁵ =
(2⁵)⁻⁵ = 2⁻²⁵
32⁻⁵ = (2³ ×
2²)⁻⁵, since 32 = 2⁵
32⁻⁵ = (4³ ×
2⁻¹)⁻⁵
Ganita
Prakash Class 8 Maths Chapter 2 Solutions
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 a re also square numbers.
See
SolutionsOnly Sometimes True
Reason:
A cube
number is of the form 𝑛³ and a square number is of the form 𝑚².
A number
that is both a perfect square and a perfect cube must have 6 power, i.e., 𝑥⁶
Example: 64
= 4³ = 8² is both.
But 8 = 2³
is not a square.
So, only
some cube numbers are also square numbers.
(ii) Fourth
powers are also square numbers.
See
SolutionsAlways True
Reason:
Any number
raised to the power 4 is also a square because:
𝑛⁴ = (𝑛²)²
Hence, all
fourth powers are perfect squares.
(iii) The
fifth power of a number is divisible by the cube of that number.
See
SolutionsAlways True
Reason:
Any number 𝑎⁵ is clearly divisible by 𝑎³, because:
𝑎⁵ = 𝑎³⋅𝑎²
So, the cube
𝑎³ is a factor of 𝑎⁵ for all 𝑎 ≠ 0.
(iv) The
product of two cube numbers is a cube number.
See
SolutionsAlways True
Reason:
Let two cube
numbers be 𝑎³ and 𝑏³.
Then, 𝑎³⋅𝑏³ = (𝑎𝑏)³
So, the
product is again a perfect cube.
(v) 𝑞⁴⁶ is both a 4th power and a 6th power (where 𝑞 is a prime number).
See
SolutionsNever True
Reason:
A number is
both a 4th and a 6th power if its exponent is a multiple of LCM(4, 6) = 12.
Since, 46 =
2 × 23 (not divisible by 4 or 6)
But the
question says 𝑞⁴⁶ is both a 4th and 6th power.
That’s impossible unless 46 is divisible by both 4 and 6.
So the
correct answer is: Never True
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
See
Solutions(i) 10⁻² × 10⁻⁵ = 10⁻²⁻⁵ = 10⁻⁷
(ii) 5⁷ ÷ 5⁴
= 5⁷⁻⁴ = 5³
(iii) 9⁻⁷ ÷
9⁴ = 9⁻⁷⁻⁴ = 9⁻¹¹
(iv)
(13⁻²)⁻³ = 13⁶
(v)
m⁵n¹²(mn)⁹ = m⁵n¹².m⁹.n⁹ = m⁵⁺⁹n¹²⁺⁹ = m¹⁴n²¹
Class 8
Maths Chapter 2 Exercises
6. If 12² =
144 what is
(i) (1.2)²
See
SolutionsGiven that 12² = 144
(i) (1.2)²
= (12/10)² =
12²/10²
= 144/100 =
1.44
(ii) (0.12)²
See
Solutions(ii) (0.12)²
= (12/100)²
= 12²/100²
= 144/10000
= 0.0144
(iii)
(0.012)²
See
Solutions(iii) (0.012)²
= (12/1000)²
= 12²/1000000²
=
144/1000000 = 0.000144
(iv) 120²
See
Solutions(iv) 120²
= (12 × 10)²
= 12² × 10²
= 144 × 100
= 14400
7. Circle
the numbers that are the same —
• 2⁴ × 3⁶
• 6⁴ × 3²
• 6¹⁰
• 18² × 6²
• 6²⁴
See
SolutionsSimplifying all the numbers:
• 2⁴ × 3⁶
• 6⁴ × 3² =
(2 × 3)⁴ × 3² = 2⁴ × 3⁴ × 3² = 2⁴ × 3⁶
• 6¹⁰ = (2 ×
3)¹⁰ = 2¹⁰ × 3¹⁰
• 18² × 6² =
(2×3²)² × (2×3)² = 2²×3⁴ × 2²×3² = 2⁴ × 3⁶
• 6²⁴ = (2 ×
3)²⁴ = 2²⁴ × 3²⁴
After
simplification, we can see that (2⁴ × 3⁶), (6⁴ × 3²) and (18² × 6²) are same.
8.
Identify the greater number in each of the following —
(i) 4³ or 3⁴
See
Solutions(i) 4³ or 3⁴
4³ = 64
3⁴ = 81
So, 3⁴ is
greater.
(ii) 2⁸ or
8²
See
Solutions(ii) 2⁸ or 8²
2⁸ = 256
8² = 64
So, 2⁸ is
greater.
(iii) 100²
or 2¹⁰⁰
See
Solutions(iii) 100² or 2¹⁰⁰
100² = 10000
2¹⁰⁰ =
(2¹⁰)¹⁰ = (1024)¹⁰ [Since, 2¹⁰ = 1024]
So, 2¹⁰⁰ is
a huge number, thus it is greater.
Class 8
Maths Chapter 2 All Questions
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?
See
SolutionsA dairy plans to produce 8.5 billion = 8.5 × 10⁹ packets.
Each packet
needs a unique numeric code using digits 0–9.
The number
of unique codes that can be made with n digits = (10)^𝑛
We must find
the smallest n such that: 10^𝑛 ≥ 8.5 × 10⁹
If we put 𝑛 = 9, we get:
10⁹ =
1,000,000,000 less than 8.5 billion.
If we put 𝑛 = 10, we get:
10¹⁰ =
10,000,000,000 more than 8.5 billion.
So, The code
must have at least 10 digits, since 10¹⁰ is the smallest power of 10 greater
than 8.5 billion.
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?
See
SolutionsThe numbers that are both squares and cubes:
• 1 = 1² =
1³
• 64 = 8² =
4³
• 729 = 27²
= 9³
• 4096 = 64²
= 16³
• 15625 =
125² = 25³
The way to
describe such numbers in general:
All numbers
that are sixth powers (like 1⁶, 2⁶, 3⁶, …) are both perfect squares and perfect
cubes.
So the
general form is:
Numbers of
the form 𝑛 = 𝑘⁶
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?
See
SolutionsIt is given that each passcode is 5 characters long.
Each
character can be alphanumeric, i.e., A–Z (26 letters) and 0–9 (10 digits)
So, total
choices per position = 26 + 10 = 36
Now, each of
the 5 positions in the code can be filled in 36 ways.
So, total
number of such codes = 36 × 36 × 36 × 36 × 36 = 36⁵
Thus, 36⁵ =
60,466,176
Therefore,
60,466,176 codes are possible.
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?
(ii) 20⁹
(ii) 10¹¹ (iii) 10¹⁰
(iv) 10¹⁸
(v) 2 × 10⁹ (vi) 10⁹ + 10⁹
See
SolutionsGiven that:
Population
of sheep = 10⁹
Population
of goat = 10⁹
So, the
total population of sheep and goat:
= 10⁹ + 10⁹
= 2 × 10⁹
Hence, the
obtions (v) and (vi) are correct.
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.
See
Solutions(i) World population ≈ 8 × 10⁹ (approx.)
Each person
has 30 clothes
So, total
clothes:
30 × 8 × 10⁹
= 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.
See
Solutions(ii) Given that 100 million bee colonies, 50,000 bees per colony.
So, the
number of bee colonies
= 100
million
= 1 × 10⁸
Each colony
has bees = 50,000
= 5 × 10⁴
Therefore,
the total bees:
(Number of
bee colonies)×(Number of bees in each colony)
= (1 × 10⁸)
× (5 × 10⁴)
= 5 × 10⁸ ×
10⁴
= 5 × 10¹²
bees
(iii) The
human body has about 38 trillion bacterial cells. Find the bacterial population
residing in all humans in the world.
See
Solutions(iii) World population ≈ 8 × 10⁹ (approx.)
The human
body has bacterial cells = 38 trillion
38 trillion
= 3.8 × 10¹³
So, total
Bacteria = (8 × 10⁹) × (3.8 × 10¹³)
= 30.4 × 10⁹
× 10¹³
= 30.4 ×
10²²
= 3.04 ×
10²³
(iv) Total
time spent eating in a lifetime in seconds.
See
Solutions(iv) Average eating time per day = 1.5 hours
= 1.5 × 60 ×
60 = 5400 seconds
Average
lifespan = 70 years
Number of
days in 70 years = 70 × 365 = 25,550 days
Total eating
time in lifespan:
= (5400) ×
25,550
=
137,970,000 seconds
= 1.3797 ×
10⁸ seconds
14. What
was the date 1 arab/1 billion seconds ago?
See
SolutionsIn the Indian number system:
1 arab =
1,00,00,00,000 = 10⁹ = 1 billion seconds
Now, convert
10⁹ seconds into years.
We know
that:
1 minute =
60 seconds
1 hour = 60
minutes = 3600 seconds
1 day = 24
hours = 86,400 seconds
1 year ≈
365.25 days (to include leap years)
So, seconds
in a year
= 365.25 ×
24 × 60 × 60
= 31,557,600
Therefore,
the number of years in 10⁹ seconds
=
10⁹/31,557,600
=
1,00,00,00,000/31,557,600
= 31.7 years
Let’s assume
today is July 29, 2025.
Go back 31.7
years
⇒ approximately 31 years and 8.5
months
So, subtract
31 years
⇒ July 29, 1994
Go back
approx. 8.5 months
⇒ Around mid-November 1993
That means 1
arab seconds ago, it was around November 1993.
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