NCERT Solutions: Exponents & Powers- 1

# NCERT Solutions for Class 8 Maths - Exponents and Powers- 1

### Exercise 11.1

Q1: Find the value of:

(i) 2

(ii) 9

(iii) 11

(iv) 5

Ans:

(i) 26 = 2 x 2 x 2 x 2 x 2 x 2 = 64
(ii) 93 = 9 x 9 x 9 = 729
(iii) 11= 11 x 11= 121
(iv) 54 = 5 x 5 x 5 x 5 = 625

Q2: Express the following in exponential form:

(i) 6 x 6 x 6 x 6
(ii) t x t
(iii) b x b x b x b
(iv) 5 x 5 x 7 x 7 x 7
(v) 2 x 2 x a x a
(vi) a x a x a x c x c x c x c x d

Ans:

(i) 6 x 6 x 6 x 6 = 64
(ii) t x t = t2
(iii) b x b x b x b = b4
(iv) 5 x 5 x 7 x 7 x 7 = 52 x 73
(v) 2 x 2 x a x a = 2x a2
(vi) a x a x a x c x c x c x c x d = a3 x c4 x d

Q3: Express each of the following numbers using exponential notation:

(i) 512
(ii) 343
(iii) 729
(iv) 3125

Ans:

(i) 512

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 29

 2            512 2 256 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1

(ii) 343
343 = 7 x 7 x 7 = 73

 7            343 7 49 7 7 1

(iii) 729
729 = 3 x 3 x 3 x 3 x 3 x 3 = 36

 3           729 3 243 3 81 3 27 3 9 3 3 1

(iv) 3125
3125 = 5 x 5 x 5 x 5 x 5

 5 3125 5 625 5 125 5 25 5 5 1

Q4: Identify the greater number, wherever possible, in each of the following:

(i)  43 and 34
(ii)  53 or 35
(iii) 28 or 82
(iv) 100or 2100
(v) 210 or 102

Ans:

(i) 43 = 4 x 4 x 4 = 64
34 = 3 x 3 x 3 x 3 = 81
Since 64 < 81
Thus, 34 is greater than 43.

(ii) 53 = 5 x 5 x 5 = 125
35 = 3 x 3 x 3 x 3 x 3 = 243
Since, 125 < 243
Thus, 34 is greater than 53.

(iii) 28= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256
82 = 8x8 = 64
Since, 256 > 64
Thus, 28 is greater than 82.

(iv) 1002 = 100 x 100 = 10,000
2100 = 2 x 2 x 2 x 2 x 2 x.......14 times x....x 2 = 16,384 x....x 2
Since, 10,000 < 16,384 x....x2
Thus, 2100 is greater than 1002.

(v)  210 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1,024
102 = 10x10 = 100
Since, 1,024 > 100
Thus, 210 >102

Q5: Express each of the following as product of powers of their prime factors:

(i) 648

(ii) 405

(iii) 540

(iv) 3,600

Ans:

(i)    648 = 23 x 34

 2 648 2 324 2 162 3 81 3 27 3 9 3 3 1

(ii) 405 = 5 x 34

 5 405 3 81 3 27 3 9 3 3 1

(iii) 540 = 22 x 33 x 5

 2 540 2 270 3 135 3 45 3 15 5 5 1

(iv) 3,600 = 24 x 3x 52

 2 3600 2 1800 2 900 2 450 3 225 3 75 5 25 5 5 1

Q6: Simplify:

(i)  2 x 103
(ii) 72 x 22
(iii) 23 x 5
(iv) 3 x 44
(v) 0 x 102
(vi) 5x 33
(vii) 24 x 32
(viii) 32 x 104

Ans:

(i)  2 x 103  = 2 x 10 x 10 x 10  = 2,000
(ii)  72 x 22 = 7 x 7 x 2 x 2  = 196
(iii) 23 x 5 = 2 x 2 x 2 x 5  = 40
(iv) 3 x 44 = 3 x 4 x 4 x 4 x 4  = 768
(v) 0 x 102 = 0 x 10 x 10  = 0
(vi) 53 x 33 = 5 x 5 x 3 x 3 x 3  = 675
(vii) 2x 32  = 2 x 2 x 2 x 2 x 3 x 3  = 144
(viii) 32 x 104 = 3 x 3x 10x10x 10x10  =  90,000

Q7: Simplify:

(i) (-4)3
(ii) (-3)x(-2)3
(iii) (-3)x (-5)2
(iv) (-2)3 x (-10)3

Ans:

(i)  (-4)3 = (-4) x (-4) x (-4) = -64
(ii) (-3) x (-2)3 =(-3) x (-2) x (-2) x (-2) = 24
(iii) (-3)2 x (-5)2 = (-3) x (-3) x (-5) x (-5) = 225
(iv) (-2)3x (-10)3 =(-2) x (-2) x (-2) x (-10) x (-10) x (-10)

Question 8: Compare the following numbers:

(i) 2.7 x 1012; 1.5 x 108

(ii) 4 x 1014; 3 x 1017

(i) 2.7 x 1012 and 1.5 x 108
On comparing the exponents of base 10,
2.7 x 1012 > 1.5x108

(ii) 4 x 1014 and 3 x 1017
On comparing the exponents of base 10,
4 x 1014 < 3 x 1017

## Exercise 11.2

Question 1: Using laws of exponents, simplify and write the answer in exponential form:

(i) 32 x 34 x 38
(ii) 615 ÷ 610
(iii) a3 x a
(iv) 7x x 72
(v) (52)2 ÷ 53
(vi) 25 x 55
(vii) a4 x b4
(viii) (34)3
(ix) [220 ÷ 215) x 23

(x) 8t x 82

(i)
(ii)
(iii)
(iv)
(v)

(vi)
(vii)
(viii)
(ix)

(x)

Q2: Simplify and express each of the following in exponential form:

(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)

Ans:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(i) 10 x 1011 = 10011
(ii)  23 > 52
(iii) 23x 32 = 6s
(iv) 30 = (1000)0

Ans:

(i) 10x1011 = 10011
L.H.S. 101+11 = 1012    and R.H.S. (102)11=1022
Since, L.H.S. ≠ R.H.S.
Therefore, it is false.

(ii) 23 > 52
L.H.S 23 = 8    and R.H.S. 52 = 25
Since, L.H,S. is not greater than R.H.S.
Therefore, it is false,

(iii) 23 x32 = 65
L.H.S. 23 x 32 = 8 x 9 = 72    and R.H.S. 65 =7.776
Since, L.H.S. ≠ R.H.S.
Therefore, it is false.

(iv) 30= (1000)0
L.H.S. 30 = 1    and R.H.S. (1000)0 = 1
Since, L.H.S. = R.H.S.
Therefore, it is true.

Q4: Express each of the following as a product of prime factors only in exponential form:
(i) 108 x 192
(ii) 270
(iii) 729 x 64
(iv) 768
Ans:

Question 5: Simplify:

(i)
(ii)
(iii)

(i)

(ii)

(iii)

## Question 1: Write the following numbers in the expanded form:

279404,

3006194,

2806196,

120719,

20068

(i) 2,79,404

= 2,00,000 + 70,000 + 9,000 + 400 + 00 + 4
= 2x100000 + 7x 10000+ 9x1000 + 4 x 100+ 0x 10 + 4x1
= 2x105+7x104 +9x103 + 4 x102+ 0x101+4x100

(ii) 30,06,194

= 30,00,000 + 0 + 0 + 6,000 + 100 + 90 + 4
= 3 x 1000000 + 0 x 100000 + 0 x 10000 + 6 x 1000 + 1x100 + 9x10 + 4x1
= 3x106 + 0x105 +0x104 +6 x103+1x102 +9x10+4x100

(iii) 28,06,196

= 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6
= 2x1000000 + 8x 100000+ 0x 10000 +6 x 1000 + 1 x 100 + 9 x 10 + 6 x 1
= 2 x106+8x105+0x104 + 6x103+1x102+9x10+6x100

(iv) 1,20,719

= 1,00,000 + 20,000 + 0 + 700 + 10 + 9
= 1 x 100000 + 2 x 10000 + 0x 1000 + 7x100 + 1x 10+9x1
= 1 x 105 + 2 x 10+ 0 x 103 + 7 x 102 + 1 x 101 + 9 x 100

(v) 20,068

= 20,000 + 00 + 00 + 60 + 8

= 2 x 10000 + 0 x 1000 + 0 x 100 + 6 x 10 + 8 x 1

Question 2:

Find the number from each of the following expanded forms:

(a) 8 x 104 + 6 x 103 + 0 x 102 + 4 x 101 + 5 x 100

(b) 4 x 105 + 5 x 103 + 3 x 102 + 2 x 100

(c) 3 x 104 + 7 x 102 + 5 x 100 (d) 9 x 105 + 2 x 102 + 3 x 101

(a) 8 x 104 + 6 x 103 + 0 x 102 + 4 x 101 + 5 x 100
= 8 x 10000 + 6x 1000 + 0 x 100 + 4 x 10 + 5 x 1
= 80000 + 6000 + 0 + 40+5
= 86,045

(b) 4 x 105 + 5 x 103 + 3 x 102 + 2 x 100
= 4 x 100000 + 0 x 10000 + 5 x 1000 + 3 x 100 + 0 x 10 + 2 x 1
= 400000 + 0 + 5000 + 3000 + 0 + 2
= 4,05,302

(c) 3 x 104 + 7 x 10z + 5 x 100
= 3 x 10000 + 0 x 1000 + 7 x 100 + 0 x 10 + 5 x 1
= 30000 + 0 + 700 + 0 + 5
= 30,705

(d) 9x 105 + 2 x 102 + 3 x 101
= 9 x 100000 + 0 x 10000 + 0 x 1000 + 2 x 100 + 3 x 10 + 0 x 1
= 900000 + 0 + 0 + 200 + 30 + 0
= 9,00,230

Question 3:

Express the following numbers in standard form:

(i) 5,00,00,000

(ii) 70,00,000

(iii) 3,18,65,00,000

(iv) 3,90,878

(v) 39087.8

(vi) 3908.78

(i) 5,00,00,000 = 5 x 1,00,00,000 = 5 x 107

(ii) 70,00,000 = 7 x 10,00,000 = 7 x 106

(iii) 3,18,65,00,000 = 31865 x 100000

= 3.1865 x 10000 x 100000 = 3.1865 x 109

(iv) 3,90,878 = 3.90878 x 100000 = 3.90878 x 105

(v) 39087.8 = 3.90878 x 10000 = 3.90878 x 104

(vi) 3908.78 = 3.90878 x 1000 = 3.90878 x 103

Question 4:

Express the number appearing in the following statements in standard form:

(a) The distance between Earth and Moon is 384,000,000 m.

(b) Speed of light in vacuum is 300,000,000 m/s.

(c) Diameter of Earth id 1,27,56,000 m.

(d) Diameter of the Sun is 1,400,000,000 m.

(e) In a galaxy there are on an average 100,000,000,0000 stars.

(f) The universe is estimated to be about 12,000,000,000 years old.

(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.

(i) The Earth has 1,353,000,000 cubic km of sea water.

(j) The population of India was about 1,027,000,000 in march, 2001.

(a) The distance between Earth and Moon = 384,000,000 m
= 384x 1000000 m
= 3.84 x 100 x 1000000
= 3.84x108 m

(b) Speed of light in vacuum = 300,000,000 m/s
= 3 x 100000000 m/s
= 3x108 m/s

(c) Diameter of the Earth = 1,27,50,000 m
= 12756 x 1000 m
= 1.2756 x 10000 x 1000 m
= 1.2756x107 m

(d) Diameter of the Sun = 1,400,000,000 m
= 14 x 100,000,000 m
= 1.4 x 10 x 100,000,000 m
= 1.4 x 109 m

(e) Average of Stars = 100, 000, 000,000
= 1 x 100,000,000,000
= 1x1011

(f) Years of Universe = 12,000,000,000 years
= 12 x 1000,000,000 years
= 1.2 x 10 x 1000,000,000 years
= 1.2 x 1010 years

(g) Distance of tlie Sun from the centre of the Milky Way Galaxy
= 300,000,000,000,000,000,000 m
= 3 x 100,000,000,000,000,000,000 m
= 3 x 1020 m

(h) Number of molecules in a drop of water weighing 1.8 gm
= 60,230,000,000,000,000,000,000
= 6023 x 10,000,000,000,000,000,000
= 6,023 x 1000 x 10,000,000,000,000,000,000
= 6.023 x1022

(i) The Earth has Sea water = 1,353,000,000 km3
= 1,353 x 1000000 km3
= 1,353 x 1000 x 1000,000 km3
= 1.353x109 km3

(j) The population of India = 1,027,000,000
= 1027x1000000 = 1,027x1000x1000000
= 1.027x109

The document NCERT Solutions for Class 8 Maths - Exponents and Powers- 1 is a part of the Class 7 Course Mathematics (Maths) Class 7.
All you need of Class 7 at this link: Class 7

## Mathematics (Maths) Class 7

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## FAQs on NCERT Solutions for Class 8 Maths - Exponents and Powers- 1

 1. What is the expanded form of a number?
Ans. The expanded form of a number is a way to write a number by showing the value of each digit in the number. For example, the number 345 in expanded form is 300 + 40 + 5.
 2. How do you write a number in expanded form?
Ans. To write a number in expanded form, you break down the number into its individual place values (hundreds, tens, ones) and write each digit with its corresponding place value. For example, the number 456 in expanded form is 400 + 50 + 6.
 3. Why is it important to understand expanded form?
Ans. Understanding expanded form helps in developing a deeper understanding of place value and how numbers are composed of different place values. It also helps in performing operations like addition and subtraction more efficiently.
 4. Can you convert any number into expanded form?
Ans. Yes, any whole number can be converted into expanded form by breaking it down into its place values. This is a fundamental concept in mathematics and is applicable to numbers of any size.
 5. How can expanded form help in simplifying mathematical calculations?
Ans. By writing numbers in expanded form, it becomes easier to perform operations like addition and subtraction as you can clearly see the value of each digit. This can help in avoiding errors and making calculations more straightforward.

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