NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

 

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

 

Introduction:

In this chapter, Exponents and Power we will learn that very large numbers are difficult to read, understand and compare, we use exponents.

We can write large numbers in a shorter form using Exponents.

For example - 10,000 = 10×10×10×10 = 104

The short notation  104 stands for the product 10×10×10×10. Here, '10' is called the base and '4' the exponent. The number 104 is read as 10 raised to the power of 4 or simply as fourth power of 10. is called the exponential form of 10,000.

 

NCERT Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

NCERT Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.2

NCERT Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.3

Class 7 Maths Exercise 11.1 (Page-173)

 

Q1. Find the value of :

(i) \(\displaystyle {{2}^{6}}\)

(ii) 9³

(iii) 11²

(iv) \(\displaystyle {{5}^{4}}\)

Solution: 

(i) 2×2×2×2×2×2 = 64

(ii) 9×9×9 = 729

(iii) 11×11 = 121

(iv) 5×5×5×5 = 625

Q2. Exress the following in exponential form:

(i) 6 × 6 × 6 × 6

(ii) t × t

(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) \(\displaystyle {{6}^{4}}\)

(ii) \(\displaystyle {{t}^{2}}\)

(iii) \(\displaystyle {{b}^{4}}\)

(iv) \(\displaystyle {{5}^{2}}\) × \(\displaystyle {{7}^{3}}\)

(v) \(\displaystyle {{2}^{2}}\) × \(\displaystyle {{a}^{2}}\)

(vi) \(\displaystyle {{a}^{3}}\) × \(\displaystyle {{c}^{4}}\) × d¹

 

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

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Solution: (i) 

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

2 ×2×2×2×2×2×2×2×2 = \(\displaystyle {{2}^{9}}\)

(ii)      NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

7 × 7× 7 = 7³

(iii)  NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

3×3×3×3×3×3 = \(\displaystyle {{3}^{6}}\)

 

(iv) NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

5×5×5×5×5 = \(\displaystyle {{5}^{5}}\)

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

(i) 43 or 34

(ii) 53  or 35

(iii)  28 or 82

(iv) 100² or 2100

(v)210  or 102

Solution: 

(i) 4³ = 4×4×4 = 64

34= 3×3×3×3= 81

= 64< 81. Hence, 3is a greater number.

ii) 5³ = 5×5×5 = 125

35 = 3×3×3×3×3 = 243

= 125< 243 Hence, 35  is greater number.

iii) 28 = 2×2×2×2×2×2×2×2 = 256

82 = 8×8 = 64

= 256> 64

Hence,  28 is greater number.

(iv) 100² = 100×100 = 10000

2100= splitting 2100  into 210
210 = 2×2×2×2×2×2×2×2×2×2= 1024

s0, 2100  = 1024×1024×1024×1024×1024×1024×1024×1024×1024×1024

so,  100²<2100  

Hence, 2100   is greater number.

(v) 210= 2×2×2×2×2×2×2×2×2×2= 1024

10²= 10×10= 100

1024> 100

hence, 210 is greater number.

 

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

(i) 648

(ii) 405

(iii) 540

(iv) 3600

Solution: 

(i) 648

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

648 = 2×2×2×3×3×3×3

= 2³× \(\displaystyle {{3}^{4}}\)

ii) 405

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

405 = 3×3×3×3×5

405 = \(\displaystyle {{3}^{4}}\)

iii) 540

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

540 = 2×2×3×3×3×5

540= 2² × 3³ × 5

iv) 3600

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

3600 = 2×2×2×2×3×3×5×5

3600= \(\displaystyle {{2}^{4}}\) × 3² × 5²

 

Q6. Simplify:

(i) 2 × 10³

(ii) 7² × 2²

(iii) 2³ × 5

(iv) 3 × \(\displaystyle {{4}^{4}}\)

(v) 0 ×10²

(vi) 5² × 3³

(vii) \(\displaystyle {{2}^{4}}\) × 3²

(viii) 3² × \(\displaystyle {{10}^{4}}\)

Solution: 

i) 2 × 10× 10× 10= 2× 1000

= 2000

ii) 7×7×2×2

= 49×4

= 196

iii) 2×2×2×5

= 8×5

= 40

iv) 3×4×4×4×4

= 3×16×16

= 48×16

= 768

v) 0( if we multiply any number to zero answer will remain zero)

vi) 5×5×3×3×3

= 25×27

= 675

vii) 2×2×2×2×3×3

= 4×4×9

= 16×9

= 144

viii) 3×3×10×10×10×10

= 9× 10000

= 90000

Q7. Simplify:

(i) (-4)³

(ii) (-3)×(-2)³

(iii) (-3)²× (-5)²

(iv) (-2)³× (-10)³

Solution: 

i)  -4×-4×-4

= 16 ×-4

= - 64

ii) (-3) × -2 ×-2×-2

= (-3 )×(-8)

= 24

iii)  -3×-3×-5×-5

= 9×25

= 225

(iv) -2×-2×-2×-10×-10×-10

= -8)×(-1000)

= 8000

 

Q8. Compare the following numbers:

(i) 2.7 × 1012 ; 1.5× 108

(ii) 4× 1014; 3× 1017

Solution: 

i)  1012 >108

So, 2.7 × 1012 > 1.5× 108

ii)  1014< 1017

So, 4× 1014 <3× 1017

 

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