NCERT Exemplar Class 8 Maths Solutions for Chapter 8 - Exponents And Powers

NCERT Exemplar Class 8 Maths Chapter 8 Exponents and Powers, made available here for students to prepare for exams. These exemplars solutions are designed by our experts in accordance with CBSE syllabus(2018-2019), which covers all the topics of standard 8, Maths chapter 8.

Exponents and powers, is one of the most important chapters for class 8 CBSE. In this chapter, the students will learn about the laws of exponents and powers of negative exponents. To understand the chapter in a better way students are advised to solve the exemplars.

Class 8 Maths NCERT Exemplars For Exponents and Powers

To understand the concepts present in each chapter of Maths as well as Science subject, solve NCERT exemplars for class 8 on all the topics covered under syllabus. Go through the topics based on which exemplars are given for chapter 8.

  • Powers with Negative Exponents
  • Laws of Exponents
  • Expressing small numbers in standard form, using Exponents

Students are also made available with exemplar books, NCERT solutions for 8th standard Maths and question papers herewith BYJU’S to prepare for exams. It is advisable to solve sample papers and previous year question papers which gives an idea of types of questions asked in the board exam from chapter exponents and powers.

NCERT Exemplar Class 8 Maths Solutions Chapter 8 Exponents And Powers:-Download PDF Here

NCERT Exemplar Solutions Class 8 Maths Chapter 8
NCERT Exemplar Solutions Class 8 Maths Chapter 8 1
NCERT Exemplar Solutions Class 8 Maths Chapter 8 2
NCERT Exemplar Solutions Class 8 Maths Chapter 8 3
NCERT Exemplar Solutions Class 8 Maths Chapter 8 4
NCERT Exemplar Solutions Class 8 Maths Chapter 8 5
NCERT Exemplar Solutions Class 8 Maths Chapter 8 6
NCERT Exemplar Solutions Class 8 Maths Chapter 8 7

Access NCERT Exemplar Class 8 Maths Chapter 8 Solutions

In questions 1 to 33, out of the four options, only one is correct. Write the correct answer.

1. In 2n, n is known as:

(a) Base (b) Constant (c) exponent (d) Variable

Solution:

(c) Exponent

Explanation: 2 is the rational number which is the base here and n is the power of 2. Hence, it is an exponent.

2. For a fixed base, if the exponent decreases by 1, the number becomes:

(a) One-tenth of the previous number.

(b) Ten times of the previous number.

(c) Hundredth of the previous number.

(d) Hundred times of the previous number.

Solution:

(a) One-tenth of the previous number

Explanation: Suppose for 106, when the exponent is decreased by 1, it becomes 105. Hence, 105/106 = 1/10.

3. 3-2 can be written as:

(a) 32 (b) 1/32 (c) 1/3-2 (d) -2/3

Solution:

(b) 1/32

Explanation: By the law of exponent we know: a-n = 1/an.

Hence, 3-2=1/32

4. The value of 1/(4)-2 is:

(a) 16 (b) 8 (c) 1/16 (d) 1/8

Solution:

(a) 16

Explanation: 1/(4)-2 = 1/(1/42) = 42 = 16

5. The value of 35 ÷ 3-6 is:

(a) 35 (b) 3-6 (c) 311 (d) 3-11

Solution:

(c) 311

Explanation: By the law of exponents, we know,

am/an=am-n

Hence, 35 ÷ 3-6 = 35-(-6) = 311

6. The value of (2/5)-2 is:

(a) 4/5 (b) 4/25 (c) 25/4 (d) 5/2

Solution:

(c) 25/4

Explanation: By the law of exponent we know: a-n = 1/an.

Hence, (2/5)-2 = 1/(2/5)2 = 1/(4/25) = 25/4

7. The reciprocal of (2/5)-1 is:

(a) 2/5 (b) 5/2 (c) –5/2 (d) –2/5

Solution:

(b) 5/2

Explanation: By the law of exponent we know: a-n = 1/an.

Hence, (2/5)-1=1/(2/5)=5/2

8. The multiplicative inverse of 10-100 is

(a) 10 (b) 100 (c) 10100 (d) 10-100

Solution:

(c) 10100

Explanation: By the law of exponent we know: a-n = 1/an.

So, 10-100 = 1/10100

The multiplicative inverse for any integer a is 1/a, such that;

a x 1/a = 1

Hence, the multiplicative inverse for 1/10100 is 10100

as, 1/10100 x 10100 = 1

9. The value of (–2)2×3-1 is

(a) 32 (b) 64 (c) – 32 (d) – 64

Solution:

(c) – 32

Explanation: (–2)2×3-1=(-2)6-1=(-2)5=-32

10. The value of (-2/3)4 is equal to:

(a) 16/81 (b) 81/16 (c) -16/81 (d) 81/ −16

Solution:

(a) 16/81

Explanation: (-2/3)4 = (-2/3)(-2/3)(-2/3)(-2/3) = 16/81

11. The multiplicative inverse of (-5/9)-99 is:

(a) (-5/9)99 (b) (5/9)99 (c) (9/-5)99 (d) (9/5)99

Solution:

(-5/9)99

Explanation: Take the reference of Q.8 mentioned above.

12. If x be any non-zero integer and m, n be negative integers, then xm × xn is equal to:

(a) xm (b) xm+n (c) xn (d) xm-n

Solution:

(b) xm+n (By the law of exponents)

13. If y be any non-zero integer, then y0 is equal to:

(a) 1 (b) 0 (c) – 1 (c) Not defined

Solution:

(a) 1 (By the law of exponent)

14. If x be any non-zero integer, then x-1 is equal to

(a) x (b) 1/x (c) – x (c) -1/x

Solution:

(b) 1/x (By the law of exponents)

15. If x be any integer different from zero and m be any positive integer, then x-m is equal to:

(a) xm (b) –xm (c) 1/xm (d) -1/xm

Solution:

(c) 1/xm (By the law of exponents)

16. If x be any integer different from zero and m, n be any integers, then (xm)n is equal to:

(a) xm+n (b) xmn (c) xm/n (d) xm-n

Solution:

(b) xmn (By the law of exponents)

17. Which of the following is equal to (-3/4)-3?

(a) (3/4)-3 (b) – (3/4)-3 (c) (4/3)3 (d) (-4/3)3

Solution:

(d) (-4/3)3

Explanation: (-3/4)-3 = 1/(-3/4)3 = (-4/3)3

(By the law of exponents: a-n = 1/an)

18. (-5/7)-5 is equal to:

(a) (5/7)-5 (b) (5/7)5 (c) (7/5)5 (d) (-7/5)5

Solution:

(d) (-7/5)5

Explanation: (-5/7)-5=1/(-5/7)5=(-7/5)5

(By the law of exponents: a-n = 1/an)

19. (-7/5)-1 is equal to:

(a) 5/7 (b) – 5/7 (c) 7/5 (d) -7/5

Solution:

(b) – 5/7

Explanation: (-7/5)-1= 1/(-7/5) = -5/7

20. (–9)3 ÷ (–9)8 is equal to:

(a) (9)5 (b) (9)-5 (c) (– 9)5 (d) (– 9)-5

Solution:

(d) (– 9)-5

Explanation: (–9)3 ÷ (–9)8 = (-9)3-8 = (-9)-5

(By the law of exponents: am ÷ an=am-n)

21. For a non-zero integer x, x7 ÷ x12 is equal to:

(a) x5 (b) x19 (c) x-5 (d) x-19

Solution:

(c) x-5

Explanation: x7 ÷ x12 = x7-12 = x-5

(By the law of exponents: am ÷ an=am-n)

22. For a non-zero integer x, (x4)-3 is equal to:

(a) x12 (b) x-12 (c) x64 (d) x-64

Solution:

(b) x-12

Explanation: (x4)-3 = x4×(-3) = x-12

(By the law of exponents: (am)n=amn)

23. The value of (7-1 – 8-1) -1 – (3-1 – 4-1)-1 is:

(a) 44 (b) 56 (c) 68 (d) 12

Solution:

(a) 44

Explanation: (7-1 – 8-1) -1 – (3-1 – 4-1)-1

= (1/7-1/8) -1 – (1/3-1/4)-1

= (1/56) -1 – (1/12) -1

= 56 – 12 = 44

24. The standard form for 0.000064 is

(a) 64 × 104 (b) 64 × 10-4 (c) 6.4 × 105 (d) 6.4 × 10-5

Solution:

(d) 6.4 × 10-5

25. The standard form for 234000000 is

(a) 2.34 × 108 (b) 0.234 × 109 (c) 2.34 × 10-8 (d) 0.234×10-9

Solution:

(a) 2.34 × 108

Explanation: 234000000 = 234 × 106 = 2.34 × 102 × 106 = 2.34 × 108

26. The usual form for 2.03 × 10-5

(a) 0.203 (b) 0.00203 (c) 203000 (d) 0.0000203

Solution:

(d) 0.0000203

27. (1/10)0 is equal to

(a) 0 (b) 1/10 (c) 1 (d) 10

Solution:

(c) 1 Since, a0 = 1 (by law of exponent)

28. (3/4)5 ÷(5/3)5 is equal to

(a) (3/4÷5/3)5 (b) (3/4 ÷ 5/3)1 (c) (3/4 ÷ 5/3)0 (d) (3/4 ÷ 5/3)10

Solution:

(a) (3/4÷5/3)5

(By law of exponent: (a)m÷(b)m = (a÷b)m

29. For any two non-zero rational numbers x and y, x4 ÷ y4 is equal to

(a) (x ÷ y)0 (b) (x ÷ y)1 (c) (x ÷ y)4 (d) (x ÷ y)8

Solution:

(c) (x ÷ y)4

(By law of exponent: (a)m÷(b)m = (a÷b)m)

30. For a non-zero rational number p, p13 ÷ p8 is equal to

(a) p5 (b) p21 (c) p-5 (d) p-19

Solution:

(a) p5

(By law of exponent: (a)m÷(a)n = (a)m-n)

31. For a non-zero rational number z, (z-2)3 equal to

(a) z6 (b) z-6 (c)z1 (d) z4

Solution:

(b) z-6

(By the law of exponents: (am)n=amn)

32. Cube of -1/2 is

(a) 1/8 (b) 1/16 (c) -1/8 (d) -1/16

Solution:

(c) -1/8

Explanation: Cube of -1/2 = (-1/2)3

= (-1/2) × (-1/2) × (-1/2) = -1/8

33. Which of the following is not the reciprocal of (2/3)4?

(a) (3/2)4 (b) (3/2)-4 (c) (2/3)-4 (d) 34/24

Solution:

(b) (3/2)-4

Explanation: (2/3)4 = 1/(2/3)-4 = (3/2) -4

In questions 34 to 50, fill in the blanks to make the statements true.

34. The multiplicative inverse of 1010 is 10-10

35. a3 × a-10 = a3+(-10) = a3-10 = a-7

36. 50 = 1

37. 55 × 5-5 = 55+(-5) = 55-5 = 50 = 1

38. The value of (1/23)2 equal to (1/26).

Explanation: (1/23)2 = (1/2)3×2 = (1/2)6

39. The expression for 8-2 as a power with the base 2 is (2)-6

Explanation: 8-2 = (2 × 2 × 2)-2 = (23)-2

40. Very small numbers can be expressed in standard form by using negative exponents.

41. Very large numbers can be expressed in standard form by using positive exponents.

42. By multiplying (10)5 by (10)-10 we get 10-5

Explanation: (10)5 × (10)-10 = 105+(-10) = 105-10 = 10-5

43. [(2/13)-6÷(2/13)3]3 × (2/13)-9 = (2/13)-36

Explanation: [(2/13)-6÷(2/13)3]3 × (2/13)-9

= [(2/13)-6-3]3 × (2/13)-9

= [(2/13)-9]3 × (2/13)-9

= (2/13)-9×3 × (2/13)-9

= (2/13)-27 × (2/13)-9

= (2/13)-27-9

= (2/13)-36

44. Find the value [4-1 +3-1 + 6-2]-1

Solution: [4-1 +3-1 + 6-2]-1

= (1/4+1/3+1/62)-1

= [(9+12+1)/36]-1

= (22/36)-1

= (36/22)

45. [2-1 + 3-1 + 4-1]0 = 1 (Using law of exponent, a0=1)

46. The standard form of (1/100000000) is 1.0 × 10-8

Explanation: (1/100000000) = 1/1×108 = 1.0 × 10-8

47. The standard form of 12340000 is 1.234 × 107

Explanation: 12340000 = 1234 × 104 = 1.234 × 10 3 × 104 = 1.234 × 107

48. The usual form of 3.41 × 106 is 3410000.

Explanation: 3.41 × 106 = 3.41 × 10 × 10 × 10 × 10 × 10 × 10

= 341 × 10 × 10 × 10 × 10

= 3410000

49. The usual form of 2.39461 × 106 is 2394610.

Explanation: 2.39461 × 106 = 2.39461 × 10 × 10 × 10 × 10 × 10 × 10

= 239461 × 10

= 2394610

50. If 36 = 6 × 6 = 62, then 1/36 expressed as a power with the base 6 is 6-2.

Explanation: 36 = 6 × 6 = 62

1/36 = 1/62 = 6-2

 

Download exemplars solutions for all chapters of Maths covered in 8th class by clicking here. Also, download BYJU’S-The Learning App and get personalized videos, explaining the concepts of exponents and powers and many maths-related topics, experiencing a new way of learning to understand the concepts easily.

Also Check

Leave a Comment

Your email address will not be published. Required fields are marked *