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

NCERT Exemplar Solutions Class 8 Maths Chapter 8 – Free PDF Download

The NCERT Exemplar Class 8 Maths Chapter 8 Exponents and Powers is available here for students to prepare for exams. These Exemplars solutions are designed by our subject experts in accordance with the CBSE syllabus (2023-2024), which covers all the topics of Class 8 Maths Chapter 8. Exponents and Powers is one of the most important chapters for Class 8 CBSE. In this chapter, students will learn about the laws of exponents and the powers of negative exponents. To understand the chapter in a better way, students are advised to solve the exemplars. To understand the concepts present in each chapter of Maths, as well as Science subjects, solve NCERT Exemplars for Class 8 on all the topics covered under the syllabus. Go through the topics based on which the Exemplars are given for Chapter 8.

• Powers with negative exponents
• Laws of exponents
• Expressing small numbers in standard form, using exponents

Students can also access exemplar books, NCERT Solutions for 8th standard Maths, and question papers from BYJU’S to prepare for exams. It is advisable to solve sample papers and previous years’ question papers to get an idea of the types of questions asked in the board exam from the chapter Exponents and Powers.

NCERT Exemplar Class 8 Maths Solutions for Chapter 8 Exponents and Powers:-Download the PDF Here

Access Answers to NCERT Exemplar Solutions for Class 8 Maths Chapter 8

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,

a^m/aⁿ=a^m-n

Hence, 3⁵ ÷ 3^-6 = 3^5-(-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 x^m × xⁿ is equal to:

(a) x^m (b) x^m+n (c) xⁿ (d) x^m-n

Solution:

(b) x^m+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) x^m (b) –x^m (c) 1/x^m (d) -1/x^m

Solution:

(c) 1/x^m (By the law of exponents)

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

(a) x^m+n (b) x^mn (c) x^m/n (d) x^m-n

Solution:

(b) x^mn (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: a^m ÷ aⁿ=a^m-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 = x^7-12 = x^-5

(By the law of exponents: a^m ÷ aⁿ=a^m-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 = x^4×(-3) = x^-12

(By the law of exponents: (a^m)ⁿ=a^mn)

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: (a^m)ⁿ=a^mn)

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 = a^3+(-10) = a^3-10 = a^-7

36. 5⁰ = 1

37. 5⁵ × 5^-5 = 5^5+(-5) = 5^5-5 = 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 = 10^5+(-10) = 10^5-10 = 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 Class 8 by clicking here. Also, download BYJU’S – The Learning App and get personalised video lessons explaining the concepts of exponents and powers and other Maths-related topics, experiencing a new way of learning to understand the concepts easily.

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