NCERT Exemplar Solutions for Class 7 Maths Chapter 11 Exponents and Powers

NCERT Exemplar Solutions for Class 7 Maths Chapter 11 Exponents and Powers are the most preferred study materials due to their easy-to-understand explanation of the concepts. Chapter 11 introduces the concepts related to exponents and powers. Referring to NCERT Exemplar Solutions for Class 7 Maths improves students’ conceptual understanding.

Exponents are the product of rational numbers multiplied several times by themselves. In this chapter, students will learn some important concepts, and they are

• Laws of Exponents
• Multiplying Powers with the Same Base
• Dividing Powers with the Same Base
• Multiplying Powers with the Same Exponents
• Dividing Powers with the Same Exponents

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Exercise Page: 337

In questions 1 to 22, there are four options, out of which one is correct. Write the correct one.

1. [(-3)²]³ is equal to

(a) (-3)⁸ (b) (-3)⁶ (c) (-3)⁵ (d) (-3)²³

Solution:-

(b) (-3)⁶

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

(a^m)ⁿ = a^mn

= -3⁶

2. For a non-zero rational number x, x⁸ ÷ x² is equal to

(a) x⁴ (b) x⁶ (c) x¹⁰ (d) x¹⁶

Solution:-

(b) x⁶

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ aⁿ = a^m–n , m>n

so, x⁸ ÷ x²

= x⁸/x²

= x^{8 – 2}

= x⁶

3. x is a non-zero rational number. Product of the square of x with the cube of x is equal to the

(a) second power of x (b) third power of x

(c) fifth power of x (d) sixth power of x

Solution:-

(c) fifth power of x

From the question, it is given that,

The square of x = x²

The cube of x = x³

As per the condition given in the question, x² × x³

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ = a^m+n

x² × x³ = x²⁺³

= x⁵

Therefore, fifth power of x.

4. For any two non-zero rational numbers x and y, x⁵ ÷ y⁵ is equal to

(a) (x÷y)¹ (b) (x÷y)⁰ (c) (x÷y)⁵ (d) (x÷y)¹⁰

Solution:-

(c) (x÷y)⁵

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ b^m = (a/b)^m

x⁵ ÷ y⁵ = (x/y)⁵ = (x ÷ y)⁵

5. a^m × aⁿ is equal to

(a) (a²)^mn (b) a^m–n (c) a^m+n (d) a^mn

Solution:-

(c) a^m+n

a^m × aⁿ is equal to a^{m + n}

6. (1⁰ + 2⁰ + 3⁰) is equal to

(a) 0 (b) 1 (c) 3 (d) 6

Solution:-

(c) 3

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a⁰ = 1

So, 1⁰ = 1, 2⁰ = 1, 3^o = 1

Then, (1 + 1 + 1)

= 3

7. Value of (10²² + 10²⁰)/10²⁰ is

(a) 10 (b) 10⁴² (c) 101 (d) 10²²

Solution:-

(c) 101

By splitting, (10²² + 10²⁰)/10²⁰ we get,

= (10²²/10²⁰) + (10²⁰/10²⁰)

By using the rule a^m ÷ aⁿ = a^m–n, m>n

= 10^{(22 – 20)} + 1

= 10² + 1

= 100 + 1

= 101

8. The standard form of the number 12345 is

(a) 1234.5 × 10¹ (b) 123.45 × 10²

(c) 12.345 × 10³ (d) 1.2345 × 10⁴

Solution:-

(d) 1.2345 × 10⁴

Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation.

9. If 2¹⁹⁹⁸ – 2¹⁹⁹⁷ – 2¹⁹⁹⁶ + 2¹⁹⁹⁵ = K2¹⁹⁹⁵, then the value of K is

(a) 1 (b) 2 (c) 3 (d) 4

Solution:-

(c) 3

2¹⁹⁹⁸ – 2¹⁹⁹⁷ – 2¹⁹⁹⁶ + 2¹⁹⁹⁵ = K2¹⁹⁹⁵

By cross multiplication we get,

(2¹⁹⁹⁸/2¹⁹⁹⁵) – (2¹⁹⁹⁷/2¹⁹⁹⁵) – (2¹⁹⁹⁶/2¹⁹⁹⁵) + (2¹⁹⁹⁵/2¹⁹⁹⁵) = K

By the rule, a^m ÷ aⁿ = a^m–n , m>n

K = (2^{1998 – 1995}) – (2^{1997 – 1995}) – (2^{1996 – 1995}) + (2^{1995 – 1995})

K = 2³ – 2² – 2¹ + 2⁰

K = 8 – 4 – 2 + 1

K = 3

10. Which of the following is equal to 1?

(a) 2⁰ + 3⁰ + 4⁰ (b) 2⁰ × 3⁰ × 4⁰

(c) (3⁰ – 2⁰) × 4⁰ (d) (3⁰ – 2⁰) × (3⁰ +2⁰)

Solution:-

(b) 2⁰ × 3⁰ × 4⁰

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a⁰ = 1

So, 2⁰ = 1, 3⁰ = 1, 4^o = 1

Then, (1 × 1 × 1)

= 1

11. In standard form, the number 72105.4 is written as 7.21054 × 10ⁿ where n is equal to

(a) 2 (b) 3 (c) 4 (d) 5

Solution:-

(c) 4

Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation.

72105.4 is written as 7.21054 × 10⁴

12. Square of (-2/3) is

(a) -2/3 (b) 2/3 (c) -4/9 (d) 4/9

Solution:-

(d) 4/9

Square of (-2/3) is = -2²/3²

= (-2 × -2)/(3 × 3)

= 4/9

13. cube of (-1/4) is

(a) -1/12 (b) 1/16 (c) -1/64 (d) 1/64

Solution:-

(c) -1/64

Cube of (-1/4) is = -1³/4³

= (-1 × -1 × -1)/(4 × 4 × 4)

= -1/64

14. Which of the following is not equal to (-5/4)⁴?

(a) (-5)⁴/4⁴ (b) 5⁴/(-4)⁴

(c) –(5⁴/4⁴) (d) (-5/4) × (-5/4) × (-5/4) × (-5/4)

Solution:-

(c) –(5⁴/4⁴)

15. Which of the following is not equal to 1 ?

(a) (2³ × 3³)/(4 × 18) (b) [(-2)³ × (-2)⁴] ÷ (-2)²

(c) (3⁰ × 5³)/(5 × 25) (d) 2⁴/(7⁰ + 3⁰)³

Solution:-

(d) 2⁴/(7⁰ + 3⁰)³

= 2⁴/(1 + 1)³

= 2⁴/2³

16. (2/3)³ × (5/7)³ is equal to

(a) ((2/3) × (5/7))⁹ (b) ((2/3) × (5/7))⁶

(c) ((2/3) × (5/7))³ (d) ((2/3) × (5/7))⁰

Solution:-

(c) ((2/3) × (5/7))³

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n.

a^m × b^m = (ab)^m

17. In standard form, the number 829030000 is written as K × 10⁸ where K is equal to (a) 82903 (b) 829.03 (c) 82.903 (d) 8.2903

Solution:-

(d) 8.2903

Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation.

829030000 is written as 8.2903 × 10⁸

18. Which of the following has the largest value?

(a) 0.0001 (b) 1/10000 (c) 1/10⁶ (d) (1/10⁶) ÷ 0.1

Solution:-

(a) 0.0001 and (b) 1/10000 = 0.0001

19. In standard form 72 crore is written as

(a) 72 × 10⁷ (b) 72 × 10⁸ (c) 7.2 × 10⁸ (d) 7.2 × 10⁷

Solution:-

(c) 7.2 × 10⁸

Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation.

72 crore is written as = 72,00,00,000 = 7.2 × 10⁸

20. For non-zero numbers a and b, (a/b)^m ÷ (a/b)ⁿ, where m>n, is equal to

(a) (a/b)^mn (b) (a/b)^{m + n} (c) (a/b)^{m – n} (d) ((a/b)^m)ⁿ

Solution:-

(c) (a/b)^{m – n}

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ aⁿ = a^m–n , m>n

21. Which of the following is not true?

(a) 3² > 2³ (b) 4³ = 2⁶ (c) 3³ = 9 (d) 25 > 5

Solution:-

(c) 3³ = 9

3 × 3 × 3 = 27

Therefore, 27 > 9

22. Which power of 8 is equal to 2⁶?

(a) 3 (b) 2 (c) 1 (d) 4

Solution:-

(b) 2

2⁶ = 2 × 2 × 2 × 2 × 2 × 2

= 64

8² = 8 × 8 = 64

In questions 23 to 39, fill in the blanks to make the statements true.

23. (-2)³¹ × (-2)¹³ = (-2)

Solution:-

(-2)³¹ × (-2)¹³ = (-2)⁴⁴

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ = a^m+n

(-2)³¹ × (-2)¹³ = (-2)^{31 + 13}

= -2⁴⁴

24. (-3)⁸ ÷ (-3)⁵ = (-3)

Solution:-

(-3)⁸ ÷ (-3)⁵ = (-3)³

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ aⁿ = a^m–n , m>n

(-3)⁸ ÷ (-3)⁵ = (-3)^8-5

= -3³

25. (11/15)⁴ × ( )⁵ = (11/15)⁹

Solution:-

(11/15)⁴ × (11/15)⁵ = (11/15)⁹

We know that,

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ = a^m+n

Therefore, (11/15)⁴ × (11/15)⁵ = (11/15)⁹

26. (-1/4)³ × (-1/4) = (-1/4)¹¹

Solution:-

(-1/4)³ × (-1/4)⁸ = (-1/4)¹¹

We know that,

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ = a^m+n

Let us assume missing power be y,

Then, (-1/4)³ × (-1/4)^y = (-1/4)¹¹

(-1/4)^y = (-1/4)¹¹/(-1/4)³

(-1/4)^y = (-1/4)^{11 – 3}

(-1/4)^y = (-1/4)⁸

27. [(7/11)³]⁴ = (7/11)

Solution:-

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

(a^m)ⁿ = a^mn

= (7/11)¹²

28. (6/13)¹⁰ ÷ [(6/13)⁵]² = (6/13)

Solution:-

(6/13)¹⁰ ÷ [(6/13)⁵]² = (6/13)⁰

In the given question first consider [(6/13)⁵]²

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

(a^m)ⁿ = a^mn

= (6/13)¹⁰

Then, (6/13)¹⁰ ÷ (6/13)¹⁰

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ aⁿ = a^m–n , m>n

(6/13)¹⁰ ÷ (6/13)¹⁰ = (6/13)^{10 – 10}

= (6/13)⁰

29. [(-1/4)¹⁶]² = (-1/4)

Solution:-

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

(a^m)ⁿ = a^mn

= (-1/4)³²

30. (13/14)⁵ ÷ ( )² = (13/14)³

Solution:-

(13/14)⁵ ÷ (13/14)² = (13/14)³

We know that,

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ aⁿ = a^m–n , m>n

Therefore, (13/14)⁵ ÷ (13/14)²= (13/14)^{5 – 2}

= (13/14)³

31. a⁶ × a⁵ × a⁰ = a

Solution:-

a⁶ × a⁵ × a⁰ = a¹¹

We know that,

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ × a^o= a^m+n+o

a⁶ × a⁵ × a⁰ = a^{5 + 6 + 0}

= a¹¹

32. 1 lakh = 10

Solution: –

1 lakh = 10⁵

Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation.

1 lakh = 1,00,000

= 10⁵

33. 1 million = 10

Solution:-

1 million = 10⁶

Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation.

1 million = 10,00,000

= 10⁶

34. 729 = 3

Solution: –

729 = 3⁶

Now, we have to find out the factors of 729.

Therefore, factors of 729 are = 3 × 3 × 3 × 3 × 3 × 3

729 = 3⁶

35. 432 = 2⁴ × 3

Solution:-

432 = 2⁴ × 3³

Now, we have to find out the factors of 432.

Therefore, factors of 432 are = 2 × 2 × 2 × 2 × 3 × 3 × 3

432 = 2⁴ × 3³

36. 53700000 = ––– × 10⁷

Solution:-

53700000 = 5.37 × 10⁷

= (53700000/10000000) × 10000000

= 5.37 × 10⁷

Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation

37. 88880000000 = ––– × 10¹⁰

Solution:-

88880000000 = 8.888 × 10¹⁰

= (88880000000/10000000000) × 10000000000

= 8.888 × 10¹⁰

38. 27500000 = 2.75 × 10––––

Solution:-

27500000 = 2.75 × 10⁷

= (27500000/10000000) × 10000000

= 2.75 × 10⁷

39. 340900000 = 3.409 × 10––––

Solution:-

340900000 = 3.409 × 10⁸

= (340900000/100000000) × 100000000

= 3.409 × 10⁸

40. Fill in the blanks with <, > or = sign.

(a) 3² ______15

Solution:-

3² < 15

Where, 3² = 3 × 3 = 9

Therefore, 9 < 15

(b) 2³ ______ 3²

Solution:-

2³ < 3²

Where, 2³ = 2 × 2 × 2 = 8

3² = 3 × 3 = 9

Therefore, 8 < 9

(c) 7⁴ ______5⁴

Solution:-

7⁴ > 5⁴

Where, 7⁴ = 7 × 7 × 7 × 7 = 2401

5 × 5 × 5 × 5 = 625

Therefore, 2401 > 625

(d) 10,000 ______ 10⁵

Solution:-

10,000 < 100000

Where, 10 × 10 × 10 × 10 × 10= 100000

Therefore, 10,000 < 100000

(e) 6³ _____4⁴

Solution:-

6³ < 4⁴

Where, 6³ = 6 × 6 × 6 = 216

4⁴ = 4 × 4 × 4 × 4 = 256

Therefore, 216 < 256

In questions 41 to 50, state whether the given statements are True or False.

41. One million = 10⁷

Solution:-

False.

We know that, 1 million = 10,00,000

= 10⁶

42. One hour = 60² seconds

Solution:-

True.

We know that, 1 hour = 60 minutes

1 minutes = 60 seconds

Therefore, 1 hour = 60 × 60 = 3600 seconds

43. 1⁰ × 0¹ = 1

Solution:-

False

1⁰ = 1

0¹ = 0

So, 1 × 0 = 0

44. (-3)⁴ = – 12

Solution:-

False.

(-3)⁴ = -3 × -3 × -3 × -3 = 81

So, 81 ≠ – 12

45. 3⁴ > 4³

Solution:-

True.

3⁴ = 3 × 3 × 3 × 3 = 81

4³ = 4 × 4 × 4 = 64

Therefore, 81 > 64

46. (-3/5)¹⁰⁰ = (-3¹⁰⁰/-5¹⁰⁰)

Solution:-

True.

Consider the LHS = (-3/5)¹⁰⁰

-3 can be written as = (-1 × 3)

So, ((-1 × 3)/5)¹⁰⁰

= (-1¹⁰⁰ × 3¹⁰⁰)/5¹⁰⁰

= (1 × 3¹⁰⁰)/5¹⁰⁰

= 3¹⁰⁰/5¹⁰⁰

Then, consider the RHS = -3¹⁰⁰/-5¹⁰⁰

= 3¹⁰⁰/5¹⁰⁰

By comparing LHS and RHS,

LHS = RHS

3¹⁰⁰/5¹⁰⁰ = 3¹⁰⁰/5¹⁰⁰

Therefore, (-3/5)¹⁰⁰ = (-3¹⁰⁰/-5¹⁰⁰)

47. (10 + 10)¹⁰ = 10¹⁰ + 10¹⁰

Solution:-

False.

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ = a^m+n

48. x⁰ × x⁰ = x⁰ ÷ x⁰ is true for all non-zero values of x.

Solution:-

True.

we know that, x⁰ = 1

so, x⁰ × x⁰ = 1 × 1 = 1

x⁰ ÷ x⁰ = 1 ÷ 1 = 1

Therefore, x⁰ × x⁰ = x⁰ ÷ x⁰

1 = 1

49. In the standard form, a large number can be expressed as a decimal number between 0 and 1, multiplied by a power of 10.

Solution:-

False

Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation.

50. 4² is greater than 2⁴.

Solution:-

False.

4² = 4 × 4 = 16

2⁴ = 2 × 2 × 2 × 2 = 16

Therefore, 4² = 2⁴