Chapter 11 Exponents & Powers- 1 Solutions

NCERT Solutions: Exponents & Powers- 1

Exercise 11.1

Q1: Find the value of:

(i) 2⁶

(ii) 9³

(iii) 11²

(iv) 5⁴

Ans:

(i) 2⁶ = 2 x 2 x 2 x 2 x 2 x 2 = 64
(ii) 9³ = 9 x 9 x 9 = 729
(iii) 11²= 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 = 2²x 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
NCERT Solutions: Exponents & Powers- 1

(ii) 343
343 = 7 x 7 x 7 = 7³

(iii) 729
729 = 3 x 3 x 3 x 3 x 3 x 3 = 3⁶ NCERT Solutions: Exponents & Powers- 1

(iv) 3125
3125 = 5 x 5 x 5 x 5 x 5 = 5⁵ NCERT Solutions: Exponents & Powers- 1

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

(i) 4³ and3⁴
(ii) 5³ or 3⁵
(iii) 2⁸ or 8²
(iv) 100²or 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, 3⁵ is greater than 5³.

(iii) 2⁸= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256
8² = 8 x 8 = 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 = 10×10 = 100
Since, 1,024 > 100
Thus, 2¹⁰ greater than 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:

648 = 2³ x 3⁴

405 = 5 x 3⁴

540 = 2² x 3³ x 5

3,600 = 2⁴ x 3²x 5²

Q6: Simplify:

(i) 2 x 10³
(ii) 7² x 2²
(iii) 2³ x 5
(iv) 3 x 4⁴
(v) 0 x 10²
(vi) 5²x 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) 5² x 3³ = 5 x 5 x 3 x 3 x 3 = 675
(vii) 2⁴x 3² = 2 x 2 x 2 x 2 x 3 x 3 = 144
(viii) 3² x 10⁴ = 3 x 3 x 10 x 10 x 10 x 10 = 90,000

Q7: Simplify:

(i) (-4)³
(ii) (-3)x(-2)³
(iii) (-3)²x (-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) = 8,000

Question 8: Compare the following numbers:

(i) 2.7 x 10¹²; 1.5 x 10⁸

(ii) 4 x 10¹⁴; 3 x 10¹⁷

Answer 8:

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

(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 a²
(iv) 7^x x 7²
(v) (5²)² ÷ 5³
(vi) 2⁵ x 5⁵
(vii) a⁴ x b⁴
(viii) (3⁴)³
(ix) [2²⁰ ÷ 2¹⁵) x 2³
(x) 8^t x 8²

Answer 1:

(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) NCERT Solutions: Exponents & Powers- 1

(ii)

(iii)

(iv)

(v) NCERT Solutions: Exponents & Powers- 1

(vi)

(vii)

(viii)

(ix) NCERT Solutions: Exponents & Powers- 1

(x) NCERT Solutions: Exponents & Powers- 1

(xi)

(xii)

Q3: Say true or false and justify your answer:

(i) 10 x 10¹¹= 100¹¹
(ii) 2³ > 5²
(iii) 2³x 3² = 6⁵
(iv) 3⁰ = (1000)⁰

Ans:

(i) 10×10¹¹ = 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 =7776
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:
(i) 108 × 192
= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 3)
= (2² × 3³) × (2⁶ × 3)
= 2⁸ × 3⁴
(ii) 270 = 2 × 3 × 3 × 3 × 3 × 5 = 2 × 3³ × 5
(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2)
= 3⁶ × 2⁶
(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3

Question 5: Simplify:

(i) NCERT Solutions: Exponents & Powers- 1
(ii)
(iii)

Answer 5:

(i)
NCERT Solutions: Exponents & Powers- 1

(ii)
NCERT Solutions: Exponents & Powers- 1

(iii)
NCERT Solutions: Exponents & Powers- 1

Exercise 11.3

Q1: Write the following numbers in the expanded form:

(i) 279404

(ii) 3006194

(iii) 2806196

(iv) 120719

(v) 20068

Ans:

(i) 2,79,404

= 2,00,000 + 70,000 + 9,000 + 400 + 00 + 4
= 2×100000 + 7x 10000+ 9×1000 + 4 x 100+ 0x 10 + 4×1
= 2×10⁵+7×10⁴ +9×10³ + 4 x10²+ 0x10¹+4×10⁰

(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 + 1×100 + 9×10 + 4×1
= 3×10⁶ + 0x10⁵ +0x10⁴ +6 x10³+1×10² +9×10+4×10⁰

(iii) 28,06,196

= 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6
= 2×1000000 + 8x 100000+ 0x 10000 +6 x 1000 + 1 x 100 + 9 x 10 + 6 x 1
= 2 x10⁶+8×10⁵+0x10⁴ + 6×10³+1×10²+9×10+6×10⁰

(iv) 1,20,719

= 1,00,000 + 20,000 + 0 + 700 + 10 + 9
= 1 x 100000 + 2 x 10000 + 0x 1000 + 7×100 + 1x 10+9×1
= 1 x 10⁵ + 2 x 10⁴+ 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

Q2: 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

Ans:

(a) 8 x 10⁴ + 6 x 10³ + 0 x 10² + 4 x 10¹ + 5 x 10⁰
= 8 x 10000 + 6 x 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 10² + 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

Q3: 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

Answer 3:

(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

Q4: 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 is 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.

Ans :

(a) The distance between Earth and Moon = 384,000,000 m
= 384x 1000000 m
= 3.84 x 100 x 1000000
= 3.84×10⁸ m

(b) Speed of light in vacuum = 300,000,000 m/s
= 3 x 100000000 m/s
= 3×10⁸ m/s

(c) Diameter of the Earth = 1,27,56,000 m
= 12756 x 1000 m
= 1.2756 x 10000 x 1000 m
= 1.2756 x 10⁷ 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 number of stars in a galaxy = 100, 000, 000,000
= 1 x 100,000,000,000
= 1×10¹¹

(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 the 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
= 1.353×10⁹ km³

(j) The population of India = 1,027,000,000
= 1027×1000000 = 1.027x1000x1000000
= 1.027×10⁹

Chapter 11 Exponents & Powers- 1 Solutions

Chapter 10 Algebraic Expressions Solutions

Chapter 13 Visualising Solid Shapes Solutions