Mathematics - Real Life | Maths 7th STd

Exercise 11.1 Solutions

Question 1 Find the value of: (i) 2⁶ (ii) 9³ (iii) 11² (iv) 5⁴

Answer:

(a) 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

(b) 9³ = 9 × 9 × 9 = 729

(c) 11² = 11 × 11 = 121

(d) 5⁴ = 5 × 5 × 5 × 5 = 625

Question 2 Express 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

Answer:

(a) 6 × 6 × 6 × 6 = 6⁴

(b) t × t = t²

(c) b × b × b × b = b⁴

(d) 5 × 5 × 7 × 7 × 7 = 5² × 7³

(e) 2 × 2 × a × a = 2² × a²

(f) a × a × a × c × c × c × c × d = a³ × c⁴ × d

Question 3 Express each of the following numbers using exponential notation: (i) 512 (ii) 343 (iii) 729 (iv) 3125

Answer:

(a) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹

(b) 343 = 7 × 7 × 7 = 7³

(c) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶

(d) 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

Question 4 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

Answer:

(a) 4³ or 3⁴: 4³ = 4 × 4 × 4 = 64 and 3⁴ = 3 × 3 × 3 × 3 = 81. Since 81 > 64, therefore 3⁴ is greater than 4³.

(b) 5³ or 3⁵: 5³ = 5 × 5 × 5 = 125 and 3⁵ = 3 × 3 × 3 × 3 × 3 = 243. Since 243 > 125, therefore 3⁵ is greater than 5³.

(c) 2⁸ or 8²: 2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 and 8² = 8 × 8 = 64. Since 256 > 64, therefore 2⁸ is greater than 8².

(d) 100² or 2¹⁰⁰: 2¹⁰⁰ is much greater than 100². As we know 2¹⁰ = 1024, which is greater than 100² = 10,000. Therefore, 2¹⁰⁰ would be significantly larger.

(e) 2¹⁰ or 10²: 2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024 and 10² = 10 × 10 = 100. Since 1024 > 100, therefore 2¹⁰ is greater than 10².

Question 5 Express each of the following as a product of powers of their prime factors: (i) 648 (ii) 405 (iii) 540 (iv) 3,600

Answer:

(a) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2³ × 3⁴

(b) 405 = 3 × 3 × 3 × 3 × 5 = 3⁴ × 5

(c) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5

(d) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴ × 3² × 5²

Question 6 Simplify: (i) 2 × 10³ (ii) 7² × 2² (iii) 2³ × 5 (iv) 3 × 4⁴ (v) 0 × 10² (vi) 5² × 3³ (vii) 2⁴ × 3² (viii) 3² × 10⁴

Answer:

(a) 2 × 10³ = 2 × 1000 = 2000

(b) 7² × 2² = 49 × 4 = 196

(c) 2³ × 5 = 8 × 5 = 40

(d) 3 × 4⁴ = 3 × 256 = 768

(e) 0 × 10² = 0 × 100 = 0

(f) 5² × 3³ = 25 × 27 = 675

(g) 2⁴ × 3² = 16 × 9 = 144

3² × 10⁴ = 9 × 10,000 = 90,000

Question 7 Simplify: (i) (– 4)³ (ii) (–3) × (–2)³ (iii) (–3)² × (–5)² (iv) (–2)³ × (–10)³

Answer:

(a) (– 4)³ = (-4) × (-4) × (-4) = -64

(b) (–3) × (–2)³ = (-3) × (-8) = 24

(c) (–3)² × (–5)² = 9 × 25 = 225

(d) (–2)³ × (–10)³ = (-8) × (-1000) = 8000

Question 8 Compare the following numbers: (i) 2.7 × 10¹² ; 1.5 × 10⁸ (ii) 4 × 10¹⁴ ; 3 × 10¹⁷

Answer:

(a) 2.7 × 10¹² ; 1.5 × 10⁸: Comparing the exponents, 12 > 8, therefore 2.7 × 10¹² is greater than 1.5 × 10⁸.

(b) 4 × 10¹⁴ ; 3 × 10¹⁷: Comparing the exponents, 17 > 14, therefore 3 × 10¹⁷ is greater than 4 × 10¹⁴.

Exercise 11.2 Solutions

Question 1 Using laws of exponents, simplify and write the answer in exponential form:

(i) 3² × 3⁴ × 3⁸ (ii) 6¹⁵ ÷ 6¹⁰ (iii) a³ × a² (iv) 7x × 7² (v) (5²)³ ÷ 5⁵ (vi) 2⁵ × 5⁵ (vii) a⁴ × b⁴ (viii) (3⁴)³ (ix) (2²⁰ ÷ 2¹⁵) × 3² × 2² (x) 8t ÷ 8²

Answer:

(a) 3² × 3⁴ × 3⁸ = 3²⁺⁴⁺⁸ = 3¹⁴ [Using aᵐ × aⁿ = aᵐ⁺ⁿ]

(b) 6¹⁵ ÷ 6¹⁰ = 6¹⁵⁻¹⁰ = 6⁵ [Using aᵐ ÷ aⁿ = aᵐ⁻ⁿ]

(c) a³ × a² = a³⁺² = a⁵ [Using aᵐ × aⁿ = aᵐ⁺ⁿ]

(d) 7x × 7² = 7¹⁺² × x = 7³x [Using aᵐ × aⁿ = aᵐ⁺ⁿ]

(e) (5²)³ ÷ 5⁵ = 5⁶ ÷ 5⁵ = 5⁶⁻⁵ = 5¹ = 5 [Using (aᵐ)ⁿ= aᵐⁿ and aᵐ ÷ aⁿ = aᵐ⁻ⁿ]

(f) 2⁵ × 5⁵ = (2 × 5)⁵ = 10⁵ [Using aᵐ × bᵐ = (a × b)ᵐ]

(g) a⁴ × b⁴ = (a × b)⁴ [Using aᵐ × bᵐ = (a × b)ᵐ]

(3⁴)³ = 3⁴ˣ³ = 3¹² [Using (aᵐ)ⁿ= aᵐⁿ]

(i) (2²⁰ ÷ 2¹⁵) × 3² × 2² = (2²⁰⁻¹⁵) × 3² × 2² = 2⁵ × 3² × 2² = 2⁵⁺² × 3² = 2⁷ × 3² [Using aᵐ ÷ aⁿ = aᵐ⁻ⁿ and aᵐ × aⁿ = aᵐ⁺ⁿ]

(j) 8t ÷ 8² = (2³)ᵗ ÷ 2⁶ = 2³ᵗ ÷ 2⁶ = 2³ᵗ⁻⁶ [Using (aᵐ)ⁿ= aᵐⁿ and aᵐ ÷ aⁿ = aᵐ⁻ⁿ]

Question 2 Simplify and express each of the following in exponential form:

(i) (2³/3⁴) × (3²/2⁴) × (2/3) (ii) (5⁷/5²) × (5³/5⁴) (iii) (4⁵ ÷ 2⁵) ÷ 5 (iv) (3/2)⁷ × (2¹¹/3⁷) × (2/3)⁴ (v) (3/7)⁴ × (3/7)³ (vi) 2⁰ + 3⁰ + 4⁰ (vii) 2⁰ × 3⁰ × 4⁰ (viii) (3⁰ + 2⁰) × 5⁰ (ix) (3/2)⁵ × (a/a⁵) × (a⁸/a³)

Answer:

(a) (2³/3⁴) × (3²/2⁴) × (2/3) = (2³⁺¹ / 3⁴⁺¹) × (3² / 2⁴) = (2⁴ / 3⁵) × (3² / 2⁴) = 2⁴⁻⁴ × 3²⁻⁵ = 2⁰ × 3⁻³ = 3⁻³ [Using aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ and a⁰ = 1]

(b) (5⁷/5²) × (5³/5⁴) = 5⁷⁻² × 5³⁻⁴ = 5⁵ × 5⁻¹ = 5⁵⁻¹ = 5⁴ [Using aᵐ × aⁿ = aᵐ⁺ⁿ and aᵐ ÷ aⁿ = aᵐ⁻ⁿ]

(c) (4⁵ ÷ 2⁵) ÷ 5 = ((2²)⁵ ÷ 2⁵) ÷ 5 = (2¹⁰ ÷ 2⁵) ÷ 5 = 2¹⁰⁻⁵ ÷ 5 = 2⁵ ÷ 5 [Using (aᵐ)ⁿ= aᵐⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ]

(d) (3/2)⁷ × (2¹¹/3⁷) × (2/3)⁴ = 3⁷⁻⁷ × 2¹¹⁻⁷⁺⁴ = 3⁰ × 2⁸ = 2⁸ [Using aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ and a⁰ = 1]

(e) (3/7)⁴ × (3/7)³ = (3/7)⁴⁺³ = (3/7)⁷ [Using aᵐ × aⁿ = aᵐ⁺ⁿ]

(f) 2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3 [Using a⁰ = 1]

(g) 2⁰ × 3⁰ × 4⁰ = 1 × 1 × 1 = 1 [Using a⁰ = 1]

(3⁰ + 2⁰) × 5⁰ = (1 + 1) × 1 = 2 × 1 = 2 [Using a⁰ = 1]

(i) (3/2)⁵ × (a/a⁵) × (a⁸/a³) = (3/2)⁵ × a¹⁻⁵ × a⁸⁻³ = (3/2)⁵ × a⁻⁴ × a⁵ = (3/2)⁵ × a⁻⁴⁺⁵ = (3/2)⁵ × a [Using aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ]

Question 3 Say true or false and justify your answer:

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

Answer:

(a) 10 × 10¹¹= 10⁰¹¹ : False. 10 × 10¹¹ = 10¹⁺¹¹ = 10¹²

(b) 2³ > 5²: False. 2³ = 8 and 5² = 25. So, 5² > 2³.

(c) 2³ × 3² = 6⁵: False. 2³ × 3² = 8 × 9 = 72, while 6⁵ would be a much larger number.

(d) 3⁰ = (1000)⁰: True. Any number (except 0) raised to the power 0 is 1.

Question 4 Express each of the following as a product of prime factors only in exponential form: (i) 108 × 192 (ii) 270 (iii) 729 × 64 (iv) 768

Answer:

(a) 108 × 192 = (2² × 3³) × (2⁶ × 3) = 2⁸ × 3⁴ [Using aᵐ × aⁿ = aᵐ⁺ⁿ]

(b) 270 = 2 × 3³ × 5

(c) 729 × 64 = 3⁶ × 2⁶ [Using aᵐ × bᵐ = (a × b)ᵐ]

(d) 768 = 2⁸ × 3

Question 5 Simplify:

(i) (2/3)² × (8/7)³ × (3/4)² (ii) (10²/2⁵) × (3/4)³ × (5/3)⁴ (iii) (5⁵/7⁵) × (7/5)³ × (10/25)⁵ × (5/6)⁵

Answer:

(a) (2/3)² × (8/7)³ × (3/4)² = (2²/3²) × (2³/7³) × (3²/2⁴) = 2⁶⁻⁴ × 3⁴⁻² × 7⁻³ = 2² × 3² × 7⁻³ [Using aᵐ × aⁿ = aᵐ⁺ⁿ and aᵐ ÷ aⁿ = aᵐ⁻ⁿ]

(b) (10²/2⁵) × (3/4)³ × (5/3)⁴ = (2² × 5²)/2⁵ × (3³/4³) × (5⁴/3⁴) = 2⁴⁻⁵ × 3³⁻⁴ × 5⁶⁻⁵ = 2⁻¹ × 3⁻¹ × 5 = 5/(2 × 3) = 5/6 [Using aᵐ × aⁿ = aᵐ⁺ⁿ and aᵐ ÷ aⁿ = aᵐ⁻ⁿ]

(c) (5⁵/7⁵) × (7/5)³ × (10/25)⁵ × (5/6)⁵ = (5⁵/7⁵) × (7³/5³) × (2⁵ × 5⁵)/(5⁵ × 5⁵) × (5⁵/2⁵ × 3⁵) = 2⁵⁻⁵ × 3⁻⁵ × 5⁵⁻⁵⁺⁵⁻⁵ × 7³⁻⁵ = 2⁰ × 3⁻⁵ × 5⁰ × 7⁻² = 1/(3⁵ × 7²) [Using aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ and a⁰ = 1]

Exercise 11.3 Solutions

Question 1 Write the following numbers in the expanded forms: 279404, 3006194, 2806196, 120719, 20068

Answer:

• 279404 = 2 × 10⁵ + 7 × 10⁴ + 9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰

• 3006194 = 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰

• 2806196 = 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰

• 120719 = 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰

• 20068 = 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

Question 2 Find the number from each of the following expanded forms:

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰ (b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰ (c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰ (d) 9 × 10⁵ + 2 × 10² + 3 × 10¹

Answer:

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰ = 86045

(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰ = 405302

(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰ = 30705

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹ = 900230

Question 3 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:

(a) 5,00,00,000 = 5 × 10⁷

(b) 70,00,000 = 7 × 10⁶

(c) 3,18,65,00,000 = 3.1865 × 10⁹

(d) 3,90,878 = 3.90878 × 10⁵

(e) 39087.8 = 3.90878 × 10⁴

(f) 3908.78 = 3.90878 × 10³

Question 4 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 the 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 100,000,000,000 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. 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.

Answer:

(a) The distance between Earth and Moon is 384,000,000 m = 3.84 × 10⁸ m.

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

(c) Diameter of the Earth is 1,27,56,000 m = 1.2756 × 10⁷ m.

(d) Diameter of the Sun is 1,400,000,000 m = 1.4 × 10⁹ m.

(e) In a galaxy there are on an average 100,000,000,000 stars = 1 × 10¹¹ stars.

(f) The universe is estimated to be about 12,000,000,000 years old = 1.2 × 10¹⁰ 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 = 3 × 10²⁰ m.

60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm = 6.023 × 10²² molecules.

(i) The earth has 1,353,000,000 cubic km of sea water = 1.353 × 10⁹ cubic km.

(j) The population of India was about 1,027,000,000 in March, 2001 = 1.027 × 10⁹.