What is the significance of negative exponents?

In Class 7 Maths Chapter 11, the significance of negative exponents lies in their ability to represent reciprocals concisely, allowing for easier simplification of expressions. They help students connect positive exponents to division concepts, enhancing their overall understanding of exponents. Additionally, negative exponents are crucial in scientific notation for expressing very small values, preparing students for future math and science topics.

Why is it important to learn the laws of exponents?

The laws of exponents make calculations involving large numbers much simpler and are essential for higher-level mathematics, including algebra, calculus, and beyond.

How does practicing exponents help in real-life situations?

Understanding exponents is important in fields like physics, finance, and computer science, where exponential growth, decay, and large numbers are commonly encountered.

NCERT Solutions

Class 7

Maths

Chapter 11 Exponents and Powers

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NCERT Solutions Class 7 Maths Chapter 11 - Exponents and Powers

In NCERT class 7 maths exponents and powers, students will learn how to represent numbers in standard form, understand the laws of exponents, and practice various operations like multiplication and division involving exponents. The solutions help students to understand different aspects of exponents and powers thus making it easier for the students to understand it.

These solutions are designed in a step-by-step manner to simplify intricate problems of chapter exponents and powers class 7, enabling students to understand complex calculations efficiently. Overall, NCERT Solutions for Class 7 Maths chapter 11 exponents and powers help students develop a strong understanding of exponents and powers, enhance their problem-solving abilities, and prepare them effectively for exams. The inclusion of examples, detailed explanations, and practical applications ensures that students can apply these concepts in real-life situations and mathematical problems.

1.0Download NCERT Solutions Class 7 Maths Chapter 11 : Free PDF

Students can Download NCERT Solutions Class 7 Maths Chapter 11 from the below link and Practice from anywhere anytime.

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

2.0NCERT Solutions Class 6 Maths Chapter 11 Exponents and Powers

These exercises prove to be quite helpful for students as it is often seen that questions can be repeated from the NCERT books during final exams and also if one practices these questions daily, he or she can figure out that they were not able to answer well in which section and gradually give more time to that particular area.

Exercises	Total Number of Questions
Exponents and powers class 7 ex 11.1	8
Exponents and powers class 7 ex 11.2	5
Exponents and powers class 7 ex 11.3	4

3.0NCERT Questions with Solutions for Class 7 Maths Chapter 11 - Detailed Solutions

Exercise: 11.1

Find the value of: (i) $2^{6}$ (ii) $9^{3}$ (iii) $1 1^{2}$ (iv) $5^{4}$ Sol. (i) $2^{6} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$ (ii) $9^{3} = 9 \times 9 \times 9 = 729$ (iii) $1 1^{2} = 11 \times 11 = 121$ (iv) $5^{4} = 5 \times 5 \times 5 \times 5 = 625$
Express the following in exponential form : (i) $6 \times 6 \times 6 \times 6$ (ii) $t \times t$ (iii) $b \times b \times b \times b$ (iv) $5 \times 5 \times 7 \times 7 \times 7$ (v) $2 \times 2 \times a \times a$ (vi) $a \times a \times a \times c \times c \times c \times c \times d$ Sol. (i) $6 \times 6 \times 6 \times 6 = 6^{4}$ (ii) $t \times t = t^{2}$ (iii) $b \times b \times b \times b = b^{4}$ (iv) $5 \times 5 \times 7 \times 7 \times 7 = 5^{2} \times 7^{3}$ (v) $2 \times 2 \times a \times a = 2^{2} \times a^{2}$ (vi) $a \times a \times a \times c \times c \times c \times c \times d = a^{3} \times c^{4} \times d$
Express each of the following numbers using exponential notation: (i) 512 (ii) 343 (iii) 729 (iv) 3125 Sol. (i) 512 $= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{9}$

2	512
2	256
2	128
2	64
2	32
2	16
2	8
2	4
2	2
	1

(ii) 343 $= 7 \times 7 \times 7 = 7^{3}$

7	343
7	49
7	7
	1

(iii) 729 $= 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^{6}$

3	729
3	243
3	81
3	27
3	9
3	3
	1

(iv) 3125 $= 5 \times 5 \times 5 \times 5 \times 5 = 5^{5}$

5	3125
5	625
5	125
5	25
5	5
	1

Identify the greater number, wherever possible, in each of the following ? (i) $4^{3}$ or $3^{4}$ (ii) $5^{3}$ or $3^{5}$ (iii) $2^{8}$ or $8^{2}$ (iv) $10 0^{2}$ or $2^{100}$ (v) $2^{10}$ or $1 0^{2}$ Sol. (i) $4^{3} = 4 \times 4 \times 4 = 64$ $3^{4} = 3 \times 3 \times 3 \times 3 = 81$ Since $64 < 81$ Thus, $3^{4}$ is greater than $4^{3}$ . (ii) $5^{3} = 5 \times 5 \times 5 = 125$ $3^{5} = 3 \times 3 \times 3 \times 3 \times 3 = 243$ Since, $125 < 243$ Thus, $3^{5}$ is greater than $5^{3}$ . (iii) $2^{8} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 256$ $8^{2} = 8 \times 8 = 64$ Since, $256 > 64$ Thus, $2^{8}$ is greater than $8^{2}$ . (iv) $10 0^{2} = 100 \times 100 = 10, 000$ $2^{100} = 2 \times 2 \times 2 \times 2 \times 2 \times \dots 14$ times $\times \dots \times 2$ $= 16, 384 \times ..... \times 2$ Since, $10, 000 < 16, 384 \times$ $\times 2$ Thus, $2^{100}$ is greater than $10 0^{2}$ . (v) $2^{10} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ $= 1, 024$ $1 0^{2} = 10 \times 10 = 100$ Since, $1024 > 100$ Thus, $2^{10} > 1 0^{2}$
Express each of the following as product of powers of their prime factors : (i) 648 (ii) 405 (iii) 540 (iv) 3,600 Sol. (i) $648 = 2^{3} \times 3^{4}$

2	648
2	324
2	162
3	81
3	27
3	9
3	3
	1

(ii) $405 = 5 \times 3^{4}$

3	405
3	135
3	45
3	15
5	5
	1

(iii) $540 = 2^{2} \times 3^{3} \times 5$

2	540
2	270
3	135
3	45
3	15
5	5
	1

(iv) $3, 600 = 2^{4} \times 3^{2} \times 5^{2}$

2	3600
2	1800
2	900
2	450
3	225
3	75
5	25
5	5
	1

Simplify : (i) $2 \times 1 0^{3}$ (ii) $7^{2} \times 2^{2}$ (iii) $2^{3} \times 5$ (iv) $3 \times 4^{4}$ (v) $0 \times 1 0^{2}$ (vi) $5^{2} \times 3^{3}$ (vii) $2^{4} \times 3^{2}$ (viii) $3^{2} \times 1 0^{4}$ Sol. (i) $2 \times 1 0^{3} = 2 \times 10 \times 10 \times 10 = 2, 000$ (ii) $7^{2} \times 2^{2} = 7 \times 7 \times 2 \times 2 = 196$ (iii) $2^{3} \times 5 = 2 \times 2 \times 2 \times 5 = 40$ (iv) $3 \times 4^{4} = 3 \times 4 \times 4 \times 4 \times 4 = 768$ (v) $0 \times 1 0^{2} = 0 \times 10 \times 10 = 0$ (vi) $5^{2} \times 3^{3} = 5 \times 5 \times 3 \times 3 \times 3 = 675$ (vii) $2^{4} \times 3^{2} = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 144$ (viii) $3^{2} \times 1 0^{4} = 3 \times 3 \times 10 \times 10 \times 10 \times 10 = 90, 000$
Simplify : (i) $(- 4)^{3}$ (ii) $(- 3) \times (- 2)^{3}$ (iii) $(- 3)^{2} \times (- 5)^{2}$ (iv) $(- 2)^{3} \times (- 10)^{3}$ Sol. (i) $(- 4)^{3} = (- 4) \times (- 4) \times (- 4) = - 64$ (ii) $(- 3) \times (- 2)^{3} = (- 3) \times (- 2) \times (- 2) \times (- 2) = 24$ (iii) $(- 3)^{2} \times (- 5)^{2} = (- 3) \times (- 3) \times (- 5) \times (- 5) = 225$ (iv) $(- 2)^{3} \times (- 10)^{3}$ $= (- 2) \times (- 2) \times (- 2) \times (- 10) \times (- 10) \times (- 10)$ $= 8, 000$
Compare the following numbers : (i) $2.7 \times 1 0^{12}; 1.5 \times 1 0^{8}$ (ii) $4 \times 1 0^{14}; 3 \times 1 0^{17}$ Sol. (i) $2.7 \times 1 0^{12}$ and $1.5 \times 1 0^{8}$ On comparing the exponents of base 10 , $2.7 \times 1 0^{12} > 1.5 \times 1 0^{8}$ (ii) $4 \times 1 0^{14}$ and $3 \times 1 0^{17}$ On comparing the exponents of base 10 , $4 \times 1 0^{14} < 3 \times 1 0^{17}$

Exercise: 11.2

Using laws of exponents, simplify and write the answer in exponential form : (i) $3^{2} \times 3^{4} \times 3^{8}$ (ii) $6^{15} \div 6^{10}$ (iii) $a^{3} \times a^{2}$ (iv) $7 \times 7^{2}$ (v) $(5^{2})^{3} \div 5^{3}$ (vi) $2^{5} \times 5^{5}$ (vii) $a^{4} \times b^{4}$ (viii) $(3^{4})^{3}$ (ix) $(2^{20} \div 2^{15}) \times 2^{3}$ (x) $8^{t} \div 8^{2}$ Sol. (i) $3^{2} \times 3^{4} \times 3^{8}$ $= 3^{(2 + 4 + 8)} = 3^{14} [∵ a^{m \times a^{n} = a^{m + n}]}$ (ii) $6^{15} \div 6^{10} = 6^{15 - 10} = 6^{5} [∵ a^{m} \div a^{n} = a^{m - n}]$ (iii) $a^{3} \times a^{2} = a^{3 + 2} = a^{5} [∵ a^{m} \times a^{n} = a^{m + n}]$ (iv) $7^{x} \times 7^{2} = 7^{x + 2} [∵ a^{m} \times a^{n} = a^{m + n}]$ (v) $(5^{2})^{3} \div 5^{3} = 5^{2 \times 3} \div 5^{3} [∵ (a^{m})^{n} = a^{m \times n}]$ $= 5^{6} \div 5^{3} = 5^{6 - 3} = 5^{3} [∵ a^{m} \div a^{n} = a^{m - n}]$ (vi) $2^{5} \times 5^{5} = (2 \times 5)^{5} = 1 0^{5} [∵ a^{m} \times b^{m} = (a \times b)^{m}]$ (vii) $a^{4} \times b^{4} = (a \times b)^{4} [∵ a^{m} \times b^{m} = (a \times b)^{m}]$ (viii) $(3^{4})^{3} = 3^{4 \times 3} = 3^{12} [∵ (a^{m})^{n} = a^{m \times n}]$ (ix) $(2^{20} \div 2^{15}) \times 2^{3}$

$= (2^{20 - 15}) \times 2^{3} [∵ a^{m} \div a^{n} = a^{m - n}] = 2^{5} \times 2^{3} = 2^{5 + 3} = 2^{8} [∵ a^{m} \times a^{n} = a^{m + n}]$ (x) $8^{t} \div 8^{2} = 8^{t - 2} [∵ a^{m} \div a^{n} = a^{m - n}]$

Simplify and express each of the following in exponential form : (i) $\frac{2 ^{3} \times 3 ^{4} \times 4}{3 \times 32}$ (ii) $[(5^{2})^{3} \times 5^{4}] \div 5^{7}$ (iii) $2 5^{4} \div 5^{3}$ (iv) $\frac{3 \times 7 ^{2} \times 1 1 ^{8}}{21 \times 1 1 ^{3}}$ (v) $\frac{3 ^{7}}{3 ^{4} \times 3 ^{3}}$ (vi) $2^{0} + 3^{0} + 4^{0}$ (vii) $2^{0} \times 3^{0} \times 4^{0}$ (viii) $(3^{0} + 2^{0}) \times 5^{0}$ (ix) $\frac{2 ^{8} \times a ^{5}}{4 ^{3} \times a ^{3}}$ (x) $(\frac{a ^{5}}{a ^{3}}) \times a^{8}$ (xi) $\frac{4 ^{5} \times a ^{8} b ^{3}}{4 ^{5} \times a ^{5} b ^{2}}$ (xii) $(2^{3} \times 2)^{2}$ Sol. (i) $\frac{2 ^{3} \times 3 ^{4} \times 4}{3 \times 32} = \frac{2 ^{3} \times 3 ^{4} \times 2 ^{2}}{3 \times 2 ^{5}} = \frac{2 ^{3 + 2} \times 3 ^{4}}{3 \times 2 ^{5}}$ $[∵ a^{m} \times a^{n} = a^{m + n}]$ $= \frac{2 ^{5} \times 3 ^{4}}{3 \times 2 ^{5}} = 2^{5 - 5} \times 3^{4 - 1} [∵ a^{m} \div a^{n} = a^{m - n}]$ $= 2^{0} \times 3^{3} = 1 \times 3^{3} = 3^{3}$ (ii) $[(5^{2})^{3} \times 5^{4}] \div 5^{7}$ $= [5^{6} \times 5^{4}] \div 5^{7} [∵ (a^{m})^{n} = a^{m \times n}] = [5^{6 + 4}] \div 5^{7} = 5^{10} \div 5^{7} [∵ a^{m \times} \times a^{n} = a^{m + n}] = 5^{10 - 7} = 5^{3} [∵ a^{m} \div a^{n} = a^{m - n}]$ (iii) $2 5^{4} \div 5^{3}$ $= (5^{2})^{4} \div 5^{3} = 5^{8} \div 5^{3} [∵ (a^{m})^{n} = a^{m \times n}]$ $= 5^{8 - 3} = 5^{5} [∵ a^{m} \div a^{n} = a^{m - n}]$ (iv) $\frac{3 \times 7 ^{2} \times 1 1 ^{8}}{21 \times 1 1 ^{3}} = \frac{3 \times 7 ^{2} \times 1 1 ^{8}}{3 \times 7 \times 1 1 ^{3}}$ $= 3^{1 - 1} \times 7^{2 - 1} \times 1 1^{8 - 3} [∵ a^{m} \div a^{n} = a^{m - n}]$ $= 3^{0} \times 7^{1} \times 1 1^{5} = 7 \times 1 1^{5}$

$= 3^{7 - 7} = 3^{0} = 1 [∵ a^{m} \div a^{n} = a^{m - n}]$ (vi) $2^{0} + 3^{0} + 4^{0} = 1 + 1 + 1 = 3 [∵ a^{0} = 1]$ (vii) $2^{0} \times 3^{0} \times 4^{0} = 1 \times 1 \times 1 = 1 [∵ a^{0} = 1]$ (viii) $(3^{0} + 2^{0}) \times 5^{0} = (1 + 1) \times 1$ $= 2 \times 1 = 2 [∵ a^{0} = 1]$ (ix) $\frac{2 ^{8} \times a ^{5}}{4 ^{3} \times a ^{3}} = \frac{2 ^{8} \times a ^{5}}{( 2 ^{2} ) ^{3} \times a ^{3}} = \frac{2 ^{8} \times a ^{5}}{2 ^{6} \times a ^{3}}$ $[∵ (a^{m})^{n} = a^{m \times n}]$ $= 2^{8 - 6} \times a^{5 - 3} = 2^{2} \times a^{2} [∵ a^{m} \div a^{n} = a^{m - n}]$ $= (2 a)^{2} [∵ a^{m} \times b^{m} = (a \times b)^{m}]$ (x) $(\frac{a ^{5}}{a ^{3}}) \times a^{8} = (a^{5 - 3}) \times a^{8} = a^{2} \times a^{8}$ $[∵ a^{m} \div a^{n} = a^{m - n}]$ $= a^{2 + 8} = a^{10} [∵ a^{m} \times a^{n} = a^{m + n}]$ (xi) $\frac{4 ^{5} \times a ^{8} b ^{3}}{4 ^{5} \times a ^{5} b ^{2}} = 4^{5 - 5} \times a^{8 - 5} \times b^{3 - 2} = 40 \times a^{3} \times b$ $[∵ a^{m} \div a^{n} = a^{m - n}]$ $= 1 \times a^{3} \times b = a^{3} b [∵ a^{0} = 1]$ (xii) $(2^{3} \times 2)^{2} = (2^{3 + 1})^{2} = (2^{4})^{2}$ $[∵ a^{m} \times a^{n} = a^{m + n}]$ $= 2^{4 \times 2} = 2^{8} [∵ (a^{m})^{n} = a^{m \times n}]$
Say true or false and justify your answer : (i) $10 \times 1 0^{11} = 10 0^{11}$ (ii) $2^{3} > 5^{2}$ (iii) $2^{3} \times 3^{2} = 6^{5}$ (iv) $3^{0} = (1000)^{0}$ Sol. (i) $10 \times 1 0^{11} = 10 0^{11}$ L.H.S. $1 0^{1 + 11} = 1 0^{12}$ and R.H.S. $(1 0^{2})^{11} = 1 0^{22}$ Since, L.H.S. $\neq =$ R.H.S. Therefore, it is false. (ii) $2^{3} > 5^{2}$ L.H.S. $2^{3} = 8$ and R.H.S. $5^{2} = 25$ Since, L.H.S. is not greater than R.H.S. Therefore, it is false. (iii) $2^{3} \times 3^{2} = 6^{5}$ L.H.S. $2^{3} \times 3^{2} = 8 \times 9 = 72$ and R.H.S. $6^{5} = 7, 776$ Since, L.H.S. $\neq =$ R.H.S. Therefore, it is false. (iv) $3^{0} = (1000)^{0}$ L.H.S. $3^{0} = 1$ and R.H.S. $(1000)^{0} = 1$ Since, L.H.S. = R.H.S. Therefore, it is true.
Express each of the following as a product of prime factors only in exponential form : (i) $108 \times 192$ (ii) 270 (iii) $729 \times 64$ (iv) 768 Sol. (i) $108 \times 192$ $= (2^{2} \times 3^{3}) \times (2^{6} \times 3) = 2^{2 + 6} \times 3^{3 + 1} = 2^{8} \times 3^{4}$

2	108
2	54
3	27
3	9
3	3
	1

2	192
2	96
2	48
2	24
2	12
2	6
3	3
	1

(ii) $270 = 2 \times 3^{3} \times 5$

2	270
3	135
3	45
3	15
5	5
	1

(iii) $729 \times 64 = 36 \times 2^{6}$

3	729
3	243
3	81
3	27
3	9
3	3
	1

2	64
2	32
2	16
2	8
2	4
2	2
	1

(iv) $768 = 2^{8} \times 3$

2	768
2	384
2	192
2	96
2	48
2	24
2	12
2	6
3	3
	1

Simplify : (i) $\frac{( 2 ^{5} ) ^{2} \times 7 ^{3}}{8 ^{3} \times 7}$ (ii) $\frac{25 \times 5 ^{2} \times t ^{8}}{1 0 ^{3} \times t ^{4}}$ (iii) $\frac{3 ^{5} \times 1 0 ^{5} \times 25}{5 ^{7} \times 6 ^{5}}$ Sol. (i) $\frac{( 2 ^{5} ) ^{2} \times 7 ^{3}}{8 ^{3} \times 7} = \frac{2 ^{5 \times 2} \times 7 ^{3}}{( 2 ^{3} ) ^{3} \times 7} = \frac{2 ^{10} \times 7 ^{3}}{2 ^{9} \times 7}$ $= 2^{10 - 9} \times 7^{3 - 1} = 2 \times 7^{2} = 2 \times 49 = 98$ (ii) $\frac{25 \times 5 ^{2} \times t ^{8}}{1 0 ^{3} \times t ^{4}} = \frac{5 ^{2} \times 5 ^{2} \times t ^{8}}{( 5 \times 2 ) ^{3} \times t ^{4}}$ $= \frac{5 ^{2 + 2} \times t ^{8 - 4}}{2 ^{3} \times 5 ^{3}} = \frac{5 ^{4} \times t ^{4}}{2 ^{3} \times 5 ^{3}}$ $= \frac{5 ^{4 - 3} \times t ^{4}}{2 ^{3}} = \frac{5 t ^{4}}{8}$ (iii) $\frac{3 ^{5} \times 1 0 ^{5} \times 25}{5 ^{7} \times 6 ^{5}} = \frac{3 ^{5} \times ( 2 \times 5 ) ^{5} \times 5 ^{2}}{5 ^{7} \times ( 2 \times 3 ) ^{5}}$ $= \frac{3 ^{5} \times 2 ^{5} \times 5 ^{5} \times 5 ^{2}}{5 ^{7} \times 2 ^{5} \times 3 ^{5}} = \frac{3 ^{5} \times 2 ^{5} \times 5 ^{5 + 2}}{5 ^{7} \times 2 ^{5} \times 3 ^{5}} = \frac{3 ^{5} \times 2 ^{5} \times 5 ^{7}}{5 ^{7} \times 2 ^{5} \times 3 ^{5}} = 2^{5 - 5} \times 3^{5 - 5} \times 5^{7 - 7} = 2^{0} \times 3^{0} \times 5^{0} = 1 \times 1 \times 1 = 1$

Exercise: 11.3

Write the following numbers in the expanded forms : 279404, 3006194, 2806196, 120719, 20068 Sol. (i) 2,79,404 $2, 00, 000 + 70, 000 + 9, 000 + 400 + 0 + 4 = 2 \times 100000 + 7 \times 10000 + 9 \times 1000 + 4 \times 100 + 0 \times 10 + 4 \times 1 = 2 \times 1 0^{5} + 7 \times 1 0^{4} + 9 \times 1 0^{3} + 4 \times 1 0^{2} + 0 \times 1 0^{1} + 4 \times 1 0^{0}$ (ii) $30, 06, 194$ $= 30, 00, 000 + 0 + 0 + 6, 000 + 100 + 90 + 4 = 3 \times 1000000 + 0 \times 100000 + 0 \times 10000 + 6 \times 1000 + 1 \times 100 + 9 \times 10 + 4 \times 1 = 3 \times 1 0^{6} + 0 \times 1 0^{5} + 0 \times 1 0^{4} + 6 \times 1 0^{3} + 1 \times 1 0^{2} + 9 \times 10 + 4 \times 1 0^{0}$ (iii) $28, 06, 196$ $= 20, 00, 000 + 8, 00, 000 + 0 + 6, 000 + 100 + 90 + 6 = 2 \times 1000000 + 8 \times 100000 + 0 \times 10000 + 6 \times 1000 + 1 \times 100 + 9 \times 10 + 6 \times 1 = 2 \times 1 0^{6} + 8 \times 1 0^{5} + 0 \times 1 0^{4} + 6 \times 1 0^{3} + 1 \times 1 0^{2} + 9 \times 10 + 6 \times 1 0^{0}$ (iv) $1, 20, 719$ $= 1, 00, 000 + 20, 000 + 0 + 700 + 10 + 9$ $= 1 \times 100000 + 2 \times 10000 + 0 \times 1000$ $+ 7 \times 100 + 1 \times 10 + 9 \times 1$ $= 1 \times 1 0^{5} + 2 \times 1 0^{4} + 0 \times 1 0^{3} + 7 \times 1 0^{2}$ $+ 1 \times 1 0^{1} + 9 \times 1 0^{0}$ (v) $20, 068 = 20, 000 + 0 + 0 + 60 + 8$ $= 2 \times 10000 + 0 \times 1000 + 0 \times 100 + 6$ $\times 10 + 8 \times 1$ $= 2 \times 1 0^{4} + 0 \times 1 0^{3} + 0 \times 1 0^{2} + 6 \times 1 0^{1}$ $+ 8 \times 1 0^{0}$
Find the number from each of the following expanded forms: (a) $8 \times 1 0^{4} + 6 \times 1 0^{3} + 0 \times 1 0^{2} + 4 \times 1 0^{1} + 5 \times 1 0^{0}$ (b) $4 \times 1 0^{5} + 5 \times 1 0^{3} + 3 \times 1 0^{2} + 2 \times 1 0^{0}$ (c) $3 \times 1 0^{4} + 7 \times 1 0^{2} + 5 \times 1 0^{0}$ (d) $9 \times 1 0^{5} + 2 \times 1 0^{2} + 3 \times 1 0^{1}$ Sol. (a) $8 \times 1 0^{4} + 6 \times 1 0^{3} + 0 \times 1 0^{2} + 4 \times 1 0^{1} +$ $5 \times 1 0^{0}$ $= 8 \times 10000 + 6 \times 1000 + 0 \times 100 + 4$ $\times 10 + 5 \times 1$ $= 80000 + 6000 + 0 + 40 + 5$ $= 86, 045$ (b) $4 \times 1 0^{5} + 5 \times 1 0^{3} + 3 \times 1 0^{2} + 2 \times 1 0^{0}$ $= 4 \times 100000 + 0 \times 10000 + 5 \times 1000$ $+ 3 \times 100 + 0 \times 10 + 2 \times 1$ $= 400000 + 0 + 5000 + 300 + 0 + 2$ $= 4, 05, 302$ (c) $3 \times 1 0^{4} + 7 \times 1 0^{2} + 5 \times 1 0^{0}$ $= 3 \times 10000 + 0 \times 1000 + 7 \times 100 + 0$ $\times 10 + 5 \times 1$ $= 30000 + 0 + 700 + 0 + 5$ $= 30, 705$ (d) $9 \times 1 0^{5} + 2 \times 1 0^{2} + 3 \times 1 0^{1}$ $= 9 \times 100000 + 0 \times 10000 + 0 \times 1000$ $+ 2 \times 100 + 3 \times 10 + 0 \times 1$ $= 900000 + 0 + 0 + 200 + 30 + 0$ $= 9, 00, 230$
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 Sol. (i) $5, 00, 00, 000 = 5 \times 1, 00, 00, 000 = 5 \times 1 0^{7}$ (ii) $70, 00, 000 = 7 \times 10, 00, 000 = 7 \times 1 0^{6}$ (iii) $3, 18, 65, 00, 000 = 31865 \times 100000$ $= 3.1865 \times 10000 \times 100000$ $= 3.1865 \times 1 0^{9}$ (iv) $3, 90, 878 = 3.90878 \times 100000$ $= 3.90878 \times 1 0^{5}$ (v) $39087.8 = 3.90878 \times 10000$ $= 3.90878 \times 1 0^{4}$ (vi) $3908.78 = 3.90878 \times 1000 = 3.90878 \times 1 0^{3}$
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 Sun from the centre of 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. Sol. (a) The distance between Earth and Moon $= 384, 000, 000 m$ $= 3.84 \times 1 0^{8} m$ (b) Speed of light in vacuum $= 300, 000, 000 m / s$ $= 3 \times 100000000 m / s$ $= 3 \times 1 0^{8} m / s$ (c) Diameter of the Earth $= 1, 27, 56, 000 m$ $= 12756 \times 1000 m$ $= 1.2756 \times 10000 \times 1000 m$ $= 1.2756 \times 1 0^{7} m$ (d) Diameter of the Sun $= 1, 400, 000, 000 m$ $= 14 \times 100, 000, 000 m$ $= 1.4 \times 10 \times 1, 00, 000, 000 m$ $= 1.4 \times 1 0^{9} m$ (e) Average of Stars $= 100, 000, 000, 000$ $= 1 \times 100, 000, 000, 000$ $= 1 \times 1 0^{11}$ (f) Years of Universe $= 12, 000, 000, 000$ years $= 12 \times 1, 000, 000, 000$ years $= 1.2 \times 10 \times 1, 000, 000, 000$ years $= 1.2 \times 1 0^{10}$ years (g) Distance of the Sun from the centre of the Milky Way Galaxy $= 300, 000, 000, 000, 000, 000, 000 m$ $= 3 \times 100, 000, 000, 000, 000, 000, 000 m$ $= 3 \times 1 0^{20} m$ (h) Number of molecules in a drop of water weighing 1.8 gm $= 60, 230, 000, 000, 000, 000, 000, 000$ $= 6023 \times 10, 000, 000, 000, 000, 000, 000$ $= 6.023 \times 1000 \times 10, 000, 000, 000, 000, 000, 000$ $= 6.023 \times 1 0^{22}$ (i) The Earth has sea water $= 1, 353, 000, 000 km^{3}$ $= 1, 353 \times 1000000 km^{3}$ $= 1.353 \times 1000 \times 1000, 000 km^{3}$ $= 1.353 \times 1 0^{9} km^{3}$ (j) The population of India = 1,027,000,000 $= 1027 \times 1000000$ $= 1.027 \times 1000 \times 1000000$ $= 1.027 \times 1 0^{9}$

4.0What Will Students Learn in Chapter 11: Exponents and Powers?

Introduction to representation of large numbers in a simpler form by using exponentiation.
The fundamental operation principles and laws of exponents including, multiplication, division of powers, raising power of a power and more.
Defining what negative exponents mean and the connection between exponents and reciprocals.
How to demonstrate very large or very small numbers using powers of 10.
Several examples of using exponents in real life based on mathematical and scientific problems.

NCERT Solutions for Class 7 Maths Other Chapters:-

Chapter 1: Integers

Chapter 2: Fractions and Decimals

Chapter 3: Data Handling

Chapter 4: Simple Equations

Chapter 5: Lines and Angles

Chapter 6: The Triangle and its Properties

Chapter 7: Comparing Quantities

Chapter 8: Rational Numbers

Chapter 9: Perimeter and Area

Chapter 10: Algebraic Expressions

Chapter 11: Exponents and Powers