NCERT Solutions Class 7 Maths Chapter 11 - Exponents and Powers

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) $11^2$ (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) $11^2=11\times 11=121$ (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) $\mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{c}\times \mathrm{c}\times \mathrm{c}\times \mathrm{c}\times \mathrm{d}$ Sol. (i) $6\times 6\times 6\times 6=6^4$ (ii) $t\times t=t^2$ (iii) $\mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}={\mathrm{b}}^4$ (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 $$\begin{gathered} =2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\ =2^9 \end{gathered}$$

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

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

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

• 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) $100^2$ or $2^{100}$ (v) $2^{10}$ or $10^2$ 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) $100^2=100\times 100=10,000$ $2^{100}=2\times 2\times 2\times 2\times 2\times \dots 14$ times $\times \dots \times 2$ $=16,384\times .....\times 2$ Since, $10,000<16,384\times$ $\times 2$ Thus, $2^{100}$ is greater than $100^2$. (v) $2^{10}=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$ $=1,024$ $10^2=10\times 10=100$ Since, $1024>100$ Thus, $2^{10}>10^2$
• Express each of the following as product of powers of their prime factors : (i) 648 (ii) 405 (iii) 540 (iv) 3,600 Sol. (i) $648=2^3\times 3^4$

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

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

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

• Simplify : (i) $2\times 10^3$ (ii) $7^2\times 2^2$ (iii) $2^3\times 5$ (iv) $3\times 4^4$ (v) $0\times 10^2$ (vi) $5^2\times 3^3$ (vii) $2^4\times 3^2$ (viii) $3^2\times 10^4$ Sol. (i) $2\times 10^3=2\times 10\times 10\times 10=2,000$ (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) $0\times 10^2=0\times 10\times 10=0$ (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) $3^2\times 10^4=3\times 3\times 10\times 10\times 10\times 10=90,000$
• 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) $2.7\times 10^{12};1.5\times 10^8$ (ii) $4\times 10^{14};3\times 10^{17}$ Sol. (i) $2.7\times 10^{12}$ and $1.5\times 10^8$ On comparing the exponents of base 10 , $2.7\times 10^{12}>1.5\times 10^8$ (ii) $4\times 10^{14}$ and $3\times 10^{17}$ On comparing the exponents of base 10 , $4\times 10^{14}<3\times 10^{17}$

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) $6^{15}÷6^{10}$ (iii) ${\mathrm{a}}^3\times {\mathrm{a}}^2$ (iv) $7\times 7^2$ (v) ${\left(5^2\right)}^3÷5^3$ (vi) $2^5\times 5^5$ (vii) ${\mathrm{a}}^4\times {\mathrm{b}}^4$ (viii) ${\left(3^4\right)}^3$ (ix) $\left(2^{20}÷2^{15}\right)\times 2^3$ (x) $8^t÷8^2$ Sol. (i) $3^2\times 3^4\times 3^8$ $=3^{(2+4+8)}=3^{14}[\because {\mathrm{a}}^{\mathrm{m}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}]}$ (ii) $6^{15}÷6^{10}=6^{15-10}=6^5\left[\because a^m÷a^n=a^{m-n}\right]$ (iii) $a^3\times a^2=a^{3+2}=a^5\left[\because a^m\times a^n=a^{m+n}\right]$ (iv) $7^{\mathrm{x}}\times 7^2=7^{\mathrm{x}+2}\left[\because {\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right]$ (v) ${\left(5^2\right)}^3÷5^3=5^{2\times 3}÷5^3\left[\because {\left(a^m\right)}^n=a^{m\times n}\right]$ $=5^6÷5^3=5^{6-3}=5^3\left[\because a^m÷a^n=a^{m-n}\right]$ (vi) $2^5\times 5^5=(2\times 5{)}^5=10^5\left[\because a^m\times b^m=(a\times b{)}^m\right]$ (vii) ${\mathrm{a}}^4\times {\mathrm{b}}^4=(\mathrm{a}\times \mathrm{b}{)}^4\left[\because {\mathrm{a}}^{\mathrm{m}}\times {\mathrm{b}}^{\mathrm{m}}=(\mathrm{a}\times \mathrm{b}{)}^{\mathrm{m}}\right]$ (viii) ${\left(3^4\right)}^3=3^{4\times 3}=3^{12}\left[\because {\left(a^m\right)}^n=a^{m\times n}\right]$ (ix) $\left(2^{20}÷2^{15}\right)\times 2^3$

$$\begin{gathered} =\left(2^{20-15}\right)\times 2^3\left[\because a^m÷a^n=a^{m-n}\right] \\ =2^5\times 2^3=2^{5+3}=2^8\left[\because a^m\times a^n=a^{m+n}\right] \end{gathered}$$ (x) $8^t÷8^2=8^{\mathrm{t}-2}\left[\because {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$

• Simplify and express each of the following in exponential form : (i) $\frac{2^3\times 3^4\times 4}{3\times 32}$ (ii) $\left[{\left(5^2\right)}^3\times 5^4\right]÷5^7$ (iii) $25^4÷5^3$ (iv) $\frac{3\times 7^2\times 11^8}{21\times 11^3}$ (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) $\left(3^0+2^0\right)\times 5^0$ (ix) $\frac{2^8\times a^5}{4^3\times a^3}$ (x) $\left(\frac{a^5}{a^3}\right)\times a^8$ (xi) $\frac{4^5\times a^8{b}^3}{4^5\times a^5{b}^2}$ (xii) ${\left(2^3\times 2\right)}^2$ Sol. (i) $\frac{2^3\times 3^4\times 4}{3\times 32}=\frac{2^3\times 3^4\times 2^2}{3\times 2^5}=\frac{2^{3+2}\times 3^4}{3\times 2^5}$ $\left[\because {\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right]$ $=\frac{2^5\times 3^4}{3\times 2^5}=2^{5-5}\times 3^{4-1}\left[\because {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$ $=2^0\times 3^3=1\times 3^3=3^3$ (ii) $\left[{\left(5^2\right)}^3\times 5^4\right]÷5^7$ $$\begin{gathered} =\left[5^6\times 5^4\right]÷5^7\left[\because {\left(a^m\right)}^n=a^{m\times n}\right] \\ =\left[5^{6+4}\right]÷5^7=5^{10}÷5^7\left[\because a^{m\times }\times a^n=a^{m+n}\right] \\ =5^{10-7}=5^3\left[\because a^m÷a^n=a^{m-n}\right] \end{gathered}$$ (iii) $25^4÷5^3$ $={\left(5^2\right)}^4÷5^3=5^8÷5^3\left[\because {\left(a^m\right)}^n=a^{m\times n}\right]$ $=5^{8-3}=5^5\left[\because {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$ (iv) $\frac{3\times 7^2\times 11^8}{21\times 11^3}=\frac{3\times 7^2\times 11^8}{3\times 7\times 11^3}$ $=3^{1-1}\times 7^{2-1}\times 11^{8-3}\left[\because {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$ $=3^0\times 7^1\times 11^5=7\times 11^5$

• $=3^{7-7}=3^0=1\left[\because a^m÷a^n=a^{m-n}\right]$ (vi) $2^0+3^0+4^0=1+1+1=3\left[\because a^0=1\right]$ (vii) $2^0\times 3^0\times 4^0=1\times 1\times 1=1\left[\because a^0=1\right]$ (viii) $\left(3^0+2^0\right)\times 5^0=(1+1)\times 1$ $=2\times 1=2\left[\because a^0=1\right]$ (ix) $\frac{2^8\times a^5}{4^3\times a^3}=\frac{2^8\times a^5}{{\left(2^2\right)}^3\times a^3}=\frac{2^8\times a^5}{2^6\times a^3}$ $\left[\because {\left(a^m\right)}^n=a^{m\times n}\right]$ $=2^{8-6}\times a^{5-3}=2^2\times a^2\left[\because a^m÷a^n=a^{m-n}\right]$ $=(2\mathrm{a}{)}^2\left[\because {\mathrm{a}}^{\mathrm{m}}\times {\mathrm{b}}^{\mathrm{m}}=(\mathrm{a}\times \mathrm{b}{)}^{\mathrm{m}}\right]$ (x) $\left(\frac{a^5}{a^3}\right)\times a^8=\left(a^{5-3}\right)\times a^8=a^2\times a^8$ $\left[\because {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$ $={\mathrm{a}}^{2+8}={\mathrm{a}}^{10}\left[\because {\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right]$ (xi) $\frac{4^5\times {\mathrm{a}}^8{\mathrm{b}}^3}{4^5\times {\mathrm{a}}^5{\mathrm{b}}^2}=4^{5-5}\times {\mathrm{a}}^{8-5}\times {\mathrm{b}}^{3-2}=40\times {\mathrm{a}}^3\times \mathrm{b}$ $\left[\because {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]$ $=1\times a^3\times b=a^{3}b\left[\because a^0=1\right]$ (xii) ${\left(2^3\times 2\right)}^2={\left(2^{3+1}\right)}^2={\left(2^4\right)}^2$ $\left[\because {\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right]$ $=2^{4\times 2}=2^8\left[\because {\left(a^m\right)}^n=a^{m\times n}\right]$
• Say true or false and justify your answer : (i) $10\times 10^{11}=100^{11}$ (ii) $2^3>5^2$ (iii) $2^3\times 3^2=6^5$ (iv) $3^0=(1000{)}^0$ Sol. (i) $10\times 10^{11}=100^{11}$ L.H.S. $10^{1+11}=10^{12}$ and R.H.S. ${\left(10^2\right)}^{11}=10^{22}$ Since, L.H.S. $\ne$ 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. $\ne$ 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) $108\times 192$ $$\begin{gathered} =\left(2^2\times 3^3\right)\times \left(2^6\times 3\right) \\ =2^{2+6}\times 3^{3+1} \\ =2^8\times 3^4 \end{gathered}$$

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

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

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

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

Exercise: 11.3

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