Exercise 11.3 (Revised) - Chapter 13 - Exponents & Powers - Ncert Solutions class 7 - Maths

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Chapter 11 - Exponents & Powers - NCERT Solutions Class 7 Maths

Ex 11.3 Question 1.

Write the following numbers in the expanded form:
279404, 3006194, 2806196, 120719, 20068

Answer:

$
\begin{aligned}
& \text {(i) } 2,79,404=2,00,000+70,000+9,000+400+00+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{aligned}
$
$
\begin{aligned}
& \text { (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 10^6+0 \times 10^5+0 \times 10^4+6 \times 10^3+1 \times 10^2+9 \times 10^1+4 \times 10^0
\end{aligned}
$
$
\begin{aligned}
& \text { (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 10^6+8 \times 10^5+0 \times 10^4+6 \times 10^3+1 \times 10^2+9 \times 10^1+6 \times 10^0
\end{aligned}
$
$
\begin{aligned}
& \text { (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
\end{aligned}
$
$
\begin{aligned}
& \text { (v) } 20,068=20,000+00+00+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
\end{aligned}
$

Ex 11.3 Question 2.

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) $9 \times 10^5+2 \times 10^2+3 \times 10^1$

Answer:
$
\begin{aligned}
& \text {(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
\end{aligned}
$
$
\begin{aligned}
& \text { (b) } 4 \times 10^5+5 \times 10^3+3 \times 10^2+2 \times 10^0 \\
& =4 \times 100000+5 \times 1000+3 \times 100+2 \times 1 \\
& =400000+5000+300+2=4,05,302
\end{aligned}
$
$
\begin{aligned}
& \text { (c) } 3 \times 10^4+7 \times 10^2+5 \times 10^0 \\
& =3 \times 10000+7 \times 100+5 \times 1 \\
& =30000+700+5=30,705
\end{aligned}
$
$
\begin{aligned}
& \text { (d) } 9 \times 10^5+2 \times 10^2+3 \times 10^1 \\
& =9 \times 100000+2 \times 100+3 \times 10 \\
& =900000+200+30=9,00,230
\end{aligned}
$

Ex 11.3 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:

(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) $3908.78=3.90878 \times 1000=3.90878 \times 10^3$

Ex 11.3 Question 4.

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 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 $100,000,000,0000$ 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 $300,000,000,000,000,000,000 \mathrm{~m}$.
(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 $\mathrm{km}$ 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 $=384,000,000 \mathrm{~m}$
$
=384 \times 1000000 \mathrm{~m}=3.84 \times 100 \times 1000000=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 100,000,000 \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
$=12 \times 1000,000,000$ years
$=1.2 \times 10 \times 1000,000,000$ years $=1.2 \times 10^{10}$ years

(g) Distance of the Sun from the $=300,000,000,000,000,000,000 \mathrm{~m}$
centre of the Milky Way Galaxy $=3 \times 100,000,000,000,000,000,000 \mathrm{~m}$ $=3 \times 10^{20} \mathrm{~m}$
(h) Number of molecules in a drop $=60,230,000,000,000,000,000,000$
of water weighing $1.8 \mathrm{gm}=6023 \times 10,000,000,000,000,000,000$
$
=6.023 \times 1000 \times 10,000,000,000,000,000,000=6.023 \times 10^{22}
$
(i) The Earth has Sea water $=1,353,000,000 \mathrm{~km}^3$
$
\begin{aligned}
& =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
\end{aligned}
$
$
\begin{aligned}
& \text { (j) The population of India }=1,027,000,000 \\
& =1027 \times 1000000 \\
& =1.027 \times 1000 \times 1000000 \\
& =1.027 \times 10^9
\end{aligned}
$