Ncert Solutions For Class 7 Maths Ex 13.3

Ncert Solutions For Class 7 Maths Chapter 13 Ex 13.3

Q.1. Write the following numbers in the expanded form:

(i)279404

(ii)3006194

(iii)2806196

(iv)120719

(v)20068

Solution:

(i)279404=2,00,000+70,000+9,000+400+00+4

=2×100000+7×10000+9×1000+4×100+0×10+4×1

=2×105+7×104+9×103+4×102+0×101+4×100

 

(ii)3006194=30,00,000+0+0+6,000+100+90+4

=3×1000000+0×100000+0×10000+6×1000+1×100+9×10+4×1=3×106+0×105+0×104+6×103+1×102+9×101+4×100

 

(iii)2806196=20,00,000+8,00,000+0+6,000+100+90+6

=2×1000000+8×100000+0×10000+6×1000+1×100+9×10+6×1=2×106+8×105+0×104+6×103+1×102+9×101+6×100

 

(iv)120719=1,00,000+20,000+0+700+10+9

==1×100000+2×10000+0×1000+7×100+1×10+9×1=1×105+2×104+0×103+7×102+1×101+9×100

 

(v)20068=20,000+00+00+60+8

==2×10000+0×1000+0×100+6×10+8×1=2×104+0×103+0×102+6×101+8×100

 

Q.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

Solution:

(a)8×104+6×103+0×102+4×101+5×100

8×10000+6×1000+0×100+4×10+5×1=80000+6000+0+40+5=86,045

 

(b)4×105+5×103+3×102+2×100

4×100000+5×1000+3×100+2×1=400000+5000+300+2=4,05,302

 

(c) 3×104+7×102+5×100

3×10000+7×100+5×1=30000+700+5=30,705

 

(d)9×105+2×102+3×101

9×100000+2×100+3×10=900000+200+30=9,00,230

 

Q.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

Solution:

(i)5,00,00,000 =5×1,00,00,000=5×107

(ii)70,00,000=7×10,00,000=7×106

(iii)3,18,65,00,000= 31865×100000=3.1865×10000×100000=3.1865×109

(iv)3,90,878 =3.90878×100000=3.90878×105

(v)39087.8 =3.90878×10000=3.90878×104

(vi)3908.78=3.90878×1000=3.90878×103

 

Q.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 Earth id 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.

Solution:

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

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

(iii)Diameter of the Earth=1,27,56,000m=12756×1000m=1.2756×10000×1000m=1.2756×107m

(iv)Diameter of the Sun=1,400,000,000m=14×100,000,000m=1.4×10×100,000,000m=1.4×109m

(v)Average of Stars=100,000,000,000=1×100,000,000,000=1×1011

(vi)Years of Universe=12,000,000,000years=12×1000,000,000years=1.2×10×1000,000,000years=1.2×1010years

(vii)Distance of the Sun from the centre of the Milky Way Galaxy=300,000,000,000,000,000,000m=3×100,000,000,000,000,000,000m=3×1020m

(viii)Number of molecules in a drop of water weighing 1.8 gm=60,230,000,000,000,000,000,000=6023×10,000,000,000,000,000,000=6.023×1000×10,000,000,000,000,000,000=6.023×1022

(ix)The Earth has Sea water=1,353,000,000km33=1,353×1000000km3=1.353×1000×1000,000km3=1.353×109km3

(x) The population of India=1,027,000,000=1027×1000000=1.027×1000×1000000=1.027×109

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