Class 8 Maths Chapter 12 Exponents and Powers
NCERT Solutions For Class 8 Maths Chapter 12 Exponents And Powers, (Maths) exam are Students are taught thru NCERT books in some of state board and CBSE Schools. As the chapter involves an end, there is an exercise provided to assist students prepare for evaluation. Students need to clear up those exercises very well because the questions withinside the very last asked from those.
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NCERT Solutions For Class 8 Maths Chapter 12 Exponents and Powers
Class 8 Maths Chapter 12 Exponents and Powers
Exercise 12.1
Page No: 197
1. Evaluate:
(i) 3-2 (ii) (-4)-2 (iii) (1/2)-5
Solution:
(i) 3-2 = (1/3)2
= 1/9
(ii) (-4)-2 = (1/-4)2
= 1/16
(iii) (1/2)-5 = (2/1)5
= 25
= 32
2. Simplify and express the result in power notation with positive exponent:
(i) (-4)4 ÷(-4)8
(ii) (1/23)2
(iii) -(3)4×(5/3)4
(iv) (3-7÷3-10)×3-5
(v) 2-3×(-7)-3
Solution:
(i)
= (-4)5/(-4)8
= (-4)5-8
= 1/(-4)3
(ii) (1/23)2
= 12/(23)2
= 1/23×2 = 1/26
(iii) -(3)4×(5/3)4
= (-1)4×34×(54/34 )
= 3(4-4)×54
= 30×54 = 54
(iv)
= (3-7/3-10)× 3-5
= 3-7 – (-10) × 3-5
= 3(-7+10)×3-5
= 33×3-5
= 3(3+-5)
= 3-2
=1/32
(v) 2-3×(-7) – 3
= (2×-7)-3
(Because am×bm = (ab)m)
= 1/(2×-7)3
= 1/(-14)3
3. Find the value of :
(i) (30+4-1)×22
(ii) (2-1×4-1)÷2 – 2
(iii) (1/2)-2+(1/3)-2+(1/4)-2
(iv) (3-1+4-1+5-1)0
(v) {(-2/3)-2}2
Solution:
(i)(30+4– 1)×22 = (1+(1/4))×22
= ((4+1)/4 )×22
= (5/4)×22
= (5/22)×22
= 5×2(2-2)
= 5×20
= 5×1 = 5
(ii)(2-1×4-1)÷2-2
= [(1/2)×(1/4)] ÷(1/4)
= (1/2×1/22 )÷ 1/4
= 1/23÷1/4
= (1/8)×(4)
= 1/2
(iii) (1/2)-2+(1/3)-2+(1/4)-2
= (2-1)-2+(3-1)-2+(4-1)-2
= 2(-1×-2)+3(-1×-2)+4(-1×-2)
= 22+32+42
= 4+9+16
=29
(iv) (3-1+4-1+5-1)0
= 1
(v) {(-2/3)-2}2 = (-2/3)-2×2
= (-2/3)-4
= (-3/2)4
= 81/16
4. Evaluate
(i) (8-1×53)/2-4
(ii) (5-1×2-2)×6-1
Solution:
(i) (8-1×53)/2-4
= 2×125 = 250
(ii) (5-1×2-2)×6-1
= (1/10)×1/6
= 1/60
5. Find the value of m for which 5m ÷ 5-3 = 55
Solution:
5m ÷ 5-3 = 55
5(m-(-3) ) = 55
5m+3 =55
Comparing exponents both sides, we get
m+3 = 5
m = 5-3
m = 2
6. Evaluate
(i)
(ii)
Solution:
(i)
= 3-4
= -1
(ii)
= 512/125
7. Simplify.
(i)
(ii)
Solution:
(i)
=
=
(ii)
= 1×1×3125
= 3125
Exercise 12.2
Page No: 200
1. Express the following numbers in standard form.
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000
Solution:
(i) 0.0000000000085 = 0.0000000000085×(1012/1012) = 8.5 ×10-12
(ii) 0.00000000000942 = 0.00000000000942×(1012/1012) = 9.42×10-12
(iii) 6020000000000000 = 6020000000000000×(1015/1015) = 6.02×1015
(iv) 0.00000000837 = 0.00000000837×(109/109) = 8.37×10-9
(v) 31860000000 = 31860000000×(1010/1010) = 3.186×1010
2.Express the following numbers in usual form.
(i) 3.02×10-6
(ii) 4.5×104
(iii)3×10-8
(iv)1.0001×109
(v) 5.8×1012
(vi)3.61492×106
Solution:
(i) 3.02×10-6 = 3.02/106 = 0 .00000302
(ii) 4.5×104 = 4.5×10000 = 45000
(iii) 3×10-8 = 3/108 = 0.00000003
(iv) 1.0001×109 = 1000100000
(v) 5.8×1012 = 5.8×1000000000000 = 5800000000000
(vi) 3.61492×106 = 3.61492×1000000 = 3614920
3. Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 1/1000000 m.
(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.
(iii) Size of bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm
Solution :
(i) 1 micron = 1/1000000
= 1/106
= 1×10-6
(ii) Charge of an electron is 0.00000000000000000016 coulombs.
= 0.00000000000000000016×1019/1019
= 1.6×10-19 coulomb
(iii) Size of bacteria = 0.0000005
= 5/10000000 = 5/107 = 5×10-7 m
(iv) Size of a plant cell is 0.00001275 m
= 0.00001275×105/105
= 1.275×10-5m
(v) Thickness of a thick paper = 0.07 mm
0.07 mm = 7/100 mm = 7/102 = 7×10-2 mm
4. In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?
Solution:
Thickness of one book = 20 mm
Thickness of 5 books = 20×5 = 100 mm
Thickness of one paper = 0.016 mm
Thickness of 5 papers = 0.016×5 = 0.08 mm
Total thickness of a stack = 100+0.08 = 100.08 mm
= 100.08×102/102 mm
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