NCERT solutions for class 8 maths chapter 12 Exponents and Powers
The NCERT Solutions for Class 8 Mathematics Chapter 12, focusing on Exponents and Powers, provide invaluable assistance to students aiming for excellence in their examinations. These solutions, aligned with the CBSE curriculum, offer a comprehensive understanding of the concepts presented in the chapter.
Access Answers to NCERT solutions for class 8 maths chapter 12 Exponents and Powers
Students can access the NCERT solutions for class 8 maths chapter 12 Exponents and Powers. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 12.1
Evaluate:
(i) 3-2
(ii) (-4)-2
(iii) (1/2)-5
(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
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
(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
Evaluate:
(i) (8-1×53)/2-4
(ii) (5-1×2-2)×6-1
(i) (8-1×53)/2-4
=
= 2×125 = 250
(ii) (5-1×2-2)×6-1
= (1/10)×1/6
= 1/60
Find the value of m for which 5m ÷ 5-3 = 55
5m ÷ 5-3 = 55
5(m-(-3) ) = 55
5m+3 =55
Comparing exponents on both sides, we get
m+3 = 5
m = 5-3
m = 2
Simplify the following:
(i)
(ii)
(i)
=
(ii)
=
=
=
= 1×1×3125
= 3125
Simplify and express the result in power notation with a 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
(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
Evaluate:
(i)
(ii)
(i)
= 3-4
= -1
(ii)
=
=
= 512/125
Exercise 12.2
Express the following numbers in standard form.
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000
(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
Express the following numbers in the 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
(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
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
(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
In a stack, there are 5 books, each having a thickness of 20 mm and 5 paper sheets, each having a thickness of 0.016 mm. What is the total thickness of the stack?
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
