An expression that represents repeated multiplication of the same factor is called a power. For example, in the case of 42, the number 4 is called the base, and the number 2 is the exponent. An exponent corresponds to the number of times the base is utilised as a factor in an expression.
——————————Exercise 10.1——————————-
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 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
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
NCERT Solutions for Class 8 Maths Chapter 9 Mensuration
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 on 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 the following:
(i)
(ii)
Solution:
(i)
=
(ii)
=
=
=
= 1×1×3125
= 3125
——————————Exercise 10.2——————————-
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 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
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
GLOBAL FOOTBALL FEVER: UNVEILING THE TOP 10 SOCCER LEAGUES AROUND THE WORLD
3. Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 1/1000000 m.
(ii) The charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.
(iii) Size of bacteria is 0.0000005 m
(iv) The size of a plant cell is 0.00001275 m
(v) The 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 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?
Solution:
The thickness of one book = 20 mm
Thickness of 5 books = 20×5 = 100 mm
The 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
