Indices Exponents Class 9 OP Malhotra Exe-6B ICSE Maths Solutions Ch-6. We Provide Step by Step Solutions / Answer of Questions on Indices Exponents OP Malhotra Maths. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.
Indices Exponents Class 9 OP Malhotra Exe-6B ICSE Maths Solutions Ch-6
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 9th |
| Chapter-6 | Indices Exponents |
| Writer | OP Malhotra |
| Exe-6B | Problems on Indices Exponents |
| Edition | 2025-2026 |
Indices Exponents Exercise- 6B Class 9 OP Malhotra Maths
Que-1: Evaluate:
(i) 5°
(ii) 2¯³
(iii) 10^-4
(iv) (- 2)¯²
(v) (2/3)°
(vi) (3/4)¯³
(vii) (−1/2)¯²
(viii) 2¯²/5¯²
(ix) {(3x^0)−1}/{(3x^0)+1}
Sol: (i) 5°
= 1
(ii) 2¯³
= 1/2³
= 1/8
(iii) 10^-4
= (1/10)^4
= 1/10000
(iv) (- 2)¯²
= (-1/2)²
= 1/4
(v) (2/3)°
= 1
(vi) (3/4)¯³
= (4/3)³
= 64/27
(vii) (−1/2)¯²
= (-2)²
= 4
(viii) 2¯²/5¯²
= 5²/2²
= 25/8
(ix) {(3x^0)−1}/{(3x^0)+1}
= {3×1-1}/{3×1+1}
= (3-1)/(3+1)
= 2/4
= 1/2
Que-2: Evaluate:
(i) 4^(1/2)
(ii) 8^(1/3)
(iii) 16^(1/4)
(iv) -27^(1/3)
(v) 32^(3/5)
(vi) (125)^(−2/3)
(vii) (8/125)^(1/3)
(viii) (1/216)^(−2/3)
(ix) -(- 27)^(-4/3)
(x) (27/8)^(−2/3)
(xi) (0.0625)^(3/4)
(xii) {12*(19/27)}^(1/3)
Sol: (i) 4^(1/2) = (2×2)^(1/2)
= (2^2)^(1/2) = 2^(2×1/2)
= 2^1 = 2
(ii) 8^(1/3) = (2×2×2)^(1/3)
= (2^3)^(1/3)
= 2^(3×1/3)
= 2^1 = 2
(iii) 16^(1/4) = (2×2×2×2)^(1/4)
= (2^4)^(1/4)
= 2^(4×1/4)
= 2^1 = 2
(iv) -27^(1/3) = [(−3)×(−3)×(−3)]^(1/3)
= [(−3)^3](1/3)
= (−3)^(3×1/3)
= (-3)^1 = – 3
(v) 32^(3/5) = (2×2×2×2×2)^(3/5)
= (2^5)^(3/5)
= 2^(5×3/5)
= 2^3
= 2 x 2 x 2 = 8
(vi) (125)^(−2/3) = (5×5×5)^(−2/3)
= (5^3)^(−2/3)
= 5^(3×(−2/3))
= 5^−2
= 1/(5)² = 1/(5×5)
= 1/25
(vii) (8/125)^(1/3)
= {(2/5)^3}^(1/3)
= (2/5)^(3×1/3)
= (2/5)^1 = 2/5.
(viii) (1/216)^(−2/3)
= {(1/6)^3}^(-2/3)
= (1/6)^(3×(-2/3))
= (1/6)^-2
= 6^2
= 36.
(ix) -(- 27)^(-4/3)
= {-(-3)^3}^(-4/3)
= -3^(3×(-4/3))
= -3^-4
= -1/81.
(x) (27/8)^(−2/3)
= {(3/2)^3}^(-2/3)
= (3/2)^-2
= (2/3)^2
= 4/9
(xi) (0.0625)^(3/4)
= {(0.5)^4}^(3/4)
= (0.5)^{4×(3/4)}
= (0.5)^3
= 0.125
(xii) {12*(19/27)}^(1/3)
= (343/27)^(1/3)
= {(7/3)^3}^(1/3)
= (7/3)^{3×(1/3)}
= 7/3 = 2*(1/3).
— : End of Indices Exponents Class 9 OP Malhotra Exe-6B ICSE Maths Ch-6. Step by Step Solutions :–
Return to :– OP Malhotra S Chand Solutions for ICSE Class-9 Maths
Thanks
Please Share with Your Friends
