Indices Exponents Class 9 OP Malhotra Exe-6A ICSE Maths Solutions

Indices Exponents Class 9 OP Malhotra Exe-6A 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-6A ICSE Maths Solutions Ch-6

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-6	Indices Exponents
Writer	OP Malhotra
Exe-6A	Problems on Indices Exponents
Edition	2025-2026

Problems on Indices Exponents with Solutions

 Class 9 OP Malhotra Exe-6A ICSE Maths Solutions Ch-6

Que-1: Write the products in the exponential form:

(i) (x . x . x . x. x) (x. x)
(ii) -1 (n . n. n) (n . n)
(iii) y⁸ x y⁵ x y²
(iv) – x (- x⁴)
(v) (3a⁷b⁸c⁹) x (5a²⁷b¹⁶c⁸)
(vi) (x + 2)² . (x + 2)⁴
(vii) (2m – 3n)^3a-2b x (2m – 3n)^6a+10b
(viii) x^2a+b-c.x^2c+a-b.x^2b+c-a

Sol:  (i) (x . x . x . x. x) (x. x)
= (x^5) . (x²)
= x^(5+2)      [x^m . x^n = x^(m+n)]
= x^7.

(ii)-1 (n . n. n) (n . n)
= -1(n³) . (n²)
= -1{n^(3+2)
= -n^5.

(iii) y⁸ x y⁵ x y²
= y^(8+5+2)
= y^15.

(iv) – x (- x⁴)
= x^1 . x^5
= x^(1+5)
= x^6.

(v) (3a⁷b⁸c⁹) x (5a²⁷b¹⁶c⁸)
= 3×5×a^(7+27).b^(8+16).c^(9+8)
= 15a^34.b^24.c^17

(vi) (x + 2)² . (x + 2)⁴
= (x+2)^(2+4)
= (x+2)^6.

(vii) (2m – 3n)^3a-2b x (2m – 3n)^6a+10b
= (2m-3n)^(3a-2b+6a+10b)
= (2m-3n)^(9a+8b)

(viii) x^2a+b-c.x^2c+a-b.x^2b+c-a
= x^(2a+b-c+2c+a-b+2b+c-a)
= x^(2a+2b+2c).

Que-2: Write each expression in the simpler form:
(i) (xy)³
(ii) (-x)^5
(iii) (-2xy)^4
(iv) 1/2
(v) (x²)^5
(vi) (7³)^8
(vii) (6a²)³
(viii) (-x²y³)³
(ix) (p²)^5 x (p³)²
(x) 3 (x^4.y³)^10 x 5(x²y²)³
(xi) (c³/d²)^7
(xii) (3p²/4q²)^n
(xiii) (a²b²/x²y³)^m

Sol: (i) (xy)³
= x³.y³

(ii) (-x)^5
= -x×-x×-x×-x×-x
= -x^5

(iii) (-2xy)^4
= (-2xy)×(-2xy)×(-2xy)×(-2xy)
= 16x^4.y^4

(iv) 1/2

(v) (x²)^5
= x²×x²×x²×x²×x²
= x^10.

(vi) (7³)^8
= 7³×7³×7³×7³×7³×7³×7³×7³
= 7^24.

(vii) (6a²)³
= 6a²×6a²×6a²
= 216a^6

(viii) (-x²y³)³
= -x²y³×(-x²y³)×(-x²y³)
= -x^6.y^9.

(ix) (p²)^5 x (p³)²
= p^(2×5) × p^(3×2)
= p^10 × p^6
= p^(10+6)
= p^16.

(x) 3 (x^4.y³)^10 x 5(x²y²)³
= 3x^(4×10).y^(3×10) × 5x^(2×3).y^(2×3)
= 3x^40. y^30 × 5x^6. y^6
= 15x^(40+6). y^(30+6)
= 15x^46. y^36

(xi) (c³/d²)^7
= {c^3(×7)}/{d^(2×7)}
= (c^21)/(d^14)

(xii) (3p²/4q²)^n
= {3^n. p^(n×2)}/{4^n. q^(2×n)}
= {3^n. p^2n}/{4^n. q^2n}

(xiii) (a²b²/x²y³)^m
= {a^(2×m). b^(2×m)}/{x^(2×m). y^(3×m)}
= {a^2m. b^2m}/{x^2m. y^3m}.

Que-3: Find the quotient:
(i) x^6 ÷ x²
(ii) x^(2a) + x^4
(iii) (p^5.q³)/(p³q²)
(iv) (−8x^27.y^21) ÷ (−16x^6.y^17)
(v) {4pq²(−5pq³)}/10p²q²
(vi) {(−4ab²)²}/16ab
(vii) {x^(a−b).y^(c−d)}/{x^(2b−a).y^c
(viii) [{m^(3n−9)}^6]/m^(2n−4)
(ix) [{(x^(2a−4))²}/x^(a+5)]³

Sol:  (i) x^6 ÷ x²
= x^(6-2)
= x^4

(ii) x^(2a) + x^a
= x^(2a-a)
= x^a

(iii) (p^5.q³)/(p³q²)
= {p^(5-3). q^(3-1)}
= p².q¹ = p².q

(iv) (−8x^27.y^21) ÷ (−16x^6.y^17)
= {-8/16} x^(27-6). y^(21-17)
= (-1/2) x^21. y^14.

(v) {4pq²(−5pq³)}/10p²q²

= -2p^(2-2). q^(5-2)
= -2p^0. q³
= -2q³

(vi) {(−4ab²)²}/16ab

= a^(2-1). b^(4-1)
= a.b³

(vii) {x^(a−b).y^(c−d)}/{x^(2b−a).y^c

= x^(a-b-2b+a). y^(c-d-c)
= x^(2a-3b). y^(-d)

(viii) [{m^(3n−9)}^6]/m^(2n−4)

= m^(16n-50).

(ix) [{(x^(2a−4))²}/x^(a+5)]³

= [{x^4a-8}/{x^(a+5)}]³
= [x^{4a-8-a-5}]³
= x^(3a-13)³
= x^(9a-39).

— : End of Indices Exponents Class 9 OP Malhotra Exe-6A ICSE Maths Ch-6 Step by Step Solutions :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-9 Maths

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