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Maths Ace Class 8 Solutions Chapter 6 Algebraic Expressions
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Ace Prime Class 8 Math Book, Chapter 6, Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions, Exercise 6.1, 6.2, 6.3 and 6.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 6 solutions.
Algebraic Expressions Exercise 6.1 Solution :
Question no – (1)
Solution :
(a) 8ab, – 10ba
= It’s a Like term.
(b) 23a2b, ab2
= It’s a Unlike term.
(c) xyz, -xyz
= It’s a Like term.
(d) 1/3 st, – 1/√3 st
= It’s a Like term.
Question no – (2)
Solution :
(a) 3x + 5
= Yes, it is a Polynomial.
(b) 3xy-2 + 7y + 2
= No, it is not a Polynomial.
(c) 5
= Yes, it is a Polynomial.
(d) 3√x + 5
= No, it is not a Polynomial.
(e) 5/(x + 1)
= No, it is not a Polynomial.
(f) 5xy2 + 3x2y – 2x2y2
= Yes, it is a Polynomial.
Question no – (3)
Solution :
(a) x2 + y2 + z2
= Trinomial
(b) 100
= Monomial
(c) xyz2 + x2yz
= Binomial
(d) a2 + b2 + c2
= Trinomial
(e) -9x – 2 + 3x
= Binomial
(f) p + 2
= Binomial
Question no – (4)
Solution :
(a) m3 + 3m2 – 18m, 23m – 4m2 and 5m3 – 1
= m3 + 3m2 – 18m + 23m – 4m2 + 5m3 – 1
= 6m3 – m2 + 5m – 1
(b) 1 – x – x2, x + 2 and 3x2
= (1 – x – x2) + (x + 2) + 3x2
= 1 – x – x2 + x + 2 + 3x2
= 2x2 + 3
(c) x2y2 – xy2 + 4xy – 6x2y and 7x2y + 9xy
= (x2y2 – xy2 + 4xy – 6x2y) + (7x2y + 9xy)
= x2y2 – xy2 + 4xy – 6x2y + 7x2y + 9xy
= x2y2 – xy2 + 13xy + x2y
Question no – (5)
Solution :
(a) 4z2xy + x2zy – 5xyz + 2xy2z from x2zy + xyz + xy2z – z2xy
∴ (x2zy + xyz + xy2z – z2xy) – (4z2xy + x2zy – 5xyz + 2xy2z)
= x2zy + xyz + xy2z – z2xy – 4z2xy – x2zy + 5xyz – 2xy2z
= 6xyz – xy2z – 5z2xy
(b) 3m2 – 2 + 11m from m3 – 9
Now, = m3 – 9 – (3m2 – 2 + 11m)
= m3 – 9 – 3m2 + 2 – 11m
= m3 – 3m2 – 11m – 7
Question no – (6)
Solution :
= 9a2 – 14a – (a2 + 7a)
= 9a2 – 14a – a2 – 7a
= 8a2 – 21a
Therefore, Number, should be added 8a2 – 21a
Question no – (7)
Solution :
= (ab + 1) – (5ab – 5)
= ab + 1 – 5ab + 5
= – 4ab + 6
Thus, – 4ab + 6 should be subtracted.
Question no – (8)
Solution :
As per the question,
Edward bought a packet of chips for = Rs (3x2 + y)
And a chocolate for = Rs (y2 + 2y).
∴ Total pay,
= (3x2 + y + y2 + 2y)
= (3x2 + y2 + 3y) Rs.
Question no – (9)
Solution :
A rectangle has length (x2 + 4y3) units
Breadth (-2x2 + y3) units.
∴ Perimeter,
= 2{(x2 + 4y3) + (- 2x2 + y3)}
= 2(x2 + 4y3 – 2x2 + y3)
= 2(5y3 – x2)
Therefore, its perimeter will be 2(5y3 – x2)
Algebraic Expressions Exercise 6.2 Solution :
Question no – (1)
Solution :
(a) ab × bc
= ab2c
(b) 2dc2 × (-6dc)
= -12d2c3
(c) str × 7r2s2
= 7r3s3t
(e) 99pqr × p3q2r
= 99 p4 q3 r2
(f) x × x × x
= x3
(g) 2ad × 5bd
= 10abd2
(h) (1/8x3y) (5/6x2y2)
= (1/8 × 5/6) (x5 y3)
= 5/48 x5 y3
(i) (3xy12)(-xy2)
= -3 x2 y14
(j) 9abc(3/8a2b2c2)
= (9 × 3/8) a3b3c3
= 27/8 a3b3c3
Question no – (3)
Solution :
(a) (5 – d2 + 2d)(2d – 1)
= (10d – 5 – 2d3 + d2 + 4d2 – 2d)
= -2d3 + 5d2 + 8d – 5
(b) pq(p2 – q2 + 1)
= p3q – pq3 + pq
(c) (x2 – y2)(4x3 – y3)
= 4x5 – x2y3 – 4x3y + y5
Question no – (4)
Solution :
As per the question,
Rectangle length is = (2x + 7) units
And breadth is = (x – 5) units.
Area of rectangle,
= (2x + 7) (x – 5)
= 2x2 – 10x + 7x – 35
= 2x2 – 3x – 35
Hence, the area of a rectangle will be 2x2 – 3x – 35
Question no – (5)
Solution :
According to the given question,
Cuboid length, breadth and height are 3ax units, (2x – a) units and 5a units,
∴ Volume of Cube,
= (3ax) (2x – a) (5a)
= (6ax2 – 3a2x) (5a)
= 30a2x2 – 15a3x
= 30a2x2 – 15a3x
Therefore, the volume of a cuboid will be 30a2x2 – 15a3x.
Algebraic Expressions Exercise 6.3 Solution :
Question no – (1)
Solution :
(a) 33x2y2z2 ÷ 11x2y2z
= 3z
(b) 119xyz2 ÷ 17z2
= 7 xy
(c) 57xz ÷ xz
= 57
(d) x3y4z6 ÷ yz
= x3y3z5
(e) –x2yz ÷ x2z
= – x2yz/x2z
= -y
(f) (12x2 + 3xy) ÷ 3x
= (4 + y)
(g) (45x2y2z2 + 36y2x2z2 + 18z2y2x2) ÷ 9xyz
= 5 xyz + 4yxz + 2 zyx
= 11 xyz
(h) (-2x2y2 + 4y2x2 – 2xy) ÷ 2xy
= -xy + 2xy – 1
= xy – 1
Question no – (2)
Solution :
(a) 9x3y3 – 14xy + 20) ÷ (xy + 5)
= 9x2y2 – 45xy + 211
(b) (x2 + 17x – 9) ÷ (x + 3)
= x + 14
(d) (a2 + 2ab + b2) ÷ (a + b)
= a2 + 2ab + b2/a + b
= (a + b)2/a + b)
= a + b
(e) (x2 + 15x + 56) ÷ (x + 8)
= x2 + 15x + 56/x + 8
= x2 + 7x + 8x + 56/x + 8
= x(x + 7) + 8(x + 7)/(x + 8)
= (x + 8) (x + 7)/(x + 8)
= x + 7
Algebraic Expressions Exercise 6.4 Solution :
Question no – (1)
Solution :
(a) (x + 13) (x + 20)
= x2 + 20x + 13x + 260
= x2 + 33x + 260
(b) (x + 30) (x + 30)
= (x + 30)2
= x2 + 2.x. 30 + (30)2
= x2 + 60x + 900
(c) (x – 7) (x + 1) (x – 1)
= (x – 7)2
= x2 – 2.x.7 + 49
= x2 – 14x + 49
(d) (x + 1) (x – 1)
= x2 – 12
= x2 – 1
(e) (2m + 1/4) (2m + 1/4)
= (2m + 1/4)2
= (2m)2 + 2.2m. 1/4 + 1/16
= 4m2 + m + 1/16
(f) (3a + 1.5) (3a + 1.5)
= (3a)2 – (1.5)2
= 9a2 – 1.25
Question no – (2)
Solution :
(a) (109)2 + (91)2
= {(100 + 9)2 + (100 – 9)2}
= {(100)2 2.100.9 + 81} + {(100)2 – 2.100.0 + 81}
= (10,000 + 1800 + 81) + (10,000 – 1800 + 81)
= 11881 + 8281
= 20162
(b) (200)2 – (100)2
= (200 + 100) (200 – 100)
= 300 × 100
= 30000
(c) (1025)2 – (975)2
= (1025 + 975) (1025 – 978)
= 2000 × 50
= 10,0000
(d) (10)2 + (20)2
= 100 + 400
= 500
Question no – (3)
Solution :
(a) (210)2
= (200 + 10)2
= (200)2 + 2.200.10 + 100
= 40000 + 4000 + 100
= 44100
(b) (190)2
= (200 – 10)2
= (200)2 – 2.200.10 + 100
= 40,000 – 4000 + 100
= 36100
(c) (44)2
= (40 + 4)2
= 9400)2 + 2.40.4 + 16
= 1600 + 320 + 16
= 1936
Question no – (4)
Solution :
As we know, a = 3 and b = 4.
(a) (a + b)2 = a2 + 2ab + b2
L.H.S, (a + b)2
= (3 + 4)2
= (7)2
= 49
R.H.S, a2 + 2.ab + b2
= 32 + 2 × 3 × 4 + 16
= 9 + 24 + 16
= 49
∴ L.H.S = R.H.S …(Proved)
(b) (a – b)2 = a2 – 2ab + b2
L.H.S, (a – b)2
= (3 – 4)2
= (- 1)2
= 1
R.H.S, a2 – 2ab + b2
= 32 – 2.3 × 4 + 16
= 9 – 24 + 16
= 1
∴ L.H.S = R.H.S …(Proved)
(c) (a + b) (a – b) = a2 – b2
L.H.S, (a + b) (a – b)
= (3 + 4) (3 – 4)
= 7 × (- 1)
= – 7
R.H.S, a2 – b2
= 32 – 42
= 9 – 16
= – 7
∴ L.H.S = R.H.S …(Proved)
Next Chapter Solution :
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