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Mathsight Class 7 Solutions Chapter 6 Algebraic Expressions
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students to solve MathSight Class 7 Mathematics Book, Chapter 6, Algebraic Expressions. Here students can easily get all the exercise questions solution for Chapter 6, Algebraic Expressions Exercise 6.1, 6.2, 6.3, and 6.4
Algebraic Expressions Exercise 6.1 Solution :
Question no – (1)
Solution :
(d) x2y2
Question no – (2)
Solution :
(a) -5
(b) 3
(c) 7
(d) 9
Question no – (3)
Solution :
(a) 7
(b) 3
(c) 0
(d) 9
Question no – (4)
Solution :
(a) -5, 3
(b) 7, 3
(c) -4
(d) -1
Question no – (5)
Solution :
(a) -3y
(b) 18a2c
(c) -37b2y
(d) 59
Question no – (6)
Solution :
Like terms :
5x2, 17x2y’
– 4xy2, -4xy2’
-7a, 14a, 55a
45x2y2, 3x2y2
Question no – (7)
Solution :
(a) Degree = 4
(b) Degree =1
(c) Degree =1
(d) Degree =4
(e) Degree =5
Question no – (8)
Solution :
(a) It is a Binomial
(b) It is a Monomial
(c) It is a Trinomial
(d) It is a Trinomial
(e) It is a Trinomial
(f) It is a Trinomial
(g) It is a Binomial
(h) It is a Monomial
Question no – (9)
Solution :
| Term | Degree | |
| (a) 2a2 + b2c2 + 7 → | Trinomial | 2 |
| (b) 9a2bc → | Monomial | 4 |
| (c) a2 + 2a + 3 → | Trinomial | 2 |
| (d) 47x2 + 34y → | Binomial | 2 |
| (e) p3 + 3pq + 4q → | Trinomial | 3 |
| (f) 3a + 7 → | Binomial | 1 |
Question no – (10)
Solution :
(a) 5x2y + 7 → 5x2y, 7
(b) 7xy2z + 3x → 7xy2z, 3x
(c) -3a2b2c → -3,a2, b2, c
(d) pq2 + 3pqr → pq2, 3pqr
Question no – (11)
Solution :
A2b2 + ab + 5
Question no – (12)
Solution :
Y = x/6
Or, y + x/6 = 0
Or, 6y + x/6 = 0
Or, x + 6y = 0
Algebraic Expressions Exercise 6.2 Solution :
Question no – (1)
Solution :
(a) (4x2+7x2)
=11x2, -11x.112x – 11x
(b) (9x2y+7x2y)
= 16x2y , -15x3y + 4xy2 + 16x2y
(c) (17a + 15a)
= 32a , 9 – 20b + 2b)
= – 18b , 32a -18b
(d) (7xy + 14xy)
= 21xy , (-8y2 – 4y2)
= 12y2 , 21 xy – 12y2
Question no – (2)
Solution :
(a) 15x2 – x2 + 17x2 – 3x2
= 28 x2
(b) -2a + 10a + 6a – 40a
= -44a
(c) a + b – a + b + a – b – a – b
= 0
(d) 8a – 7b + 4a + 3b + 14b – 2a
= 10a +10b
Question no – (3)
Solution :
(a) a4 + b4 + 2a2b2
2a4 – 7b4 – 4a2b2
+ 2a4 – 3b4
——————————
5a4 – 9b4 – 2a2b2
(b) -56a2b
-318a2b
+ 29a2b
112a2b
—————————
54a2b
(c) 7a + 5b – 3c
+ 6a – 2b + 2c
5a + 4b – 4c
—————————
18a – 7b – 5c
(d) – x2 – y2 – z2
2x2 + 3y2 + 4z2
-x2 -z2
—————————
0 + 2y2 – 2z2
Question no – (4)
Solution :
(a) 2x – 8x2 – 9 + 5x2 + 3x + 4
= 5x – 3x2 – 5
(b) 7 + 4a – 7b + 2b + 4a – 6
= 13a – 5b + 1
(c) -7c + 4a – 2b + 2a – 7b + 4c
= 6a – 9b – 3c
(d) -4x2 + 7x + 1 + 3x2 + 10
= x2 + 7x + 11
Question no – (5)
Solution :
A + 2B + C + 2D
= 5a – 7b + 7 + 2(-4a – 7 + 4b) + 9a – 7 – 2b + 2(14b – 2a + 4)
= 5a – 7b+ 7 – 8a -14 + 8b + x – 7 – 2b + 28b – 4a + 8
= 2a + 27b – 6
Question no – (6)
Solution :
Given, 2p + 3q
= 2(4a – 4b – 4c) + 3(7a + 7b – c)
= 8a – 8b – 8c + 21a + 21b – 3c
= 29a + 13b – 11c
Question no – (7)
Solution :
(a) 2x + 2x + 2x + 5 + 2x + 5
= 8x + 10
(b) x + x + x + 1 + x + 1 + x + 5 + x + 4
= 6x + 11
Question no – (8)
Solution :
Cheetah run,
= 10 × 10
= 100 m.
Zebra run = 10k.
∴ Expression = 100 – 10k
Algebraic Expressions Exercise 6.3 Solution :
Question no – (1)
Solution :
(a) 17x2 – 15x2
= 2x2
(b) – 40a – 20
= – 60 a
(c) a – b – a – b
= 2b
(d) 14b – 2a – 11a – 7b
= -13a
= 7b
Question no – (2)
Solution :
(a) 2a4 – 7b4 – 4a2b2 – a4 – b4 – 2a2b2
= a4 – 6b4 – 6a2b2
(b) 29a2b2 + 56a2b2
= 85a2b2
(c) 0 – 7a + 5b-3c
= -7a + 5b-3c
(d) 1 – x2 – y2 – z2
(e) 5x2 + 3x + 4 – 2x + 8x2 + 9
= 13x2 + x + 13
(f) 2b + 4a – 6 – 7 – 9a + 7b
= -5a + 9b – 13
(g) 2a – 7b + 4c + 7c – 4a + 2b
= -2a – 5b + 11c
(h) -3x2 + 10 + 4x2 – 7x – 1
= x2 + 3x – 1
Question no – (3)
Solution :
(5x2 + x + 2 – 2x + 3x2 + 4) – x – x2 – 1
= 8x2 – x + 6 – x – x2 – 1
= 7x2 – 2x + 5
Question no – (4)
Solution :
3a2 – 4b2 + 6ab + 8b2 + 4a2 – 2ab
= 7a2 + 4b2 + 4ab
Question no – (5)
Solution :
7a + 7b – c-4a + 4b + 4c
= 3a + 11b + 3c
Question no – (6)
Solution :
10x – 8 – 7x + 5
= 3x – 3
Question no – (7)
Solution :
A – 2B
= 2x2 + 4x + 3 – 2(5x2 + 3)
= 2x2 + 4x + 3 – 10x2 – 6
= -8x2 + 4x – 3
Question no – (8)
Solution :
3x + 4 – 5x + 3x – 7 – 5
= x – 14
Or, x – 8
= x – 14
Question no – (9)
Solution :
1 week both a/c
= 10x + 120 + 50x + 35
= 60x + 155
(a) 5 week both a/c
= 5 (60x + 155)
= 300x + 775
Algebraic Expressions Exercise 6.4 Solution :
Question no – (1)
Solution :
(a) 5x + (3x – 2y + 5)
= 8x – 2y + 5
(b) 4x – {3x – (2y – x) + 4}
= 4x – 3x – 2y + x + 4
(c) a – [5b – {3b – (4 – 2a) + 7}]
= a – 5b – 3b – 4 – 2a + 7
= -a – 8b + 3
(d) 3 + [x – {4y – (5x + 2y – 5) + 2x2} – (3x2 + 6y)-5]
= 3 + x – 4y – 5x + 2y – 5 + 2x2 – 3x2 – 6y – 5
= -7 – 4x – 8y – x2
= -x2 – 4x – 8y – 7
(e) 10x + [x3 + 3x2 – {2x2 – (7 – 2x – 9x3) – 5x3} – 3x2 + 5]
= 10x + x3 + 3x2 – 2x2 – 7 + 2x + 9x3 – 5x3 – 3x2 + 5
= 12x + 5x3 – 2x2 – 2
(f) ab [bc – ca – {ba – (3b – ac) – (ab – cb)}]
= ab[bc – ca – {ba – 3b + ac – ab + cb}]
= ab [bc – ca + 3b – ac – cb]
= 3ab2
Question no – (2)
Solution :
(a) 5a2 + 6ab – 2a2 + 2ab – 10
= 5(-2)2 + 6(-2).3 – 2(-2)2 + 2.3.(-2) – 10
= 20 – 36 – 8 – 12 – 10
= -46
(b) 2x2 + 7yx + 5x2y + 6xy – 10
= 2(1)2 + 7.(1)1 + 5.12.(-1) + 6.10(-1) – 10
= 2 – 7 – 5 – 6 – 10
= -26
(c) 13a2b + 5a2 + 3ab
= 13(3)2.(1) + 5(3)2 + 3.3.(-1)
= -117 + 45 – 9
= -81
(d) 15a – 2(3a + 8) – 4(5a – 3)
= 15.3 – 2(3.3 + 8) – 4(5.3 – 3)
= 45 – 2(9 + 8) – 4(15 – 3)
= 45 – 2 × 17 – 4 × 12
= 45 – 34 – 48
= -37
Question no – (3)
Solution :
(a) Perimeter of square,
= 2(2.5 + x)
= (5 + 2x)m.
(b) Perimeter of required,
= 2(10 + x + 7 – y)
= 2(17 + x – y)
= 2x – 2y + 34
Question no – (4)
Solution :
(a) Area,
= 1/2 × 3.5 × 10
= 7 cm2
(b) Area,
= 5.58 × 4
= 22 cm2
Question no – (5)
Solution :
Surface area = 6x2
= 6 × 42
= 96 m2
Volume = x3
= 43
= 64 m3
Question no – (7)
Solution :
5x2 + 2
= 5(5)2 + 2
= 5 × 25 + 2
= 125 + 2
= 127
Question no – (8)
Solution :
Let, Number of students before trip = x
Number of kulche = 4x
Extra study add = x + 10
Now, according to question,
= 3(x + 10) = 4x
Or, 3x + 30
= 4x
Or, 4x – 3x = 30
Or, x = 30
∴ Number of student of trip
= 30 + 10
= 40
(a) 240x + 150y + 110z
(b) 240(2) + 150(3) + 110(1)
= 480 + 300 + 110
= 890 Rs
(c) 240(4) + 150(4)
= 960+600
= 1560 Rs
Therefore, Seema spent more then Rohan.
Previous Chapter Solution :
