# Mathsight Class 7 Solutions Chapter 6

## 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) pq+ 3pqr → pq2, 3pqr

Question no – (11)

Solution :

A2b+ 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 + 4xy+ 16x2y

(c) (17a + 15a)

= 32a , 9 – 20b + 2b)

= – 18b , 32a -18b

(d) (7xy + 14xy)

= 21xy , (-8y– 4y2)

= 12y2 , 21 xy – 12y2

Question no – (2)

Solution :

(a) 15x– x+ 17x– 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) a+ b+ 2a2b2

2a– 7b– 4a2b2

+ 2a– 3b4
——————————
5a– 9b– 2a2b2

(b) -56a2b

-318a2b

+  29a2b

112a2b
—————————
54a2b

(c) 7a + 5b – 3c

+ 6a – 2b + 2c

5a + 4b – 4c
—————————
18a – 7b – 5c

(d) – x– y– z2

2x+ 3y+ 4z2

-x2 -z2
—————————
0 + 2y– 2z2

Question no – (4)

Solution :

(a) 2x – 8x– 9 + 5x+ 3x + 4

=  5x – 3x– 5

(b) 7 + 4a – 7b + 2b + 4a – 6

= 13a – 5b + 1

(c) -7c + 4a – 2b + 2a – 7b + 4c

= 6a – 9b – 3c

(d) -4x+ 7x + 1 + 3x+ 10

= x+ 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) 17x– 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) 2a– 7b– 4a2b– a– b– 2a2b2

= a– 6b– 6a2b2

(b) 29a2b+ 56a2b2

= 85a2b2

(c) 0 – 7a + 5b-3c

= -7a + 5b-3c

(d) 1 – x– y– z2

(e) 5x+ 3x + 4 – 2x + 8x+ 9

= 13x+ x + 13

(f) 2b + 4a – 6 – 7 – 9a + 7b

= -5a + 9b – 13

(g) 2a – 7b + 4c + 7c – 4a + 2b

= -2a – 5b + 11c

(h) -3x+ 10 + 4x– 7x – 1

= x+ 3x – 1

Question no – (3)

Solution :

(5x+ x + 2 – 2x + 3x+ 4) – x – x– 1

= 8x– x + 6 – x – x– 1

= 7x– 2x + 5

Question no – (4)

Solution :

3a– 4b+ 6ab + 8b+ 4a– 2ab

= 7a+ 4b+ 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

= 2x+ 4x + 3 – 2(5x+ 3)

= 2x+ 4x + 3 – 10x– 6

= -8x+ 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} – (3x+ 6y)-5]

= 3 + x – 4y – 5x + 2y – 5 + 2x– 3x– 6y – 5

= -7 – 4x – 8y – x2

= -x– 4x – 8y – 7

(e) 10x + [x+ 3x– {2x– (7 – 2x – 9x3) – 5x3} – 3x+ 5]

= 10x + x+ 3x– 2x– 7 + 2x + 9x– 5x– 3x+ 5

= 12x + 5x– 2x– 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)+ 6(-2).3 – 2(-2)+ 2.3.(-2) – 10

= 20 – 36 – 8 – 12 – 10

= -46

(b) 2x+ 7yx + 5x2y + 6xy – 10

= 2(1)+ 7.(1)1 + 5.12.(-1) + 6.10(-1) – 10

= 2 – 7 – 5 – 6 – 10

= -26

(c) 13a2b + 5a+ 3ab

= 13(3)2.(1) + 5(3)+ 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 :

5x+ 2

= 5(5)+ 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 :

Updated: May 27, 2023 — 6:18 am