# Maths Ace Class 7 Solutions Chapter 6

## Maths Ace Class 7 Solutions Chapter 6 Algebraic Expressions

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Maths Ace Prime Class 7 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, 6.4 and 6.5 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 6.2 Solution :

Question no – (1)

Solution :

(a) Subtract the product of 2 and x from 7.

= 7 – 2x

(b) The product of 3x and y divided by 5.

= 3xy/5

(c) The sum of squares of x and y.

= x2 + y2

(d) 7 times m is added to 5 times n.

= 7m + 5x

(e) Take away 6 from the sum of p and q.

= (p + q) – 6

Question no – (3)

Solution :

According to the question,

(a) 5p + 3

(b) 7pq + p2

(c) 6pq – 3pr + 6pq2

 Term of contain P Coefficients of P 5p 5 7pq 7q 6 pq, – 3pq, 6pq2 6q, – 3p, 6q2

Question no – (4)

Solution :

As per the question,

(a) pq – 1

(b) 2y+ 6

(c) – xy + 3x2y2

 Terms Numerical Coefficient (a) pq – 1 1 (b) 2y2 + 6 2 (c) – xy + 3x2y2 – 1, 3

Question no – (5)

Solution :

Like terms are,

= (8p– p), ) (3q2, 2q2, – q2), (- 2r, r)

= (7pq, – 2pq, 2pq), (- 3t, t/2)

Question no – (6)

Solution :

(a) 5a, 2a, 6a

= Like term

(b) – 5xy, – 5x2

= Unlike terms

(c) – 2pq, 1/2qp

= Like terms

(d) 2x, 1/2y

= Unlike terms

Question no – (7)

Solution :

(a) -4xyz + 3y2

= binomial

(b) 5pqr

= monomial

(c) –x + y – 2z

= trinomial

(d) 71x2 + 75y2 – 15z2

= trinomial

(e) 3x

= monomial

(f) – 6xy

= monomial

(g) a/3 + b/4

= binomial

(h) 2x2zy

= monomial

(i) 10

= monomial

(j) l2m2 + m2n2

= binomial

Question no – (8)

Solution :

(a) 3x2 + 1 – 4x4 + 4x3

= – 4x4 + 4x3 + 3x2 + 1

= degree = 4

(b) 5p3 + 6p7 – 3p2 – 1

= 6p7 + 5p3 – 3p2 – 1

= degree = 7

(c) – 2x2y2 + 3x3y2 + 5xy – 4x3y3

= – 4xy3y3 + 3x3y2 + 5xy

= degree = 6

(d) – 3a2b + 5ab + 5a2b2 – 4ab3 + 6a7b2

= 6a7b2 + 5a2b2 – 4ab3 – 3a2b + 5ab

= degree = 9

(e) 2/3 x2 + 4/5xy – 1/5x2yz

= – 1/5x2yz + 2/5x2 + 4/5xy

= degree = 4

Algebraic Expressions Exercise 6.3 Solution :

Question no – (1)

Solution :

(a) 13x, – 27x and – 39x

= 13x + (- 27x) + (- 39k)

= 53x

(b) 19a, – 55a and 14a

= 19a + (- 55a) + 14a

= – 22a

(c) 7xyz and – 4xyz

= 7xyz + (- 4xyz)

= 3xyz

(d) – 15pq, 12qp, – 23pq and 7pq

= 15pq + 12pq + (- 23pq) + 7pq

= – 19pq

Question no – (2)

Solution :

(a) 5xy + 2x, – 2xy + 3y and 8xy – 3x + 4y

= 5xy + 2x + (- 2xy + 3y) + (8x – 3x + 4y)

= 5xy + 2x – 2xy + 3y + 8x – 3x + 4y

= 3xy + 7x + 7y

(b) m + 1 and m2 + m

= (m + 1) + m2 + m

= m2 + 2m + 1

(c) 4t4 + 2t + 11, – 3t2 + t and t4 – 2

= 4t4 + 2t + 11+ (- 3t2 + t) + (t4 – 2)

= 4t4 + 2t + 11 – 3t2 + t + t4 – 2

= 5t4 – 3t2 + 3t + 9

(d) 2a + 3b – 4c, 8c – 1 and 5b + 4

= 2a + 3b – 4c + 8c – 1 + 5b + 4

= 2a + 8b + 4c + 3

Question no – (3)

Solution :

(a) 2x3 – 7×4 + 7x5 and 11x3 – 4x4 – x

= 2x3 + 11x3 – 7x4 – 4x4

= 13x3 – 11x4 + 7×5 – x

(b)  -2p – 3p2 and 1 – 5p + 8p2

= – 2p + (- 5p) + (- 3p)+ 8p2 + 1

= – 7p + 5p2 + 1

= 5p2 – 7p + 1

(c) – y + 5, 2y2 + 4 and 6t – 2y2

= – y + 6y + 2y2 – 2y2 + 4

= 5y + 4

(d) x2 – y2 – 1, y2 – 1 – x2 and 1 – x2 – y2

= x2 – x2 – x2 + y2 – y2 – y2 + 1 – 1 – 1

= – x2 – y2 – 1

Question no – (4)

Solution :

(a) 12a + 2b2 + 3a3 + (- 3a) + 17b2 + 75a

= 3a3 + 19b2 + 84a

(b) 23b + 127b3 + 5b2 + 13b – 7b2 + 5b

= 127b3 – 2b2 + 41b

(c) 17x + 5x + 7x2 + (- 19x2) + 33x

= – 12x2 + 55x

(d) 3m + 6n – 3m – 2m

= 4x

Question no – (5)

Solution :

If A = 3x2 – 7x – 8 and B = x2 – 5x – 4, find 2A + B.

= 2A + B = 2 (3x2 – 7x – 8) + (x2 – 5x – 4)

= 6x2 – 14x – 16x + x2 – 5x – 4

= 5x2 – 19x – 20

Question no – (6)

Solution :

Rohan spends = (3a + 4b + 2c) + (7a + 5b – c)

= 3a – 4b + 2c + 7a + 5b – c

= 10a + b + c

Therefore, He spend in all 10a + b + c

Question no – (7)

Solution :

Given sides are – (x + 4)

Perimeter of square

= 4 (x + 4)

= (4x + 16) cm.

Thus, the perimeter of the given square is (4x + 16) cm.

Algebraic Expressions Exercise 6.4 Solution :

Question no – (1)

Solution :

(a) 6yz from – 2yz

= 6yz + 2yz

= 8yz

(b) 7p2 from 3p2

= 4p2

(c) (9y7 – 4y – 1) from (- y7 + 10y + 7)

= 9y– 4y – 1 + y7 – 10y – 7

= 8y7 – 14y – 8

(d) (2a3 – 3a2 + a + 1) from (2a2 – 3a2 + a)

= 2a3 – 3a2 + a + 1 – 2a3 + 3a2 – a

= 1

(e) 7x + 77y – 777z from – 77y – 7z

= 7x + 77y – 777x – 777x + 77y + 7x

= 770x + 154y – 770z

Question no – (2)

Solution :

(a) Given, (2h4 – 2h + 1) from (3h4 – 8h – 7)

= (2h4 – 2h + 1) – (3h4 – 8h – 7)

= 2h4 – 2h + 1 – 3h4 + 8h + 7

= – h4 + 6h + 8

(b) (- p5 + 4p4 – 3p – 3) from (p4 + 3p3 – 2p – 19)

= (- p5 + 4p4 – 3p – 3) – (p4 + 3p3 – 2p – 19)

= – 5p + 4p4 – 3p – 3 – p4 – 3p3 + 2p + 19

= – p5 – 3p4 – 3p3 – p + 16

Question no – (3)

Solution :

(a) (a – 4) – (a + 9) – (a – 3) + (a + 11)

= a – 4 + a + 9 – a + 3 + a + 1

= – 1 + 1

= 0

(b) 4(2p – 3q – 9) – 2 (3p – 5q)

= 8p – 12p – 36 – 6p + 10q

= 2p – 2q – 36

(c) 6 – {2x – (4x – 2)}

= 6 {2x – 4x – 2}

= 6 – 2x – 2

= 4 – 2x

(d) (3x2 + 5xy – 7y2) + (11x2 – 6xy + 2y2) – (x2 + 2xy – 3x2)

= 3x2 + 5xy – 7y2 + 11x2 – 6xy + 2y2 – x2 + 2xy + 3x2

= 16x2 – 5y2 + xy

Question no – (4)

Solution :

Left of the part ribbon

= (5x + 5) – (3x + 3)

= 5x + 5 – 3x – 3

= 2x + 2

Question no – (5)

Solution :

= (- 2x2 + 3xy – 5) – (4x2 – 5xy + 9)

= – 2x2 + 3xy – 5 – 4x2 + 5xy – 9

= – 6x2 + 8xy – 14

Question no – (6)

Solution :

Need to subtract,

= (7x3 + 11x2 + 5x – 9) + (3x2 + 15x + 7)

= 7x3 + 11x2 + 5x – 9 + 3x2 + 15x + 7

= 7x3 + 14x2 + 20x – 2

Algebraic Expressions Exercise 6.5 Solution :

Question no – (1)

Solution :

(a) x + 6 when x = – 2

= – 2 + 6 [∵ x = – 2]

= 4

(b) 1 – a – b when a = – a and b = 2

= 1 – (- 1) – (2) [∵ – a = – 1 – b = – 2]

= 1 + 1 – 2

= 2 – 2

= 0

(c) x3 + x2 + x + 5 when x = – 3

= (- 3)3 + (- 3)2 + (- 3) + 5 [∵ x = – 3]

= – 27 + 9 – 3 + 5

= – 16

(d) 4x – 5 (3x + 1) when x = 2

= 4x – 15x + 1

= 4 × 2 – 15 × 2 + 1 [∵ x = 2]

= 8 – 30 + 1

= – 2

Question no – (2)

Solution :

(a) 2p + q – 3pq

= 2.3 + (- 3) – (3) (- 3)

= 6 – 3 – 27

= 3 – 27

= – 24

Question no – (3)

Solution :

Given,

(a) y = – 4

(b) y = 0

(c) y = 6

(d) y = 1/6

Now,

(a) 6x2y – 3 (2x2y – 4y)

= 6x2y – 6x2y + 12y

= 12 × (- 4) [∵ y = – 4]

= – 48

(b) 6x2y – 3 (2x2y – 4y)

= 6x2y – 6x2y + 12y

= 12 × 1/6 [ y = 1/6]

= 2

(c) 6x2y – 3 (2x2y – 4y)

= 6x2y – 6x2y + 12y

= 12 × 6 = 72 [∵ y = 6]

(d) 6x2y – 3 (2x2y – 4y)

= 6x2y – 6x2y + 12y

= 12 × 1/2 [∵ y = 1/6]

= 2

Question no – (5)

Solution :

Given, 8x2 – 7x + a = 0

or, 8 (- 1)2 – 7 (- 1) + a = 0 [∵ x = – 1]

or, 8 + 7 + a = 0

or, 15 + a = 0

or, a = – 15

Question no – (6)

Solution :

(a) Number of meters (n) in p kilometres

= n = 1000P

(b) Area of a triangle (A) is half of the product of its base (b) and height (h)

= A = 1/2 (b × h)

(c) Loss (L) in a transaction is the deduction of selling price (s) from cost price (c)

= L = c – s

(d) The speed (s) of a vehicle is obtained by dividing the distance traveled (d) by the time taken (t)

= S = d/t

(e) Perimeter of a regular pentagon (P) is 5 times the length of its side (l).

= p = 5L

Next Chapter Solution :

Updated: June 17, 2023 — 4:59 am