Collins Maths Solutions Class 7 Chapter 11

Collins Maths Solutions Class 7 Chapter 11 Operations on Algebraic Expressions

Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Collins Maths Class 7 Mathematics, Chapter 11, Operations on Algebraic Expressions. Here students can easily find chapter 7 solutions with exercise wise explanation. Students will find proper solutions for Exercise 11.1, 11.2 and 11.3 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.

Operations on Algebraic Expressions Exercise 11.1 Solution :

Question no – (1) 

Solution : 

(a) 36x2y + 28xy + 10

= 36x2y = 2 × 2 × 3 × 3 × x  × x × y

= 28xy = 2 × 2 × 8 × x × y

= 10 = 2 × 5

(b) 2x5 – 6x3 + 28x2 + x  + 9

= 2x= 2 × x × x ×x × x × x

= 6x3 = 2 × 3 × x × x × x

= 28x2 = 2 × 2 × 7 ×  x  × x

= x = x

= 9 =  3 ×  3

(c) 5abc + 25a2bc

= 5abc =  5 × a × b × c

= 25a2bc = 5 × 5 × a × a × b × c

Question no – (2)

Solution : 

Variable term Numerical coefficient of variable term Constant term
45hx, 32y, 45, 32 -6
3t3, 2t2 3, 2 +10
h/4, 7h 1/4, 7 +21

Question no – (3)

Solution : 

(a) Degree,

= (2 + 1 + 1)

=  4

(b) Degree,

= (1 + 1)

= 2

(c) Degree,

= (3 + 2 + 1)

=  6

(d) Degree,

= (3 + 3 + 2)

=  8

(e) Degree,

= (2 + 3)

= 5

Question no – (4)

Solution : 

(a) Degree,

= 3

(b) Degree,

= 3

(c) Degree,

= 9

Question no – (5) 

Solution : 

(a) Like terms

(b) Unlike terms.

(c) Unlike terms

(d) Like terms.

Question no – (6)

Solution : 

(a) Trinomial

(b) Monomial

(c) Binomial

(d) Trinomial

(e) Trinomial

(f) Polynomial

Operations on Algebraic Expressions Exercise 11.2 Solution :

Question no – (1) 

Solution : 

(a) Let, father age = x year.

Ravi’s age = ( x – 32) year.

(b) Let, lions = x

tigers = (x – 10)

(c) Let, last year =  x

Now, (x – 20)% rainfall in this year.

(d) Toms brother age = x year

Tom’s age

= (x – 4) years.

(e) Algebraic expression = 300d

(f) Amount of money of each employee get,

= 10000/x

[x = total number of employee]

Operations on Algebraic Expressions Exercise 11.3 Solution :

Question no – (1) 

Solution : 

(a) (5a – 9b) + (-16a + 19b)

= 5a – 9b – 16a + 19b

= – 11a + 10b

(b) (41h + 15g) + (7g – 13h)

= 41h  + 15g + 7g – 13h

= 28h + 22g

(c) (5p3 – 3p2 – 6 p – 3) + (3p3 + 7b – 10)

= 5p3 – 3p2 – 6p – 3 + 3p3 + 7p – 10

= 8p3 – 3p2 + p – 13

(d) (-14a2 + 21b2 + 9ab) + (33a2 – 25b2 + ab)

= -14a2 + 21b2 + 9ab + 33a2 – 25b2 + ab

= 19a2 – 4b2 + 10ab

Question no – (2)

Solution : 

(a) (7b + 16c) – (52c – 12b)

= 7b + 16c – 52c  + 12b

= 19b – 36c

(b) (19l6 + 15l2 -12l) – (19l6 + 15l2 – 12l)

= 19l6 + 15l2 – 12l 19l6 – 15l2 + 12l

= 0

(c) (12k2 + 2k – 9) – (9k2 – 14k + 6)

= 12k2 + 2k – 9 – 9k2 + 14k – 6

= 3k2 + 16k  – 15

(d) (17bc + 19ca – 12ab) – (-2ab + 5bc + 3ca)

= 17bc + 19ca – 12ab – 2ab – 5bc – 3ca

= 12bc + 16ca – 10ab

Question no – (3)

Solution : 

(a) A + A + C

=  -4x2 + 3y2 + 14xy + (-4x2 + 3y2 + 14xy) + (22x2 – 25y2 + 17xy)

= -4x2 + 3y2 + 14xy – 4x2 + 3y2 + 14xy + 22x2 – 25y2 + 17xy

= 14x2 – 19y2 + 45xy

(b) A + C – B

= – 4x2 + 3y2 + 14xy + 22x2 – 25xy + 17xy – (21x2 – 15xy + 9y2)

= – 4x2 + 3y2 + 14xy + 22x2 – 25y2 + 17xy – 21x2 + 15xy – 9y2

= – 3x2 – 31y2 + 46xy

(c) C – A + B

= 22x2 – 25y2 + 17xy – (-4x2 + 3y2 + 14xy) + 21x2 – 15xy + 9y2

= 22x2  – 25y2  + 17xy + 4x2 – 3y2 – 14xy  + 21x2 – 15xy + 9x2

= 47x2 – 19y2 – 12xy

Question no – (4) 

Solution : 

Sum = (7a – 2b + 3c) + (5a + 6b – 9c)

= 7a -2b  +3c + 5a + 6b – 9c

= 12a + 4b – 6c

Now, (14a – 25b + 30c) – (12 a 4b – 6c)

=  14a – 25b + 30c – 12a 4b + 6c

=  2a –  29b + 36c

Question no – (5)

Solution : 

To be added,

= (54x8 – 40x3 + 72x2 – 98) – (13×8 – 12x3 + x2 – 19x + 49)

= 54x8 – 40x3 + 72x2 – 98 + 12x3 – x2 + 19x – 49

= 41x8 – 28x3 + 71x2 + 19x – 147

Question no – (6) 

Solution :

Perimeter,

= 2 [(-4x + 6y + 3z) + (12x – 4y + 9z)

= 2 [-4x + 6y + 3z +12x – 4y + 9z]

= 2 [8x + 2y + 12z]

= 16x + 4y + 24z

Question no – (7) 

Solution : 

Third side,

= 29a + 9b – 6c – {(16a + 5b – 2c) – (-6a + 7b – (2c)}

= 29a + 9b – 6c – {16a + 5b – 2c + 6a – 7b + 12c}

= 29a + 9b – 6c – 16a – 5b + 2c – 6a + 7b – 12c

=  7a + 11b – 16c

Next Chapter Solution : 

👉 Chapter 12 👈

Updated: June 15, 2023 — 9:18 am

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