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Number Line Prime Class 7 Solutions Chapter 12 Algebraic Expressions
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Number Line Prime Class 7 Math Book, Chapter 12, Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions, Exercise 12A, 12B, 12C and 12D 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 12 solutions. Here in this post all the solutions are based on latest Syllabus.
Algebraic Expressions Exercise 12A Solution :
Question no – (1)
Solution :
| Number of 7s | 1 | 2 | 3 | x |
| Number of match sticks | 6 | 11 | 16 | y |
| Pattern | 5 × 1 + 1 | 5 × 2 + 1 | 5 × 3 + 1 | y = 5x + 1 |
Question no – (2)
Solution :
| Number of box | 1 | 2 | 3 | x |
| Number of line | 4 | 7 | 10 | x |
| Pattern | 3 × 1 + 1 | 3 × 2 + 1 | 3 × 3 × 1 | y = 3x + 1 |
Algebraic Expressions Exercise 12B Solution :
Question no – (1)
Solution :
(a) Given, 7x2 + y3 – 3xy
= Trinomials
(b) Given, – 27
= Monomials
(c) Given, xy3 + x2/2 + 3x
= Trinomials
(d) Given, 4x + 4 + 4
= 4x + 8
= Binomials
Question no – (2)
Solution :
| Numerical coefficient : | Literal coefficient : | |
| (a) – 36x3y/11 | – 36/11 | x3y |
| (b) 33y3z2y2 | 33 | x2y3z2 |
| (c) – 28x6/15 | -28/15 | x6 |
| (d) 12z6y3 | 12 | z6y3 |
Question no – (3)
Solution :
(a) As per the question – 32yz, – 6z2, – 17yz, 96z2, 23zy
= 32yz, – 17yz, 23zy are like terms
= – 6yz2, 96z2 are like terms
(b) As per the question – 18x3y, 6x3, 27x3y, x3, yx2
= 18x3y, 27x3y – are like terms
= – 6x3, x3 – are like terms
(c) Given, 35pq2, – 16p2q, 56pq2, – 16pq2, – 18p2q, – 18pq2, p2q2, 5p2q2
= 35pq2, 56pq2,- 16pq2,- 18pq2, – are like terms
= -16p2q, -18p2q, – are like terms
= 5p2q2, p2q2 – are like terms
Question no – (4)
Solution :
(a) Given, x4 -2x6 -4
= Degree is 6
(b) As per the question, 5xy + 8x2y + 7x4y4 + 9y6
= Degree is (4 + 4)
= 8
(c) Given, 16p2q + 32pq2 + 8p2q2
= Degree (2 + 2)
= 4
(d) 4/3m3 + 1/2n3 -8/26 m3n3
= Degree (3 + 3)
= 6
Question no – (5)
Solution :
(a) Given, x in -5x2yz3
= -5xyz2
(b) As per the question, xy2 in x2y2z2
= xz2
(c) Given, -2a in -2/7 a2bc4
= 1/z abc4
(d) Given, x3 in -x3
= -1
Question no – (6)
Solution :
(a) x + y/3
(b) xy + (x + y)
(c) 2/3x – 3
Algebraic Expressions Exercise 12C Solution :
Question no – (1)
Solution :
(a) Given, 13x, -27x and – 39x
= 13x + (- 27x) + (- 39x)
= 13x – 27x – 39x
= 13x – 66x
= – 53x
(b) Given, 19x, – 55x and 2/3 x
= 19x – 55x + 2/3 x
= – 36x + 2/3 x
= -108x + 2x /3
= -106x/3
(c) Given, 7xyz and -4xyz
= 7xyz – 4xyz
= 3xyz
(d) As per the question, – 15pq, 12qp, – 23pq and 7pq
= – 15pq + 12pq – 23pq + 7pq
= – 3pq – 16pq
= – 19pq
Question no – (2)
Solution :
(a) Given, 5xy + 2x, – 2xy + 3y and 8xy – 3x + 4y
5xy + 2
+ – 2xy + 3y
8xy – 3x + 4y
————————————
+ 11xy – x = 7y
(b) As per the question, m + 1 and m2 + m
m + 1
+ m2 + m
————————————
M2 + 2m + 1
(c) Given, 4t4 + 2t + 11, – 3t2 + t and t4 – 2
4t4 + 2t + 11
3t2 + t
t4 – 2
————————————
5t4 – 3t2 = 3t + 9
(d) Given, 2a + 3b – 4c, 8c – 1 and 5b + 4
2a + 3b – 4c
8c – 1
+ + 5b + 4
————————————
2a + 8b + 4c + 3
Question no – (3)
Solution :
(a) As per the question, 2x3 – 7x4 + 7x5 and 11x3 – 4x4 – x
= 2x3 – 7x4 + 7x5 + 11x3 – 4x4 – x
= (2x3 +11x3) + (-7x4 – 4x4) + 7x5 – x
= 7x5 – 11x4 + 13x3 – x
(b) As per the question, – 2p – 3p2 and 1 – 5p + 8p2
= (- 2p – 3p2) + (1 – 5p + 8p2)
= 1 + (-2p – 5p) + (- 3p2 + 8p2)
= 1 – 7p + 5p2
(c) As per the question, – y + 5, 2y2 + 4 and 6y – 2y2
= – y + 5, 2y2 + 4 + 6y – 2y2
= 5y + 9
(d) As per the question, x2 – y2 -1, y2 – 1 – x and 1 – x2 – y2
= x2 – y2 – 1, y2 – 1 – x + 1 – x2 – y2
= – 1 – x – y2
= – (1 + x + y2)
Question no – (4)
Solution :
(a) Given, 12a + 2b2 + 3a3 + (- 3a) + 17b2 + 75a
= 9a + 19b2 + 3a3 + 75a
= 3a3 + 19b2 + 84a
(b) Given, 23b + 127b3 + 5b2 + 13b – 7b2 + 5b
= 41b – 2b2 + 127b3
(c) Given, 17x + 5x + 7x2 + (-19x2) + 33x
= 55x – 12x2
(d) Given, 3m + 6n – 3m – 2n
= 4n
Question no – (5)
Solution :
As per the question, A = 3x2 – 7x – 8, B = x2 – 5x – 4
∴ 2A + B = 2 (3x2 – 7x – 8) + x2 – 5x – 4
= 6x2 – 14x – 16 + x2 – 5x – 4
= 7x2 – 19x – 20
Algebraic Expressions Exercise 12D Solution :
Question no – (1)
Solution :
(a) Given, 6yz from – 2yz
= 6yz + (- 2yz)
= – 6yz – 2yz
= – 8yz
(b) As per the question, (9y7 – 4y – 1) from (- y7 + 10y + 7)
= (9y7 – 4y – 1) + (- y7 + 10y + 7)
= – 9y7 – 4y – 1 – y7 + 10y + 7
= -10y4 + 14y + 8
(c) Given, (2a3 – 3a2 + a + 1) from (2a3 – 3a2 + a)
= – 2a3 – 3a2 + a + 1 +2a3 – 3a2 + a
= -1
(d) Given, 7x + 77y – 777z from 777x – 77y – 7z
= – 7x + 77y – 777z + 777x – 77y – 7z
= + 70x – 154y + 770z
Question no – (2)
Solution :
(a) Given, (2h4 – 2h + 1) from (3h4 – 8h – 7)
= 2h4 – 2h – 1 + 3h4 – 8h – 7
= + h4 – 6hb – 8
(b) Given, (- p5 + 4p4 – 3p – 3 ) from ( p4 + 3p3 – 2p – 19)
= + p5 – 4p4 + 3p + 3 + p4 + 3p3 – 2p – 19
= + p5 – 3p4 + 3p3 + p – 16
(c) Given, (x2 + 4) from (- 2x2 + 9x + 1)
= – 2x2 + 9x + 1 – x2 + 4
= x2 – 2x2 + 9x – 3
= – (+ 3x2 – 9x + 3)
(d) As per the question, (11x + 7y – 3) from (9x – 21y – 3)
= 9x – 21y – 3 – 11x + 7y – 3
= – 2x – 28y
= – 2 (x + 14y)
Question no – (3)
Solution :
(a) As per the question, (a – 4) – (a + 9) – (a – 3) + ( a + 11)
= (a – 4) – (a + 9) – (a – 3) + ( a + 11)
= a – 4 – a – 9 – a + 3 + a + 11
= 10
(b) As per the question, (m – 14) – (m + 5) + (m – 16) + (m + 9)
= m – 14 – m – 5 + m – 16 + m + 9
= 2m – 26
(c) Given, (r + 13) – (r + 15) – (r – 17) – (r – 19)
= r + 13 – r – 15 – r + 17 – r + 19
= – 2r + 34
Question no – (4)
Solution :
∴ Ribbon is left = (5x + 5) – (13x + 3)
= 5x + 5 – 3x – 3
= (2x + 2)m
Question no – (5)
Solution :
∴ We should add = (- 2x2 + 3xy – 5) – ( 4x2 – 5xy + 9)
= – 2x2 + 3xy – 5 -4x2 + 5xy – 9
= – 6x2 + 8xy – 14
Question no – (6)
Solution :
Given in the question, P = 3, q = -3
(a) Given, 2p + q – 3pq
= 2p + q – 3pq
= 2 × 3 + (- 3) – 3 (3) (- 3)
= 6 – 3 + 27
= 30
(b) As per the question, p2q2 + p3 – q3 + pq2 + p2q
= p2q2 + p3 – q3 + pq2 + p2q
= p3 – q3 + pq (pq + q + p)
= 33 – (- 3)3 + 3(- 3) [3 × (- 3) + 3 – 3]
= 27 + 27 – 9 [- 9 + 0]
= 27 + 27 + 81
= 135
Question no – (7)
Solution :
As per the given question, x = -1, y = 2
As per the question, 2x2 – 3y + 3x2 + 5y2 + 5x – 7 – 7x + 2y
= 5x2 + 5y2 – 2x – y – 7
= 5(x2 + y2) – 2x – y -7
= 5(1 + 4) – 2 (-1) – 2 – 7
= 25 + 2 – 9
= 18
Next Chapter Solution :
