NCERT Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions

NCERT Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 7 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 7 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 7 Mathematics are an important part of exams for Class 7 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 7 Mathematics and also download more latest study material for all subjects. Chapter 12 Algebraic Expressions is an important topic in Class 7, please refer to answers provided below to help you score better in exams

Chapter 12 Algebraic Expressions Class 7 Mathematics NCERT Solutions

Class 7 Mathematics students should refer to the following NCERT questions with answers for Chapter 12 Algebraic Expressions in Class 7. These NCERT Solutions with answers for Class 7 Mathematics will come in exams and help you to score good marks

Chapter 12 Algebraic Expressions NCERT Solutions Class 7 Mathematics

Exercise 12.1

Q.1) Get the algebraic expressions in the following cases using variables, constants and arithmetic operations:
(i) Subtraction of z from y.
(ii) One-half of the sum of numbers x and y.
(iii) The number z multiplied by itself.
(iv) One-fourth of the product of numbers p and q.
(v) Numbers x and y both squared and added.
(vi) Number 5 added to three times the product of m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
Sol.1) i) 𝑦 βˆ’ 𝑧      ii) π‘₯+𝑦/2
iii) 𝑧2                      iv) π‘π‘ž/4
v) π‘₯2 + 𝑦2             vi) 3π‘šπ‘› + 5         
vii) 10 βˆ’ 𝑦𝑧        viii) π‘Žπ‘ βˆ’ (π‘Ž + 𝑏)

Q.2) (i) Identify the terms and their factors in the following expressions, show the terms and factors by tree diagram:
(a) π‘₯ -3                  (b) 1 + π‘₯ + π‘₯2             (c) 𝑦 βˆ’ 𝑦3
(d) 5π‘₯𝑦2 + 7π‘₯2𝑦    
(e) βˆ’π‘Žπ‘ + 2𝑏2 βˆ’ 3π‘Ž2
(ii) Identify the terms and factors in the expressions given below:
(a) -4π‘₯ + 5    (b) βˆ’4π‘₯ + 5𝑦                  (c) 5𝑦 + 3𝑦2        (d) π‘₯𝑦 + 2π‘₯2𝑦2
(e) π‘π‘ž + π‘ž     (f) 1. π‘Žπ‘ βˆ’ 2.4𝑏 + 3.6π‘Ž   (g) 3/4 π‘₯ + 1/4   (h) 0.1𝑝2 + 0.2π‘ž
Sol.2)

""NCERT-Solutions-Class-7-Mathematics-Algebraic-Expressions-4

""NCERT-Solutions-Class-7-Mathematics-Algebraic-Expressions-5

ii) (a) -4π‘₯ + 5                    (b) βˆ’4π‘₯ + 5𝑦
Terms βˆ’4π‘₯, 5                        Terms: βˆ’4π‘₯, 5𝑦
Factors: βˆ’4, π‘₯;                      5 factors: βˆ’4, π‘₯; 5, 𝑦

(c) 5𝑦 + 3𝑦                    (d) π‘₯𝑦 + 2π‘₯2𝑦2
Terms: 5𝑦, 3𝑦                    Terms: π‘₯𝑦, 2π‘₯2𝑦2
Factors: 5, 𝑦; 3, 𝑦, 𝑦             Factors: π‘₯, 𝑦 ; 2π‘₯, π‘₯, 𝑦, 𝑦

(e) π‘π‘ž + π‘ž                         (f) 1.2π‘Žπ‘ βˆ’ 2.4𝑏 + 3.6π‘Ž
Terms: π‘π‘ž, π‘ž                        Terms: 1.2π‘Žπ‘, βˆ’2.4𝑏, 3.6π‘Ž
Factors: 𝑝, π‘ž ; π‘ž                    Factors: 1.2, π‘Ž , 𝑏 ; βˆ’2.4, 𝑏 ; 3.6, π‘Ž

(g) 3/4 π‘₯ + 1/4                  (h) 0.1𝑝2 + 0.2π‘ž2
                                           Terms: 3/4 π‘₯, 1/4
                                           Terms: 0.1𝑝2, 0.2π‘ž2
                                           Factors: 3/4, π‘₯ ; 1/4
                                           Factors: 0.1, 𝑝, 𝑝; 0.2, π‘ž, π‘ž
 

Q.3) Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 βˆ’ 3𝑑                (ii) 1 + 𝑑 + 𝑑2 + 𝑑3   (iii) π‘₯ + 2π‘₯𝑦 + 3𝑦
(iv) 100π‘š + 1000𝑛   (v) βˆ’p2q2 + 7π‘π‘ž       (vi) 1.2π‘Ž + 0.8𝑏
(vii) 3.14π‘Ÿ2              (viii) 2(𝑙 + 𝑏)            (ix) 0.1𝑦 + 0.01𝑦2
Sol.3)

""NCERT-Solutions-Class-7-Mathematics-Algebraic-Expressions-6

Q.4) (a) Identify terms which contain π‘₯ and give the coefficient of π‘₯.
(i) 𝑦2π‘₯ + 𝑦        (ii) 13𝑦2 βˆ’ 8𝑦π‘₯      (iii) π‘₯ + 𝑦 + 2   (iv) 5 + 𝑧 + zx
(v) 1 + π‘₯ + π‘₯𝑦   (vi) 12π‘₯𝑦2 + 25    (vii) 7π‘₯ + π‘₯𝑦2
(b) Identify terms which contain 𝑦2 and give the coefficient of 𝑦2 .
(i) 8 βˆ’ π‘₯𝑦2  (ii) 5𝑦2 + 7π‘₯   (iii) 2π‘₯2 βˆ’ 15π‘₯𝑦2 + 7𝑦2
Sol.4)

""NCERT-Solutions-Class-7-Mathematics-Algebraic-Expressions-7

Q.5) Classify into monomials, binomials and trinomials:
(i) 4𝑦 βˆ’ 7π‘₯ (ii) 𝑦2 (iii) π‘₯ + 𝑦 βˆ’ π‘₯𝑦 (iv) 100
(v) π‘Žπ‘ βˆ’ π‘Ž βˆ’ 𝑏 (vi) 5 βˆ’ 3𝑑 (vii) 4𝑝2π‘ž βˆ’ 4π‘π‘ž2 (viii) 7π‘šπ‘›
(ix) 𝑧2 βˆ’ 3𝑧 + 8 (x) π‘Ž2 + 𝑏2 (xi) 𝑧2 + 𝑧 (xii) 1 + π‘₯ + π‘₯2
Sol.5) Type of Polynomial
(i) 4𝑦 βˆ’ 7π‘₯                          Binomial
(ii) 𝑦2                                 Monomial
(iii) π‘₯ + 𝑦 βˆ’ π‘₯𝑦                    Trinomial
(iv) 100                             Monomial
(v) π‘Žπ‘ βˆ’ π‘Ž βˆ’ 𝑏                    Trinomial
(vi) 5 βˆ’ 3𝑑                          Binomial
(vii) 4𝑝2π‘ž βˆ’ 4π‘π‘ž2               Binomial
(viii) 7π‘šπ‘›                          Monomial
(ix) 𝑧2 βˆ’ 3𝑧 + 8                 Trinomial
(x) π‘Ž2 + 𝑏2                        Binomial
(xi) 𝑧2 + 𝑧                         Binomial
(xii) 1 + π‘₯ + π‘₯                 Trinomial

Q.6) State whether a given pair of terms is of like or unlike terms:
(i) 1, 100 (ii) βˆ’7π‘₯, (5/2) π‘₯ (iii) βˆ’29π‘₯, βˆ’29𝑦
(iv) 14π‘₯𝑦, 42𝑦π‘₯ (v) 4π‘š2𝑝, 4π‘šπ‘2 (vi) 12π‘₯𝑧, 12π‘₯2𝑧2
Sol.6)

""NCERT-Solutions-Class-7-Mathematics-Algebraic-Expressions-8

The terms which have the same algebraic factors are called like terms and when the terms have different algebraic factors, they are called unlike terms.

Q.7) Identify like terms in the following:
a) βˆ’π‘₯2𝑦, βˆ’4𝑦π‘₯2, 8π‘₯2, 2π‘₯𝑦2, 7𝑦, βˆ’11π‘₯2 βˆ’ 100π‘₯, βˆ’11𝑦π‘₯, 20π‘₯2𝑦, βˆ’6π‘₯2, 𝑦 , 2π‘₯𝑦 , 3π‘₯
b) 10π‘π‘ž, 7𝑝, 8𝑝, βˆ’π‘2π‘ž2, βˆ’7π‘π‘ž , βˆ’100π‘ž , βˆ’23, 12𝑝2π‘ž2, βˆ’5𝑝2, 41, 2405𝑝,
78π‘žπ‘, 13𝑝2π‘ž, π‘žπ‘2701𝑝2
Sol.7) a) Like terms
i) βˆ’π‘₯𝑦2, 2π‘₯𝑦2 ii) βˆ’4𝑦π‘₯2, 20π‘₯2𝑦 iii) 8π‘₯2, βˆ’11π‘₯2, βˆ’6π‘₯2
iv) 7𝑦, 𝑦 v) βˆ’100π‘₯, 3π‘₯ vi) βˆ’11𝑦π‘₯, 2π‘₯𝑦
b) Like terms are:
i) 10π‘π‘ž, βˆ’7π‘π‘ž , 78π‘π‘ž ii) 7𝑝, 2405𝑝 iii) 8π‘ž, βˆ’100π‘ž
iv) βˆ’π‘2π‘ž2, 12𝑝2π‘ž2 v) βˆ’12, 41 vi) βˆ’5𝑝2, 701𝑝2
vii) 13𝑝2π‘ž, π‘žπ‘2

Exercise 12.2

Q.1) Simplify combining like terms:
i) 21𝑏 βˆ’ 32 + 7𝑏 βˆ’ 20𝑏 ii) βˆ’π‘§2 + 13𝑧2 βˆ’ 5π‘₯ + 7𝑧3 βˆ’ 15𝑧
iii) 𝑝 βˆ’ (𝑝 βˆ’ π‘ž) βˆ’ π‘ž βˆ’ (π‘ž βˆ’ 𝑝) iv) 3π‘Ž βˆ’ 2𝑏 βˆ’ π‘Žπ‘ βˆ’ (π‘Ž βˆ’ 𝑏 + π‘Žπ‘) + 3π‘Žπ‘ + 𝑏 βˆ’π‘Ž
v) 5π‘₯2𝑦 βˆ’ 5π‘₯2 + 3𝑦π‘₯2 βˆ’ 3𝑦2 + π‘₯2 βˆ’ π‘¦2 + 8π‘₯𝑦2 βˆ’ 3𝑦2
vi) (3𝑦2 + 5𝑦 βˆ’ 4) βˆ’ (8𝑦 βˆ’ π‘¦2 βˆ’ 4)
Sol.1) i) 21𝑏 βˆ’ 32 + 7𝑏 βˆ’ 20𝑏
= 21𝑏 + 7𝑏 βˆ’ 20𝑏 βˆ’ 32
= 28𝑏 βˆ’ 20𝑏 βˆ’ 32 = 8𝑏 βˆ’ 32

ii) βˆ’z2 + 13z2 βˆ’ 5π‘₯ + 7z3 βˆ’ 15𝑧
= 7z3 + (βˆ’z2 + 13z2) βˆ’ (5𝑧 + 15𝑧)
= 7z3 + 12z2 βˆ’ 20𝑧

iii) 𝑝 βˆ’ (𝑝 βˆ’ π‘ž) βˆ’ π‘ž βˆ’ (π‘ž βˆ’ 𝑝)
= 𝑝 βˆ’ 𝑝 + π‘ž βˆ’ π‘ž βˆ’ π‘ž + 𝑝
= 𝑝 βˆ’ 𝑝 + 𝑝 + π‘ž βˆ’ π‘ž βˆ’ π‘ž = 𝑝 βˆ’ π‘ž

iv) 3π‘Ž βˆ’ 2𝑏 βˆ’ π‘Žπ‘ βˆ’ (π‘Ž βˆ’ 𝑏 + π‘Žπ‘) + 3π‘Žπ‘ + 𝑏 βˆ’ π‘Ž
= 3π‘Ž βˆ’ 2𝑏 βˆ’ π‘Žπ‘ βˆ’ π‘Ž + 𝑏 βˆ’ π‘Žπ‘ + 3π‘Žπ‘ + 𝑏 βˆ’ π‘Ž
= 3π‘Ž βˆ’ π‘Ž βˆ’ π‘Ž βˆ’ 2𝑏 + 𝑏 + 𝑏 βˆ’ π‘Žπ‘ βˆ’ π‘Žπ‘ + 3π‘Žπ‘
= (3π‘Ž βˆ’ π‘Ž βˆ’ π‘Ž) βˆ’ (2𝑏 βˆ’ 𝑏 βˆ’ 𝑏) βˆ’ (π‘Žπ‘ + π‘Žπ‘ βˆ’ 3π‘Žπ‘)
= π‘Ž βˆ’ 0 βˆ’ (π‘Ž βˆ’ 𝑏)
= π‘Ž + π‘Žπ‘

v) 5π‘₯2𝑦 βˆ’ 5π‘₯2 + 3𝑦π‘₯2 βˆ’ 3𝑦2 + π‘₯2 βˆ’ π‘¦2 + 8π‘₯𝑦2 βˆ’ 3𝑦2
= 5π‘₯2𝑦 + 3𝑦π‘₯2 + 8π‘₯𝑦2 βˆ’ 5π‘₯2 + π‘₯2 βˆ’ 3𝑦2 βˆ’ π‘¦2 βˆ’ 3𝑦2
= (5π‘₯2𝑦 + 3π‘₯2𝑦) + 8π‘₯𝑦2 βˆ’ (5π‘₯2 βˆ’ π‘₯2) βˆ’ (3𝑦2 + π‘¦2 + 3𝑦2)
= 8π‘₯2𝑦 + 8π‘₯𝑦2 βˆ’ 4π‘₯2 βˆ’ 7𝑦2

vi) (3𝑦2 + 5𝑦 βˆ’ 4) βˆ’ (8𝑦 βˆ’ 𝑦2 βˆ’ 4)
= 3𝑦2 + 5𝑦 βˆ’ 4 βˆ’ 8𝑦 + 𝑦2 + 4
= (3𝑦2 + π‘¦2) + (5𝑦 βˆ’ 8𝑦) βˆ’ (4 βˆ’ 4)
= 4𝑦2 βˆ’ 3𝑦 βˆ’ 0 = 4𝑦2 βˆ’ 3𝑦

Q.2) Add:
i) 3π‘šπ‘›, βˆ’5π‘šπ‘›, 8π‘šπ‘› βˆ’ 4π‘šπ‘›
ii) 𝑑 βˆ’ 8𝑑𝑧, 3𝑑𝑧 βˆ’ 𝑧, 𝑧 βˆ’ 𝑑
iii) βˆ’7π‘šπ‘› + 5, 12π‘šπ‘› + 2, 9π‘šπ‘› βˆ’ 8, βˆ’2π‘šπ‘› βˆ’ 3
iv) π‘Ž + 𝑏 βˆ’ 3, 𝑏 βˆ’ π‘Ž + 3, π‘Ž βˆ’ 𝑏 + 3
v) 14π‘₯ + 10𝑦 βˆ’ 12π‘₯𝑦 βˆ’ 13, 18 βˆ’ 7π‘₯ βˆ’ 10𝑦 + 8π‘₯𝑦, 4π‘₯𝑦
vi) 5π‘š βˆ’ 7𝑛, 3𝑛 βˆ’ 4π‘š + 2, 2π‘š βˆ’ 3π‘šπ‘› βˆ’ 5
vii) 4π‘₯2𝑦, βˆ’3π‘₯𝑦2, βˆ’5π‘₯𝑦2, 5π‘₯2𝑦
viii) 3𝑝2π‘ž2 βˆ’ 4π‘π‘ž + 5, βˆ’10𝑝2π‘ž2, 15 + 9π‘π‘ž + 7𝑝2π‘ž2
ix) π‘Žπ‘ βˆ’ 4π‘Ž, 4𝑏 βˆ’ π‘Žπ‘, 4π‘Ž βˆ’ 4𝑏
x) π‘₯2 βˆ’ π‘¦2 βˆ’ 1, π‘¦2 βˆ’ 1 βˆ’ π‘₯2 , 1 βˆ’ π‘₯2 βˆ’ π‘¦2
Sol.2) i) 3π‘šπ‘›, βˆ’5π‘šπ‘›, 8π‘šπ‘› βˆ’ 4π‘šπ‘›
= 3π‘šπ‘› + (βˆ’5π‘šπ‘›) + 8π‘šπ‘› + (βˆ’4π‘šπ‘›)
= (3 βˆ’ 5 + 8 βˆ’ 4)π‘šπ‘› = 2π‘šπ‘›

ii) 𝑑 βˆ’ 8𝑑𝑧, 3𝑑𝑧 βˆ’ 𝑧, 𝑧 βˆ’ 𝑑
= 𝑑 βˆ’ 8𝑑𝑧 + 3𝑑𝑧 βˆ’ 𝑧 + 𝑧 βˆ’ 𝑑
= 𝑑 βˆ’ 𝑑 βˆ’ 8𝑑𝑧 + 3𝑑𝑧 βˆ’ 𝑧 + 𝑧
= (1 βˆ’ 1)𝑑 + (βˆ’8 + 3)𝑑𝑧 + (βˆ’1 + 1)𝑧
= 0 βˆ’ 5𝑑𝑧 + 0 = βˆ’5𝑑𝑧

iii) βˆ’7π‘šπ‘› + 5, 12π‘šπ‘› + 2, 9π‘šπ‘› βˆ’ 8, βˆ’2π‘šπ‘› βˆ’ 3
= βˆ’7π‘šπ‘› + 5 + 12π‘šπ‘› + 2 + 9π‘šπ‘› βˆ’ 8 + (βˆ’2π‘šπ‘›) βˆ’ 3
= βˆ’7π‘šπ‘› + 12π‘šπ‘› + 9π‘šπ‘› βˆ’ 2π‘šπ‘› + 5 + 2 βˆ’ 8 βˆ’ 3
= (βˆ’7 + 12 + 9 βˆ’ 2)π‘šπ‘› + 7 βˆ’ 11
= 12π‘šπ‘› βˆ’ 4

iv) π‘Ž + 𝑏 βˆ’ 3, 𝑏 βˆ’ π‘Ž + 3, π‘Ž βˆ’ 𝑏 + 3
= π‘Ž + 𝑏 βˆ’ 3 + 𝑏 βˆ’ π‘Ž + 3 + π‘Ž βˆ’ 𝑏 + 3
= (π‘Ž βˆ’ π‘Ž + π‘Ž) + (𝑏 + 𝑏 βˆ’ 𝑏) βˆ’ 3 + 3 + 3
= π‘Ž + 𝑏 + 3

v) 14π‘₯ + 10𝑦 βˆ’ 12π‘₯𝑦 βˆ’ 13, 18 βˆ’ 7π‘₯ βˆ’ 10𝑦 + 8π‘₯𝑦, 4π‘₯𝑦
= 14π‘₯ + 10𝑦 βˆ’ 12π‘₯𝑦 βˆ’ 13 + 18 βˆ’ 7π‘₯ βˆ’ 10𝑦 + 8π‘₯𝑦 + 4π‘₯𝑦
= 14π‘₯ βˆ’ 7π‘₯ + 10𝑦 βˆ’ 10𝑦 βˆ’ 12π‘₯𝑦 + 8π‘₯𝑦 + 4π‘₯𝑦 βˆ’ 13 + 18
= 7π‘₯ + 0𝑦 + 0π‘₯𝑦 + 5 = 7π‘₯ + 5

vi) 5π‘š βˆ’ 7𝑛, 3𝑛 βˆ’ 4π‘š + 2, 2π‘š βˆ’ 3π‘šπ‘› βˆ’ 5
= 5π‘š βˆ’ 7𝑛 + 3𝑛 βˆ’ 4π‘š + 2 + 2π‘š βˆ’ 3π‘šπ‘› βˆ’ 5
= 5π‘š βˆ’ 4π‘š + 2π‘š βˆ’ 7𝑛 + 3𝑛 βˆ’ 3π‘šπ‘› + 2 βˆ’ 5
= (5 βˆ’ 4 + 2)π‘š + (βˆ’7 + 3)𝑛 βˆ’ 3π‘šπ‘› βˆ’ 3
= 3π‘š βˆ’ 4𝑛 + 3π‘šπ‘› βˆ’ 3

vii) 4π‘₯2𝑦, βˆ’3π‘₯𝑦2, βˆ’5π‘₯𝑦2, 5π‘₯2𝑦
= 4π‘₯2𝑦 + (βˆ’3π‘₯𝑦2) + (βˆ’5π‘₯𝑦2) + 5π‘₯2𝑦
= 4π‘₯2𝑦 + 5π‘₯2𝑦 βˆ’ 3π‘₯𝑦2 βˆ’ 5π‘₯𝑦2
= 9π‘₯2𝑦 βˆ’ 8π‘₯𝑦2

viii) 3𝑝2π‘ž2 βˆ’ 4π‘π‘ž + 5, βˆ’10𝑝2π‘ž2, 15 + 9π‘π‘ž + 7𝑝2π‘ž2
= 3𝑝2π‘ž2 βˆ’ 4π‘π‘ž + 5 + (βˆ’10𝑝2π‘ž2) + 15 + 9π‘π‘ž + 7𝑝2π‘ž2
= 3𝑝2π‘ž2 βˆ’ 10𝑝2π‘ž2 + 7𝑝2π‘ž2 + 4π‘π‘ž + 9π‘π‘ž + 5 + 15
= (3 βˆ’ 10 + 7)𝑝2π‘ž2 + (βˆ’4 + 9)π‘π‘ž + 20
= 0𝑝2π‘ž2 + 5π‘π‘ž + 20 = 5π‘π‘ž + 20

ix) π‘Žπ‘ βˆ’ 4π‘Ž, 4𝑏 βˆ’ π‘Žπ‘, 4π‘Ž βˆ’ 4𝑏
= π‘Žπ‘ βˆ’ 4π‘Ž + 4𝑏 βˆ’ π‘Žπ‘ + 4π‘Ž βˆ’ 4𝑏
= βˆ’4π‘Ž + 4π‘Ž + 4𝑏 βˆ’ 4𝑏 + π‘Žπ‘ βˆ’ π‘Žπ‘
= 0 + 0 + 0 = 0

x) π‘₯2 βˆ’ π‘¦2 βˆ’ 1, π‘¦2 βˆ’ 1 βˆ’ π‘₯2 , 1 βˆ’ π‘₯2 βˆ’ π‘¦2
= π‘₯2 βˆ’ π‘¦2 βˆ’ 1 + π‘¦2 βˆ’ 1 βˆ’ π‘₯2 + 1 βˆ’ π‘₯2 βˆ’ 𝑦^2
= π‘₯2 βˆ’ π‘₯2 βˆ’ π‘₯2 βˆ’ π‘¦2 + π‘¦2 βˆ’ π‘¦2 βˆ’ 1 βˆ’ 1 + 1
= (1 βˆ’ 1 βˆ’ 1)π‘₯2 + (βˆ’1 + 1 βˆ’ 1)𝑦2 βˆ’ 1 βˆ’ 1 + 1
= βˆ’π‘₯2 βˆ’ π‘¦2 βˆ’ 1

Q.3) Subtract:
i) βˆ’5𝑦2 π‘“π‘Ÿπ‘œπ‘š π‘¦2
ii) 6π‘₯𝑦 π‘“π‘Ÿπ‘œπ‘š βˆ’ 12π‘₯𝑦
iii) (π‘Ž βˆ’ 𝑏)π‘“π‘Ÿπ‘œπ‘š (π‘Ž + 𝑏)

iv) π‘Ž(𝑏 βˆ’ 5)π‘“π‘Ÿπ‘œπ‘š 𝑏(5 βˆ’ π‘Ž)
v) βˆ’π‘š2 + 5π‘šπ‘› π‘“π‘Ÿπ‘œπ‘š 4π‘š2 βˆ’ 3π‘šπ‘› + 8
vi) βˆ’π‘₯2 + 10π‘₯ βˆ’ 5 π‘“π‘Ÿπ‘œπ‘š 5π‘₯ βˆ’ 10
vii) 5π‘Ž2 βˆ’ 7π‘Žπ‘ + 5𝑏2 π‘“π‘Ÿπ‘œπ‘š 3π‘Žπ‘ βˆ’ 2π‘Ž2 βˆ’ 2𝑏2
viii) 4π‘π‘ž βˆ’ 5π‘ž2 βˆ’ 3𝑝2 π‘“π‘Ÿπ‘œπ‘š 5𝑝2 + 3π‘ž2 βˆ’ π‘π‘ž
Sol.3) i) 𝑦2 βˆ’ (βˆ’5𝑦2) = 𝑦2 + 5𝑦2
= 6𝑦2

ii) βˆ’12π‘₯𝑦 βˆ’ (6π‘₯𝑦) = βˆ’12π‘₯𝑦 βˆ’ 6π‘₯𝑦
= βˆ’18π‘₯𝑦

iii) (π‘Ž + 𝑏) βˆ’ (π‘Ž βˆ’ 𝑏) = π‘Ž + 𝑏 βˆ’ π‘Ž + 𝑏
= π‘Ž βˆ’ π‘Ž + 𝑏 + 𝑏 = 2𝑏

iv) 𝑏(5 βˆ’ π‘Ž) βˆ’ π‘Ž(𝑏 βˆ’ 5) = 5𝑏 βˆ’ π‘Žπ‘ βˆ’ π‘Žπ‘ + 5π‘Ž
= 5𝑏 βˆ’ 2π‘Žπ‘ + 5π‘Ž
= 5π‘Ž + 5𝑏 βˆ’ 2π‘Žπ‘

v) 4π‘š2 βˆ’ 3π‘šπ‘› + 8 βˆ’ (βˆ’π‘š2 + 5π‘šπ‘›)
= 4π‘š2 βˆ’ 3π‘šπ‘› + 8 + π‘š2 βˆ’ 5π‘šπ‘›
= 4π‘š2 + π‘š2 βˆ’ 3π‘šπ‘› βˆ’ 5π‘šπ‘› + 8
= 5π‘š2 βˆ’ 8π‘šπ‘› + 8

vi) 5π‘₯ βˆ’ 10 βˆ’ (βˆ’π‘₯2 + 10π‘₯ βˆ’ 5)
= 5π‘₯ βˆ’ 10 + π‘₯2 βˆ’ 10π‘₯ + 5
= π‘₯2 + 5π‘₯ βˆ’ 10π‘₯ βˆ’ 10 + 5
= π‘₯2 βˆ’ 5π‘₯ βˆ’ 5

vii) 3π‘Žπ‘ βˆ’ 2π‘Ž2 βˆ’ 2𝑏2 βˆ’ (5π‘Ž2 βˆ’ 7π‘Žπ‘ + 5𝑏2)
= 3π‘Žπ‘ βˆ’ 2π‘Ž2 βˆ’ 2𝑏2 βˆ’ 5π‘Ž2 βˆ’ 7π‘Žπ‘ βˆ’ 5𝑏2
= 3π‘Žπ‘ + 7π‘Žπ‘ βˆ’ 2π‘Ž2 βˆ’ 5π‘Ž2 βˆ’ 2𝑏2 βˆ’ 5𝑏2
= 10π‘Žπ‘ βˆ’ 7π‘Ž2 βˆ’ 7𝑏2
= βˆ’7π‘Ž2 βˆ’ 7𝑏+ 10π‘Žπ‘

viii) 5𝑝2 + 3π‘ž2 βˆ’ π‘π‘ž βˆ’ (4π‘ž βˆ’ 5π‘ž2 βˆ’ 3𝑝2)
= 5𝑝2 + 3π‘ž2 βˆ’ π‘π‘ž βˆ’ 4π‘π‘ž + 5π‘ž2 + 3𝑝2
= 5𝑝2 + 3𝑝2 + 3π‘ž2 + 5π‘ž2 βˆ’ π‘π‘ž βˆ’ 4π‘π‘ž
= 8𝑝2 + 8π‘ž2 βˆ’ 5π‘π‘ž

Q.4) (a) What should be added to π‘₯2 + π‘₯𝑦 + 𝑦to obtain 2π‘₯2 + 3π‘₯𝑦 ?
(b) What should be subtracted from 2π‘Ž + 8𝑏 + 10 to get βˆ’3π‘Ž + 7𝑏 + 16 ?
Sol.4) A) Let 𝑝 should be added
Then according to the question,
π‘₯2 + π‘₯𝑦 + 𝑦2 + 𝑝 = 2π‘₯2 + 3π‘₯𝑦
β‡’ 𝑝 = 2π‘₯2 + 3π‘₯𝑦 βˆ’ (π‘₯2 + π‘₯𝑦 + 𝑦2)
β‡’ 𝑝 = 2π‘₯2 + 3π‘₯𝑦 βˆ’ π‘₯2 βˆ’ π‘₯𝑦 βˆ’ 𝑦2
β‡’ 𝑝 = 2π‘₯2 βˆ’ π‘₯2 βˆ’ 𝑦2 + 3π‘₯𝑦 βˆ’ π‘₯𝑦
β‡’ 𝑝 = π‘₯2 βˆ’ 𝑦2 + 2π‘₯𝑦
Hence, π‘₯2 βˆ’ 𝑦2 + 2π‘₯𝑦 should be added

B) let π‘ž should be subtracted
Then according to question,
2π‘Ž + 8𝑏 + 10 βˆ’ π‘ž = βˆ’3π‘Ž + 7𝑏 + 16
β‡’ βˆ’π‘ž = βˆ’3π‘Ž + 7𝑏 + 16 βˆ’ (2π‘Ž + 8𝑏 + 10)
β‡’ βˆ’π‘ž = βˆ’3π‘Ž + 7𝑏 + 16 βˆ’ 2π‘Ž βˆ’ 8𝑏 βˆ’ 10
β‡’ βˆ’π‘ž = βˆ’3π‘Ž βˆ’ 2π‘Ž + 7𝑏 βˆ’ 8𝑏 + 16 βˆ’ 10
β‡’ βˆ’π‘ž = βˆ’5π‘Ž βˆ’ 𝑏 + 6
β‡’ π‘ž = βˆ’(βˆ’5π‘Ž βˆ’ 𝑏 + 6)
β‡’ π‘ž = 5π‘Ž + 𝑏 βˆ’ 6

Q.5) What should be taken away from 3π‘₯2 βˆ’ 4𝑦2 + 5π‘₯𝑦 + 20 to obtain βˆ’π‘₯2 βˆ’ 𝑦2 + 6π‘₯𝑦 + 20 ?
Sol.5) Let π‘ž should be subtracted
Then according to question,
3π‘₯2 βˆ’ 4𝑦2 + 5π‘₯𝑦 + 20 βˆ’ π‘ž = βˆ’π‘₯2 βˆ’ π‘¦2 + 6π‘₯𝑦 + 20
β‡’ π‘ž = 3π‘₯2 βˆ’ 4𝑦2 + 5π‘₯𝑦 + 20 βˆ’ (βˆ’π‘₯2 βˆ’ π‘¦2 + 6π‘₯𝑦 + 20)
β‡’ π‘ž = 3π‘₯2 βˆ’ 4𝑦2 + 5π‘₯𝑦 + 20 + π‘₯2 + π‘¦2 βˆ’ 6π‘₯𝑦 βˆ’ 20
β‡’ π‘ž = 3π‘₯2 + π‘₯2 βˆ’ 4𝑦2 + π‘¦2 + 5π‘₯𝑦 βˆ’ 6π‘₯𝑦 + 20 βˆ’ 20
β‡’ π‘ž = 4π‘₯2 βˆ’ 3𝑦2 βˆ’ π‘₯𝑦 + 0
Hence, 4π‘₯2 βˆ’ 3𝑦2 βˆ’ π‘₯𝑦 should be subtracted.

Q.6) (a) From the sum of 3π‘₯ – 𝑦 + 11 and – 𝑦 – 11, subtract 3π‘₯ – 𝑦 – 11.
(b) From the sum of 4 + 3π‘₯ and 5 – 4π‘₯ + 2π‘₯2, subtract the sum of 3π‘₯2 β€“ 5π‘₯ and – π‘₯2 + 2π‘₯ + 5.
Sol.6) a) According to question,
(3π‘₯ βˆ’ 𝑦 + 11) + (βˆ’π‘¦ βˆ’ 11) βˆ’ (3π‘₯ βˆ’ 𝑦 βˆ’ 11) = 3π‘₯ βˆ’ 𝑦 + 11 βˆ’ 𝑦 βˆ’ 11 βˆ’ 3π‘₯ + 𝑦 + 11
= 3π‘₯ βˆ’ 3π‘₯ βˆ’ 𝑦 + 𝑦 + 11 βˆ’ 11 + 11
= (3 βˆ’ 3)π‘₯ βˆ’ (1 + 1 βˆ’ 1)𝑦 + 11 + 11 βˆ’ 11
= 0π‘₯ βˆ’ 𝑦 + 11 = βˆ’π‘¦ + 11

b) According to the question,
[(4 + 3π‘₯) + (5 βˆ’ 4π‘₯ + 2π‘₯2)] βˆ’ [(3π‘₯2 βˆ’ 5π‘₯) + (βˆ’π‘₯2 + 2π‘₯ + 5)]
= [4π‘₯ + 3π‘₯ + 5 βˆ’ 4π‘₯ + 2π‘₯2] βˆ’ [3π‘₯2 βˆ’ 5π‘₯ βˆ’ π‘₯2 + 2π‘₯ + 5]
= [2π‘₯2 + 3π‘₯ βˆ’ 4π‘₯ + 5 + 4] βˆ’ [3π‘₯2 βˆ’ π‘₯2 + 2π‘₯ βˆ’ 5π‘₯ + 5]
= [2π‘₯2 βˆ’ π‘₯ + 9] βˆ’ [2π‘₯2 βˆ’ 3π‘₯ + 5]
= 2π‘₯2 βˆ’ π‘₯ + 9 βˆ’ 2π‘₯2 + 3π‘₯ βˆ’ 5
= 2π‘₯2 βˆ’ 2π‘₯2 βˆ’ π‘₯ + 3π‘₯ + 9 βˆ’ 5
= 2π‘₯ + 4

Exercise 12.3

Q.1) If m = 2, find the value of:
i) π‘š βˆ’ 2 ii) 3π‘š βˆ’ 5 iii) 9 βˆ’ 5π‘š iv) 3π‘š2 βˆ’ 2π‘š βˆ’ 7 v) 5π‘š/2 βˆ’ 4
Sol.1) i) π‘š βˆ’ 2
= 2 βˆ’ 2 = 0                          [putting π‘š = 2]
ii) 3π‘š βˆ’ 5
= 3 Γ— 2 βˆ’ 5                          [putting π‘š = 2]
= 6 βˆ’ 5 = 1
iii) 9 βˆ’ 5π‘š
= 9 βˆ’ 5 Γ— 2                          [putting π‘š = 2]
= 9 βˆ’ 10 = βˆ’1

iv) 3π‘š2 βˆ’ 2π‘š βˆ’ 7
= 3(2)2 βˆ’ 2(2) βˆ’ 7               [putting π‘š = 2]
= 3 Γ— 4 βˆ’ 2 Γ— 2 βˆ’ 7
= 12 βˆ’ 4 βˆ’ 7
= 12 βˆ’ 11 = 1

v) (5π‘š/2) βˆ’ 4
= (5Γ—2/2) βˆ’ 4                       [putting π‘š = 2]
= (10/2) βˆ’ 4
= 5 βˆ’ 4 = 1

Q.2) If π‘₯ = βˆ’2, find
(i) 4𝑝 + 7
(ii) βˆ’3𝑝2 + 4𝑝 + 7
(iii) βˆ’2𝑝3 βˆ’ 3𝑝2 + 4𝑝 + 7
Sol.2) (i) 4𝑝 + 7
= 4(βˆ’2) + 7                                         [putting 𝑝 = βˆ’2]
= βˆ’8 + 7 = βˆ’1

(ii) βˆ’3𝑝2 + 4𝑝 + 7
= βˆ’3(βˆ’2)2 + 4(βˆ’2) + 7                        [putting 𝑝 = βˆ’2]
= βˆ’3 Γ— 4 βˆ’ 8 + 7
= βˆ’12 βˆ’ 8 + 7
= βˆ’20 + 7 = βˆ’13

(iii) βˆ’2𝑝3 βˆ’ 3𝑝2 + 4𝑝 + 7
= βˆ’2(βˆ’2)3 βˆ’ 3(βˆ’2)2 + 4(βˆ’2) + 7         [putting 𝑝 = βˆ’2]
= βˆ’2 Γ— (βˆ’8) βˆ’ 3 Γ— 4 βˆ’ 8 + 7
= 16 βˆ’ 12 βˆ’ 8 + 7
= βˆ’20 + 23 = 3

Q.3) Find the value of the following expressions, when π‘₯ =- 1:
(i) 2π‘₯ βˆ’ 7 ii) βˆ’π‘₯ + 2 iii) π‘₯2 + 2π‘₯ + 1 iv) 2π‘₯2 βˆ’ π‘₯ βˆ’ 2
Sol.3) i) 2π‘₯ βˆ’ 7
= 2(βˆ’1) βˆ’ 7                             [putting π‘š = 2]
= βˆ’2 βˆ’ 7 = βˆ’9

ii) βˆ’π‘₯ + 2
= βˆ’(βˆ’1) + 2                             [putting π‘š = 2]
= 1 + 2 = 3

iii) π‘₯2 + 2π‘₯ + 1
= (βˆ’1)2 + 2(βˆ’1) + 1                 [putting π‘š = 2]
= 1 βˆ’ 2 + 1
= 2 βˆ’ 2 = 0

iv) 2π‘₯2 βˆ’ π‘₯ βˆ’ 2
= 2(βˆ’1)2 βˆ’ (βˆ’1) βˆ’ 2                 [putting π‘š = 2]
= 2 Γ— 1 + 1 βˆ’ 2
= 2 + 1 βˆ’ 2
= 3 βˆ’ 2 = 1

Q.4) If π‘Ž = 2, 𝑏 = βˆ’2, find the value of:
i) π‘Ž2 + 𝑏2 ii) π‘Ž2 + π‘Žπ‘ + 𝑏2 iii) π‘Ž2 βˆ’ 𝑏2

Sol.4) i) π‘Ž2 + 𝑏2
= (2)2 + (2)2
= 4 + 4 = 8

ii) π‘Ž2 + π‘Žπ‘ + 𝑏2
= (2)2 + (2)(βˆ’2) + (βˆ’2)2
= 4 βˆ’ 4 + 4 = 4

iii) π‘Ž2 βˆ’ 𝑏2
= (2)2 βˆ’ (βˆ’2)2
= 4 βˆ’ 4 = 0

Q.5) When π‘Ž = 0, 𝑏 = βˆ’1 find the value of the given expressions:
(i) 2π‘Ž + 2𝑏 ii) 2π‘Ž2 + 𝑏2 + 1 iii) 2π‘Ž2𝑏 + 2π‘Žπ‘2 + π‘Žπ‘ iv) π‘Ž2 + π‘Žπ‘ + 2
Sol.5) i) 2π‘Ž + 2𝑏
= 2(0) + 2(βˆ’1)
= 0 βˆ’ 2 = βˆ’2

ii) 2π‘Ž2 + 𝑏2 + 1
= 2(0)2 + (βˆ’1)2 + 1
= 2 Γ— 0 + 1 + 1 = 0 + 2 = 0            [putting π‘Ž = 0, 𝑏 = βˆ’1]

iii) 2π‘Ž2𝑏 + 2π‘Žπ‘2 + π‘Žπ‘
= 2(0)2(βˆ’1) + 2(0)(βˆ’1)2 + (0)(βˆ’1)
= 0 + 0 + 0 = 0

iv) π‘Ž2 + π‘Žπ‘ + 2
= (0)2 + (0)(βˆ’1) + 2
= 0 + 0 + 2 = 2

Q.6) Simplify the following expressions and find the value at π‘₯ = 2:
(i) π‘₯ + 7 + 4(π‘₯ βˆ’ 5)
(ii) 3(π‘₯ + 2) + 5π‘₯ βˆ’ 7
(iii) 10π‘₯ + 4(π‘₯ βˆ’ 2)
(iv) 5(3π‘₯ βˆ’ 2) + 4π‘₯ + 8
Sol.6) (i) π‘₯ + 7 + 4(π‘₯ βˆ’ 5)
= π‘₯ + 7 + 4π‘₯ βˆ’ 20
= 4π‘₯ + π‘₯ + 7 βˆ’ 20 = 5π‘₯ βˆ’ 13
= 5(2) βˆ’ 13 = 10 βˆ’ 13                    [𝑃𝑒𝑑𝑑𝑖𝑛𝑔 π‘₯ = 2 ]
= βˆ’3

(ii) 3(π‘₯ + 2) + 5π‘₯ βˆ’ 7
= 3π‘₯ + 6 + 5π‘₯ βˆ’ 7
= 3π‘₯ + 5π‘₯ + 6 βˆ’ 7 = 8π‘₯ βˆ’ 1
= 8( 2) – 1                                    [𝑃𝑒𝑑𝑑𝑖𝑛𝑔 π‘₯ = 2 ]
= 16 – 1 = 15

(iii) 10π‘₯ + 4(π‘₯ βˆ’ 2)
= 10π‘₯ + 4π‘₯ βˆ’ 8
= 14π‘₯ βˆ’ 8
= 14( 2) – 8                                   [𝑃𝑒𝑑𝑑𝑖𝑛𝑔 π‘₯ = 2 ]
28 – 8 = 20

(iv) 5(3π‘₯ βˆ’ 2) + 4π‘₯ + 8
= 15π‘Ž βˆ’ 10 + 4π‘₯ + 8
= 15π‘Ž + 4π‘Ž βˆ’ 10 + 8 = 19π‘Ž βˆ’ 2 [𝑃𝑒𝑑𝑑𝑖𝑛𝑔 = 2 ]
= 19(2) βˆ’ 2 = 38 βˆ’ 2
= 36

Q.7) Simplify these expressions and find their values if π‘₯ = 3, π‘Ž =– 1, 𝑏 =– 2.
i) 3π‘₯ – 5 – π‘₯ + 9 ii) 2– 8π‘₯ + 4π‘₯ + 4 iii) 3π‘Ž + 5 – 8π‘Ž + 1
iv) 10– 3𝑏– 4– 5𝑏 v) 2π‘Žβ€“ 2𝑏– 4– 5 + π‘Ž
Sol.7) Putting values π‘₯ = 3, π‘Ž =– 1, 𝑏 =– 2.

i) 3π‘₯ – 5 – π‘₯ + 9
= 3π‘₯ βˆ’ π‘₯ βˆ’ 5 + 9 = 2π‘₯ + 4
= 2 Γ— 3 + 4
= 6 + 4 = 10

ii) 2– 8π‘₯ + 4π‘₯ + 4
= βˆ’8π‘₯ + 4π‘₯ + 2 + 4 = βˆ’4π‘₯ βˆ’ 6
= βˆ’4 Γ— 3 + 6
= βˆ’12 + 6 = βˆ’12

iii) 3π‘Ž + 5 – 8π‘Ž + 1
= 3π‘Ž βˆ’ 8π‘Ž + 5 + 1 = βˆ’5π‘Ž + 6
= βˆ’5(βˆ’1) + 6
= 5 + 6 = 11

iv) 10– 3𝑏– 4– 5𝑏
= βˆ’3𝑏 βˆ’ 5𝑏 + 10 βˆ’ 4 = βˆ’8𝑏 + 6
= βˆ’8(βˆ’2) + 6
= 16 + 6 = 22

v) 2π‘Žβ€“ 2𝑏– 4– 5 + π‘Ž
= 2π‘Ž + π‘Ž βˆ’ 2𝑏 βˆ’ 4 βˆ’ 5
= 3π‘Ž βˆ’ 2𝑏 βˆ’ 9 = 3(βˆ’1) βˆ’ 2(βˆ’2) βˆ’ 9
= βˆ’3 + 4 βˆ’ 9 = βˆ’8

Q.8) (i) If 𝑧 = 10, find the value of 𝑧3 βˆ’ 3(𝑧 βˆ’ 10) .
(ii) 𝐼𝑓 𝑝 = -10, find the value of 𝑝2 βˆ’ 2𝑝 βˆ’ 100.
Sol.8) i) 𝑧3 βˆ’ 3(𝑧 βˆ’ 10)
= 𝑧3 βˆ’ 3𝑧 + 30                                        [putting 𝑧 = 10]
= (10 Γ— 10 Γ— 10) βˆ’ (3 Γ— 10) + 30
= 1000 βˆ’ 30 + 30 = 1000
ii) 𝑝2 βˆ’ 2𝑝 βˆ’ 100
= (βˆ’10) Γ— (βˆ’10) βˆ’ 2(βˆ’10) βˆ’ 100              [putting 𝑝 = βˆ’10]
= 100 + 20 βˆ’ 100 = 20

Q.9) What should be the value of π‘Ž if the value of 2π‘₯2 + π‘₯ – π‘Ž equals to 5, when π‘₯ = 0?
Sol.9) 2π‘₯2 + π‘₯ – π‘Ž = 5 , when π‘₯ = 0
(2 Γ— 0) + 0 βˆ’ π‘Ž = 5
= 0 βˆ’ π‘Ž = 5
= π‘Ž = βˆ’5

Q.10) Simplify the expression and find its value when π‘Ž = 5 and 𝑏 =– 3.
2(π‘Ž2 + π‘Žπ‘) + 3– π‘Žπ‘
Sol.
10) 2(π‘Ž2 + π‘Žπ‘) + 3– π‘Žπ‘ , where π‘Ž = 5 & 𝑏 = βˆ’3
= 2π‘Ž2 + 2π‘Žπ‘ + 3 βˆ’ π‘Žπ‘
= 2π‘Ž2 + 2π‘Žπ‘ βˆ’ π‘Žπ‘ + 3
= 2π‘Ž2 + π‘Žπ‘ + 3
Now, putting the values of π‘Ž π‘Žπ‘›π‘‘ 𝑏
= 2 Γ— (5 Γ— 5) + 5 Γ— (βˆ’3) + 3
= 50 βˆ’ 15 + 3 = 38

Exercise 12.4

Q.1) Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

""NCERT-Solutions-Class-7-Mathematics-Algebraic-Expressions-3

Sol.1)

""NCERT-Solutions-Class-7-Mathematics-Algebraic-Expressions-2

(i) 5𝑛 + 1
Putting 𝑛 = 5,               5 Γ— 5 + 1 = 25 + 1 = 26
Putting 𝑛 = 10,             5 Γ— 10 + 1 = 50 + 1 = 51
Putting 𝑛 =100,            5 Γ— 100 + 1 = 500 + 1 = 501

(ii) 3𝑛 + 1
Putting 𝑛 = 5,               3 Γ— 5 + 1 = 15 + 1 = 16
Putting 𝑛 = 10,             3 Γ— 10 + 1 = 30 + 1 = 31
Putting 𝑛 =100,            3 Γ— 100 + 1 = 300 + 1 = 301

(iii) 5𝑛 + 2
Putting 𝑛 = 5,               5 Γ— 5 + 2 = 25 + 2 = 27
Putting 𝑛 = 10,             5 Γ— 10 + 2 = 50 + 2 = 52
Putting 𝑛 =100,            5 Γ— 100 + 2 = 500 + 2 = 502

Q.2) Use the given algebraic expression to complete the table of number patterns.

""NCERT-Solutions-Class-7-Mathematics-Algebraic-Expressions-1

Sol.2) (i) 2𝑛 βˆ’ 1
Putting 𝑛 =100,              2 Γ— 100 – 1 = 200 – 1 = 199

(ii) 3𝑛 + 2
Putting 𝑛 = 5,                 3 Γ— 5 + 2 = 15 + 2 = 17
Putting 𝑛 = 10,               3 Γ— 10 + 2 = 30 + 2 = 32
Putting 𝑛 =100,              3 Γ— 100 + 2 = 300 + 2 = 302

(iii) 4𝑛 + 1
Putting 𝑛 = 5,                 4 Γ— 5 + 1 = 20 + 1 = 21
Putting 𝑛 = 10,               4 Γ— 10 + 1 = 40 + 1 = 41
Putting 𝑛 =100,              4 Γ— 100 + 1 = 400 + 1 = 401

(iv) 7𝑛 + 20
Putting 𝑛 = 5,                 7 Γ— 5 + 20 = 25 + 20 = 55
Putting 𝑛 = 10,               7 Γ— 10 + 20 = 70 + 20 = 90
Putting 𝑛 =100,              7 Γ— 100 + 20 = 700 + 20 = 720

(v) 𝑛2 + 1
Putting 𝑛 = 5,                 5 Γ— 5 + 1 = 25 + 1 = 26
Putting 𝑛 = 10,               10 Γ— 10 + 1 = 100 + 1 = 101
Putting 𝑛 =100,              100 Γ— 100 + 1 = 10000 + 1 = 10001

""NCERT-Solutions-Class-7-Mathematics-Algebraic-Expressions

More Study Material

NCERT Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions

NCERT Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 7 Mathematics textbook online or you can easily download them in pdf.

Chapter 12 Algebraic Expressions Class 7 Mathematics NCERT Solutions

The Class 7 Mathematics NCERT Solutions Chapter 12 Algebraic Expressions are designed in a way that will help to improve the overall understanding of students. The answers to each question in Chapter 12 Algebraic Expressions of Mathematics Class 7 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 12 Algebraic Expressions Class 7 chapter of Mathematics so that it can be easier for students to understand all answers.

NCERT Solutions Chapter 12 Algebraic Expressions Class 7 Mathematics

Class 7 Mathematics NCERT Solutions Chapter 12 Algebraic Expressions is a really good source using which the students can get more marks in exams. The same questions will be coming in your Class 7 Mathematics exam. Learn the Chapter 12 Algebraic Expressions questions and answers daily to get a higher score. Chapter 12 Algebraic Expressions of your Mathematics textbook has a lot of questions at the end of chapter to test the students understanding of the concepts taught in the chapter. Students have to solve the questions and refer to the step-by-step solutions provided by Mathematics teachers on studiestoday to get better problem-solving skills.

Chapter 12 Algebraic Expressions Class 7 NCERT Solution Mathematics

These solutions of Chapter 12 Algebraic Expressions NCERT Questions given in your textbook for Class 7 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 7.

Class 7 NCERT Solution Mathematics Chapter 12 Algebraic Expressions

NCERT Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions detailed answers are given with the objective of helping students compare their answers with the example. NCERT solutions for Class 7 Mathematics provide a strong foundation for every chapter. They ensure a smooth and easy knowledge of Revision notes for Class 7 Mathematics. As suggested by the HRD ministry, they will perform a major role in JEE. Students can easily download these solutions and use them to prepare for upcoming exams and also go through the Question Papers for Class 7 Mathematics to clarify all doubts

Where can I download latest NCERT Solutions for Class 7 Mathematics Chapter 12 Algebraic Expressions

You can download the NCERT Solutions for Class 7 Mathematics Chapter 12 Algebraic Expressions for latest session from StudiesToday.com

Can I download the NCERT Solutions of Class 7 Mathematics Chapter 12 Algebraic Expressions in Pdf

Yes, you can click on the link above and download NCERT Solutions in PDFs for Class 7 for Mathematics Chapter 12 Algebraic Expressions

Are the Class 7 Mathematics Chapter 12 Algebraic Expressions NCERT Solutions available for the latest session

Yes, the NCERT Solutions issued for Class 7 Mathematics Chapter 12 Algebraic Expressions have been made available here for latest academic session

How can I download the Chapter 12 Algebraic Expressions Class 7 Mathematics NCERT Solutions

You can easily access the links above and download the Chapter 12 Algebraic Expressions Class 7 NCERT Solutions Mathematics for each chapter

Is there any charge for the NCERT Solutions for Class 7 Mathematics Chapter 12 Algebraic Expressions

There is no charge for the NCERT Solutions for Class 7 Mathematics Chapter 12 Algebraic Expressions you can download everything free

How can I improve my scores by reading NCERT Solutions in Class 7 Mathematics Chapter 12 Algebraic Expressions

Regular revision of NCERT Solutions given on studiestoday for Class 7 subject Mathematics Chapter 12 Algebraic Expressions can help you to score better marks in exams

Are there any websites that offer free NCERT solutions for Chapter 12 Algebraic Expressions Class 7 Mathematics

Yes, studiestoday.com provides all latest NCERT Chapter 12 Algebraic Expressions Class 7 Mathematics solutions based on the latest books for the current academic session

Can NCERT solutions for Class 7 Mathematics Chapter 12 Algebraic Expressions be accessed on mobile devices

Yes, studiestoday provides NCERT solutions for Chapter 12 Algebraic Expressions Class 7 Mathematics in mobile-friendly format and can be accessed on smartphones and tablets.

Are NCERT solutions for Class 7 Chapter 12 Algebraic Expressions Mathematics available in multiple languages

Yes, NCERT solutions for Class 7 Chapter 12 Algebraic Expressions Mathematics are available in multiple languages, including English, Hindi