# NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 7 Maths Chapter 12 Algebraic Expressions 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 number m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) Y- Z
(ii) $$\frac{1}{2}$$ X+Y
(iii) Z×Z = $$Z^2$$
(iv) $$\frac{1}{4}$$pq
(v) $$x^2 + Y^2$$
(vi) 3mm + 5
(vii) 10 – yz
(viii) ab – (a + b)

Q.2 (i) Identify the terms and their factors in the following expressions show the terms and factors by tree diagrams.
(a) x – 3
(b) $$1 + x + x^2$$
(c) $$y – y^3$$
(d)$$5xy^2 + 7x^2y$$
(e) $$-ab + 2b^2 – 3a^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(a) Expression: ( x-3 )
Term I = x
Term II = -3
(b) Expression: $$1 + x + x^2$$
Term I = 1
Term II = x
Term III = $$x^2$$ = x × x
(c) Expression: $$y – y^3$$
Term I = y
Term II = $$-y^3$$ = -y × -y ×-y
(d) Expression: $$5xy^2 + 7x^2y$$
Term I = $$5xy^2$$ = 5×x×y×y
Term II = $$7x^2y$$ = 7× x × x × y
(e) Expression: $$-ab + 2b^2 – 3a^2$$
Term I = -ab = -a × b
Term II = $$2b^2$$ = 2 × b × b
Term III = $$-3a^2$$ = -3 ×a×a

Q.2 (ii) Identify terms and factors in the expression given below:
(a) -4x + 5
(b) -4x + 5y
(c) $$5y + 3y^2$$
(d)$$xy + 2x^2y^2$$
(e) pq + q
(f) 1.2ab – 2.4b + 3.6a
(g) $$\frac{3}{4} x+\frac{1}{4}$$
(h) $$0.1p^2 + 0.2q^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

Q.3 Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i)$$5 – 3t^2$$
(ii)$$1 + t + t^2 + t^3$$
(iii)$$x + 2xy + 3y$$
(iv) $$100m + 1000n$$
(v) $$– p^2q^2 + 7pq$$
(vi)$$1.2 a + 0.8 b$$
(vii)$$3.14 r^2$$
(viii)$$2 (l + b)$$
(ix)$$0.1 y + 0.01 y^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

Q.4 (a) Identify terms which contain x and give the coefficient of x.
(i)$$y^2x + y$$
(ii) $$13y^2 – 8yx$$
(iii) $$x + y + 2$$
(iv)$$5 + z + zx$$
(v) $$1 + x + xy$$
(vi)$$12 xy^2 + 25$$
(vii) $$7x + xy^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) Expression: $$y^2x + y$$
Term: $$y^2x$$
Coefficient of x = $$y^2$$
(ii) Expression: $$13y^2 – 8yx$$
Term: $$8yx$$
Coefficient of x = $$8y$$
(iii) Expression: $$x + y + 2$$
Term: $$x$$
Coefficient of x = 1
(iv) Expression:$$5 + z + zx$$
Term: $$zx$$
Coefficient of x = z
(v)Expression:$$1 + x + xy$$
Term I $$x$$
Term II xy
Coefficient of x = 1, y
(vi) Expression:$$12 xy^2 + 25$$
Term: $$12 xy^2$$
Coefficient of x = $$12 y^2$$
(vii) Expression:$$7x + xy^2$$
Term I $$7x$$
Term II $$xy^2$$
Coefficient of x = $$7 , y^2$$

Q.4 (b) Identify terms which contain $$y^2$$and give the coefficients of$$y^2.$$
(i) $$8 – xy^2$$
(ii) $$5y^2 + 7x$$
(iii) $$2x^2y – 15xy^2 + 7y^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) Expression: $$8 – xy^2$$
Term : $$-xy^2$$
Coefficient of$$y^2 =- x$$
(ii) Expression: $$5y^2 + 7x$$
Term : $$5y^2$$
Coefficient of$$y^2 = 5$$
(iii) Expression: $$2x^2y – 15xy^2 + 7y^2$$
Term : $$– 15xy^2 , 7y^2$$
Coefficient of$$y^2 = -15x , 7$$

Q.5 Classify into monomials, binomials and trinomials: (i) 4y – 7x
(ii)$$y^2$$
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) $$4p^2q – 4pq^2$$
(viii) 7mn
(ix)$$z^2 – 3z + 8$$
(x) $$a^2 + b^2$$
(xi)$$z^2 + z$$
(xii) $$1 + x + x^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) 4y – 7x __ Binomial
(ii)$$y^2$$__ Monomial
(iii) x + y – xy__ Trinomial
(iv) 100__Monomial
(v) ab – a – b__Trinomial
(vi) 5 – 3t__Binomial
(vii) $$4p^2q – 4pq^2$$__ Binomial
(viii) 7mn__Monimial
(ix)$$z^2 – 3z + 8$$__Trinomial
(x) $$a^2 + b^2$$ __ Binomial
(xi)$$z^2 + z$$__Binomial
(xii) $$1 + x + x^2$$__ Trinomial

Q.6 State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
(ii) -7x, $$\frac{5}{2}$$x
(iii) -29x, -29y
(iv) 14xy, 42yx
(v)$$4m^2p, 4mp^2$$
(vi) $$12xz, 12 x^2y^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) 1, 100 like
(ii) -7x, $$\frac{5}{2}$$x like
(iii) -29x, -29y unlike
(iv) 14xy, 42yx like
(v)$$4m^2p, 4mp^2$$ unlike
(vi) $$12xz, 12 x^2y^2$$ unlike

Q.7 Identify like terms in the following:
$$(a)-xy^2, -4yx^2, 8x^2, 2xy^2, 7y^2, -11x^2,$$
$$-100x, -11yx, 20x^2y, -6x^2, y, 2xy, 3x$$
$$(b) 10pq, 7p, 8q, -p^2q^2, -7qp, -100q, -23, 12q^2p^2,$$
$$-5p^2, 41, 2405p, 78qp, 13p^2q, qp^2, 701p^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(a) $$(i) – xy^2, 2xy^2$$
$$(ii) – 4yx^2, 20x^2y$$
$$(iii) 8x^2, – 11x^2, – 6x^2$$
$$(iv) 7y, y$$
(v) – 100x, 3x
(vi) – 11yx, 2xy (b)(i) 10pq, – 7qp, 78qp
(ii) 7p, 2405p
(iii) 8q, – 100q
(iv) $$– p^2q^2, 12q^2p^2$$
(v) – 23, 41
(vi) $$– 5p^2, 701p^2$$
(vii) $$13p^2q, qp^2$$

## NCERT solutions for class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2

Q.1 Simplify combining like terms:
(i) 21b -32 + 7b- 20b
(ii) $$-z^2 + 13z^2 -5z + 7z^3 – 15z$$
(iii)$$p – (p – q) – q – (q – p)$$
(iv) 3a – 2b — ab – (a – b + ab) + 3ab + 6 – a
$$(v) 5x^2y – 5x^2 + 3yx^2 – 3y^2 + x^2 – y^2 + 8xy^2 -3y^2$$
$$(vi) (3y^2 + 5y – 4) – (8y – y^2 – 4)$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) 21b – 32 + 7b – 20b
= 21b + 7b -20b -32
= 8b -32
(ii) $$-z^2 + 13z^2 -5z + 7z^3 – 15z$$
= $$-z^2 + 13z^2 -5z - 15z + 7z^3$$
= $$11z^2 -20z + 7z^3$$
(iii) ( p – (p – q) – q – (q – p)
= p-p +q-q-q+p
= p- q
(iv) 3a – 2b — ab – (a – b + ab) + 3ab + 6 – a
= 3a - 2b - ab - a + b - ab +3ab + 6 - a
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= a +ab
(v) $$5x^2y – 5x^2 + 3yx^2 – 3y^2 + x^2 – y^2 + 8xy^2 -3y^2$$
= $$5x^2y + 3yx^2 – 5x^2 + x^2 – 3y^2 – y^2 – 3y^2+ 8xy^2$$
= 8x^2y – 4x^2 – 7y^2 + 8xy^2
(vi) $$(3y^2 + 5y – 4) – (8y – y^2 – 4)$$
= $$3y^2 + 5y – 4 – 8y + y^2 + 4)$$
= $$3y^2 + y^2 + 5y – 8y – 4 + 4$$
= $$4y^2 – 3y$$

(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz, -z, z – t
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii)$$4x^2y, -3xy^2, -5xy^2, 5x^2y$$
$$(viii) 3p^2q^2 – 4pq + 5, -10p^2q^2, 15 + 9pq + 7p^2q^2$$
(ix) ab – 4a, 4b – ab, 4a – 4b
$$(x) x^2 – y^2 – 1, y^2 – 1 – x^2, 1 – x^2 – y^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) 3mn, -5mn, 8mn, -4mn
= (3 mn) + (-5 mn) + (8 mn) + (- 4 mn)
= 3 mn - 5 mn + 8 mn - 4 mn
= 2mn
(ii) t – 8tz, 3tz – z, z – t
= t – 8tz + 3tz – z + z – t
= t – t – 8tz + 3tz – z + z
= – 5 tz
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
= -7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn) – 3
= -7mn + 12 mn + 9mn – 2 mn + 5 + 2 – 8 – 3
= 12 mn – 4
(iv) a + b – 3, b – a + 3, a – b + 3
= a + b – 3 + b – a + 3 + a – b + 3
= a – a + a + b + b – b – 3 + 3 + 3
= a + b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
= 14x + 14y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
= -12xy + 8xy + 4xy + 14x – 7x + 10y – 10y – 13 + 18
= 0 + 7x + 0 + 5
= 7x + 5
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
= 5m -7n + 3n — 4m + 2 + 2m – 3mn – 5
= 5m – 4m + 2m – 7n + 3n- 3mn + 2 – 5
= 3m – 4n – 3mn – 3
(vii)$$4x^2y, -3xy^2, -5xy^2, 5x^2y$$
= $$4x^2y + (-3xy^2) + (-5xy^2) + 5x^2y$$
=$$4x^2y + 5x^2y – 3xy^2 – 5xy^2$$
=$$9x^2y – 8xy^2$$
$$(viii) 3p^2q^2 – 4pq + 5, -10p^2q^2, 15 + 9pq + 7p^2q^2$$
$$= 3p^2q^2 – 4pq + 5 + (- 10p^2q^2) + 15 + 9pq + 7p^2q^2$$
$$= 3p^2q2 – 10p^2q^2 + 7p^2q^2 – 4pq + 9pq + 5 + 15$$
= 5pq + 20
(ix) ab – 4a, 4b – ab, 4a – 4b
= ab – 4a + (4b – ab) + (4a – 4b)
= ab – 4a + 4b – ab + 4a – 4b
= ab – ab – 4a + 4a + 4b – 4b
= 0
$$(x) x^2 – y^2 – 1, y^2 – 1 – x^2, 1 – x^2 – y^2$$
$$= x^2 – y^2 – 1 + (y^2 – 1 – x^2) + (1 – x^2 – y^2)$$
$$= x^2 – y^2 – 1 + y^2 – 1 – x^2 + 1 – x^2 – y^2$$
$$= x^2 – x^2 – x^2 – y^2 + y^2 – y^2 – 1 – 1 + 1$$
$$= -x^2 – y^2 -1$$

Q.3 Subtract:
$$(i) -5y^2 from y^2$$
$$(ii) 6xy from -12xy$$
$$(iii) (a – b) from (a + b)$$
$$(iv) a(b – 5) from b(5 – a)$$
$$(v) -m^2 + 5mn from 4m^2 – 3mn + 8$$
$$(vi) -x^2 + 10x – 5 from 5x – 10$$
$$(vii) 5a^2 – 7ab + 5b^2 from 3ab – 2a^2 – 2b^2$$
$$(viii) 4pq – 5q^2 – 3p^2 from 5p^2 + 3q^2 – pq$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

$$(i) -5y^2 from y^2$$
$$= y^2 – (-5y^2)$$
$$= y^2 + 5y^2$$
$$= 6y^2$$
$$(ii) 6xy from -12xy$$
= -12xy – 6xy
= – 18xy
$$(iii) (a – b) from (a + b)$$
= (a + b) – (a – b)
= a + b – a + b
= a – a + b + b
= 2b
$$(iv) a(b – 5) from b(5 – a)$$
= b (5 -a) – a (b – 5)
= 5b – ab – ab + 5a
= 5a + 5b – 2ab
$$(v) -m^2 + 5mn from 4m^2 – 3mn + 8$$
$$= 4m^2 – 3mn + 8 – (- m^2 + 5mn)$$
$$= 4m^2 – 3mn + 8 + m^2 – 5mn$$
$$= 4m^2 + m^2 – 3mn – 5mn + 8$$
$$= 5m^2 – 8mn + 8$$
$$(vi) -x^2 + 10x – 5 from 5x – 10$$
$$= 5x – 10 – (-x^2 + 10x – 5)$$
$$= 5x – 10 + x^2 – 10x + 5$$
$$= x^2 + 5x – 10x – 10 + 5$$
$$= x^2 – 5x – 5$$
$$(vii) 5a^2 – 7ab + 5b^2 from 3ab – 2a^2 – 2b^2$$
$$= 3ab – 2a^2 – 2b^2 – (5a^2 – 7ab + 5b^2)$$
$$= 3ab – 2a^2 – 2b^2 – 5a^2 + 7ab – 5b^2$$
$$= 3ab + 7ab – 2a^2 – 5a^2 – 2b^2 – 5b^2$$
$$= 10ab – 7a^2 – 7b^2$$
$$(viii) 4pq – 5q^2 – 3p^2 from 5p^2 + 3q^2 – pq$$
$$= 5p^2 + 3q^2 – pq – (4pq – 5q^2 – 3p^2)$$
$$= 5p^2 + 3q^2 – pq – 4pq + 5q^2 + 3p^2$$
$$= 5p^2 + 3p^2 + 3q^2 + 5q^2 – pq – 4pq$$
$$= 8p^2 + 8q^2 – 5pq$$

Q.4 (a) What should be added to $$x^2 + xy + y^2$$ to obtain $$2x^2 + 3xy?$$
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16?

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(a) $$p + (x^2 + xy + y^2) = 2x^2 + 3xy$$
$$p = (2x^2 + 3xy) – (x^2 + xy + y^2)$$
$$p = 2x^2 + 3xy – x^2 – xy – y^2$$
$$p = 2x^2 – x^2 + 3xy – xy – y^2$$
$$p = x^2 + 2xy – y^2$$
(b) 2a + 8b + 10 – m = -3a + 7b + 16
m = (2a + 8b + 10) – (-3a + 7b + 16)
m = 2a + 8b + 10 + 3a – 7b – 16
m = 2a + 3a + 8b – 7b + 10 – 16
m = 5a + b – 6

Q.5 What should be taken away from$$3x^2 – 4y^2 + 5xy + 20$$to obtain $$– x^2 – y^2 + 6xy + 20?$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

$$3x^2 – 4y^2 + 5xy + 20 – m = -x^2 – y^2 + 6xy + 20$$
$$m= 3x^2 – 4y^2 + 5xy + 20 – (-x^2 – y^2 + 6xy + 20)$$
$$m= 3x^2 – 4y^2 + 5xy + 20 + x^2 + y^2 – 6xy – 20$$
$$m = 3x^2 + x^2 – 4y^2 + y^2 + 5xy – 6xy + 20 – 20$$
$$m = 4x^2 – 3y^2 – xy$$

Q.6 (a) From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and$$5 – 4x + 2x^2,$$subtract the sum of $$3x^2 – 5x$$ and$$-x^2 + 2x + 5.$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(a) sum of 3x – y + 11 and -y – 11
= 3x – y + 11 + (-y – 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y
Now, subtract 3x – y – 11 from 3x – 2y = 3x – 2y – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11
(b) sum of 4 + 3x and$$5 – 4x + 2x^2,$$
$$= 4 + 3x + (5 – 4x + 2x^2)$$
$$= 4 + 3x + 5 – 4x + 2x^2$$
$$= 4 + 5 + 3x – 4x + 2x^2$$
$$= 9 – x + 2x^2$$
$$= 2x^2 – x + 9$$
Now sum of $$3x^2 – 5x$$ and$$-x^2 + 2x + 5.$$
$$= 3x^2 – 5x + (-x^2 + 2x + 5)$$
$$= 3x^2 – 5x – x^2 + 2x + 5$$
$$= 3x^2 – x^2 – 5x + 2x + 5$$
$$= 2x^2 – 3x + 5$$
Now, we have to subtract $$2x^2 – 3x + 5$$ from $$2x^2 – x + 9$$
$$= 2x^2 – x + 9 – (2x^2 – 3x + 5)$$
$$= 2x^2 – x + 9 – 2x^2 + 3x – 5$$
$$= 2x^2 – 2x^2 – x + 3x + 9 – 5$$
$$= 2x + 4$$

## NCERT solutions for class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.3

Q.1 If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) $$3m^2 – 2m – 7$$
(v) $$\frac{5m}{2}-4$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) m – 2
= 2 - 2 = 0
(ii) 3m – 5
= 3 × 2 - 5 = 6 -5 = 1
(iii) 9 - 5m
= 9 - 5×2 = 9 -10 -1
(iv) $$3m^2 – 2m – 7$$
= 3×2×2 - 2×2 - 7 = 12 - 11 = 1
(v) $$\frac{5m}{2}-4$$
= $$\frac{5×2}{2}-4$$
= 5- 4 = 1

Q.2 If p = -2, find the value of:
(i) 4p + 7
$$(ii) -3p^2 + 4p + 7$$
$$(iii) -2p^3 – 3p^2 + 4p + 7$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) 4p + 7
= (4 × (-2)) + 7
= -8 + 7
= -1
$$(ii) -3p^2 + 4p + 7$$
$$= (-3 × (-2)^2) + (4 × (-2)) + 7$$
= (-3 × 4) + (-8) + 7
= -12 – 8 + 7
= -20 + 7
= -13
$$(iii) -2p^3 – 3p^2 + 4p + 7$$
$$= (-2 × (-2)^3) – (3 × (-2)^2) + (4 × (-2)) + 7$$
= (-2 × -8) – (3 × 4) + (-8) + 7
= 16 – 12 – 8 + 7
= 23 – 20
= 3

Q.3 Find the value of the following expressions, when x = –1:
(i) 2x – 7
(ii) – x + 2
$$(iii) x^2 + 2x + 1$$
$$(iv) 2x^2 – x – 2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) 2x – 7
= (2 × -1) – 7
= – 2 – 7
= – 9
(ii) – x + 2
= – (-1) + 2
= 1 + 2
= 3
$$(iii) x^2 + 2x + 1$$
$$= (-1)^2 + (2 × -1) + 1$$
= 1 – 2 + 1
= 0
$$(iv) 2x^2 – x – 2$$
$$= (2 × (-1)^2) – (-1) – 2$$
= (2 × 1) + 1 – 2
= 2 + 1 – 2
= 3 – 2
= 1

Q.4 If a = 2, b = -2, find the value of:
$$(i) a^2 + b^2$$
$$(ii) a^2 + ab + b^2$$
$$(iii) a^2 – b^2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

$$(i) a^2 + b^2$$
$$= (2)^2 + (-2)^2$$
= 4 + 4
= 8
$$(ii) a^2 + ab + b^2$$
$$= 2^2 + (2 × -2) + (-2)^2$$
= 4 + (-4) + (4)
= 4 – 4 + 4
= 4
$$(iii) a^2 – b^2$$
$$= 2^2 – (-2)^2$$ = 4 – (4)
= 4 – 4
= 0

Q.5 When a = 0, b = -1, find the value of the given expressions:
$$(i) 2a + 2b$$
$$(ii) 2a^2 + b^2 + 1$$
$$(iii) 2a^2b + 2ab^2 + ab$$
$$(iv) a^2 + ab + 2$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) 2a + 2b
= 2(0) + 2(-1)
= 0 – 2 = -2
$$(ii) 2a^2 + b^2 + 1$$
$$= 2(0)^2 + (-1)^2 + 1$$
=0 + 1 + 1 = 2
$$(iii) 2a^2b + 2ab^2 + ab$$
$$= 2(0)^2 (-1) + 2(0)(-1)^2 + (0)(-1)$$
=0 + 0 + 0 = 0
$$(iv) a^2 + ab + 2$$
$$= (0)^2 + (0)(-1) + 2$$
= 0 + 0 + 2 = 2

Q.6 Simplify the expressions and find the value if x is equal to 2.
(i) x + 7 +4(x – 5)
(ii) 3(x + 2) + 5x – 7
(iii) 6x + 5(x – 2)
(iv) 4(2x – 1) + 3x + 11

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) x + 7 + 4(x – 5)
= x + 7 + 4x – 20
= 5x – 13
Putting x = 2, we get
= 5 × 2 – 13
= 10 – 13 = -3
(ii) 3(x + 2) + 5x – 7
= 3x + 6 + 5x -7
= 8x – 1
Putting x = 2, we get
= 8 × 2 – 1
= 16 – 1 = 15
(iii) 6x + 5(x – 2)
= 6x + 5x – 10
= 11 × – 10
Putting x = 2, we get
= 11 × 2 – 10
= 22 – 10 = 12
(iv) 4(2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 11x + 7
Putting x = 2, we get
= 11 × 2 + 7
= 22+ 7 = 29

Q.7 Simplify these expressions and find their values if x = 3, a = -1, b = -2.
(i) 3x – 5 – x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1
(iv) 10 – 3b – 4 – 55
(v) 2a – 2b – 4 – 5 + a

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) 3x – 5 – x + 9 = 2x + 4
Putting x = 3, we get
2 × 3 + 4
= 6 + 4 = 10
(ii) 2 – 8x + 4x + 4
= -8x + 4x + 2 + 4
= -4x + 6
Putting x = 3, we have
= -4 × 3 + 6
= -12 + 6 =-6
(iii) 3a + 5 – 8a +1
= 3a – 8a + 5 + 1
= -5a + 6
Putting a = -1, we get
= -5(-1) + 6
= 5 + 6 = 11
(iv) 10 – 3b – 4 – 5b
= -3b – 5b + 10 – 4
= -8b + 6
Putting b = -2, we get
= -8(-2) + 6
= 16 + 6 = 22
(v) 2a – 2b – 4 – 5 + a
= 2a + a – 2b – 4 – 5
= 3a – 2b – 9
Putting a = -1 and b = -2, we get
= 3(-1) – 2(-2) – 9
= -3 + 4 – 9
= 1 – 9 = -8

Q.8 (i) If z = 10, find the value of $$z^3 – 3(z – 10).$$
(ii) If p = -10, find the value of$$p^2 -2p – 100.$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

$$(i) z^2 – 3(z – 10)$$
$$= z^2 – 3z + 30$$
Putting z = 10, we get
$$= (10)^3 – 3(10) + 30$$
= 1000 – 30 + 30 = 100
$$(ii) p^2 – 2p – 100$$
Putting p = -10, we get
$$(-10)^2 – 2(-10) – 100$$
= 100 + 20 – 100 = 20

Q.9 What should be the value of a if the value of $$2x^2 + x – a$$ equals to 5, when x = 0?

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

$$2x^2 + x – a = 5$$
Putting x = 0, we get
$$2(0)^2 + (0) – a = 5$$
0 + 0 – a = 5
-a = 5
$$\rightarrow$$ a = -5

Q.10 Simplify the expression and find its value when a = 5 and b = – 3.
$$2(a^2 + ab) + 3 – ab$$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

$$2(a^2 + ab) + 3 – ab$$
$$= 2a^2 + 2ab + 3 – ab$$
$$= 2a^2 + 2ab – ab + 3$$
$$= 2a^2 + ab + 3$$
Putting, a = 5 and b = -3, we get
$$= 2(5)^2 + (5)(-3) + 3$$
= 2 × 25 – 15 + 3
= 50 – 15 + 3
= 53 – 15 = 38

## NCERT solutions for class 7 Maths Chapter 12 Algebraic Expressions 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.

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(a) The number of line segments required to form n digits is given by the expressions. = 5n+1
For figures, the number of line segments = 5 × 5 + 1 = 25 + 1 = 26
For figures, the number of line segments = 5 × 10 + 1 = 50 + 1 = 51
For figures, the number of line segments = 5 × 100 + 1 = 500 + 1 = 501
(a) For figures, the number of line segments =3 ×5 + 1 = 15 + 1 = 16
For figures, the number of line segments = 3 × 10 + 1 = 30 + 1 = 31
For figures, the number of line segments = 3 × 100 + 1 = 300 + 1 = 301
(c) For 5 figures, the number of line numbers = 5 × 5 + 2 = 25 + 2 = 27
For 10 figures, the number of line segments = 5 × 10 + 2 = 50 + 2 = 52
For 100 figures, the number of line segments = 5 × 100 + 2 = 500 + 2 = 502

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

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) Given expression is 2n – 1
For n = 100,
2 × 100 – 1
= 200 – 1 = 199
(ii) Given expression is 3n + 2
For n = 5,
3 × 5 + 2
= 15 + 2 = 17
For n = 10,
3 × 10 + 2 = 30 + 2 = 32
For n = 100,
3 × 100 + 2 = 300 + 2 = 302
(iii) Given expression is 4n + 1
For n = 5, 4 × 5 + 1 = 20 + 1 = 21
For n = 100, 4 × 100 + 1 = 400 + 1 = 401
(iv) Given expression is 7n + 20
For n = 5, 7 × 5 + 20 = 35 + 20 = 55
For n = 10, 7 × 10 + 20 = 70 + 20 = 90
For n = 100, 7 × 100 + 20 = 700 + 20 = 720
(v) Given expression is $$n^2 + 1$$
For n = 5, $$5^2 + 1$$= 25 + 1 = 26
For n = 10,$$10^2 + 1$$= 100 + 1 = 101

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