NCERT solutions for class 7 Maths Chapter 12 Algebraic Expressions

Solution for 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.


Open in new tab link

Answer :

(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) \(-ab + 2b^2 – 3a^2 \)


Open in new tab link

Answer :

(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) \( 0.1p^2 + 0.2q^2 \)


Open in new tab link

Answer :

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)\( 0.1 y + 0.01 y^2 \)


Open in new tab link

Answer :

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) \( 7x + xy^2 \)


Open in new tab link

Answer :

(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) \(2x^2y – 15xy^2 + 7y^2 \)


Open in new tab link

Answer :

(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) \(1 + x + x^2 \)


Open in new tab link

Answer :

(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) \(12xz, 12 x^2y^2 \)


Open in new tab link

Answer :

(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:
\( (a)-xy^2, -4yx^2, 8x^2, 2xy^2, 7y^2, -11x^2,\)
\( -100x, -11yx, 20x^2y, -6x^2, y, 2xy, 3x \)
\( (b) 10pq, 7p, 8q, -p^2q^2, -7qp, -100q, -23, 12q^2p^2, \)
\( -5p^2, 41, 2405p, 78qp, 13p^2q, qp^2, 701p^2 \)


Open in new tab link

Answer :

(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) \(13p^2q, qp^2 \)

Solution for 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
\( (v) 5x^2y – 5x^2 + 3yx^2 – 3y^2 + x^2 – y^2 + 8xy^2 -3y^2 \)
\((vi) (3y^2 + 5y – 4) – (8y – y^2 – 4) \)


Open in new tab link

Answer :

(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) \( (3y^2 + 5y – 4) – (8y – y^2 – 4) \)
= \( 3y^2 + 5y – 4 – 8y + y^2 + 4) \)
= \( 3y^2 + y^2 + 5y – 8y – 4 + 4 \)
= \( 4y^2 – 3y \)

Q.2 Add:
(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
\( (x) x^2 – y^2 – 1, y^2 – 1 – x^2, 1 – x^2 – y^2 \)


Open in new tab link

Answer :

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


Open in new tab link

Answer :

\((i) -5y^2 from y^2 \)
\(= y^2 – (-5y^2) \)
\(= y^2 + 5y^2 \)
\(= 6y^2\)
\((ii) 6xy from -12xy \)
= -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?


Open in new tab link

Answer :

(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 \(– x^2 – y^2 + 6xy + 20?\)


Open in new tab link

Answer :

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

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\( -x^2 + 2x + 5. \)


Open in new tab link

Answer :

(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 \(2x^2 – x + 9\)
\(= 2x^2 – x + 9 – (2x^2 – 3x + 5)\)
\(= 2x^2 – x + 9 – 2x^2 + 3x – 5\)
\(= 2x^2 – 2x^2 – x + 3x + 9 – 5\)
\(= 2x + 4\)

Solution for 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) \(\frac{5m}{2}-4 \)


Open in new tab link

Answer :

(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
\((ii) -3p^2 + 4p + 7\)
\((iii) -2p^3 – 3p^2 + 4p + 7 \)


Open in new tab link

Answer :

(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
\((iii) x^2 + 2x + 1\)
\((iv) 2x^2 – x – 2\)


Open in new tab link

Answer :

(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:
\((i) a^2 + b^2\)
\((ii) a^2 + ab + b^2\)
\((iii) a^2 – b^2\)


Open in new tab link

Answer :

\((i) a^2 + b^2\)
\(= (2)^2 + (-2)^2\)
= 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:
\((i) 2a + 2b \)
\((ii) 2a^2 + b^2 + 1\)
\((iii) 2a^2b + 2ab^2 + ab\)
\((iv) a^2 + ab + 2\)


Open in new tab link

Answer :

(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


Open in new tab link

Answer :

(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


Open in new tab link

Answer :

(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\( p^2 -2p – 100.\)


Open in new tab link

Answer :

\((i) z^2 – 3(z – 10) \)
\(= z^2 – 3z + 30 \)
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?


Open in new tab link

Answer :

\(2x^2 + x – a = 5\)
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.
\(2(a^2 + ab) + 3 – ab\)


Open in new tab link

Answer :

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

Solution for 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


Open in new tab link

Answer :

(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.


Open in new tab link

Answer :

(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



Get it on Google Play