# NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

Written by Team Trustudies
Updated at 2021-02-11

## 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)$5x{y}^{2}+7{x}^{2}y$
(e) $?ab+2{b}^{2}–3{a}^{2}$

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: $5x{y}^{2}+7{x}^{2}y$
Term I = $5x{y}^{2}$ = 5×x×y×y
Term II = $7{x}^{2}y$ = 7× x × x × y
(e) Expression: $?ab+2{b}^{2}–3{a}^{2}$
Term I = -ab = -a × b
Term II = $2{b}^{2}$ = 2 × b × b
Term III = $?3{a}^{2}$ = -3 ×a×a

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

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)$5–3{t}^{2}$
(ii)$1+t+{t}^{2}+{t}^{3}$
(iii)$x+2xy+3y$
(iv) $100m+1000n$
(v) $–{p}^{2}{q}^{2}+7pq$
(vi)$1.2a+0.8b$
(vii)$3.14{r}^{2}$
(viii)$2\left(l+b\right)$
(ix)$0.1y+0.01{y}^{2}$

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)${y}^{2}x+y$
(ii) $13{y}^{2}–8yx$
(iii) $x+y+2$
(iv)$5+z+zx$
(v) $1+x+xy$
(vi)$12x{y}^{2}+25$
(vii) $7x+x{y}^{2}$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) Expression: ${y}^{2}x+y$
Term: ${y}^{2}x$
Coefficient of x = ${y}^{2}$
(ii) Expression: $13{y}^{2}–8yx$
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:$12x{y}^{2}+25$
Term: $12x{y}^{2}$
Coefficient of x = $12{y}^{2}$
(vii) Expression:$7x+x{y}^{2}$
Term I $7x$
Term II $x{y}^{2}$
Coefficient of x = $7,{y}^{2}$

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

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) Expression: $8–x{y}^{2}$
Term : $?x{y}^{2}$
Coefficient of${y}^{2}=?x$
(ii) Expression: $5{y}^{2}+7x$
Term : $5{y}^{2}$
Coefficient of${y}^{2}=5$
(iii) Expression: $2{x}^{2}y–15x{y}^{2}+7{y}^{2}$
Term : $–15x{y}^{2},7{y}^{2}$
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) $4{p}^{2}q–4p{q}^{2}$
(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) $4{p}^{2}q–4p{q}^{2}$__ 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)$4{m}^{2}p,4m{p}^{2}$
(vi) $12xz,12{x}^{2}{y}^{2}$

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)$4{m}^{2}p,4m{p}^{2}$ unlike
(vi) $12xz,12{x}^{2}{y}^{2}$ unlike

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

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

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

## 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) $?{z}^{2}+13{z}^{2}?5z+7{z}^{3}–15z$
(iii)$p–\left(p–q\right)–q–\left(q–p\right)$
(iv) 3a – 2b — ab – (a – b + ab) + 3ab + 6 – a
$\left(v\right)5{x}^{2}y–5{x}^{2}+3y{x}^{2}–3{y}^{2}+{x}^{2}–{y}^{2}+8x{y}^{2}?3{y}^{2}$
$\left(vi\right)\left(3{y}^{2}+5y–4\right)–\left(8y–{y}^{2}–4\right)$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

(i) 21b – 32 + 7b – 20b
= 21b + 7b -20b -32
= 8b -32
(ii) $?{z}^{2}+13{z}^{2}?5z+7{z}^{3}–15z$
= $?{z}^{2}+13{z}^{2}?5z?15z+7{z}^{3}$
= $11{z}^{2}?20z+7{z}^{3}$
(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) $5{x}^{2}y–5{x}^{2}+3y{x}^{2}–3{y}^{2}+{x}^{2}–{y}^{2}+8x{y}^{2}?3{y}^{2}$
= $5{x}^{2}y+3y{x}^{2}–5{x}^{2}+{x}^{2}–3{y}^{2}–{y}^{2}–3{y}^{2}+8x{y}^{2}$
= 8x^2y – 4x^2 – 7y^2 + 8xy^2
(vi) $\left(3{y}^{2}+5y–4\right)–\left(8y–{y}^{2}–4\right)$
= $3{y}^{2}+5y–4–8y+{y}^{2}+4\right)$
= $3{y}^{2}+{y}^{2}+5y–8y–4+4$
= $4{y}^{2}–3y$

(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)$4{x}^{2}y,?3x{y}^{2},?5x{y}^{2},5{x}^{2}y$
$\left(viii\right)3{p}^{2}{q}^{2}–4pq+5,?10{p}^{2}{q}^{2},15+9pq+7{p}^{2}{q}^{2}$
(ix) ab – 4a, 4b – ab, 4a – 4b
$\left(x\right){x}^{2}–{y}^{2}–1,{y}^{2}–1–{x}^{2},1–{x}^{2}–{y}^{2}$

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)$4{x}^{2}y,?3x{y}^{2},?5x{y}^{2},5{x}^{2}y$
= $4{x}^{2}y+\left(?3x{y}^{2}\right)+\left(?5x{y}^{2}\right)+5{x}^{2}y$
=$4{x}^{2}y+5{x}^{2}y–3x{y}^{2}–5x{y}^{2}$
=$9{x}^{2}y–8x{y}^{2}$
$\left(viii\right)3{p}^{2}{q}^{2}–4pq+5,?10{p}^{2}{q}^{2},15+9pq+7{p}^{2}{q}^{2}$
$=3{p}^{2}{q}^{2}–4pq+5+\left(?10{p}^{2}{q}^{2}\right)+15+9pq+7{p}^{2}{q}^{2}$
$=3{p}^{2}q2–10{p}^{2}{q}^{2}+7{p}^{2}{q}^{2}–4pq+9pq+5+15$
= 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
$\left(x\right){x}^{2}–{y}^{2}–1,{y}^{2}–1–{x}^{2},1–{x}^{2}–{y}^{2}$
$={x}^{2}–{y}^{2}–1+\left({y}^{2}–1–{x}^{2}\right)+\left(1–{x}^{2}–{y}^{2}\right)$
$={x}^{2}–{y}^{2}–1+{y}^{2}–1–{x}^{2}+1–{x}^{2}–{y}^{2}$
$={x}^{2}–{x}^{2}–{x}^{2}–{y}^{2}+{y}^{2}–{y}^{2}–1–1+1$
$=?{x}^{2}–{y}^{2}?1$

Q.3 Subtract:
$\left(i\right)?5{y}^{2}from{y}^{2}$
$\left(ii\right)6xyfrom?12xy$
$\left(iii\right)\left(a–b\right)from\left(a+b\right)$
$\left(iv\right)a\left(b–5\right)fromb\left(5–a\right)$
$\left(v\right)?{m}^{2}+5mnfrom4{m}^{2}–3mn+8$
$\left(vi\right)?{x}^{2}+10x–5from5x–10$
$\left(vii\right)5{a}^{2}–7ab+5{b}^{2}from3ab–2{a}^{2}–2{b}^{2}$
$\left(viii\right)4pq–5{q}^{2}–3{p}^{2}from5{p}^{2}+3{q}^{2}–pq$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

$\left(i\right)?5{y}^{2}from{y}^{2}$
$={y}^{2}–\left(?5{y}^{2}\right)$
$={y}^{2}+5{y}^{2}$
$=6{y}^{2}$
$\left(ii\right)6xyfrom?12xy$
= -12xy – 6xy
= – 18xy
$\left(iii\right)\left(a–b\right)from\left(a+b\right)$
= (a + b) – (a – b)
= a + b – a + b
= a – a + b + b
= 2b
$\left(iv\right)a\left(b–5\right)fromb\left(5–a\right)$
= b (5 -a) – a (b – 5)
= 5b – ab – ab + 5a
= 5a + 5b – 2ab
$\left(v\right)?{m}^{2}+5mnfrom4{m}^{2}–3mn+8$
$=4{m}^{2}–3mn+8–\left(?{m}^{2}+5mn\right)$
$=4{m}^{2}–3mn+8+{m}^{2}–5mn$
$=4{m}^{2}+{m}^{2}–3mn–5mn+8$
$=5{m}^{2}–8mn+8$
$\left(vi\right)?{x}^{2}+10x–5from5x–10$
$=5x–10–\left(?{x}^{2}+10x–5\right)$
$=5x–10+{x}^{2}–10x+5$
$={x}^{2}+5x–10x–10+5$
$={x}^{2}–5x–5$
$\left(vii\right)5{a}^{2}–7ab+5{b}^{2}from3ab–2{a}^{2}–2{b}^{2}$
$=3ab–2{a}^{2}–2{b}^{2}–\left(5{a}^{2}–7ab+5{b}^{2}\right)$
$=3ab–2{a}^{2}–2{b}^{2}–5{a}^{2}+7ab–5{b}^{2}$
$=3ab+7ab–2{a}^{2}–5{a}^{2}–2{b}^{2}–5{b}^{2}$
$=10ab–7{a}^{2}–7{b}^{2}$
$\left(viii\right)4pq–5{q}^{2}–3{p}^{2}from5{p}^{2}+3{q}^{2}–pq$
$=5{p}^{2}+3{q}^{2}–pq–\left(4pq–5{q}^{2}–3{p}^{2}\right)$
$=5{p}^{2}+3{q}^{2}–pq–4pq+5{q}^{2}+3{p}^{2}$
$=5{p}^{2}+3{p}^{2}+3{q}^{2}+5{q}^{2}–pq–4pq$
$=8{p}^{2}+8{q}^{2}–5pq$

Q.4 (a) What should be added to ${x}^{2}+xy+{y}^{2}$ to obtain $2{x}^{2}+3xy?$
(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) $p+\left({x}^{2}+xy+{y}^{2}\right)=2{x}^{2}+3xy$
$p=\left(2{x}^{2}+3xy\right)–\left({x}^{2}+xy+{y}^{2}\right)$
$p=2{x}^{2}+3xy–{x}^{2}–xy–{y}^{2}$
$p=2{x}^{2}–{x}^{2}+3xy–xy–{y}^{2}$
$p={x}^{2}+2xy–{y}^{2}$
(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$3{x}^{2}–4{y}^{2}+5xy+20$to obtain $–{x}^{2}–{y}^{2}+6xy+20?$

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

$3{x}^{2}–4{y}^{2}+5xy+20–m=?{x}^{2}–{y}^{2}+6xy+20$
$m=3{x}^{2}–4{y}^{2}+5xy+20–\left(?{x}^{2}–{y}^{2}+6xy+20\right)$
$m=3{x}^{2}–4{y}^{2}+5xy+20+{x}^{2}+{y}^{2}–6xy–20$
$m=3{x}^{2}+{x}^{2}–4{y}^{2}+{y}^{2}+5xy–6xy+20–20$
$m=4{x}^{2}–3{y}^{2}–xy$

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$5–4x+2{x}^{2},$subtract the sum of $3{x}^{2}–5x$ 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$5–4x+2{x}^{2},$
$=4+3x+\left(5–4x+2{x}^{2}\right)$
$=4+3x+5–4x+2{x}^{2}$
$=4+5+3x–4x+2{x}^{2}$
$=9–x+2{x}^{2}$