Mathematics – Class 7 – Chapter 12 – Algebraic Expression – Exercise 12.1 – NCERT Exercise Solution

1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.

(i) Subtraction of z from y.

Solution:

= Y – z

(ii) One-half of the sum of numbers x and y.

Solution:

= ½ (x + y)

= (x + y)/2

(iii) The number z multiplied by itself.

Solution:

= z × z

= z²

(iv) One-fourth of the product of numbers p and q.

Solution:

= ¼ (p × q)

= pq/4

(v) Numbers x and y both squared and added.

Solution:

= x² + y²

(vi) Number 5 added to three times the product of numbers m and n.

Solution:

= 3mn + 5

(vii) Product of numbers y and z subtracted from 10.

Solution:

= 10 – (y × z)

= 10 – yz

(viii) Sum of numbers a and b subtracted from their product.

Solution:

= (a × b) – (a + b)

= ab – (a + b)

2. (i) Identify the terms and their factors in the following expressions. Show the terms and factors by tree diagrams.

(a) x – 3

Solution:

Expression: x – 3

Terms: x, -3

Factors: x; -3

(b) 1 + x + x²

Solution:

Expression: 1 + x + x²

Terms: 1, x, x²

Factors: 1; x; x,x

(c) y – y³

Solution:

Expression: y – y³

Terms: y, -y³

Factors: y; -y, -y, -y

(d) 5xy² + 7x²y

Solution:

Expression: 5xy² + 7x²y

Terms: 5xy², 7x²y

Factors: 5, x, y, y; 7, x, x, y

(e) – ab + 2b² – 3a²

Solution:

Expression: -ab + 2b² – 3a²

Terms: -ab, 2b², -3a²

Factors: -a, b; 2, b, b; -3, a, a

(ii) Identify terms and factors in the expressions given below:

(a) – 4x + 5

(b) – 4x + 5y

(c) 5y + 3y² 

(d) xy + 2x²y²

(e) pq + q

(f) 1.2 ab – 2.4 b + 3.6 a

(g) ¾ x + ¼

(h) 0.1 p² + 0.2 q²

Solution:

 Expressions are defined as, numbers, symbols, and operators (such as +. –, ×, and ÷) grouped together which show the value of something.

In an algebraic expression, a term is either a single number or variable, or numbers and variables are multiplied together. Terms are separated by + or – signs or sometimes by division.

Sl.No.	Expression	Terms	Factors
(a)	– 4x + 5	-4x 5	-4, x 5
(b)	– 4x + 5y	-4x 5y	-4, x 5, y
(c)	5y + 3y2	5y 3y2	5, y 3, y, y
(d)	xy + 2x2y2	xy 2x2y2	x, y 2, x, x, y, y
(e)	pq + q	pq q	P, q Q
(f)	1.2 ab – 2.4 b + 3.6 a	1.2ab -2.4b 3.6a	1.2, a, b -2.4, b 3.6, a
(g)	¾ x + ¼	¾ x ¼	¾, x ¼
(h)	0.1 p2 + 0.2 q2	0.1p2 0.2q2	0.1, p, p 0.2, q, q

3. Identify the numerical coefficients of terms (other than constants) in the following expressions:

(i) 5 – 3t² 

(ii) 1 + t + t² + t³ 

(iii) x + 2xy + 3y

(iv) 100m + 1000n

(v) – p²q² + 7pq

(vi) 1.2 a + 0.8 b

(vii) 3.14 r² 

(viii) 2 (l + b)

(ix) 0.1 y + 0.01 y²

Solution:

Sl.No.	Expression	Terms	Coefficients
(i)	5 – 3t2	– 3t2	-3
(ii)	1 + t + t2 + t3	t t2 t3	1 1 1
(iii)	x + 2xy + 3y	x 2xy 3y	1 2 3
(iv)	100m + 1000n	100m 1000n	100 1000
(v)	– p2q2 + 7pq	-p2q2 7pq	-1 7
(vi)	1.2 a + 0.8 b	1.2a 0.8b	1.2 0.8
(vii)	3.14 r2	3.142	3.14
(viii)	2 (l + b)	2l 2b	2 2
(ix)	0.1 y + 0.01 y2	0.1y 0.01y2	0.1 0.01

4. (a) Identify terms which contain x and give the coefficient of x.

(i) y²x + y

(ii) 13y² – 8yx

(iii) x + y + 2

(iv) 5 + z + zx

(v) 1 + x + xy

(vi) 12xy² + 25

(vii) 7x + xy²

Solution:

Sl.No.	Expression	Terms	Coefficient of x
(i)	y2x + y	y2x	y2
(ii)	13y2 – 8yx	– 8yx	-8y
(iii)	x + y + 2	x	1
(iv)	5 + z + zx	x zx	1 z
(v)	1 + x + xy	xy	y
(vi)	12xy2 + 25	12xy2	12y2
(vii)	7x + xy2	7x xy2	7 y2

(b) Identify terms which contain y² and give the coefficient of y².

(i) 8 – xy² 

(ii) 5y² + 7x

(iii) 2x²y – 15xy² + 7y²

Solution:

Sl.No.	Expression	Terms	Coefficient of y2
(i)	8 – xy2	– xy2	– x
(ii)	5y2 + 7x	5y2	5
(iii)	2x2y – 15xy2 + 7y2	– 15xy2 7y2	– 15x 7

5. Classify into monomials, binomials and trinomials.

NOTE:

Monomial: An expression with only one term is called a monomial.

Binomial: An expression that contains two unlike terms is called a binomial.

Trinomial: An expression that contains three terms is called a trinomial.

(i) 4y – 7z

Solution:

Binomial.

(ii) y²

Solution:

Monomial.

(iii) x + y – xy

Solution:

Trinomial.

(iv) 100

Solution:

Monomial.

(v) ab – a – b

Solution:

Trinomial.

(vi) 5 – 3t

Solution:

Binomial.

(vii) 4p²q – 4pq²

Solution:

Binomial.

(viii) 7mn

Solution:

Monomial.

(ix) z² – 3z + 8

Solution:

Trinomial.

(x) a² + b²

Solution:

Binomial.

(xi) z² + z

Solution:

Binomial.

(xii) 1 + x + x²

Solution:

Trinomial.

6. State whether a given pair of terms is of like or unlike terms.

Like term: When term have the same algebraic factors, they are like terms.

Unlike term: The terms have different algebraic factors, they are unlike terms.

(i) 1, 100

Solution:

Like term.

(ii) –7x, (5/2)x

Solution:

Like term.

(iii) – 29x, – 29y

Solution:

Unlike terms.

(iv) 14xy, 42yx

Solution:

Like term.

(v) 4m²p, 4mp²

Solution:

Unlike terms.

(vi) 12xz, 12x²z²

Solution:

Unlike terms.

7. Identify like terms in the following:

(a) – xy², – 4yx², 8x², 2xy², 7y, – 11x², – 100x, – 11yx, 20x²y, – 6x², y, 2xy, 3x

Solution:

When term have the same algebraic factors, they are like terms.

They are,

– xy², 2xy²

– 4yx², 20x²y

8x², – 11x², – 6x²

7y, y

– 100x, 3x

– 11yx, 2xy

(b) 10pq, 7p, 8q, – p²q², – 7qp, – 100q, – 23, 12q²p², – 5p², 41, 2405p, 78qp,13p²q, qp², 701p²

Solution:

When term have the same algebraic factors, they are like terms.

They are,

10pq, – 7qp,

78qp7p, 2405p8q,

– 100q– p²q², 12q²p²– 23,

41– 5p²,701p²13p²q, qp²