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NCERT · Class 7 · Maths

Chapter 10: Algebraic Expressions

Practice each question, then open the answer or worked solution.

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Exercise 12.1

Textbook page 234 · 7 questions
Q1Written answer

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

(i) Subtraction of z from y.

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Worked solution

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)

Q2Written answer

(i) Identify the terms and their factors in the following expressions.

Show the terms and factors by tree diagrams.

(a) x – 3

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Worked solution

Solution:-

Expression: x – 3

Terms: x, -3

Factors: x; -3

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 1

(b) 1 + x + x2

Solution:-

Expression: 1 + x + x2

Terms: 1, x, x2

Factors: 1; x; x,x

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 2

(c) y – y3

Solution:-

Expression: y – y3

Terms: y, -y3

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

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 3

(d) 5xy2 + 7x2y

Solution:-

Expression: 5xy2 + 7x2y

Terms: 5xy2, 7x2y

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

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 4

(e) – ab + 2b2 – 3a2

Solution:-

Expression: -ab + 2b2 – 3a2

Terms: -ab, 2b2, -3a2

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

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 5

(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 that show the value of something.

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

Factors are defined as numbers we can multiply together to get another number.

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
Q3Written answer

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

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Worked solution

Solution:-

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

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

A coefficient is a number used to multiply a variable (2x means 2 times x, so 2 is a coefficient). Variables on their own (without a number next to them) actually have a coefficient of 1 (x is really 1x).

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
Q4Written answer

(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

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Worked solution

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
Q5Written answer

Classify into monomials, binomials and trinomials.

(i) 4y – 7z

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Worked solution

Solution:-

Binomial.

An expression which contains two unlike terms is called a binomial.

(ii) y2

Solution:-

Monomial.

An expression with only one term is called a monomial.

(iii) x + y – xy

Solution:-

Trinomial.

An expression which contains three terms is called a trinomial.

(iv) 100

Solution:-

Monomial.

An expression with only one term is called a monomial.

(v) ab – a – b

Solution:-

Trinomial.

An expression which contains three terms is called a trinomial.

(vi) 5 – 3t

Solution:-

Binomial.

An expression which contains two unlike terms is called a binomial.

 

(vii) 4p2q – 4pq2

Solution:-

Binomial.

An expression which contains two unlike terms is called a binomial.

 

(viii) 7mn

Solution:-

Monomial.

An expression with only one term is called a monomial.

(ix) z2 – 3z + 8

Solution:-

Trinomial.

An expression which contains three terms is called a trinomial.

(x) a2 + b2

Solution:-

Binomial.

An expression which contains two unlike terms is called a binomial.

(xi) z2 + z

Solution:-

Binomial.

An expression which contains two unlike terms is called a binomial.

(xii) 1 + x + x2

Solution:-

Trinomial.

An expression which contains three terms is called a trinomial.

Q6Written answer

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

(i) 1, 100

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Worked solution

Solution:-

Like term.

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

 

(ii) –7x, (5/2)x

Solution:-

Like term.

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

 

(iii) – 29x, – 29y

Solution:-

Unlike terms.

The terms have different algebraic factors, they are unlike terms.

(iv) 14xy, 42yx

Solution:-

Like term.

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

 

(v) 4m2p, 4mp2

Solution:-

Unlike terms.

The terms have different algebraic factors, they are unlike terms.

(vi) 12xz, 12x2z2

Solution:-

Unlike terms.

The terms have different algebraic factors, they are unlike terms.

Q7Written answer

Identify like terms in the following.

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

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Worked solution

Solution:-

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

They are,

10pq, – 7qp, 78qp

7p, 2405p

8q, – 100q

– p2q2, 12q2p2

– 23, 41

– 5p2, 701p2

13p2q, qp2