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:

= x+ 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.ExpressionTermsFactors
(a)– 4x + 5-4x 5-4, x 5
(b)– 4x + 5y-4x 5y-4, x 5, y
(c)5y + 3y25y 3y25, y 3, y, y
(d)xy + 2x2y2xy 2x2y2x, y 2, x, x, y, y
(e)pq + qpq qP, q Q
(f)1.2 ab – 2.4 b + 3.6 a1.2ab -2.4b 3.6a1.2, a, b -2.4, b 3.6, a
(g)¾ x + ¼¾ x ¼¾, x ¼
(h)0.1 p2 + 0.2 q20.1p2 0.2q20.1, 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.ExpressionTermsCoefficients
(i)5 – 3t2– 3t2-3
(ii)1 + t + t2 + t3t t2 t31 1 1
(iii)x + 2xy + 3yx 2xy 3y1 2 3
(iv)100m + 1000n100m 1000n100 1000
(v)– p2q2 + 7pq-p2q2 7pq-1 7
(vi)1.2 a + 0.8 b1.2a 0.8b1.2 0.8
(vii)3.14 r23.1423.14
(viii)2 (l + b)2l 2b2 2
(ix)0.1 y + 0.01 y20.1y 0.01y20.1 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.ExpressionTermsCoefficient of x
(i)y2x + yy2xy2
(ii)13y2 – 8yx– 8yx-8y
(iii)x + y + 2x1
(iv)5 + z + zxx zx1 z
(v)1 + x + xyxyy
(vi)12xy2 + 2512xy212y2
(vii)7x + xy27x xy27 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.ExpressionTermsCoefficient of y2
(i)8 – xy2– xy2– x
(ii)5y2 + 7x5y25
(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

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