# Exercise – 12.1

Q1. Get the algebraic expression in the following cases using variables, constant and arithmetic operation.

• Subtraction of z from y.

Solution:-

= Y – z

• One – half of the sum of numbers x and y.

Solution:-

= 1/2 ( x + y )

= (x + y ) / 2

• The number z multiplied by itself.

Solution:-

= z x z

= z2

• One – fourth of the product of number p and q.

Solution:-

= 1/4 ( p x q )

= pq/4

• Number x and y both squared and added .

Solution:-

= x2 + y2

• Number 5 added to three times the product of numbers m and n.

Solution:-

= 3mn +

5

• Product of numbers y and z subtracted from 10.

Solution:-

= 10 – ( y x z )

= 10 – yz

• Sum of numbers a and b subtracted from their product.

Solution:-

= ( a x b ) – ( a + b )

= ab – ( a + b )

Q2. (i) Identify the terms and their factor in the following expression

Show the terms the factors by tree diagrams.

• X – 3

Solution:-

Expression: x – 3

Terms: x, -3

Factor: x; -3

• 1 + X + X2

Solution:-

NCERT solutions for class 7 science chapter -16

Expression: 1 + x + x2

Term: 1,x , x2

Factors: 1; x; x,x

• Y – y3

Solution:-

Expression: y – y3

Terms: y, -y3

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

• 5xy2 + 7x2y

Solution:-

Expression: 5x2y + 7x2y

Terms: 5xy2 , 7x2y

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

• -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 expression gien 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) 3/4 x + 1/4

(h)  0.1 p2 + 0.2 q2

Solution:-

 S.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 1.2 ab , -2.4b , 3.6 a 1.2 , a, b, -2.4 , b 3.6 , a (g) 3/4 x + 1/4 3/4 x 1/4 3/4 x 1/4 (h) 0.1 p2 + 0.2 q2 0.1 p2,0.212 0.1, p ,p , 0.2 , q , q

Q-3 identify the numerical coefficients of terms in the following expressions :

• 5-3t2(ii) 1 + t + t2(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)
• 1 y + 0.01 y2

Solution:-

 SI No. Expression Terms Coefficients (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 + 0.8b 1.2a 0.8b 1.2 0.8 (vii) 3.14r2 3.14r2 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

Q4. (A) Identify terms which contain x and give the coefficient of x.

• Y2x + y (ii) 13y2 – 8yx (iii) x + y + 2

(iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy2 + 25 (vii) 7x + xy2

Solution:-

 SI. No. Expression Terms Coefficient of x (i) Y2x + y Y2x Y2 (ii) 13y2 – 8yx -8yz -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:-

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

Q5. Classify into monomials binomials and trinomials.

• 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) 7mm

Solution:-

Monomial.

(ix ) z2 – 3z + 8

Solution:-

Trinomial

• A2 + b2

Solution:-

BIONOMIAL

(XI) Z2 + Z

Solution:-

Binomial

(xii ) 1 + x + x2

Solution:-

Trinomial

Q6. State  whether a given pair of terms is like or  unlike terms.

• 1,100

Solution:-

Like term

• -7x , (5/2)x

Solution:-

Like term

• -29 x , – -29y

Solution:-

Like terms

• 14xy, 42yx.

Solution:-

Like term

• 4m2p , 4mp2

Solution:-

Unlike terms.

• 12xz , 12x2z2

Solution:-

Unlike term.

Q7. Identify like terms in the following :

• -xy2 , -4yx2 ,8x2, 2xy2, 7y, – 11x2 – 100x, – 11yx, 20x2y, – 6x2 , y , 2xy, 3x

Solution:-

When the term have the same algebraic factor they are like terms.

• 10pq , 7p , 8q , -p2q2, -7qp, , -100q, -23 , 12q2p2 , -5p2 , 41, 2450p , 78 qp, 13p2q , qp2, 710p2

Solution:-

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

they, are

10pq, – 7 qp , 78qp

8q, – 100 q

-p2q2 , 12q2p2

– 23,41

-13p2q, qp2

Chapter – 12 Algebraic Expression Exercise – 12.1 Q1. Get the algebraic expression in the following cases using variables, constant and…