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NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

Exercise 12.1

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 m and n.
(vii) A product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.

Answer

(i) y - z
(ii) (x + y)/2
(iii) z2
(iv) pq/4
(v) x2 + y2
(vi) 3mn + 5
(vii) 10 - yz
(viii) ab - (a + b) 

2. (i) Identify the terms and their factors in the following expressions, show the terms and factors by tree diagram: 
(a) x – 3
(b) 1 + x + x2
(c) y – y3
(d) 5xy2 + 7x2y
(e) -ab + 2b2 – 3a2

Answer


(ii) Identify the terms and factors in the expressions given below: 
(a) -4x + 5
(b) -4x + 5y 
(c) 5y + 3y2
(d) xy + 2x2y2
(e) pq + q
(f) 1.2ab - 2.4b + 3.6a
(h) 0.1 p2 + 0.2q2

Answer

(a) -4x+ 5
Terms: -4x,5
Factors: -4,x ; 5

(b) -4x + 5y
Terms: -4x, 5y
Factors: -4,x ; 5,y

(c) 5y + 3y2
Terms: 5y,3y2
Factors: 5, y ; 3,y,y

(d) xy+2x2y2
Terms: xy,2x2y2 
Factors: x,y ; 2x,x,y,y

(e) pq+q
Terms: pq,q
Factors: p,q ; q

(f) 1.2ab-2.4b+3.6a
Terms: 1,2ab.-2.4b,3 6a
Factors: 1.2.a.b ; -2.4,6 ; 3.6,a


(h) 0.1p2+0.2q2
Terms: 0.1 p2,0.2q2
Factors: 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 + t2
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) -p2q2 + 7pq
(vi) 1.2a + 0.8b
(vii) 3.14 r2
(viii) 2(l+b)
(ix) 0.1y + 0.01y2

Answer

S.No. Expression Terms Numerical Coefficient
(i) 5-3t2 -3t1 -3
(ii)1+t+t2+t3 t 1


t2 1


t3 1
(iii) x + 2xy + 3y x 1


2xy 2


3y 3
(iv) 100m+1000n 100 m 100


1000 n 1000
(v) -p2q2+7 pq -p2q2 -1


7 pq 7
(vi) 1.2a+0.8b 1.2 a 1.2


0.8b 0.8
(vii) 3.14 r2 3.14 r2 3.14
(viii) 2 (l + b) = 2l+ 2b 2l 2


2b 2
(ix) 0.1y + 0.01y2 0.1y 0.1


0.01y2 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 + x25
(vii) 7x + xy2

(b) Identify terms which contain y2 and give the coefficient of y2.
(i) 8 - xy2
(ii) 5y2 + 7x
(iii) 2x2y - 15xy2 + 7y2

Answer

S.No. Expression Term with factor x Coefficient of x
(i) y2x + y y2 x y
(ii) 13y2 -8yx -8 yx -8 y
(iii) x + y + 2 x 1
(iv) 5 + z + zx zx z
(v) 1 + x + xy x 1


xy 1
(vi) 12xy2 +25 12 xy2 12y2
(vii) 7x+xy2 xy2 y2


7x 7

S. No.  Expression  Term containing y2 Coefficient of y2
(i) 8-xy2 -xy2 -x
(ii) 5y+7x 5y2 5
(iii) 2x2y-15xy2 + 7y2 -15xy2 -15x


7y2 7

5. Classify into monomials, binomials and trinomials: 
(i) 4y - 7x
(ii) y2
(iii) x + y - xy
(iv) 100
(v) ab - a - b
(vi) 5 - 3t
(vii) 4p2q - 4pq2
(viii) 7mn
(ix) z2 - 3z + 8
(x) a2 + b2
(xi) z2 + z
(xii) 1 + x + x2

Answer

S.No. Expression Type of Polynomial
(i) 4y-7z Binomial
(ii) y2 Monomial
(iii) x+y-xy Trinomial
(iv) 100 Monomial
(v) ab-a-b Trinomial
(vi) 5-3t Binomial
(vii) 4p2q-4pq2 Binomial
(viii) 7mn Monomial
(ix) z2-3z + 8 Trinomial
(x) a2 + b2 Binomial
(xi) z2 +z Binomial
(xii) 1 + x + x2 Trinomial

6. State whether a given pair of terms is of like or unlike terms: 
(i) 1,100
(ii) 
(iii) -29x, -29y
(iv) 14xy, 42 yx
(v) 4m2p, 4mp2
(vi) 12xz, 12x2 z2

Answer

S.No. Pair of terms Like / Unlike terms
(i) 1, 100 Like terms
(ii) Like terms
(iii) -29x,-29y Unlike terms
(iv) 14xy,42yx Like terms
(v) 4m2p,4mp2 Unlike terms
(vi) 12xz,12x2z2 Unlike terms

7. Identify like terms in the following: 
(a) -xy2, -4yx2, 8x2, 2xy2, 7y,  -11x2  - 100x, - 11yx, 20x2y, -6x2, y, 2xy, 3x

Answer

(i) -xy2,2 xy2
(ii) -4yx2 , 20x2y
(iii) 8x2,-11x2,-6x2 
(iv) 7y, y
(v) -100x, 3x
(vi) -11yx, 2xy

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

Answer

(i) 10 pq - 7 pq,78 pq
(ii) 7p, 2405 p
(iii) 8q,- 100q
(iv) -p2q2, 12p2q2
(v) -12,41
(vi) -5p2,701p2 
(vii) 13 p2q,qp2

Exercise 12.2

1. Simplify combining like terms: 
(i) 21b – 32 + 7b – 20b

Answer

When term have the same algebraic factors, they are like terms.
Then,
= (21b + 7b – 20b) – 32
= b (21 + 7 – 20) – 32
= b (28 - 20) – 32
= b (8) - 32
= 8b - 32

(ii) – z2 + 13z2 – 5z + 7z3 – 15z

Answer

When term have the same algebraic factors, they are like terms.
Then,
= 7z3 + (-z2 + 13z2) + (-5z – 15z)
= 7z3 + z2 (-1 + 13) + z (-5 - 15)
= 7z3 + z2 (12) + z (-20) + 7z3 
= 7z3 + 12z2 – 20z + 7z3

(iii) p – (p – q) – q – (q – p)

Answer

When term have the same algebraic factors, they are like terms.
Then,
= p – p + q – q – q + p
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a

Answer

When term have the same algebraic factors, they are like terms.
Then,
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)
= a (1 - 2) + b (-2 + 2) + ab (-2 + 3)
= a (1) + b (0) + ab (1)
= a + ab

(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2 

Answer

When term have the same algebraic factors, they are like terms.
Then,
= 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2
= x2y (5 + 3) + x2 (- 5 + 1) + y2 (-3 – 1 -3) + 8xy2
= x2y (8) + x2 (-4) + y2 (-7) + 8xy2
= 8x2y - 4x2 – 7y2 + 8xy2 

(vi) (3y2 + 5y – 4) – (8y – y2 – 4)

Answer

When term have the same algebraic factors, they are like terms.
Then,
= 3y2 + 5y – 4 – 8y + y2 + 4
= 3y2 + y2 + 5y – 8y – 4 + 4
= y2 (3 + 1) + y (5 - 8) + (-4 + 4)
= y2 (4) + y (-3) + (0)
= 4y2 – 3y.

2. Add: 

(i) 3mn, – 5mn, 8mn, – 4mn

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 3mn + (-5mn) + 8mn + (- 4mn)
= 3mn – 5mn + 8mn – 4mn
= mn (3 – 5 + 8 - 4)
= mn (11 - 9)
= mn (2)
= 2mn

(ii) t – 8tz, 3tz – z, z – t

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= t – 8tz + (3tz - z) + (z - t)
= t – 8tz + 3tz – z + z - t
= t – t – 8tz + 3tz – z + z
= t (1 - 1) + tz (- 8 + 3) + z (-1 + 1)
= t (0) + tz (- 5) + z (0)
= - 5tz 

(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= - 7mn + 5 + 12mn + 2 + (9mn - 8) + (- 2mn - 3)
= - 7mn + 5 + 12mn + 2 + 9mn – 8 - 2mn - 3
= - 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
= mn (-7 + 12 + 9 - 2) + (5 + 2 – 8 - 3)
= mn (- 9 + 21) + (7 - 11)
= mn (12) – 4
= 12mn - 4

(iv) a + b – 3, b – a + 3, a – b + 3

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 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 (1 – 1 + 1) + b (1 + 1 - 1) + (-3 + 3 + 3)
= a (2 -1) + b (2 -1) + (-3 + 6)
= a (1) + b (1) + (3)
= a + b + 3

(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 14x + 10y – 12xy – 13 + (18 – 7x – 10y + 8xy) + 4xy
= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
= 14x – 7x + 10y– 10y – 12xy + 8xy + 4xy – 13 + 18
= x (14 - 7) + y (10 - 10) + xy(-12 + 8 + 4) + (-13 + 18)
= x (7) + y (0) + xy(0) + (5)
= 7x + 5

(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 5m – 7n + (3n – 4m + 2) + (2m – 3mn - 5)
= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
= m (5 - 4 + 2) + n (-7 + 3) – 3mn + (2 - 5)
= m (3) + n (-4) – 3mn + (-3)
= 3m – 4n – 3mn - 3

(vii) 4x2y, – 3xy2, –5xy2, 5x2y

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 4x2y + (-3xy2) + (-5xy2) + 5x2y
= 4x2y + 5x2y – 3xy2 – 5xy2
= x2y (4 + 5) + xy2 (-3 - 5)
= x2y (9) + xy2 (- 8)
= 9x2y – 8xy2

(viii) 3p2q2 – 4pq + 5, – 10 p2q2, 15 + 9pq + 7p2q2

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 3p2q2 – 4pq + 5 + (- 10p2q2) + 15 + 9pq + 7p2q2
= 3p2q2 – 10p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15
= p2q2 (3 -10 + 7) + pq (-4 + 9) + (5 + 15)
= p2q2 (0) + pq (5) + 20
= 5pq + 20

(ix) ab – 4a, 4b – ab, 4a – 4b

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= ab – 4a + (4b – ab) + (4a – 4b)
= ab – 4a + 4b – ab + 4a – 4b
= ab – ab – 4a + 4a + 4b – 4b
= ab (1 -1) + a (4 - 4) + b (4 - 4)
= ab (0) + a (0) + b (0)
= 0

(x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2 

Answer

When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= x2 – y2 – 1 + (y2 – 1 – x2) + (1 – x2 – y2)
= x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2 
= x2 – x2 – x2 – y2 + y2 – y2 – 1 – 1 + 1
= x2 (1 – 1- 1) + y2 (-1 + 1 - 1) + (-1 -1 + 1)
= x2 (1 - 2) + y2 (-2 +1) + (-2 + 1)
= x2 (-1) + y2 (-1) + (-1)
= -x2 – y2 -1

3. Subtract: 

(i) -5y2 from y2

Answer

y2 - (-5y2) = y2 + 5y2 = 6y2

(ii) 6xy from -12xy

Answer

-12xy -(6xy) = -12xy - 6xy = -18xy

(iii) (a - b) from (a + b)

Answer

(a + b)-(a -b) = a + b -a + b
= a - a + b + b = 2b

(iv) a (b - 5) from b (5 - a)

Answer

= b (5 - a)-a (b -5)
= 5b - ab - ab + 5a
= 5b - 2ab+5a
= 5a + 5b -2ab

(v) -m2 + 5mn from 4m2 - 3mn + 8

Answer

= 4m2 - 3mn + 8 - (- m2 + 5mn)
= 4m2 - 3mn + 8 + m2 - 5mn
= 4m2 + m2 - 3mn - 5mn + 8
= 5m2 - 8mn + 8

(vi) -x2 +10x - 5 from 5x-10

Answer

= 5x – 10 – (-x2 + 10x - 5)
= 5x – 10 + x2 – 10x + 5
= x2 + 5x – 10x – 10 + 5
= x2 – 5x - 5

(vii) 5a2 - 7ab + 5b2 from 3ab - 2a2 - 2b2

Answer

= 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2)
= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2 
= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2 
= 10ab – 7a2 – 7b2

(viii) 4pq - 5q2 - 3p2 from 5p2 + 3q2 - pq

Answer

= 5p2 + 3q2 – pq – (4pq – 5q2 – 3p2)
= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2 
= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq
= 8p2 + 8q2 – 5pq

4. (a) What should be added to x2 +xy+y2 to obtain 2x2 +3xy ?

Answer

Let p should be added.
Then according to question,
x2 + xy + y2 + p = 2x2 + 3xy
⇒ p = 2x2 + 3xy - (x2 + xy + y2)
⇒ p = 2x2 + 3xy - x2 - xy - y2
⇒ p = 2x2 - x2- y2 +3xy - xy
⇒ p = x2 - y2 + 2xy
Hence, x2 - y2 + 2xy should be added.

(b) What should be subtracted from 2a + 8b+10 to get -3a + 7b + 16?

Answer

Let q should be subtracted.
Then according to question, 2a + 8b + 10-q = -3a + 7b + 16
⇒ -q = -3a +7b + 16 - (2a + 8b + 10)
⇒ -q = -3a + 7b + 16 - 2a - 8b - 10
⇒ - q = -3a - 2a + 7b - 8b + 16 - 10
⇒ -q = -5a - b + 6
⇒ q = - (- 5a - b + -6)
⇒ q = 5a + b - 6

5. What should be taken away from 3x2- 4y2 + 5xy + 20 to obtain - x2 - y2 + 6xy + 20 ? 

Answer

Let q should be subtracted.
Then according to question,
3x2 - 4y2 + 5xy + 20 -q = -x2 - y2 + 6xy + 20
⇒ q = 3x2 - 4y2 + 5xy + 20 - (-x2 - y2 + 6xy + 20)
⇒ q = 3x2 - 4y2 + 5xy+ 20 + x2 + y2 - 6xy - 20
⇒ q = 3x2 + x2 - 4y2 +y2 + 5xy - 6xy + 20-20
⇒ q =  4x2 - 3y2 -  xy + 0
Hence, 4x2 -3y2 -xy should be subtracted.

6. (a) From the sum of 3x - y + 11 and - y - 11, subtract 3x - y - 11.

Answer

First we have to find out the 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) From the sum of 4 + 3x and 5 - 4x + 2x2, subtract the sum of 3x2 - 5x and -x2 + 2x + 5.

Answer

First we have to find out the sum of 4 + 3x and 5 – 4x + 2x2 
= 4 + 3x + (5 – 4x + 2x2)
= 4 + 3x + 5 – 4x + 2x2 
= 4 + 5 + 3x – 4x + 2x2 
= 9 - x + 2x2 
= 2x2 – x + 9     … [equation 1]
Then, we have to find out the sum of 3x2 – 5x and – x2 + 2x + 5.

Exercise 12.3 

1. If m = 2, find the value of: 

(i) m - 2    
(ii) 3m - 5    
(iii) 9 - 5m
(iv) 3m2 - 2m - 7
(v) 

Answer

(i) m - 2 = 2 - 2    [Putting m = 2]
= 0

(ii) 3m - 5 = 3 x 2 - 5     [Putting m = 2]
= 6 - 5 = 1

(iii) 9 - 5m = 9 - 5 x 2    [Putting m = 2]
= 9 - 10 = - 1

(iv) 3m2 - 2m - 7
= 3(2)2 - 2 (2) - 7        [Putting m = 2]
=3 × 4 - 2 × 2 - 7
 = 12-4-7
 = 12- 11 = 1

(v)  [Putting m = 2]

= 5 - 4 = 1

2. If p = -2, find the value of:
(i) 4p + 7
(ii) - 3p2 + 4p + 7
(iii) -2p3 - 3p2 +4/7 + 7

Answer

(i) 4p + 7 = 4 (- 2) + 7    [Putting p= -2]

= -8 + 7 = -1

(ii) -3p2+4p + 7
= -3 (-2)2+ 4 (-2) + 7    [Putting p = - 2]
= - 3 × 4 - 8 + 7
= - 12 - 8 + 7
= -20 + 7 = -13

(iii) - 2p3 - 3p2 +4p + 7
= - 2 (-2)3 - 3(-2)2 + 4 (-2) + 7     [Putting p = - 2]
= -2 ×(-8)-3 ×4 -8 + 7
= 16-12-8 + 7 
= -20 + 23 = 3

3. Find the value of the following expressions, when x = -1:
(i) 2x - 7
(ii) -x + 2
(iii) x2 + 2x  + 1
(iv) 2x2- x - 2

Answer

(i) 2x - 7 = 2 (-1) - 7      [Putting x= - 1]
= - 2 - 7 = - 9

(ii) - x + 2 = - (-1) + 2     [Putting x= - 1]
= 1 + 2 = 3

(iii) x2 + 2 x + 1 = (-1)2 + 2 (-1) + 1    [Putting x= - 1] 
= 1 - 2 + 1
= 2 - 2 = 0

(iv) 2x2- x - 2 = 2 (-1)2 - (-1) - 2     [Putting x= - 1] 
= 2x 1 + 1-2
= 2 + 1 - 2
= 3 - 2 = 1

4. If a = 2,b = -2, find the value of: 
(i) a2 + b2 
(ii) a2+ab + b2
(iii) a2 - b2

Answer

(i) a2 + b2 ( 2)2 + (- 2)2    [Putting a = 2. b = - 2 ]
= 4 + 4 = 8

(ii) a2+ab + b2 
= (2) + ( 2) (- 2) +(-2)2   [Putting a = 2. b = - 2 ]
= 4 - 4 + 4 = 4

(iii) a2 - b2 = (2)2 - (-2)2  [Putting a = 2,b = - 2]
= 4 - 4 = 0

5. When a = 0, b = -1, find the value of the given expressions: 
(i) 2a + 2b
(ii) 2a2+b2+1
(iii) 2a2b + 2ab2 +ab
(iv) a2+ab+2

Answer

(i) 2a + 2b = 2 (0) + 2 (-1)    [Putting a - 0,b = - 1]
= 0 - 2 = -2  

(ii) 2a2 + b2 + 1 = 2 (0)2 + (-1)2 + 1      [Putting a - 0,b = - 1]
= 2 x 0 + 1+ 1 = 0 + 2 = 2

(iii) 2a2b + 2ab2 + ab = 2(0)2 (-1) + 2 (0 )(-1)2 + (0 )(-1)     [Putting a - 0,b = - 1]
= 0 + 0 + 0 = 0

(iv) a2 +ab + 2 - (0)2 + (0) (-1) + 2   [Putting a - 0,b = - 1]
= 0 + 0 + 2 = 2

6. Simplify the expressions and find the value if x is equal to 2: 
(i) x + 7 + 4 (x- 5)
(ii) 3 (x + 2) + 5x - 7
(iii) 6x + 5 (x - 2)
(iv) 4 (2x - 1) + 3x + 11

Answer

(i) x + 7 + 4(x- 5) = x + 7 + 4x - 20 = x + 4 x + 7 - 20
= 5 x - 13 = 5 x 2 - 13                            [Putting x = 2]
= 10-13 = -3

(ii) 3 (x+ 2) + 5x - 7 = 3x + 6 + 5x -7 = 3x + 5x + 6 - 7
= 8x - 1 = 8 x 2-1                    [Putting x = -1]
= 16 - 1 = 15

(iii) 6x + 5 (x - 2) = 6x + 5x -10 = 11x - 10
= 11 x 2 - 10                      [Putting x = -1]
= 22 - 10 = 12

(iv) 4(2x - 1) + 3x + 11 = 8x - 4 + 3x +11 = 8x + 3a - 4 + 11
= 11a + 7 = 11 x 2 + 7 [Putting x = - 1]
= 22+7 = 29

7. Simplify these expressions and find their values if x = 3,a = -1, b = - 2 :
(i) 3x - 5 - x + 9
(ii) 2 - 8x + 4x + 4
(iii) 3a + 5 - 8a + 1
(iv) 10 - 3b - 4 - 5b
(v) 2a - 2b - 4 - 5 + a

Answer

(i) 3a - 5 - x + 9 = 3x - x - 5 + 9 = 2x + 4
= 2x3+4         [Putting a = 3]
= 6 + 4 = 10

(ii) 2 - 8x + 4x + 4 = - 8x + 4x + 2 + 4 = -4x + 6
= - 4 x 3 + 6     [Putting a = 3]
= -12 + 6 =12

(iii) 3a + 5 - 8a + 1 = 3a - 8a + 5 + 1 = - 5a + 6
= -5(- 1) + 6       [Putting a = - 1]
= 5 + 6 = 11

(iv) 10 - 3b - 4 - 5b = - 3b - 5b + 10 - 4 = -8b+6
= -8 (-2)+ 6    [Putting b = -2]
= 16 + 6 = 22

(v) 2a - 2b - 4 - 5 + a = 2a + a - 2b - 4 - 5
= 3a - 2b - 9 = 3 (-1)-2 (-2) -9    [Putting a = -1 , b = - 2]
= -3 + 4 -9 = -8

8. (i) If z = 10, find the value of z3 - 3 (z - 10).
(ii) If p = - 10, find the value of p2 - 2p - 100

Answer

(i) z3 -3(z-10) = (10)3-3(10 - 10)       [Putting z = 10]
= 1000 - 3 x 0 = 1000- 0
= 1000

(ii) p2 - 2p - 100 = (-10)2 - 2 (-10) - 100    (Putting p = - 10]
= 100+ 20 - 100 = 20

9. What should be the value of a if the value of 2x2 + x - a equals to 5, when x = 0 ? 

Answer

Given: 2x2 + x - a = 5
⇒ 2 (0)2 + 0 - a = 5     [Putting x = 0]
⇒ 0 + 0 - a = 5
⇒ a = -5
Hence, the value of a is -5.

10. Simplify the expression and find its value when a = 5 and b = - 3: 2 (a2 + ab) + 3 - ab

Answer

Given 2 (a2 + ab) + 3 - ab
⇒ 2a2 + 2ab + 3 - ab
⇒ 2a2 + 2ab - ab + 3
⇒ 2a2 + ab + 3
⇒ 2 (5)2 + (5) (-3) + 3   [Putting a = 5 , b = -3]
⇒ 2 x 25 - 15 + 3
⇒ 50 - 15 + 3
⇒ 38.

Go Back To NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions


Chapter 12 NCERT Solutions are accurate and detailed which will encourage students to learn new topics. Expressions that contain constants and variables, or just variables, are called algebraic expressions.

• A term that contains only a number is called a constant term.

• The constants and the variables whose product makes a term of an algebraic expression, are called the factors of the term. The factors of a constant term in an algebraic expression are not considered.

• The numerical factor of a variable term is called its coefficient. The variable factors of a term are called its algebraic factors.

There are total 4 exercises in the Chapter 12 which will help you in finding specific questions that you're looking for by visiting the link below.


NCERT Solutions will improve your retention memory and studying habits. Students can develop their understanding of the chapter and obtain maximum marks in the exams.


NCERT Solutions for Class 7 Maths Chapters:



Chapter 1 Integers
Chapter 2 Fractions and Decimals
Chapter 3 Data Handling
Chapter 4 Simple Equations
Chapter 5 Lines and Angles
Chapter 6 The Triangle and its Properties
Chapter 7 Congruence of Triangles
Chapter 8 Comparing Quantities
Chapter 9 Rational Numbers
Chapter 10 Practical Geometry
Chapter 11 Perimeter and Area
Chapter 13 Exponents and Powers
Chapter 14 Symmetry
Chapter 15 Visualising Solid Shapes



FAQ on Chapter 12 Algebraic Expressions

What are the benefits of NCERT Solutions for Chapter 12 Algebraic Expressions Class 7 NCERT Solutions?



NCERT Solutions for Class 7 can help you in figuring out the latest marking scheme and devise your own strategy. These NCERT Solutions are prepared as per the accordance of latest CBSE guidelines so you can score maximum marks.

Ritu spends ₹x daily and saves ₹ y per day. What is her income after 3 weeks?



₹21 (x + y).

Simplify the expression and find its value when a = 5 and b = –3. 2(a + ab) + 3 – ab.



2a2 + ab + 3, 38.


A bag contains 25 paise and so paise coins whose total values is ₹30. If the total number of 25 paise coins is four times that of 50 paise coins, find the number of each type of coins.



50 Paise coins = 20, 25 Paise coins = 80.
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