# NCERT Solutions for Class 7 Maths Ch 12 Algebraic Expressions| PDF Download

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Class 7 Maths NCERT Solutions will serve as beneficial tool that can be used to recall various questions any time. It can be used to enrich knowledge and make lessons for learners more exciting. These NCERT Solutions are prepared as per the accordance of latest CBSE guidelines so you can score maximum marks. 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.

(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  (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

(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

 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

 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

 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

 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

(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

(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

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

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)

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

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

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)

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.

(i) 3mn, – 5mn, 8mn, – 4mn

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

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

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

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

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

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

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

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

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

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

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

(ii) 6xy from -12xy

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

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

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

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

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

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

= 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

= 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

= 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

= 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 ?

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?

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 ?

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.

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.

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) (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

(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

(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

(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

(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

(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

(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

(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 ?

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

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.

₹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.