Home » Class 7 Maths » NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expressions

NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expressions


Exercise 12.1 : Solutions of Questions on Page Number : 234


Question 1:  Get the algebraicexpressions 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 number m and n.

(vii) 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)

(iii) z2

(iv)

(v) x2 + y2

(vi) 5 + 3 (mn)

(vii) 10 − yz

(viii) ab − (a + b)


Question 2: 

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

Show the terms and factors by tree diagrams.

(a) x − 3 (b) 1 + x + x2 (c) y − y3

(d) 

(e) − ab + 2b2 − 3a2

(ii) Identify terms and factors in the expressions given below:

(a) − 4x + 5 (b) − 4x + 5y (c) 5y + 3y2

(d) (e) pq + q

(f) 1.2 ab − 2.4 b + 3.6 a (g)

(h) 0.1p2 + 0.2 q2

Answer :

(i)

(a)

(b)


(ii)

Row Expression Terms Factors
(a) − 4x + 5 − 4x

5

− 4, x

5

(b) − 4x + 5 − 4x

5

− 4, x

5

(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) 1.2ab − 2.4b + 3.6a 1.2ab

− 2.4b

3.6a

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) 0.1p2 + 0.2q2 0.1p2

0.2q2

0.1, p,p ,  0.2, q,q

Q3 :  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) 1.2a + 0.8b
(vii) 3.14 r2 (viii) 2 (l + b) (ix) 0.1y + 0.01 y2

Answer :

 

Row 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) 1.2a +0.8b 1.2a

0.8b

1.2

0.8

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

2b

2

2

(viii) 0.1+ 0.01y2 0.1

0.01y2

 

0.1

0.01


Q4 :  (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
(b) Identify terms which contain y2 and give the coefficient of y2.
(i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y -15xy2 + 7y2

Answer :


Q5 :  Classify into monomials, binomials and trinomials.
(i) 4y – 7z (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 :
The monomials, binomials, and trinomials have 1, 2, and 3 unlike terms in it respectively.

(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


Question 6 : 

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

(i) 1, 100

(ii) 

(iii) − 29x, − 29y

(iv) 14xy, 42yx

(v)4m2p, 4mp2

(vi)12xz, 12 x2z2

 Answer : 

The terms which have the same algebraic factors are called like terms. However, when the terms have different algebraic factors, these are called unlike terms.

(i) 1, 100

Like

(ii) − 7x

Like

(iii) −29x, −29y

Unlike

(iv) 14xy, 42yx

Like

(v) 4m2p, 4mp2

Unlike

(vi) 12xz, 12x2z2

Unlike


Q7 :  Identify like terms in the following:
(a) -xy2, – 4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, -11yx, 20x2y, -6x2, y, 2xy,3x

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

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

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


Exercise 12.2 : Solutions of Questions on Page Number : 239


Q1 :  Simplify combining like terms:
(i) 21b – 32 + 7b – 20b
(ii) – z2 + 13z2 – 5z + 7z3 – 15z
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
(v) 5x2y – 5x2 + 3y x2 – 3y2 + x2 – y2 + 8xy2 -3y2
(vi) (3 y2 + 5y – 4) – (8y – y2 – 4)

Answer :
(i) 21b – 32 + 7b – 20b = 21b + 7b – 20b – 32
= b (21 + 7 – 20) -32
= 8b – 32

(ii) – z2 + 13z2 – 5z + 7z3 – 15z = 7z3 – z2 + 13z2 – 5z – 15z
= 7z3 + z2 (-1 + 13) + z (-5 – 15)
= 7z3 + 12z2 – 20z

(iii) p – (p – q) – q – (q – p) = p – p + q – q – q + p
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ba + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= a (3 – 1 – 1) + b (- 2 + 1 + 1) + ab (-1 -1 + 3)
= a + ab

(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
= 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2 + 8xy2
= x2y (5 + 3) + x2 (-5 + 1) + y2(-3 – 1 – 3) + 8xy2
= 8x2y – 4x2 – 7y2 + 8xy2

(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
= 3y2 + 5y – 4 – 8y + y2 + 4
= 3y2 + y2 + 5y – 8y – 4 + 4
= y2 (3 + 1) + y (5 – 8) + 4 (1 – 1)
= 4y2 – 3y


Q2 : Add:
(i) 3mn, – 5mn, 8mn, -4mn
(ii) t – 8tz, 3tz – z, z – t
(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x2y, – 3xy2, – 5xy2, 5x2y
(viii) 3p2q2 – 4pq + 5, – 10p2q2, 15 + 9pq + 7p2q2
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x2 – y2 – 1 , y2 – 1 – x2, 1- x2 – y2

Answer :
(i) 3mn + (-5mn) + 8mn + (-4mn) = mn (3 – 5 + 8 – 4)
= 2mn

(ii) (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)
= -5tz

(iii) (- 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)
= 12mn – 4

(iv) (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 (- 1 + 1 + 1)
= a + b + 3

(v) (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8yx) + 4xy
= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8yx + 4xy
= 14x – 7x + 10y – 10y – 12xy + 8yx + 4xy – 13 + 18
= x (14 – 7) + y (10 – 10) + xy (- 12 + 8 + 4) – 13 + 18
= 7x + 5

(vi) (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
= 3m – 4n – 3mn – 3

(vii) 4x2 y – 3xy2 – 5xy2 + 5x2y = 4x2 y + 5x2y – 3xy2 – 5xy2
= x2 y (4 + 5) + xy2 (- 3 – 5)
= 9x2y – 8xy2

(viii) (3p2q2 – 4pq + 5) + (-10 p2q2) + (15 + 9pq + 7p2q2)
= 3p2q2 – 4pq + 5 – 10 p2q2 + 15 + 9pq + 7p2q2
= 3p2q2 – 10 p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15
= p2q2 (3 – 10 + 7) + pq (- 4 + 9) + 5 + 15
= 5pq + 20

(ix) (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)
= 0

(x) (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 – y2 – 1


Q3 :  Subtract:
(i) – 5y2 from y2
(ii) 6xy from – 12xy
(iii) (a – b) from (a + b)
(iv) a (b – 5) from b (5 – a)
(v) – m2 + 5mn from 4m2 – 3mn + 8
(vi) – x2 + 10x – 5 from 5x – 10
(vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 -2b2
(viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq

Answer :

(i) y2 – (-5y2) = y2 + 5y2 = 6y2

(ii) – 12xy – (6xy) = -18xy

(iii) (a + b) – (a – b) = a + b – a + b = 2b

(iv) b (5 – a) – a (b – 5) = 5b – ab – ab + 5a
= 5a + 5b – 2ab

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

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

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


Q4 :  (a) What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?

Answer :
(a) Let a be the required term.
a + (x2 + y2 + xy) = 2x2 + 3xy
a = 2x2 + 3xy – (x2 + y2 + xy)
a = 2x2 + 3xy – x2 – y2 – xy
a = 2x2 – x2 – y2 + 3xy – xy
= x2 – y2 + 2xy

(b) Let p be the required term.
(2a + 8b + 10) – p = – 3a + 7b + 16
p = 2a + 8b + 10 – (- 3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10- 16
= 5a + b – 6


Q5 :  What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain
– x2 – y2 + 6xy + 20?

Answer :
Let p be the required term.
(3x2 – 4y2 + 5xy + 20) – p = – x2 – y2 + 6xy + 20
p = (3x2 – 4y2 + 5xy + 20) – (- x2 – y2 + 6xy + 20)
= 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20
= 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20
= 4x2 – 3y2 – xy


Q6 : (a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and
– x2 + 2x + 5.

Answer :
(a) (3x – y + 11) + (- y – 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y
(3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= – y + 11

(b) (4 + 3x) + (5 – 4x + 2x2) = 4 + 3x + 5 – 4x + 2x2
= 3x – 4x + 2x2 + 4 + 5
= – x + 2x2 + 9
(3x2 – 5x) + (- x2 + 2x + 5) = 3x2 – 5x – x2 + 2x + 5
= 3x2 – x2 – 5x + 2x + 5
= 2x2 – 3x + 5
(- x + 2x2 + 9) – (2x2 – 3x + 5)
= – x + 2x2 + 9 – 2x2 + 3x – 5
= – x + 3x + 2x2 – 2x2 + 9 – 5
= 2x + 4


Exercise 12.3 : Solutions of Questions on Page Number : 242


Question 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 = 0

(ii) 3m − 5 = (3 × 2) − 5 = 6 − 5 = 1

(iii) 9 − 5m = 9 − (5 × 2) = 9 −10 = −1

(iv) 3m2 − 2m − 7 = 3 × (2 × 2) − (2 × 2) − 7

= 12 − 4 − 7 = 1

(v)


Q2 :  If p = -2, find the value of:
(i) 4p + 7
(ii) -3p2 + 4p + 7
(iii) -2p3 – 3p2 + 4p + 7

Answer :
(i) 4p + 7 = 4 x (-2) + 7 = – 8 + 7 = -1

(ii) – 3p2 + 4p + 7 = -3 (-2) x (-2) + 4 x (-2) + 7
= – 12 – 8 + 7 = -13

(iii) -2p3 – 3p2 + 4p + 7
= -2 (-2) x (-2) x (-2) – 3 (-2) x (-2) + 4 x (-2) + 7
= 16 – 12 – 8 + 7 = 3


Q3 :  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 x (-1) – 7 = -9

(ii) – x + 2 = – (-1) + 2 = 1 + 2 = 3

(iii) x2 + 2x + 1 = (-1) x (-1) + 2 x (-1) + 1
= 1 – 2 + 1 = 0

(iv) 2x2 – x – 2 = 2 (-1) x (-1) – (-1) – 2
= 2 + 1 – 2 = 1


Q4 : 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 = 4 + 4 = 8

(ii) a2 + ab + b2
= (2 x 2) + 2 x (-2) + (-2) x (-2)
= 4 – 4 + 4 = 4

(iii) a2 – b2
= (2)2 – (-2)2 = 4 – 4 = 0


Q5 :  When a = 0, b = – 1, find the value of the given expressions:
(i) 2a + 2b (ii) 2a2 + b2 + 1
(iii) 2a2 b + 2ab2 + ab (iv) a2 + ab + 2

Answer :
(i) 2a + 2b = 2 x (0) + 2 x (-1) = 0 – 2 = -2

(ii) 2a2 + b2 + 1
= 2 x (0)2 + (-1) x (-1) + 1
= 0 + 1 + 1 = 2

(iii) 2a2b + 2ab2 + ab
= 2 x (0)2 x (-1) + 2 x (0) x (-1) x (-1) + 0 x (-1)
= 0 + 0 + 0 = 0

(iv) a2 + ab + 2
= (0)2 + 0 x (-1) + 2
= 0 + 0 + 2 = 2


Q6 :  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 + 4x + 7 – 20
= 5x – 13
= (5 x 2) – 13
= 10 – 13 = -3

(ii) 3 (x + 2) + 5x – 7 = 3x + 6 + 5x – 7
= 3x + 5x + 6 – 7 = 8x – 1
= (8 x 2) – 1 = 16 – 1 =15

(iii) 6x + 5 (x – 2) = 6x + 5x – 10
= 11x – 10
= (11 x 2) – 10 = 22 – 10 = 12

(iv) 4 (2x – 1) + 3x + 11 = 8x – 4 + 3x + 11
= 11x + 7
= (11 x 2) + 7
= 22 + 7 = 29


Q7 :  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) 3x – 5 – x + 9 = 3x – x – 5 + 9
= 2x + 4 = (2 x 3) + 4 = 10

(ii) 2 – 8x + 4x + 4 = 2 + 4 – 8x + 4x
= 6 – 4x = 6 – (4 x 3) = 6 – 12 = -6

(iii) 3a + 5 – 8a + 1 = 3a – 8a + 5 + 1
= – 5a + 6 = -5 x (-1) + 6
= 5 + 6 = 11

(iv) 10 – 3b – 4 – 5b = 10 – 4- 3b – 5b
= 6 – 8b = 6 – 8 x (-2)
= 6 + 16 = 22

(v) 2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5
= 3a – 2b – 9s
= 3 x (-1) – 2 (-2) – 9
= – 3 + 4 – 9 = -8


Q8 :  (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) = z3 – 3z + 30
= (10 x 10 x 10) – (3 x 10) + 30
= 1000 – 30 + 30 = 1000

(ii) p2 – 2p – 100
= (-10) x (-10) – 2 (-10) – 100
= 100 + 20 – 100 = 20


Q9 : What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0?

Answer :
2x2 + x – a = 5, when x = 0
(2 x 0) + 0 – a = 5
0 – a = 5
a = -5


Q10 : Simplify the expression and find its value when a = 5 and b = -3.
2 (a2 + ab) + 3 – ab

Answer :
2 (a2 + ab) + 3 – ab = 2a2 + 2ab + 3 – ab
= 2a2 + 2ab – ab + 3
= 2a2 + ab + 3
= 2 x (5 x 5) + 5 x (-3) + 3
= 50 – 15 + 3 = 38


Exercise 12.4 : Solutions of Questions on Page Number : 246


Question 1 :

Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

(a)

(b)

(c)

If the number of digits formed is taken to be n, the number of segments required to form ndigits is given by the algebraic expression appearing on the right of each pattern.

How many segments are required to form 5, 10, 100 digits of the kind −

 ,  , .

Answer :

(a) It is given that the number of segments required to form n digits of the kind

 is (5n + 1).

Number of segments required to form 5 digits = (5 × 5 + 1)

= 25 + 1 = 26

Number of segments required to form 10 digits = (5 × 10 + 1)

= 50 + 1 = 51

Number of segments required to form 100 digits = (5 × 100 + 1)

= 500 + 1 = 501

(b) It is given that the number of segments required to form n digits of the kind  is (3n + 1).

Number of segments required to form 5 digits = (3 × 5 + 1)

= 15 + 1 = 16

Number of segments required to form 10 digits = (3 × 10 + 1)

= 30 + 1 = 31

Number of segments required to form 100 digits = (3 × 100 + 1)

= 300 + 1 = 301

(c)It is given that the number of segments required to form n digits of the kind  is (5n + 2).

Number of segments required to form 5 digits = (5 × 5 + 2)

= 25 + 2 = 27

Number of segments required to form 10 digits = (5 × 10 + 2)

= 50 + 2 = 52

Number of segments required to form 100 digits = (5 × 100 + 2)

= 500 + 2 = 502