NCERT Solutions for Class 7 Maths Chapter 10 Algebraic Expressions

NCERT
SOLUTIONS
CLASS - 7TH

aglase .co

Book : Mathematics Ncert Solutions | Chapter-12 Maths

Class : 7th
Subject : Maths
Chapter : 12
Chapter Name : Algebraic Expressions

Exercise 12.1

Q1 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 numbers 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) (x + y)
1

2

(iii)z 2

1
(pq)
( iv ) 4
2 2
(v) x + y

(vi) 5 + 3(mn)

(vil) 10 − yz

(viii) ab − (a + b)

Page : 234 , Block Name : Exercise 12.1

Q2 (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 + x 2

(c) y – y 3

(d) 5xy + 7x y
2 2

(e) – ab + 2b – 3a 2 2

(ii) Identify terms and factors in the expressions given below:
(a) – 4x + 5
(b) – 4x + 5y
(c) 5y + 3y 2

(d) xy + 2x y 2 2

(e) pq + q

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

(f) 1.2 ab – 2.4 b + 3.6 a
3 1
x +
4 4

(h) 0.1 p + 0.2 q
2 2

Answer. (i)

(ii)

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Page : 234 , Block Name : Exercise 12.1

Q3 Identify the numerical coef cients of terms (other than constants) in the following expressions:
(i) 5 – 3t2

(ii) 1 + t + t + t 3
2

(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) – p q + 7pq
2 2

(vi) 1.2 a + 0.8 b
(vii) 3.14 r 2

(viii) 2 (l + b)
(ix) 0.1 y + 0.01 y
2

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Page : 235 , Block Name : Exercise 12.1

Q4 (a) Identify terms which contain x and give the coef cient of x.
(i) y x + y
2

(ii) 13y – 8yx
2

(iii) x + y + 2
(iv) 5 + z + zx
(v) 1 + x + xy
(vi) 12xy + 25 2

(vii) 7x + xy 2

(b) Identify terms which contain y and give the coef cient of y .
2 2

(i) 8 – xy 2

(ii) 5y + 7x
2

(iii) 2x y – 15xy + 7y
2 2 2

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Page : 235 , Block Name : Exercise 12.1

Q5 Classify into monomials, binomials and trinomials.
(i) 4y – 7z
(ii) y
2

(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p q – 4pq
2 2

(viii) 7mn
(ix) z – 3z + 8
2

(x) a + b
2 2

(xi) z + z
2

(xii) 1 + x + x 2

Answer. The monomials, respectively. binomials, and trinomials have 1, 2, and 3 unlike terms in it
(i) Binomial
(ii) Monomial
(iii) Trinomial
(iv) Monomial
(v) Trinomial
(vi) Binomial
(vii) Binomial
(viii) Monomial
(ix) Trinomial
(x) Binomial
(xi) Binomial
(xii) Trinomial

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Page : 235 , Block Name : Exercise 12.1

Q6 State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
(ii) −7x, x 5

2

(iii) – 29x, – 29y
(iv) 14xy, 42yx
(v) 4m p, 4mp 2 2

(vi) 12xz, 12x z 2 2

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) Like
(ii) LikeLike
(iii) Unlike
(iv) Like
(v) Unlike
(vi) Unlike

Page : 235 , Block Name : Exercise 12.1

Q7 Identify like terms in the following:
(a)
2 2 2 2 2 2
−xy , −4yx , 8x , 2xy , 7y, −11x , −100x, −11yx, 20x y

2
−6x , y, 2xy, 3x

(b)
2 2 2 2 2
10pq, 7p, 8q, −p q , −7qp, −100q, −23, 12q p , −5p , 41, 2405p, 78qp

2 2 2
13p q, qp , 701p

2 2
(a) − xy , 2xy

2 2
−4yx , 20x y

2 2 2
8x , −11x , −6x

7y, y

−100x, 3x

−11xy, 2xy

(b) 10pq, −7qp, 78qp

7p, 2405p

8q, −100q

2 2 2 2
−p q , 12p q

−23, 41

2 2
−5p , 701p
2 2
13p q, qp

Page : 235 , Block Name : Exercise 12.1

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Exercise 12.2

Q1 Simplify combining like terms:
(i) 21b − 32 + 7b − 20b

2 2 3
(ii) − z + 13z − 5z + 7z − 15z

(iii) p − (p − q) − q − (q − p)

(iv) 3a − 2b − ab − (a − b + ab) + 3ab + b − a
2 2 2 2 2 2 2 2
(v) 5x y − 5x + 3yx − 3y + x − y + 8xy − 3y

2 2
(vi) (3y + 5y − 4) − (8y − y − 4)

Answer. (i) 21b - 32 + 7b- 20b = 21b + 7b - 20b - 32
= b (21 +7 - 20) -32
= - 8b - 32
2 2 3 3 2 2
(ii) − z + 13z − 5z + 7z − 15z = 7z − z + 13z − 5z − 15z

3 2
= 7z + z (−1 + 13) + z(−5 − 15)

3 2
= 7z + 12z − 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 + 3ba + 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
2 2 2 2 2 2 2 2
(v)5x y − 5x + 3yx − 3y + x − y + 8xy − 3y

2 2 2 2 2 2 2 2
= 5x y + 3yx − 5x + x − 3y − y − 3y + 8xy

2 2 2 2
= x y(5 + 3) + x (−5 + 1) + y (−3 − 1 − 3) + 8xy

2 2 2 2
= 8x y − 4x − 7y + 8xy
2 2
(vi) (3y + 5y − 4) − (8y − y − 4)

2 2
= 3y + 5y − 4 − 8y + y + 4

2 2
= 3y + y + 5y − 8y − 4 + 4

2
= y (3 + 1) + y(5 − 8) + 4(1 − 1)

2
= 4y − 3y

Page : 239 , Block Name : Exercise 12.2

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

(iv) 14x + 10y − 12xy − 13, 18 − 7x − 10y + 8xy, 4xy

(vi) 5m − 7n, 3n − 4m + 2, 2m − 3x − 10y + 8xy, 4xy

2 2 2 2
(vi) 4x y, −3xy , −5xy , 5x y

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2 2 2 2 2 2
(viii) 3p q − 4pq + 5, −10p q , 15 + 9pq + 7p q

(ix) ab − 4a, 4b − ab, 4a − 4b

2 2 2 2 2 2
(x) x − y − 1, y − 1 − x ,1 − x − y

(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
2 2 2 2 2 2 2 2
(vii) 4x y − 3xy − 5xy + 5x y = 4x y + 5x y − 3xy − 5xy

2 2
= x y(4 + 5) + xy (−3 − 5)

2 2
= 9x y − 8xy

( 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

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2 2 2 2 2 2
(x) (x − y − 1) + (y − 1 − x ) + (1 − x − y )

2 2 2 2 2 2
= x − y − 1 + y − 1 − x + 1 − x − y

2 2 2 2 2 2
= x − x − x − y + y − y − 1 − 1 + 1

2 2
= x (1 − 1 − 1) + y (−1 + 1 − 1) + (−1 − 1 + 1)

2 2
= −x − y − 1

Page : 239 , Block Name : Exercise 12.2

Q3 Subtract:
2 2
(i) − 5y from y

(ii) 6xy from − 12xy

(iii) (a − b) from (a + b)

(iv) a(b − 5) from b(5 − a)

2 2
(v) − m + 5mn from 4m − 3mn + 8

2
(vi) − x + 10x − 5 from 5x − 10

2 2 2 2
(vi) 5a − 7ab + 5b from 3ab − 2a − 2b

2 2 2 2
(vii) 4pq − 5q − 3p from 5p + 3q − pq

2 2 2 2 2
(i) y − (−5y ) = y + 5y = 6y

(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
2 2 2 2
(v) (4m − 3mn + 8) − (−m + 5mn) = 4m − 3mn + 8 + m − 5mn

2 2
= 4m + m − 3mn − 5mn + 8

2
= 5m − 8mn + 8
2 2
( vi) (5x − 10) − (−x + 10x − 5) = 5x − 10 + x − 10x + 5

2
= x + 5x − 10x − 10 + 5

2
= x − 5x − 5
2 2 2 2
(vii) (3ab − 2a − 2b ) − (5a − 7ab + 5b )

2 2 2 2
= 3ab − 2a − 2b − 5a + 7ab − 5b

2 2 2 2
= 3ab + 7ab − 2a − 5a − 2b − 5b

2 2
= 10ab − 7a − 7b
2 2 2 2
(vili) 4pq − 5q − 3p from 5p + 3q − pq

2 2 2 2
(5p + 3q − pq) − (4pq − 5q − 3p )

2 2 2 2
= 5p + 3q − pq − 4pq + 5q + 3p

2 2 2 2
= 5p + 3p + 3q + 5q − pq − 4pq

2 2
= 8p + 8q − 5pq

Page : 240 , Block Name : Exercise 12.2

Q4 (a) What should be added to x 2
+ xy + y
2
to obtain 2x 2
+ 3xy? (b) What should be subtracted from

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2a + 8b + 10 to get – 3a + 7b + 16?

Answer. (a) Let a be the required term.
2 2 2
a + (x + y + xy) = 2x + 3xy

2 2 2
a = 2x + 3xy − (x + y + xy)

2 2 2
a = 2x + 3xy − x − y − xy

2 2 2
a = 2x − x − y + 3xy − xy

2 2
= x − y + 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

Page : 240 , Block Name : Exercise 12.2

Q5 What should be taken away from 3x 2
− 4y
2
+ 5xy + 20 to obtain −x 2
− y
2
+ 6xy + 20 ?

Answer. Let p be the required term.
2 2 2 2
(3x − 4y + 5xy + 20) − p = −x − y + 6xy + 20

2 2 2 2
p = (3x − 4y + 5xy + 20) − (−x − y + 6xy + 20)

2 2 2 2
= 3x − 4y + 5xy + 20 + x + y − 6xy − 20

2 2 2 2
= 3x + x − 4y + y + 5xy − 6xy + 20 − 20

2 2
= 4x − 3y − xy

Page : 240 , Block Name : Exercise 12.2

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 + 2x , subtract the sum of 3x – 5x and –x + 2x + 5.
2 2 2

( 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

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2 2
(b)(4 + 3x) + (5 − 4x + 2x ) = 4 + 3x + 5 − 4x + 2x

2
= 3x − 4x + 2x + 4 + 5

2
= −x + 2x + 9

2 2 2 2
(3x − 5x) + (−x + 2x + 5) = 3x − 5x − x + 2x + 5

2 2
= 3x − x − 5x + 2x + 5
2
= 2x − 3x + 5

2 2
(−x + 2x + 9) − (2x − 3x + 5)

2 2
= −x + 2x + 9 − 2x + 3x − 5

2 2
= −x + 3x + 2x − 2x + 9 − 5

= 2x + 4

Page : 240 , Block Name : Exercise 12.2

Exercise 12.3

Q1 If m = 2, nd the value of: (i) m – 2 (ii) 3m – 5 (iii) 9 – 5m
2 5m
3m − 2m − 7 (v) 4
2

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

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

= 12 − 4 − 7 = 1

5m 5×2
(iv) − 4 = ( ) − 4 = 1
2 2

Page : 242 , Block Name : Exercise 12.3

Q2 If p = – 2, nd the value of:
2 3 2
4p + 7 (ii) − 3p + 4p + 7 (iii) − 2p − 3p + 4p + 7

(i) 4p + 7 = 4 × (−2) + 7 = −8 + 7 = −1

2
(ii) − 3p + 4p + 7 = −3(−2) × (−2) + 4 × (−2) + 7

= −12 − 8 + 7 = −13
3 2
( iii) − 2p − 3p + 4p + 7

= −2(−2) × (−2) × (−2) − 3(−2) × (−2) + 4 × (−2) + 7

= 16 − 12 − 8 + 7 = 3

Page : 242 , Block Name : Exercise 12.3

Q3 Find the value of the following expressions, when x = –1:
(i) 2x – 7 (ii) – x + 2 (iii) x + 2x + 1 (iv) 2x – x – 2
2 2

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Answer. (i) 2x - 7 = 2 x (-1) - 7 = -9
(ii) − x + 2 = −(−1) + 2 = 1 + 2 = 3

2
(iii) x + 2x + 1 = (−1) × (−1) + 2 × (−1) + 1

= 1 − 2 + 1 = 0

2
( iv) 2x − x − 2 = 2(−1) × (−1) − (−1) − 2

= 2 + 1 − 2 = 1

Page : 242 , Block Name : Exercise 12.3

Q4 If a = 2, b = – 2, nd the value of:
(i) a + b (ii) a + ab + b (iii) a - b
2 2 2 2 2 2

2 2
(i)a + b

2 2
= (2) + (−2) = 4 + 4 = 8

2 2
(ii) a + ab + b

= (2 × 2) + 2 × (−2) + (−2) × (−2)

= 4 − 4 + 4 = 4

2 2
(iii) a − b

2 2
= (2) − (−2) = 4 − 4 = 0

Page : 242 , Block Name : Exercise 12.3

Q5 When a = 0, b = – 1, nd the value of the given expressions:
2 2 2 2
(i) 2a + 2b (ii) 2a + b + 1 (iii) 2a b + 2ab + ab
2
a + ab + 2

Answer. (i) 2a + 2b = 2 x (0) + 2 x (-1) = 0 - 2 = -2
2 2
( ii) 2a + b + 1

2
= 2 × (0) + (−1) × (−1) + 1

= 0 + 1 + 1 = 2
2 2
(iii) 2a b + 2ab + ab

2
= 2 × (0) × (−1) + 2 × (0) × (−1) × (−1) + 0 × (−1)

= 0 + 0 + 0 = 0
2
( iv )a + ab + 2

2
= (0) + 0 × (−1) + 2

= 0 + 0 + 2 = 2

Page : 242 , Block Name : Exercise 12.3

Q6 Simplify the expressions and nd the value if x is equal to 2
(i) x + 7 + 4 (x – 5)
(ii) 3 (x + 2) + 5x – 7

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(iii) 6x + 5 (x – 2)
(iv) 4(2x – 1) + 3x + 11

( i) x + 7 + 4(x − 5) = x + 7 + 4x − 20

= x + 4x + 7 − 20

= 5x − 13

= (5 × 2) − 13

= 10 − 13 = −3

(ii) 3(x + 2) + 5x − 7 = 3x + 6 + 5x − 7

= 3x + 5x + 6 − 7 = 8x − 1

= (8 × 2) − 1 = 16 − 1 = 15

(iii) 6x + 5(x − 2) = 6x + 5x − 10

= 11x − 10

= (11 × 2) − 10 = 22 − 10 = 12

( iv) 4(2x − 1) + 3x + 11 = 8x − 4 + 3x + 11

= 11x + 7

= (11 × 2) + 7

= 22 + 7 = 29

Page : 242 , Block Name : Exercise 12.3

Q7 Simplify these expressions and nd 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) 3x − 5 − x + 9 = 3x − x − 5 + 9

= 2x + 4 = (2 × 3) + 4 = 10

(ii) 2 − 8x + 4x + 4 = 2 + 4 − 8x + 4x

= 6 − 4x = 6 − (4 × 3) = 6 − 12 = −6

(iii) 3a + 5 − 8a + 1 = 3a − 8a + 5 + 1

= −5a + 6 = −5 × (−1) + 6

= 5 + 6 = 11

( iv) 10 − 3b − 4 − 5b = 10 − 4 − 3b − 5b

= 6 − 8b = 6 − 8 × (−2)

= 6 + 16 = 22

(v)2a − 2b − 4 − 5 + a = 2a + a − 2b − 4 − 5

= 3a − 2b − 9s

= 3 × (−1) − 2(−2) − 9

= −3 + 4 − 9 = −8

Page : 242 , Block Name : Exercise 12.3

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Q8 (i) If z = 10, nd the value of z − 3(z − 10).
3

(ii) If p = – 10, nd the value of p − 2p − 1002

3 3
(i)z − 3(z − 10) = z − 3z + 30

= (10 × 10 × 10) − (3 × 10) + 30

= 1000 − 30 + 30 = 1000

2
(ii)p − 2p − 100

= (−10) × (−10) − 2(−10) − 100

= 100 + 20 − 100 = 20

Page : 242 , Block Name : Exercise 12.3

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

2
2x + x − a = 5, when x = 0

(2 × 0) + 0 − a = 5

0 − a = 5

a = −5

Page : 242 , Block Name : Exercise 12.3

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

2 2
2 (a + ab) + 3 − ab = 2a + 2ab + 3 − ab

2
= 2a + 2ab − ab + 3

2
= 2a + ab + 3

= 2 × (5 × 5) + 5 × (−3) + 3

= 50 − 15 + 3 = 38

Page : 242 , Block Name : Exercise 12.3

Exercise 12.4

Q1 Observe the patterns of digits made from line segments of equal length. You will nd such
segmented digits on the display of electronic watches or calculators.

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

Number of segments required to form 5 digits = ( 5 x 5 + 1)
= 25 +1 = 26
Number of segments required to form 10 digits = ( 5 x 10 + 1)
= 50 +1 = 51
Number of segments required to form 100 digits = ( 5 x 100 + 1)
= 500 +1 = 501

Number of segments required to form 5 digits = ( 3 x 5 + 1)
= 15 +1 = 16
Number of segments required to form 10 digits = ( 3 x 10 + 1)
= 30 +1 = 31
Number of segments required to form 100 digits = ( 3 x 100 + 1)
= 300 +1 = 301

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Number of segments required to form 5 digits = ( 5 x 5 + 2)
= 25 +2 = 27
Number of segments required to form 10 digits = ( 5 x 10 + 2)
= 50 +2 = 52
Number of segments required to form 100 digits = ( 5 x 100 + 2)
= 500 +2 = 502

Page : 246 , Block Name : Exercise 12.4

Q2 Use the given algebraic expression to complete the table of number patterns.

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Page : 247 , Block Name : Exercise 12.4

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