NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expressions

Class 7 Maths NCERT Solutions Chapter 12 – Algebraic Expressions comprises of the 4 Exercises

• Exercise 12.1
• Exercise 12.2
• Exercise 12.3
• Exercise 12.4

This Chapter contains the Exercises relating to the following topics , which are discussed in Chapter 12 – Algebraic Expressions Class 7 NCERT book  : –

• 12.1 INTRODUCTION
• 12.2 HOW ARE EXPRESSIONS FORMED?
• 12.3 TERMS OF AN EXPRESSION
• 12.4 LIKE AND UNLIKE TERMS
• 12.5 MONOMIALS, BINOMIALS, TRINOMIALS AND POLYNOMIALS
• 12.6 ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS
• 12.7 FINDING THE VALUE OF AN EXPRESSION
• 12.8 USING ALGEBRAIC EXPRESSIONS – FORMULAS AND RULES

NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.1

NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.2

NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.3

NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.4

NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.1

1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.

a) Subtraction of z from y
b) One-half of the sum of numbers x and y
c) The number z multiplied by itself
d) One-fourth of the product of numbers p and q
e) Number x and y both squared and added
f) Number 5 added to three times the product of numbers m and n
g) Product of numbers y and z subtracted from 10
h) Sum of numbers a and b subtracted from their product

Solution:

a) y – z
b) 1/2(x + y)
c) z²
d) (1/4)pq
e) x² + y²
f) 5 + 3mn
g) 10 – yz
h) ab – (a + b)

2. Write four more rational numbers in each of the following patterns:

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) 5xy² + 7x²y
e) -ab + 2b² – 3a²

ii. Identify terms and factors in the expression given below:

a) -4x + 5
b) -4x + 5y
c) 5y + 3y²
d) xy + 2x²y²
e) pq + q
f) 1.2ab – 2.4b + 3.6a
g) (3/4)x + (1/4)
h) 1p² + 0.2q²

Solution:

i. Tree Diagrams

ii. Terms and factors

Q.3  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. 1.2 a + 0.8 b
vii. 3.14 r²
viii. 2 (l + b)
ix. 0.1 y + 0.01 y²

Solution:

Numerical coefficients of terms (other than constants)

Q.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 + 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

Solution:

a)

b)

Q.5 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

Solution:

Monomials: (ii), (iv), (viii)
Binomials: (i), (vi), (vii), (x), (xi)
Trinomials: (iii), (v), (ix), (xii)

Q.6 State whether a given pair of terms is of like or unlike terms.

i. 1, 100
ii. -7x, (5/2)x
iii. -29x, -29y
iv. 14xy, 42yx
v. 4m²p, 4mp²
vi. 12xz, 12x²z²

i. Like
ii. Like
iii. Unlike
iv. Like
v. Unlike
vi. Unlike

Q.7 Identify the 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

a) Groups of like terms are as follows

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

b) Groups of like terms are as follows

i. 10pq, -7qp, 78qp
ii. 7p, 2405p
iii. 8q, -100q
iv. -p²q², 12q²p²
v. -23, 41
vi. -5p², 701p²
vii. 13p²q, qp²

NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.2

1. Simplify by 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 + 3yx² – 3y² + x² – y² + 8xy² – 3y²
vi. (3y² + 5y – 4) – (8y – y² – 4)

i. 21b – 32 + 7b – 20b
= 21b – 20b + 7b – 32
= 8b -32

ii. – z² + 13z² – 5z + 7z³ – 15z
= 7z³ – z² + 13z² – 5z – 15z
= 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) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab
=a + ab

v. 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
= 5x²y + 3yx² + 8xy² – 5x² + x² – 3y² – y² – 3y²
= 8x²y + 8xy² – 4x² – 7y²

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

2. 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, – 10 p²q², 15 + 9pq + 7p²q²
ix. ab – 4a, 4b – ab, 4a – 4b
x. x² – y² – 1, y² – 1 – x², 1 – x² – y²

i. 3mn + (– 5mn) + 8mn + (– 4mn)
= 3mn – 5mn + 8mn – 4mn
= 2mn

ii. t – 8tz + 3tz – z + z – t
= t – t + z – z – 8tz + 3tz
= -5tz

iii. – 7mn + 5 + 12mn + 2 + 9mn – 8 + (– 2mn – 3)
= -7mn + 12 mn + 9mn – 2mn + 5 + 2 – 8 – 3
= 12mn – 4

iv. a + b – 3 + b – a + 3 + a – b + 3
= a – a + a + b + b – b – 3 + 3 + 3
= a + b + 3

v. 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
= 14x – 7x + 10y – 10y – 12xy + 8xy + 14xy – 13 + 18
= 7x + 5

vi. 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= 5m – 4m + 3m – 7n + 3n – 3mn + 2 – 5
= 3m – 4n – 3mn – 3

vii. 4x²y + (– 3xy²) + (–5xy²) + 5x²y
= 4x²y + 5x²y – 3xy² – 5xy²
= 9x²y – 8xy²

viii. 3p²q² – 4pq + 5 + – 10 p²q² + 15 + 9pq + 7p²q²
= 3p²q² – 10p²q² + 7p²q² – 4pq + 9pq + 5 + 15
= 5pq + 20

ix. ab – 4a + 4b – ab + 4a – 4b
= ab – ab – 4a + 4a + 4b – 4b
= 0

x. x² – y² – 1 + y² – 1 – x² + 1 – x² – y²
= x² – x² – x² – y² + y² – y² – 1 – 1 + 1
= -x² – y² – 1

3. Simplify by combining like terms:

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

i. 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 – 5mn
= 5m² – 8mn + 8

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

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

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

4. (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?

(a) Let K be added to x² + xy + y² to obtain 2x² + 3xy
⇒ K + x² + xy + y² = 2x² + 3xy
K = 2x² + 3xy – (x² + xy + y²)
K = 2x² + 3xy – x² – xy – y²
K = x² + 2xy – y²

(b) Let L be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16
⇒ 2a + 8b + 10 – L = – 3a + 7b + 16
2a + 8b + 10 = – 3a + 7b + 16 + L
L = 2a + 8b + 10 – (– 3a + 7b + 16)
L = 2a + 8b + 10 + 3a – 7b – 16
L = 5a + b -6

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

Solution:

Let P be taken away from 3x² – 4y² + 5xy + 20 to obtain – x² – y² + 6xy + 20
⇒ 3x² – 4y² + 5xy + 20 – P = – x² – y² + 6xy + 20
3x² – 4y² + 5xy + 20 = – x² – y² + 6xy + 20 + P
P = 3x² – 4y² + 5xy + 20 –(– x² – y² + 6xy + 20)
P = 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
P = 4x² – 3y² – xy

6. (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.

(a) The algebraic expression for the statement will be
[(3x – y + 11) + (– y – 11)] – (3x – y – 11)
Simplifying the expression
= (3x – 2y) – 3x + y + 11
= – y + 11

(b) The algebraic expression for the statement will be
[(4 + 3x) + (5 – 4x + 2x²)] – [(3x² – 5x) + (–x² + 2x + 5)]
Simplifying the expression
= (2x² – x + 9) – (2x² – 3x + 5)
= 2x² – x + 9 – 2x² + 3x – 5
= 2x + 4

NCERT Solutions for Class 7 Maths Chapter 12 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. (5m/2) – 4

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 × 2² – 2 × 2 – 7
= 12 – 4 – 7
= 1

v. (5m/2) – 4
= [(5 x 2)/2] – 4
= 5 – 4
= 1

2. If p = -2, find the value of:

i. 4p + 7
ii. -3p² + 4p + 7
iii. -2p³ – 3p² + 4p + 7

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

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

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

3. Find the value of the following expression, when x = -1:

i. 2x – 7
ii. -x + 2
iii. x² + 2x + 1
iv. 2x² – x – 2

i. 2x – 7
= 2 × (-1) – 7
= -2 – 7
= -9

ii. -x + 2
= -(-1) + 2
= 1 + 2
= 3

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

iv. 2x² – x – 2
= 2 × (-1)² – (-1) – 2
= 2 + 1 – 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
= 2² + (-2)²
= 4 + 4
= 8

ii. a² + ab + b²
= 2² + 2 × (-2) + (-2)²
= 4 – 4 + 4
= 4

iii. a² – b²
= 2² – (-2)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)
= 0 – 2
= -2

ii. 2a² + b² + 1
= 2 × (0)² + (-1)² + 1
= 0 + 1 + 1
= 2

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

iv. a² + ab + 2
= (0)² + (0) × (-1) + 2
= 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
= 5x – 13
= 5 × 2 – 13
= -3

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

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

iv. 4 (2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 11x + 7
= 11 × 2 + 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. 3x – 5 – x + 9 = 2x + 4
= 2 × 3 + 4
= 6 + 4
= 10

ii. 2 – 8x + 4x + 4 = -4x + 6
= -4 × 3 + 6
= -12 + 6
= -6

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

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

v. 2a – 2b – 4 – 5 + a
= 3a – 2b – 9
= 3 × (-1) – 2 × (-2) – 9
= -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)
= 10³ – 3 (10 – 10)
= 1000 – 0
= 1000

ii. p² – 2p – 100
= (-10)² – 2 (-10) – 100
= 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?

Solution:

Value of the expression at x = 0
2x² + x – a
= 2 × (0)² + 0 – a
= -a
According to the question, value of expression at x = 0 is 5
-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

Solution:

2 (a² + ab) + 3 – ab
= 2a² + 2ab + 3 – ab
= 2a² + ab + 3
Value of the expression when a = 5 and b = -3
2 (a² + ab) + 3 – ab
= 2a² + ab + 3
= 2 × 5² + 5 × (-3) + 3
= 50 – 15 + 3
= 38

NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.4

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.

If the number of digits formed is taken to be n, the number of segments required to form n digits 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

Solution:

Segments required

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

Solution:

Evaluating the value of expressions with respective values from the columns

With this we come to the end of NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions. We hope these helped you study your subject.

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