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

**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) z^{2 }d) (1/4)pq

e) x^{2} + y^{2 }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 + x**

^{2 }c) y – y

^{3 }d) 5xy

^{2}+ 7x

^{2}y e) -ab + 2b

^{2}– 3a

^{2}

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

**a) -4x + 5 b) -4x + 5y c) 5y + 3y**

^{2 }d) xy + 2x

^{2}y

^{2 }e) pq + q f) 1.2ab – 2.4b + 3.6a g) (3/4)x + (1/4) h) 1p

^{2}+ 0.2q

^{2}

**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 – 3t ^{2 }ii. 1 + t + t^{2} + t^{3 }iii. x + 2xy + 3y**

**iv. 100m + 1000n**

**v. – p**

^{2}q^{2}+ 7pq**vi. 1.2 a + 0.8 b**

**vii. 3.14 r**

^{2 }viii. 2 (l + b)**ix. 0.1 y + 0.01 y**

^{2}**Solution:**

Numerical coefficients of terms (other than constants)

**Q.4 (a) Identify terms which contain x and give the coefficient of x**

**i. y ^{2}x + y**

**ii. 13y**

^{2}– 8yx**iii. x + y + 2**

**iv. 5 + z + zx**

**v. 1 + x + xy**

**vi. 12xy**

^{2}+ 25**vii. 7x + xy**

^{2}**(b) Identify terms which contain y ^{2} and give the coefficient of y^{2}**

**i. 8 – xy ^{2 }ii. 5y^{2} + 7x**

**iii. 2x**

^{2}y – 15xy^{2}+ 7y^{2}**Solution:**

**a)**

**b)**

**Q.5 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**

^{2}q – 4pq^{2 }viii. 7mn**ix. z**

^{2}– 3z + 8**x. a**

^{2}+ b^{2 }xi. z^{2}+ z**xii. 1 + x + x**

^{2}**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. 4m ^{2}p, 4mp^{2 }vi. 12xz, 12x^{2}z^{2}**

i. Like

ii. Like

iii. Unlike

iv. Like

v. Unlike

vi. Unlike

**Q.7 Identify the like terms in the following:**

**a) -xy ^{2}, -4yx^{2}, 8x^{2}, 2xy^{2}, 7y, -11x^{2}, -100x, -11yx, 20x^{2}y, -6x^{2}, y, 2xy, 3x**

**b) 10pq, 7p, 8q, -p**

^{2}q^{2}, -7qp, -100q, -23, 12q^{2}p^{2}, -5p^{2}, 41, 2405p, 78qp, 13p^{2}q, qp^{2}, 701p^{2}**a)** Groups of like terms are as follows

i. -xy^{2}, 2xy^{2 }ii. -4yx^{2}, 20x^{2}y

iii. 8x^{2}, -11x^{2}, -6x^{2 }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. -p^{2}q^{2}, 12q^{2}p^{2 }v. -23, 41

vi. -5p^{2}, 701p^{2 }vii. 13p^{2}q, qp^{2}

### NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.2

**1. Simplify by combining like terms:**

**i. 21b – 32 + 7b – 20b ii. – z ^{2} + 13z^{2} – 5z + 7z^{3} – 15z iii. p – (p – q) – q – (q – p) iv. 3a – 2b – ab – (a – b + ab) + 3ab + b – a v. 5x^{2}y – 5x^{2} + 3yx^{2} – 3y^{2} + x^{2} – y^{2} + 8xy^{2} – 3y^{2 }vi. (3y^{2} + 5y – 4) – (8y – y^{2} – 4)**

**i.** 21b – 32 + 7b – 20b

= 21b – 20b + 7b – 32

= 8b -32

**ii.** – z^{2} + 13z^{2} – 5z + 7z^{3} – 15z

= 7z^{3} – z^{2} + 13z^{2} – 5z – 15z

= 7z^{3} + 12z^{2} – 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.** 5x^{2}y – 5x^{2} + 3yx^{2} – 3y^{2} + x^{2} – y^{2} + 8xy^{2} – 3y^{2 }= 5x^{2}y + 3yx^{2} + 8xy^{2} – 5x^{2} + x^{2} – 3y^{2} – y^{2} – 3y^{2 }= 8x^{2}y + 8xy^{2} – 4x^{2} – 7y^{2}

**vi.** (3y^{2} + 5y – 4) – (8y – y^{2} – 4)

= 3y^{2} + 5y – 4 – 8y + y^{2} + 4

= 3y^{2} + y^{2} + 5y – 8y – 4 + 4

= 4y^{2} – 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. 4x**

^{2}y, – 3xy

^{2}, –5xy

^{2}, 5x

^{2}y viii. 3p

^{2}q

^{2}– 4pq + 5, – 10 p

^{2}q

^{2}, 15 + 9pq + 7p

^{2}q

^{2 }ix. ab – 4a, 4b – ab, 4a – 4b x. x

^{2}– y

^{2}– 1, y

^{2}– 1 – x

^{2}, 1 – x

^{2}– y

^{2}

**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.** 4x^{2}y + (– 3xy^{2}) + (–5xy^{2}) + 5x^{2}y

= 4x^{2}y + 5x^{2}y – 3xy^{2} – 5xy^{2 }= 9x^{2}y – 8xy^{2}

**viii.** 3p^{2}q^{2} – 4pq + 5 + – 10 p^{2}q^{2} + 15 + 9pq + 7p^{2}q^{2 }= 3p^{2}q^{2} – 10p^{2}q^{2} + 7p^{2}q^{2} – 4pq + 9pq + 5 + 15

= 5pq + 20

**ix.** ab – 4a + 4b – ab + 4a – 4b

= ab – ab – 4a + 4a + 4b – 4b

= 0

**x.** x^{2} – y^{2} – 1 + y^{2} – 1 – x^{2} + 1 – x^{2} – y^{2}

= x^{2} – x^{2} – x^{2} – y^{2} + y^{2} – y^{2} – 1 – 1 + 1

= -x^{2} – y^{2} – 1

**3. Simplify by combining like terms:**

**i. -5y ^{2} from y^{2 }ii. 6xy from –12xy**

iii. (a – b) from (a + b)

iv. a (b – 5) from b (5 – a)

v. –m^{2} + 5mn from 4m^{2} – 3mn + 8

vi. – x^{2} + 10x – 5 from 5x – 10

vii. 5a^{2} – 7ab + 5b^{2} from 3ab – 2a^{2} – 2b^{2 }viii. 4pq – 5q^{2} – 3p^{2} from 5p^{2} + 3q^{2} – pq

**i.** y^{2} – (-5y^{2})

= 6y^{2}

**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.** 4m^{2} –3mn +8 – (-m^{2} + 5mn)

= 4m^{2} – 3mn + 8 + m^{2} – 5mn

= 5m^{2} – 8mn + 8

**vi.** 5x – 10 – (-x^{2} + 10x – 5)

= 5x – 10 + x^{2} – 10x + 5

= x^{2} – 5x – 5

**vii.** 3ab – 2a^{2} – 2b^{2} – (5a^{2} – 7ab + 5b^{2})

= 3ab –2a^{2} –2b^{2} –5a^{2} +7ab -5b^{2}

= 10ab – 7a^{2} – 7b^{2}

**viii.** 5p^{2} + 3q^{2} – pq – (4pq – 5q^{2} – 3p^{2})

= 5p^{2} +3q^{2} –pq –4pq +5q^{2} +3p^{2}

= 8p^{2} + 8q^{2} – 5pq

**4. (a) What should be added to x ^{2 }+ xy + y^{2} to obtain 2x^{2} + 3xy?**

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

**(a)** Let K be added to x^{2 }+ xy + y^{2} to obtain 2x^{2} + 3xy

⇒ K + x^{2 }+ xy + y^{2} = 2x^{2} + 3xy

K = 2x^{2} + 3xy – (x^{2 }+ xy + y^{2})

K = 2x^{2} + 3xy – x^{2 }– xy – y^{2 }K = x^{2} + 2xy – y^{2}

**(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 3x ^{2} – 4y^{2} + 5xy + 20 to obtain – x^{2} – y^{2} + 6xy + 20?**

**Solution:**

Let P be taken away from 3x^{2} – 4y^{2} + 5xy + 20 to obtain – x^{2} – y^{2} + 6xy + 20

⇒ 3x^{2} – 4y^{2} + 5xy + 20 – P = – x^{2} – y^{2} + 6xy + 20

3x^{2} – 4y^{2} + 5xy + 20 = – x^{2} – y^{2} + 6xy + 20 + P

P = 3x^{2} – 4y^{2} + 5xy + 20 –(– x^{2} – y^{2} + 6xy + 20)

P = 3x^{2} – 4y^{2} + 5xy + 20 + x^{2} + y^{2} – 6xy – 20

P = 4x^{2} – 3y^{2} – 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 + 2x ^{2}, subtract the sum of 3x^{2} – 5x and –x^{2} + 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 + 2x^{2})] – [(3x^{2} – 5x) + (–x^{2} + 2x + 5)]

Simplifying the expression

= (2x^{2} – x + 9) – (2x^{2} – 3x + 5)

= 2x^{2} – x + 9 – 2x^{2} + 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. 3m ^{2} – 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.** 3m^{2} – 2m – 7

= 3 × 2^{2} – 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. -3p ^{2} + 4p + 7**

**iii. -2p**

^{3}– 3p^{2}+ 4p + 7**i.** 4p + 7

= 4 × (-2) + 7

= -8 + 7

= -1

**ii.** -3p^{2} + 4p + 7

= -3 × (-2)^{2} + 4 × (-2) + 7

= -12 – 8 + 7

= -13

**iii.** -2p^{3} – 3p^{2} + 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. x ^{2} + 2x + 1**

**iv. 2x**

^{2}– x – 2**i.** 2x – 7

= 2 × (-1) – 7

= -2 – 7

= -9

**ii.** -x + 2

= -(-1) + 2

= 1 + 2

= 3

**iii.** x^{2} + 2x + 1

= (-1)^{2} + 2 × (-1) + 1

= 1 – 2 + 1

= 0

**iv.** 2x^{2} – x – 2

= 2 × (-1)^{2} – (-1) – 2

= 2 + 1 – 2

= 1

**4. If a = 2, b = -2, find the value of:**

**i. a ^{2} + b^{2 }ii. a^{2} + ab + b^{2 }iii. a^{2} – b^{2}**

**i.** a^{2} + b^{2}

= 2^{2} + (-2)^{2}

= 4 + 4

= 8

**ii.** a^{2} + ab + b^{2}

= 2^{2} + 2 × (-2) + (-2)^{2}

= 4 – 4 + 4

= 4

**iii.** a^{2} – b^{2}

= 2^{2} – (-2)^{2}

= 4 – 4

= 0

**5. When a = 0, b = – 1, find the value of the given expressions:**

**i. 2a + 2b**

**ii. 2a ^{2} + b^{2} + 1**

**iii. 2a**

^{2}b + 2ab^{2}+ ab**iv. a**

^{2}+ ab + 2**i.** 2a + 2b

= 2 × 0 + 2 × (-1)

= 0 – 2

= -2

**ii.** 2a^{2} + b^{2} + 1

= 2 × (0)^{2} + (-1)^{2} + 1

= 0 + 1 + 1

= 2

**iii.** 2a^{2}b + 2ab^{2} + ab

= 2 × (0)^{2 }× (-1) + 2 × (0) × (-1)^{2} + 0 × (-1)

= 0 + 0 + 0

= 0

**iv.** a^{2} + 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 z ^{3} – 3 (z – 10)**

**(ii) If p = -10, find the value of p**

^{2}– 2p – 100**i.** z^{3} – 3 (z – 10)

= 10^{3} – 3 (10 – 10)

= 1000 – 0

= 1000

**ii.** p^{2} – 2p – 100

= (-10)^{2} – 2 (-10) – 100

= 100 + 20 – 100

= 20

**9. What should be the value of a if the value of 2x ^{2} + x – a equals to 5, when x = 0?**

**Solution:**

Value of the expression at x = 0

2x^{2} + 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(a ^{2} + ab) + 3 – ab**

**Solution:**

2 (a^{2} + ab) + 3 – ab

= 2a^{2} + 2ab + 3 – ab

= 2a^{2} + ab + 3

Value of the expression when a = 5 and b = -3

2 (a^{2} + ab) + 3 – ab

= 2a^{2} + ab + 3

= 2 × 5^{2} + 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.**

**Download NCERT Solutions for Class 7 Maths Chapter 12 – Algebraic Expressions**