These Class 7 Maths Extra Questions Chapter 10 Algebraic Expressions will help students prepare well for the exams.

## Class 7 Maths Chapter 10 Extra Questions Algebraic Expressions

### Class 7 Maths Algebraic Expressions Extra Questions – 2 Marks

Question 1.

Write any 4 like terms to -9x^{2}y.

Solution:

4 like terms to – 9x^{2}y are

\(\frac { -1 }{ 2 }\)x^{2}y, 7x^{2}y, \(\frac { 3 }{ 4 }\)x^{2}y, 9x^{2}y, etc.

Question 2.

Find the term of the following expression.

2ab^{2} + 9 – 7pq + r

Solution:

Express the following as algebraic expression using variable, constants and arthematic operation.

Question 3.

i) 7 is subtracted from 3 times of r.

Solution:

3r – 7

ii) 9 times of product of x and y^{2}

Solution:

9xy^{2}

Question 4.

i) Twice of x is subtracted from \(\frac { 1 }{ 2 }\)

Solution:

\(\frac { 1 }{ 2 }\) – 2x

ii) Number a and b are added and \(\frac { 1 }{ 3 }\) added.

Solution:

a + b + \(\frac { 1 }{ 3 }\)

Question 5.

i) Sum of x^{2} and y^{2} is subtracted from z^{2}.

Solution:

z^{2} – (x^{2} + y^{2}) = z^{2} – x^{2} – y^{2}

ii) One fifth of product of p and q^{2} is added to 7.

Solution:

\(\frac{\mathrm{pq}^2}{5}\) + 7

Question 6.

i) Number 7 is added to three times the product of p and q.

Solution:

3pg + 7

ii) The difference of I and m is divided by 7.

Solution:

\(\frac{l-\mathrm{m}}{7}\)

Question 7.

3 – p

Solution:

Question 8.

p – p^{2}

Solution:

Question 9.

15xy^{2} + 7x

Solution:

Question 10.

9ab =l^{2}m^{2}

Solution:

Question 11.

Write any two examples of monomial.

Solution:

Examples of monomial, contain only one term are 7px, \(\frac { -9 }{ 2 }\)xy^{2}. etc.

Question 12.

Define Bionomial and given an example.

Solution:

Binomial : An expression containing only two term is called a Binomial.

Ex: a + b^{2}

Question 13.

How many terms are there is any trinomial ? Given one example.

Solution:

There will be 3 terms in a trinomial

Ex: a – b + c^{2}

Question 14.

7xy – 9p^{2}

Solution:

Question 15.

a^{2} – 7b^{2} + 6ab

Solution:

### Algebraic Expressions Extra Questions Class 7 – 3 Marks

Question 16.

3xy, x + y + \(\frac { p }{ 2 }\), 9 + ab^{2}

Solution:

Question 17.

a – b – \(\frac { c }{ 12 }\), ab^{2}c^{2}d^{2}, \(\frac { a }{ 2 }\) + \(\frac { b }{ 3 }\)

Solution:

Question 18.

7, 9 + p^{2} – q^{2}, 7 – xyz

Solution:

Question 19.

i) 2,200

Solution:

Like terms

ii) -7x, \(\frac { 5x }{ 2 }\)

Solution:

like terms

iii) 8a, 3b^{2}

Solution:

unlike terms

Question 20.

i) -11x^{2}, 10y^{2}

Solution:

unlike terms

ii) -ab, -7ab

Solution:

Like terms

iii) -7.5x, 0.5 x

Solution:

Like terms

Question 21.

i) 9 abc, 11 b^{2}c

Solution:

unlike terms

ii) \(\frac { 1 }{ 2 }\)x^{2}y, \(\frac { -3 }{ 211 }\)x^{2}y, yx^{2}

Solution:

Like terms

Question 22.

a = 2,b = -1

i) a^{2} – 2b

ii) a – 2b^{2}

iii) 2a^{2} – 2b^{2}

Solution:

Given a = 2, b = -1

i) a^{2} – 2b = 22 – 2(-1) = 4 + 2 = 6

ii) a – 2b^{2} = 2 – 2(-1)^{2} = 2 – 2 = 0

iii) 2a^{2} – 2b^{2} = 2(2)^{2} – 2(-1)^{2}

= 2 × 4 – 2 = 8 – 2 = 6

Question 23.

a = -1, b = -2 , c = 1

i) a – 2b + c

ii) 2a + 3b – 4c

Solution:

Given a = -1,b = -2, c = 1

i) a – 2b + c = -1 -2(-2) + 1 = -1 + 4 + 1 = 3 + 1 = 4

ii) 2a + 3b -4c = 2(-1) + 3(-2) – 4(1) = – 2 – 6 – 4 = – 12

Question 24.

x = \(\frac { 1 }{ 2 }\), y = \(\frac { 1 }{ 3 }\)

i) 2x + 3y

ii) 2x – 3y

iii) x + y

Solution:

Given x = \(\frac { 1 }{ 2 }\), y = \(\frac { 1 }{ 3 }\)

i) 2x + 3y = \(2\left(\frac{1}{2}\right)\) + \(3\left(\frac{1}{3}\right)\) = 1 + 1 = 2

ii) 2x – 3y = \(2\left(\frac{1}{2}\right)\) – \(3\left(\frac{1}{3}\right)\) = 1 – 1 = 0

iii) x + y = \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 3 }\) = \(\frac { 3+2 }{ 6 }\) = \(\frac { 5 }{ 6 }\)

Question 25.

2(x + 1) – (x – 3) + 7

Solution:

Given expression 2(x + 1) – (x – 3) + 7

= 2x + 2 – x + 3 + 7

= 2x – x + 2 + 3 + 7 = x + 12

Question 26.

9 – (3 – (2 – x)) + 7(x – 1)

Solution:

Given expression

9 – [3 – (2 – x)] + 7(x – 1)

= 9 – [3 – 2 + x] + 7x – 7

= 9 – [1 + x] + 7x – 7

= 9 – 1 – x + 7x – 7

= 9 – 1 – 7 – x + 7x

= 9 – 8 + 6x = 1 + 6x

Question 27.

Find the value of 5n^{2} – \(\frac { 3 }{ 2 }\)n + 1 if n = 0.

Solution:

n = 0

5n^{2} – \(\frac { 3 }{ 2 }\)n + 1 = 5(0)^{2} – \(\frac { 3 }{ 2 }\)(0) + 1 = 0 + 1 = 1

Question 28.

Find the value of (5 -x)^{2} + (8 + x)^{2} if x = -1.

Solution:

x = -1

(5 -x)^{2} + (8 + x)^{2}

= [5 -(-1)]^{2} + (8 – 1)^{2} = (6)^{2} + 7^{2} = 36 + 49 = 85

### Extra Questions of Algebraic Expressions Class 7 – 5 Marks

Question 29.

Write an example of the following.

i) a monomial

ii) a binomial

iii) a trinomlal

iv) a constant

v) a polynomial

Solution:

i) 2xy is a monomial

ii) a+ b^{2} is a binomial

iii) p^{2} – q + 2r is a trinomial

iv) 2024 is a constant

v) 3x^{2} – 3 + x^{3} + 7x is a polynomial

Question 30.

In the figure given below, find the perimeter of the following rectangle and find It at x = 4, y = 1.

Solution:

In the given rectangle

l = 2x + y; b = x + y

Perimeter of rectangle = 2(1 + b)

= 2(2x + y + x + y)

= 2(3x + 2y) = 6x + 4y

Given x = 4, y = 1

6x + 4y = 6(4) + 4(1) = 24 + 4 = 28

Question 31.

If m = -2, find the values of the following.

i) m^{2} – 2

ii) m^{3} – 8

iii) 2m^{2} + 1

iv) 6m – \(\frac { 3 }{ 2 }\)

v) m + 2

Solution:

m = -2

i) m^{2} – 2 = (-2)^{2} – 2 = 4 – 2 = 2

ii) m^{3} – 8= (-2)^{3} – 8 = – 8 – 8 = – 16

iii) 2m^{2} + 1 = 2(-2)^{2} + 1 = 2 × 4 + 1 = 8 + 1 = 9

iv) 6m – \(\frac { 3 }{ 2 }\) = 6(2) – \(\frac { 3 }{ 2 }\) = -12 – \(\frac { 3 }{ 2 }\) = \(\frac { -24-3 }{ 2 }\) = \(\frac { -27}{ 2 }\)

v) m + 2 = -2 + 2 = 0

Question 32.

Simplify the following expression and find the value of y is equal to -1 .

3y – 7 + 9(y – 1) + 3

Solution:

3y – 7 + 9(y – 1) + 3

= 3y – 7 + 9y – 9 + 3

= 3y + 9y – 7 – 9 + 3

= 12y – 16 + 3 = 12y – 13

Given y = -1

12y – 13 = 12(-1) – 13 = – 12 – 13 = – 25

Question 33.

If m = 3, n = – 2, find the value of

i) m^{n} + n^{m}

ii) 2mn – n

iii) 1 -4mn

iv) 8m – 4n

Solution:

Given m = 3, n = – 2

i) m^{n} + n^{m} = 3^{2} + (-2)^{3} = 9 – 8 = 1.

ii) 2mn – n = 2(3)(-2) – (-2) = – 12 + 2 = – 10

iii) 1 – 4mn = 1 – 4(3) (-2) = 1 + 24 = 25

iv) 8m – 4n = 8(3) – 4(2) = 24 + 8 = 32

Question 34.

If x = -1, find the following values

i) x^{3} – 1

ii) 2x^{2} – x + 1

iii) x^{2024} – 3

iv) \(\frac { x }{ 2 }\) + \(\frac { 1 }{ 2 }\)

v) 2x – 3

Solution:

Given x = -1

i) x^{3} – 1 = (-1)^{3} – 1 = – 1 – 1 = – 2

ii) 2x^{2} – x + 1 = 2(-1)^{2} – (- 1) + 1 = 2 + 1 + 1 = 4

iii) x^{2024} – 3 = (- 1)^{2024} – 3 = 1 – 3 = – 2

iv) \(\frac { x }{ 2 }\) + \(\frac { 1 }{ 2 }\) = \(\frac { -1 }{ 2 }\) + \(\frac { 1 }{ 2 }\) = 0

v) 2x – 3 = 2(-1) – 3 = – 2 – 3 = – 5

Question 35.

If p = \(\frac { -1 }{ 2 }\), q = \(\frac { 1 }{ 2 }\) find the following.

i) p^{2} – 1

ii) q^{2} – 1

iii) p + q

iv) pq – 1

v) \(\frac { p }{ q }\)

Solution:

Given p = \(\frac { -1 }{ 2 }\), q = \(\frac { 1 }{ 2 }\)

i) p^{2} – 1 = \(\left(\frac{-1}{2}\right)^2\) – 1 = \(\frac { 1 }{ 4 }\) – 1 = \(\frac { 1 -4 }{ 4 }\) = \(\frac { -3 }{ 4 }\)

ii) q^{2} – 1 = \(\left(\frac{1}{2}\right)^2\) – 1 = \(\frac { 1 }{ 4 }\) – 1 = \(\frac { 1 -4 }{ 4 }\) = \(\frac { -3 }{ 4 }\)

iii) p + q = \(\frac { -1 }{ 2 }\) + \(\frac { 1 }{ 2 }\) = 0

iv) pq – 1 = \(\frac { -1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) – 1 = \(\frac { -1 }{ 4 }\) – \(\frac { 1 }{ 1 }\) = \(\frac { -1-4 }{ 4 }\) × \(\frac { -5 }{ 4 }\)

Question 36.

Answer the following questions.

i) How many terms a trinomial contain?

ii) How many maximum number of terms In a binomial contain?

iii) How many variables are there In a constant?

iv) All polynomials are monomials? is it true Justify.

i) 3

ii) 2

iii) 0

iv No, it need not be.

3x^{2} + 7x – 1 is a polynomial but not a monomial.

Question 37.

What should be added to xy + yz + zx to get – xy – yz – zx.

Solution:

Let k is to be added.

k + (xy + yz + zx) = – xy – yz – zx

k = – xy – yz – zx – xy – yz – zx

k = – 2xy – 2yz – 2zx

Question 38.

Simplify – a – [b – (a – 2b) – a + 3a]

Solution:

– a -[b – (a – 2b) – a + 3a]

= – a -[b – a + 2b + 2a]

= – a – [3b + a]

= – a – 3b – a = – 2a – 3b

Question 39.

The length of a rectangle is 2a – 3b + c and its breadth is a – 2b – c, then find its perimeter.

Solution:

In a rectangle

l = 2a – 3b + c; b = a – 2b – c

Perimeter = 2(l + b)

= 2[2a – 3b + c + a – 2b – c]

= 2[2a + a – 3b – 2b + c – c ]

= 2[3a – 5b] = 6a – 10b

Question 40.

From the sum of x + y – 2z and 2x – 2y + 3 subtract the sum of 3x – y + z and – x – 3y – 2z.

Solution: