Algebraic Expressions Class 7 Extra Questions with Answers

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 -9x²y.
Solution:
4 like terms to – 9x²y are
\(\frac { -1 }{ 2 }\)x²y, 7x²y, \(\frac { 3 }{ 4 }\)x²y, 9x²y, etc.

Question 2.
Find the term of the following expression.
2ab² + 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 y²
Solution:
9xy²

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 x² and y² is subtracted from z².
Solution:
z² – (x² + y²) = z² – x² – y²

ii) One fifth of product of p and q² 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 – p²
Solution:

Question 9.
15xy² + 7x
Solution:

Question 10.
9ab =l²m²
Solution:

Question 11.
Write any two examples of monomial.
Solution:
Examples of monomial, contain only one term are 7px, \(\frac { -9 }{ 2 }\)xy². etc.

Question 12.
Define Bionomial and given an example.
Solution:
Binomial : An expression containing only two term is called a Binomial.
Ex: a + b²

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 + c²

Question 14.
7xy – 9p²
Solution:

Question 15.
a² – 7b² + 6ab
Solution:

Algebraic Expressions Extra Questions Class 7 – 3 Marks

Question 16.
3xy, x + y + \(\frac { p }{ 2 }\), 9 + ab²
Solution:

Question 17.
a – b – \(\frac { c }{ 12 }\), ab²c²d², \(\frac { a }{ 2 }\) + \(\frac { b }{ 3 }\)
Solution:

Question 18.
7, 9 + p² – q², 7 – xyz
Solution:

Question 19.
i) 2,200
Solution:
Like terms

ii) -7x, \(\frac { 5x }{ 2 }\)
Solution:
like terms

iii) 8a, 3b²
Solution:
unlike terms

Question 20.
i) -11x², 10y²
Solution:
unlike terms

ii) -ab, -7ab
Solution:
Like terms

iii) -7.5x, 0.5 x
Solution:
Like terms

Question 21.
i) 9 abc, 11 b²c
Solution:
unlike terms

ii) \(\frac { 1 }{ 2 }\)x²y, \(\frac { -3 }{ 211 }\)x²y, yx²
Solution:
Like terms

Question 22.
a = 2,b = -1
i) a² – 2b
ii) a – 2b²
iii) 2a² – 2b²
Solution:
Given a = 2, b = -1
i) a² – 2b = 22 – 2(-1) = 4 + 2 = 6
ii) a – 2b² = 2 – 2(-1)² = 2 – 2 = 0
iii) 2a² – 2b² = 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 5n² – \(\frac { 3 }{ 2 }\)n + 1 if n = 0.
Solution:
n = 0
5n² – \(\frac { 3 }{ 2 }\)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 + 7² = 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+ b² is a binomial
iii) p² – q + 2r is a trinomial
iv) 2024 is a constant
v) 3x² – 3 + x³ + 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) m² – 2
ii) m³ – 8
iii) 2m² + 1
iv) 6m – \(\frac { 3 }{ 2 }\)
v) m + 2
Solution:
m = -2
i) m² – 2 = (-2)² – 2 = 4 – 2 = 2
ii) m³ – 8= (-2)³ – 8 = – 8 – 8 = – 16
iii) 2m² + 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) mⁿ + n^m
ii) 2mn – n
iii) 1 -4mn
iv) 8m – 4n
Solution:
Given m = 3, n = – 2
i) mⁿ + n^m = 3² + (-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) x³ – 1
ii) 2x² – x + 1
iii) x²⁰²⁴ – 3
iv) \(\frac { x }{ 2 }\) + \(\frac { 1 }{ 2 }\)
v) 2x – 3
Solution:
Given x = -1
i) x³ – 1 = (-1)³ – 1 = – 1 – 1 = – 2
ii) 2x² – x + 1 = 2(-1)² – (- 1) + 1 = 2 + 1 + 1 = 4
iii) x²⁰²⁴ – 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) p² – 1
ii) q² – 1
iii) p + q
iv) pq – 1
v) \(\frac { p }{ q }\)
Solution:
Given p = \(\frac { -1 }{ 2 }\), q = \(\frac { 1 }{ 2 }\)
i) p² – 1 = \(\left(\frac{-1}{2}\right)^2\) – 1 = \(\frac { 1 }{ 4 }\) – 1 = \(\frac { 1 -4 }{ 4 }\) = \(\frac { -3 }{ 4 }\)

ii) q² – 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.
3x² + 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: