RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions

RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 8
Subject Maths
Chapter Chapter 9
Chapter Name Algebraic Expressions
Exercise Additional Questions
Number of Questions 43
Category RBSE Solutions

Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions

I. Objective Type Questions

Question 1.
The product of xy, yz and zx is
(a) 2xyz
(b) x²y²z²
(c) xy + yz + zx
(d) x3y3z3

Question 2.
Coefficient of x5 in \(\frac { 8 }{ 5 } { x }^{ 5 }\) is
(a) \(\frac { 8 }{ 5 }\)
(b) 5
(c) 8
(d) \(\frac { 5 }{ 8 }\)

Question 3.
Coefficient of yz in \(\frac { 15 }{ 16 }\) xyz is
RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions img-1

Question 4.
The number of terms in the expression 8x² + 2xy + 3x² + 2y² + 2x² is
(a) 3
(b) 2
(c) 5
(d) 4

RBSE Solutions

Question 5.
The sum of 4xy, 3.5xy, – 2xy, 2xz is
(a) 2.5xy + 5.2xz
(b) 5.5xy + 2xz
(c) 2.5xy + 5xz
(d) 5.5xy + 2xy

Question 6.
The product of 3x (x + y) is
(a) 3x² + 3xy
(b) 3x² + y
(c) x² + 3xy
(d) 3x + 3y

Question 7.
Subtracting 2a + 3b from 5a + 2b, we get
(a) 3a + 2b
(b) 2a + 3b
(c) 3a – b
(d) 5a – 3b

Question 8.
The product of pq + qr + 2p and 0 is
(a) 0
(b) 1
(c) pqr
(d) pq + qr + rp

RBSE Solutions

Question 9.
Like term of expression 7x²y is
(a) 7xy
(b) – 10x²y
(c) 7
(d) 7x²

Answers
1. (b)
2. (a)
3. (b)
4. (a)
5. (b)
6. (a)
7. (c)
8. (a)
9. (b)

RBSE Solutions

II. Fill in the blanks

Question 1.
The coefficient of x4 in \(\frac { 2 }{ 8 } { x }^{ 4 }\) is ___

Question 2.
The solution of 8xyz – 5xyz is ___

Question 3.
The sum of 8x² + 2x and 4x + 2 is ___

Question 4.
The product(RBSESolutions.com)of 2x, 4x and \(\frac { 2 }{ 3 }x\) is ___

RBSE Solutions

Question 5.
x² + (a + b) x + ab = (x + a) ___

Answers
1. \(\frac { 2 }{ 8 }\)
2. 3xyz
3. 8x² + 6x + 2
4. \(\frac { 16{ x }^{ 3 } }{ 3 } \)
5. (x + b)

III. True/False Type Questions

Question 1.
4y – 7z is a trinomial.

Question 2.
x + \(\frac { 1 }{ x }\) is a polynomial.

Question 3.
(x + a) (x + b) = x² + (a + b) x + ab is a identity.

Question 4.
1 and 100 are homogeneous expressions.

Answers
1. False
2. False
3. True
4. True.

RBSE Solutions

IV. Matching Type Questions

Part 1 Part 2
1. 4x x 5y x 72 = (a) 140xyz
2. Coefficient of x in 1 + x + x² (b) 0
3. Sum of ab – bc, bc – ca, ca – ab (c) a² – b²
4. (a + b) (a – b) = (d) 1

Answers
1. ⇔ (a)
2. ⇔ (d)
3. ⇔ (b)
4. ⇔ (c)

RBSE Solutions

V. Very Short Answer Type Questions

Question 1.
Find the coefficient of x² in \(-\frac { 3 }{ 7 }\)x²y² .
Solution
The coefficient of x² in \(-\frac { 3 }{ 7 }\)x²y²
= \(-\frac { 3 }{ 7 }\)y²

Question 2.
Write the number of terms in 5xy + 2xz + 3xy + x² + y².
Solution:
5xy + 2xz + 3xy + x² + y²
= 5xy + 3xy + 2xz + x² + y²
= 8xy + 2xz + x² + y²
Hence the number of terms = 4

Question 3.
Find the sum \(\frac { 5 }{ 6 } x+\frac { 7 }{ 6 } x+\frac { 1 }{ 6 } x\)
Solution:
RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions img-2

RBSE Solutions

Question 4.
Find the sum of (2x + 7), (4x – 2) land (6x + 4).
Solution:
RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions img-3

Question 5.
Find the product of (x² + 2x) and (2x + 3).
Solution:
(x² + 2x) (2x + 3)
= x² (2x + 3) + 2x (2x + 3)
= 2x³ + 3x² + 4x² + 6x .
= 2x³ + 7x² + 6x

Question 6.
Find the(RBSESolutions.com)value of 3x² + 4xy + 2y² if x = 5 and y = 2.
Solution:
3x² + 4xy + 2y²
= 3 (5)² + 4(5) (2) + 2(2)²
= 75 + 40 + 8
= 123

RBSE Solutions

Question 7.
Find the product :
(3x + 5) (5x – 3)
Solution
(3x + 5) (5x – 3)
= 3x(5x – 3) + 5(5x – 3)
= (3x) (5x) + (3x) (- 3) + (5) (5x) + (5)(- 3)
= 15x² – 9x + 25x – 15
= 15x² + 16x – 15

Question 8.
If x = \(\frac { 1 }{ 2 }\),y = \(\frac { 2 }{ 3 }\) and z = \(\frac { 1 }{ 3 }\) then find the value of \(\frac { 1 }{ 8 }xyz\)
Solution
\(\frac { 1 }{ 8 }xyz\)
RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions img-4

VI. Short Answer Type Questions

Question 1.
Simplify
RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions img-5
Solution
RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions img-6

RBSE Solutions

Question 2.
If \(x+\frac { 1 }{ x }=7\), the find the value of \({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \).
Solution
\(x+\frac { 1 }{ x }=7\)
Squaring both sides
RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions img-7

RBSE Solutions

Question 3.
If 2(a² + b²) = (a + b)² then prove that a = b.
Solution
2 (a² + b²) = (a + b)²
⇒ 2a² + 2b² = a² + 2ab + b²
(Using identity I)
⇒ a² – 2ab + b² = 0
⇒ (a – b)² = 0 (Using identity II)
⇒ a – b = 0
⇒ a = b

Question 4.
What is the value of a if 2x² + x – a is equal to 5 and x = 0.
Solution
According(RBSESolutions.com)to question,
2(0)² + (0) – a = 5
⇒ – a = 5
⇒ a = – 5

RBSE Solutions

Question 5.
Simplify the expression 2(a² + ab) + 3 – ab and find its value when a = 5 and b = – 3.
Solution
2(a² + ab) + 3 – ab
= 2a² + 2ab + 3 – ab
= 2a² + ab + 3
= 2(5)² + (5) (- 3) + 3
When a = 5 and b = – 3
= 50 – 15 + 3
= 38

Question 6.
Using proper identity, find the value of (1.2)² – (0.8)².
Solution
(1.2)² – (0.8)²
= (1.2 + 0.8) (1.2 – 0.8)
(Using identity III)
= (2) (0.4)
= 0.8

RBSE Solutions

Question 7.
Using identity (x + a) (x + b) = x² + (a + b) x + ab, find the value of following – 201 x 202
Solution
201 x 202
= (200 + 1) x (200 + 2)
= (200)² + (1 + 2) (200) + 1 x 2 (Using given identity)
= 40000 + 600 + 2
= 40602

Question 8.
To get – x² – y² + 6xy + 20, what is to be subtracted from 3x² – 4y² + 5xy + 20?
Solution
Required expression
= (3x² – 4y² + 5xy + 20) – (- x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= 4x² – 3y² – xy

Question 9.
Simplify
(a + b) (a – b) – (a² + b²)
Solution
(a + b) (a – b) – (a² + b²)
= a (a – b) + b (a – b) – (a² + b²)
= a² – ab + ba – b² – a² – b²
= a² – ab + ab – b² – a² – b²
= a² – a² – ab + ab – b² – b²
= 0 – 0 – 2b²
= – 2b²

RBSE Solutions

Question 10.
Using identity a² – b² = (a + b) (a – b) find the product
(i) (2a + 7) (2a – 7)
(ii) (p² + q²) (p² – q²)
Solution
(i) (2a + 7) (2a – 7) = (2a)² – (1)²
= 2²a² – 7²
= 4a² – 49
(ii) (p² + q²) (p² – q²)
= (p²)² – (q²)²
= p4 – q4

Question 11.
Find the product using suitable identity
(i) (5x – 3) (5x + 3)
(ii) 103 x 99
Solution
(i) (5x – 3) (5x + 3)
= (5x)² – (3)² (Using identity 3)
= 25x² – 9
(ii) 103 x 99
(100 + 3) x (100 – 1)
(100 + 3) x {100 + (- 1)}
= (100)² + {3 + (-1)} 100 + (3) (-1) (Using given identity)
= 1000 + 200 – 3
= 10197

RBSE Solutions

Question 12.
Simplify
(x – 5)² + 10x.
Solution
(x – 5)² + 10x = x² – 2 × x × 5 + (5)² + 10x
[Formula : (a – b)² = a² – 2ab + b²]
= x² – 10x + 25 + 10x
= x² + 25

Question 13.
Using suitable identities, evaluate the 72 x 68.
Solution
72 x 68
= (70 + 2) (70 – 2)
= (70)² – (2)²
[Using in identity (a + b) (a – b) = a² – b²]
= 4900 – 4
= 4896

RBSE Solutions

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