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

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

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

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)

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 ___

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.

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)

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:

Question 4.
Find the sum of (2x + 7), (4x – 2) land (6x + 4).
Solution:

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

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
\(\frac { 1 }{ 8 }xyz\)

VI. Short Answer Type Questions

Question 1.
Simplify

Solution

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

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

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

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

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

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

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