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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