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) x^{3}y^{3}z^{3}

Question 2.

Coefficient of x^{5} 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 x^{4} 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

= p^{4} – q^{4}

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

We hope the given RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions, drop a comment below and we will get back to you at the earliest.

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