RBSE Solutions for Class 8 Maths Chapter 11 Linear Equations with One Variable Additional Questions is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Linear Equations with One Variable |

Exercise |
Additional Questions |

Number of Questions |
35 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable Additional Questions

**I. Objective Type Questions**

Question 1.

The necessary condition for a linear equation is

(a) highest power of variable is 1.

(b) highest power of variable is 2.

(c) highest power of variable is 3.

(d) None of the above.

Question 2.

On transposing the term with sign changed

(a) in ‘×’

(b) in ‘÷‘

(c) in ‘-‘

(d) remains same

Question 3.

The value of x in 3x = 21 is

(a) 7

(b) 24

(c) 18

(d) 63

Question 4.

The solution(RBSESolutions.com)of a linear equation may be

(a) any natural number

(b) any rational number

(c) any real number

(d) any whole number

Question 5.

The solution of x – 4 = 5 is

(a)4×5

(b)5×4

(c)5-4

(d)5+4

Question 6.

In the following the linear equation is

(a) \(\frac { x }{ 4 } =\frac { 4 }{ x } \)

(b) \(\frac { 1 }{ x } +\frac { 1 }{ x-1 } =1\)

(c) \(\frac { x }{ 2 } +\frac { x }{ 3 } =\frac { 1 }{ 4 } \)

(d) x² + 2x + 3 = 0

Question 7.

The degree of (x – 1)² = x² – 3 is

(a) 1

(b) 2

(c) 0

(d) 3

Question 8.

The solution of \(\frac { 5 }{ x }=2\) is

(a) 10

(b) \(\frac { 2 }{ 5 }\)

(c) \(\frac { 5 }{ 2 }\)

(d) \(\frac { 1 }{ 10 }\)

Answers

1. (a)

2. (c)

3. (a)

4. (c)

5. (d)

6. (c)

7. (a)

8. (c)

**II. Fill in the blanks**

Question 1.

Sign___is always used in equations.

Question 2.

Variables can be___from one side to another side as numbers.

Question 3.

An equation involving only___polynomials is called a linear equation.

Question 4.

A value of the(RBSESolutions.com)variable which makes the equation a true statement, is called a___or a___of the equation.

Answers

1. ‘=’

2. transposed

3. linear

4. solution, root.

**III. True/False Type Questions**

Question 1.

The root of z ÷ 4 = – 8 is 32.

Question 2.

The root of 3x = \(\frac { 20 }{ 7 }\) – x is \(\frac { 5 }{ 7 }\)

Question 3.

The root of 2x + 3 = 2(x – 4) does not exist.

Question 4.

The largest number is 3 consecutive numbers is x + 1 and smallest is x.

Answers

1. False

2. True

3. True

4. False.

**IV. Matching Type Questions**

Part 1 |
Part 2 |

1. One’s digit is 1 and ten’s digit is 2 | (a) 0 |

2. Degree of linear equation | (b) \(\frac { 3 }{ 5 }\) |

3. The maximum solutions of a linear equation | (c) 21 |

4. If numerator is 3 and denominator is 5 then the fraction is | (d) 1 |

Answers

1. ⇔ (c)

2. ⇔ (d)

3. ⇔ (a)

4. ⇔ (b)

**V. Very Short Answer Type Questions**

Question 1.

What do you mean by LHS in algebraic equations?

Answer:

LHS means left hand(RBSESolutions.com)side expression in a equation.

Question 2.

Which type of equations can be solved by linear equations?

Answer:

Age problems, number(RBSESolutions.com)problems, Area problems etc. can be solved by linear equations.

Question 3.

Solve 2x – 3 = 7.

Solution

2x – 3 = 7

⇒ 2x = 7 + 3

⇒ 2x = 10

⇒ \(x=\frac { 10 }{ 2 }=5\)

Question 4.

Solve

2x – 3 = x + 2

Solution:

2x – 3 = x + 2

⇒ 2x – x = 2 + 3

⇒ x = 5

Question 5.

Solve the equation \(\frac { x }{ 3 } +1=\frac { 7 }{ 15 } \)

Solution:

\(\frac { x }{ 3 } +1=\frac { 7 }{ 15 } \)

Question 6.

Check the equation \(\frac { 15 }{ 4 }-7x=9\) at \(x=-\frac { 3 }{ 4 }\)

Solution:

LHS = \(\frac { 15 }{ 4 }-7x\)

Hence solution is right

**VI. Short Answer Type Questions**

Question 1.

Solve the equation 4x – [2 + {x – (3 – x)}] = 3x + 6

Solution:

4x – [2 + {x – (3 – x)}] = 3x + 6

⇒ 4x – [2 + (x – 3 + x)] = 3x + 6

⇒ 4x – [2 + 2x – 3] = 3x + 6

⇒ 4x – (2x – 1) = 3x + 6

⇒ 4x – 2x + 1 = 3x + 6

⇒ 2x + 1 = 3x + 6

⇒ 2x – 3x = 6 – 1

⇒ – x = 5

⇒ x = – 5

Question 2.

Solve the equation

√3x – 2 = 2√3 + 4

Solution:

⇒ √3x – 2 = 2√3 + 4

⇒ √3x = 2 + 2√3 + 4

⇒ √3x = 2√3 + 6

⇒ √3x = 2√3(1+√3)

⇒ \(x=\frac { 2\sqrt { 3 } (1+\sqrt { 3 } ) }{ \sqrt { 3 } } \)

⇒ x = 2(1 + √3)

Question 3.

Solve the equation \(\frac { 8x+3 }{ 2x-4 } =\frac { 4x }{ x-5 } \)

Solution:

⇒ \(\frac { 8x+3 }{ 2x-4 } =\frac { 4x }{ x-5 } \)

⇒ (8x + 3) (x – 5) = 4x (2x – x)

⇒ 8x² – 40x + 3x – 15 = 8x² – 16x

⇒ – 40x + 3x + 16x = 15

⇒ – 21x = 15

⇒ \(x=-\frac { 15 }{ 21 }\)

⇒ \(x=-\frac { 5 }{ 7 }\)

Question 4.

A bag contains 15 coins of two-rupee and five-rupee. The total amount of the coins is Rs 45. How many coins of each find are there in the bag.

Solution:

Let there may be x two-rupee coins in the bag.

Then No. of 5-rupee coins = 15 – x

Value of x two rupee coins = Rs 2x

Value of 5-rupee(RBSESolutions.com)coins = Rs 5(15 – x)

According to question

⇒ 2x + 5 (15 – x) = 45

⇒ 2x + 75 – 5x = 45

⇒ 2x – 5x + 75 = 45

⇒ – 3x + 75 = 45

⇒ – 3x = 45 – 75

⇒ – 3x = – 30

\(x=\frac { -30 }{ -3 }\)

⇒ x = 10

Hence, No. of 2-rupee coins = 10

and No. of 5-rupee coins = 15 – 10 = 5

Question 5.

Find the three consecutive numbers where the sum of twice of first number, thrice of second number and four times of third number is 182.

Solution:

Let three consecutive numbers be x, x + 1 and x + 2.

Then according to question 2x + 3(x + 1) + 4 (x + 2) = 182

⇒ 2x + 3x + 3 + 4x + 8 – 182

⇒ 9x + 11 = 182

⇒ 9x = 182 – 11

⇒ 9x = 171

⇒ x = \(\frac { 171 }{ 9 }\) = 19

Hence, the first number = 19

second number = 19 + 1 = 20

and third number = 19 + 2 = 21

Question 6.

Ramesh’s father is 27 years older than Ramesh. After 5 years, the ratio of ages of Ramesh and his father would be 2 : 3. Find their present age.

Solution:

Let age of Ramesh = x years

Then father’s age = (x + 27) years

After 5 years from now, age of Ramesh = (x + 5) years

After 5 years(RBSESolutions.com)from now, age of father = x+ 27 + 5 = (x+ 32) years

According to question, \(\frac { x+5 }{ x+32 } =\frac { 2 }{ 3 } \)

or 3(x + 5) = 2(x + 32)

or 3x + 15 = 2x + 64

or 3x – 2x = 64 – 15

or x = 49

Ramesh’s age x = 49 years

Father’s age x + 27 = 49 + 27 = 76 years

Question 7.

Solve

\(\frac { { 6x }^{ 2 }+13x-4 }{ 2x+5 } =\frac { { 12x }^{ 2 }+5x-2 }{ 4x+3 } \)

Solution:

Given that

\(\frac { { 6x }^{ 2 }+13x-4 }{ 2x+5 } =\frac { { 12x }^{ 2 }+5x-2 }{ 4x+3 } \)

⇒ (6x² + 13x – 4) (4x + 3) = (12x² + 5x – 2) (2x + 5) (Cross multiplication)

⇒ (6x² + 13x – 4) × 4x + (6x² + 13x – 4) × 3 = (12x² + 5x – 2) × 2x + (12x² + 5x – 2) × 5

⇒ 24x³ + 52x² – 16x + 18x² + 39x – 12 = 24x³ + 10x² – 4x + 60x² + 25x – 10

⇒ 24x³ + 70x² + 23x – 12 = 24x³ + 70x² + 21x – 10

⇒ 24x³ + 70x² + 23x – 24x³ – 70x² – 21x = – 10 + 12

⇒ 2x = 2

⇒ x = 1

Hence, x = 1 is the required solution of given equation.

Question 8.

Denominator of a rational number exceeds its numerator by 5. If 2 is added to its numerator and denominator then we get 1/2. Find the rational number.

Solution:

Let the numerator be x.

According to question, denominator of a rational number exceeds its numerator by 5,

hence the rational(RBSESolutions.com)number is \(\frac { x }{ x+5 }\)

Now by adding 2 to both numerator and denominator.

\(\frac { x+2 }{ x+5+2 }\) = \(\frac { 1 }{ 2 }\)

⇒ 2(x + 2) = x + 7

⇒ 2x + 4 = x + 7

⇒ 2x – x = 7 – 4

⇒ x = 3

Hence, Numerator = 3

Denominator = 3 + 5 = 8

Required rational number = \(\frac { 3 }{ 8 }\)

Question 9.

Length of a rectangle exceeds its breadth by 3 meter. If its perimeter is 54 meter then find its length and breadth.

Solution:

Let the breadth of rectangle be x metre.

∴ Length of rectangle = (3 + x) metre

∵ Perimeter = 2 (L + B)

∴ Perimeter = 2{(3 + x) + x}

= 2(2x + 3) m

According to question,

2(2x + 3) = 54

Therefore 2x + 3 = \(\frac { 54 }{ 2 }\)

2x + 3 = 27

2x = 27 – 3 = 24

\(x=\frac { 24 }{ 2 }=12\)

Hence the breadth(RBSESolutions.com)of rectangle =12 meter and length of rectangle = 12 + 3 = 15 meter

Question 10.

Find three consecutive multiples of 11, whose sum is 363.

Solution:

Let three consecutive(RBSESolutions.com)multiples of 11 be 11 x, 11(x + 1) and 11(x + 2)

According to question,

11 x + 11(x + 1) + 11(x + 2) = 363

⇒ 11 x + 11 x + 11 + 11 x + 22 = 363

33 x = 363 – 33

⇒ 33 x = 330

\(x=\frac { 330 }{ 33 }\)

⇒ x = 10

Three consecutive multiple numbers

11x = 11 x 10 = 110

11(x + 1) = 11(10 + 1) = 11 × 11 = 121

11(x + 2) = 11(10 + 2) = 11 × 12 = 132

Three consecutive multiple numbers are 110, 121 and 132

We hope the given RBSE Solutions for Class 8 Maths Chapter 11 Linear Equations with One Variable Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable Additional Questions, drop a comment below and we will get back to you at the earliest.

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