RBSE Solutions for Class 8 Maths Chapter 11 Linear Equations with One Variable Ex 11.2 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 Exercise 11.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Linear Equations with One Variable |

Exercise |
Exercise 11.2 |

Number of Questions |
10 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable Ex 11.1

Question 1.

The numerator of a rational number is less than its denominator by 3. If 5 is added to both i.e., its numerator and its denominator then it becomes \(\frac { 3 }{ 4 }\). Find the numbers.

Solution:

Let the denominator of rational number be x.

Then, according to(RBSESolutions.com)question, numerator is = x – 3

Hence, the rational number = \(\frac { x-3 }{ x }\).

Now, 5 is added to numerator and denominator, we get (according to question)

⇒ 4(x + 2) = 3(x + 5)

⇒ 4x + 8 = 3x + 15

⇒ 4x – 3x = 15 – 8

⇒ x = 7

Hence, denominator = 7

and numerator = 7 – 3 = 4

∴ Required rational number = \(\frac { 4 }{ 7 }\)

Question 2.

What should be added in numerator and denominator of fraction \(\frac { 5 }{ 13 }\) so that the fraction become \(\frac { 3 }{ 5 }\) ?

Solution:

Let the required number be x.

Then, according(RBSESolutions.com)to question,

\(\frac { 5+x }{ 13+x } =\frac { 3 }{ 5 } \)

⇒ 5(5 + x) = 3(13 + x)

⇒ 25 + 5x = 39 + 3x

⇒ 5x – 3x = 39 – 25

⇒ 2x = 14

\(x=\frac { 14 }{ 2 }\)

⇒ x = 7

Hence, the required number be 7.

Question 3.

What should be subtracted from numerator and denominator of fraction \(\frac { 15 }{ 19 }\) so that the fraction becomes \(\frac { 5 }{ 7 }\) ?

Solution:

Let the required number be x.

Then, according(RBSESolutions.com)to question,

\(\frac { 15-x }{ 19-x } =\frac { 5 }{ 7 } \)

⇒ 7(15 – x) = 5(19 – x)

⇒ 105 – 7x = 95 – 5x

⇒ – 7x + 5x = 95 – 105

⇒ – 2x = – 10

\(x=\frac { -10 }{ -2 }\)

⇒ x = 5

Hence, the required number is 5.

Question 4.

Ramesh distributed his capital, half of the Capital to his wife, one third to his son and remaining 50,000/- to his daughter. Find the total amount of his capital.

Solution:

Let x be the total money.

Then, according to question,

Money given to wife = \(\frac { x }{ 2 }\)

Money(RBSESolutions.com)given to son = \(\frac { x }{ 3 }\)

Money given to daughter = Rs 50,000

∴From question,

\(\frac { x }{ 2 } +\frac { x }{ 3 } +50,000=x\)

⇒ 3x + 2x + 50,000 x 6 = 6x [Multiplying by 6]

⇒ 5x + 3,00,000 = 6x

⇒ 3,00,000 = 6x – 5x

⇒ 3,00,000 = x

⇒ x = 3,00,000

Hence, the required money is Rs 3,00,000.

Question 5.

5 times of any number is 48 more than its double. Find the number.

Solution:

Let the required number be x.

Then, according to question,

Five times of number = 5x

Twice the number = 2x

Then,(RBSESolutions.com)according to question,

⇒ 5x = 2x + 48

⇒ 5x – 2x = 48

⇒ 3x = 48

⇒ \(x=\frac { 48 }{ 3 }\)

⇒ x = 16

Hence, the required number is 16.

Question 6.

Distribute 45 in this way that one part is 7 less than three times of another part.

Solution:

Let one part be x.

Then according to question second part = 45 – x.

Then,(RBSESolutions.com)according to question x = 3(45 – x) – 7

⇒ x = 135 – 3x – 7

⇒ x = – 3x + 135 – 7

⇒ x = – 3x + 128

⇒ x + 3x = 128

⇒ 4x = 128

⇒ \(x=\frac { 128 }{ 4 }\)

⇒ x = 32

Hence, one part = 32

Hence, second part = 45 – 32 = 13

Hence, the required part are 32 and 13.

Question 7.

Age of Ranu is three times of Sujal’s age. After 4 years, sum of their age will be 40 years. Find their present age.

Solution:

Let x years be the age of Sujal.

Then 3x will be the age of Ranu.

After 4 years,

Age of Sujal = (x + 4) years

Age of Ranu = (3x + 4) years

According to question,

(x + 4) + (3x + 4) = 40

⇒ x+3x + 4 + 4 = 40

⇒ 4x + 8 = 40

⇒ 4x = 40 – 8

⇒ 4x = 32

⇒ x = \(\frac { 32 }{ 8 }\)

⇒ x = 8

Hence the present age of Sujal = 8 years and the present age of Ranu = 3 x 8 years

= 24 years.

Question 8.

Length of a rectangle exceeds its breadth by 6 meter. If its perimeter is 64 meter then find its length and breadth.

Solution:

Let x meter be the breadth of rectangle.

Then, length(RBSESolutions.com)of rectangle = (x + 6) meter

∴Perimeter = 2 (Length + Breadth)

= 2 {x + (x + 6)}

= 2 (2x + 6) meter

According to question,

2(2x + 6) = 64

⇒ 2x + 6 = \(\frac { 64 }{ 2 }\)

⇒ 2x + 6 = 32

⇒ 2x = 32 – 6

⇒ 2x = 26

⇒ \(x=\frac { 26 }{ 2 }\)

⇒ x = 13

Hence, the breadth of rectangle = 13 meter length of rectangle = (13 + 6) meter = 19 meter

Question 9.

Sum of the digits of a two digit number is 12. New number formed by reversing the digit is greater than the original number by 54. Find the original number.

Solution:

Let one’s digit be x.

Since the sum(RBSESolutions.com)of digits is 12.

Therefore, ten’s digit = (12 – x) ….(i)

∴Number = 10 x ten’s digit + One’s digit

= 10 (12 – x) + x

= 120 – 10x + x

= 120 – 9x

Now, if digits are reversed, then

One’s digit = 12 – x

and ten’s digit = x

∴ New number = 10 x ten’s digit + one’s digit

= 10(x) + 12 – x

= 10x + 12 – x

= 10x – x + 12

= 9x + 12

According to question,

⇒ 9x + 12 = (120 – 9x) + 54

⇒ 9x + 12 = 120 + 54 – 9x

⇒ 9x + 12 = 174 – 9x

⇒ 9x + 9x = 174 – 12

⇒ 18x = 162

⇒ \(x=\frac { 162 }{ 18 }\)

⇒ x = 9

One’s digit = 9

and ten’s digit = 12 – 9 = 3

Hence, the required number is = 39.

Question 10.

In two-digit number, first digit is four times of second digit. Adding this to the new number formed by reversing the digits, 110 is obtained. Find the numbers.

Solution:

Let the one’s digit be x.

∴ Ten’s digit is 4x.

∴ Number = 10 x Ten’s digit + One’s digit

= 10 x (4x) + x

= 40x + x

= 41x

Now the digits are reversed, one’s digit = 4x, and ten’s digit = x

∴ Number = 10 x ten’s digit + one’s digit

= 10 x x + 4x

= 10x + 4x

= 14x

According to question

41x + 14x = 110

⇒ 55x = 110

\(x=\frac { 110 }{ 55 }\)

⇒ x = 2

Hence, One’s digit = 2

and Ten’s digit =4 x 2 = 8

Number = 82

Again, if one’s digit is 4 times of ten’s digit, then number = 28

Hence, the required(RBSESolutions.com)number is 82 or 28.

We hope the given RBSE Solutions for Class 8 Maths Chapter 11 Linear Equations with One Variable Ex 11.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable Exercise 11.2, drop a comment below and we will get back to you at the earliest.

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