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

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.