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

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.
Answer: Let's say the denominator of the rational number is \( x \). This means the numerator is \( x - 3 \). So, the original rational number is \( \frac {x-3}{x} \). If we add 5 to both the numerator and the denominator, the new fraction becomes \( \frac {x-3+5}{x+5} \). We are told this new fraction is equal to \( \frac {3}{4} \). We can set up the equation to find \( x \).
\( \frac {x-3+5}{x+5} = \frac {3}{4} \)
\( \implies \frac {x+2}{x+5} = \frac {3}{4} \)
\( \implies 4(x + 2) = 3(x + 5) \) (Cross-multiplying helps solve for \( x \)).
\( \implies 4x + 8 = 3x + 15 \)
\( \implies 4x - 3x = 15 - 8 \)
\( \implies x = 7 \)
Now that we know \( x = 7 \), the denominator is 7. The numerator is \( 7 - 3 = 4 \). So, the required rational number is \( \frac {4}{7} \).
In simple words: We find a number for the denominator, and then subtract 3 to get the numerator. When we add 5 to both parts, the new fraction becomes \( \frac{3}{4} \). By solving, we find the number is \( \frac{4}{7} \).

🎯 Exam Tip: Always verify your answer by plugging the calculated number back into the original problem statement to ensure it satisfies all conditions.

Question 2. What should be added in numerator and denominator of fraction \( \frac {5}{13} \) so that the fraction become \( \frac {3}{5} \)?
Answer: Let the number that needs to be added to both the numerator and denominator be \( x \). So, if we add \( x \) to 5 and to 13, the new fraction should be \( \frac {3}{5} \).
\( \frac {5+x}{13+x} = \frac {3}{5} \)
\( \implies 5(5 + x) = 3(13 + x) \) (Cross-multiplication is a simple way to solve for \( x \)).
\( \implies 25 + 5x = 39 + 3x \)
\( \implies 5x - 3x = 39 - 25 \)
\( \implies 2x = 14 \)
\( \implies x = \frac {14}{2} \)
\( \implies x = 7 \)
Therefore, the required number is 7.
In simple words: We need to add 7 to both the top and bottom of the fraction \( \frac {5}{13} \) to change it into \( \frac {3}{5} \).

🎯 Exam Tip: When dealing with fractions in equations, cross-multiplication is usually the quickest way to remove the denominators and simplify the problem.

Question 3. What should be subtracted from numerator and denominator of fraction \( \frac {15}{19} \) so that the fraction becomes \( \frac {5}{7} \)?
Answer: Let the number that needs to be subtracted from both the numerator and denominator be \( x \). When we subtract \( x \) from 15 and from 19, the new fraction should be \( \frac {5}{7} \).
\( \frac {15-x}{19-x} = \frac {5}{7} \)
\( \implies 7(15 - x) = 5(19 - x) \) (Cross-multiplying both sides helps remove the fractions).
\( \implies 105 - 7x = 95 - 5x \)
\( \implies -7x + 5x = 95 - 105 \)
\( \implies -2x = -10 \)
\( \implies x = \frac {-10}{-2} \)
\( \implies x = 5 \)
So, the required number is 5.
In simple words: We need to take away 5 from both the top and bottom of the fraction \( \frac {15}{19} \) to make it equal to \( \frac {5}{7} \).

🎯 Exam Tip: Be careful with negative signs when rearranging terms in the equation; a common mistake is to forget to change the sign when moving a term to the other side.

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.
Answer: Let \( x \) be the total capital (money) Ramesh has. He gave half of his capital to his wife: \( \frac {x}{2} \). He gave one-third of his capital to his son: \( \frac {x}{3} \). The remaining amount, Rs 50,000, was given to his daughter. The sum of all these parts must be equal to his total capital \( x \).
\( \frac {x}{2} + \frac {x}{3} + 50000 = x \)
To solve this, we can multiply the entire equation by the least common multiple of the denominators (2 and 3), which is 6. This clears the fractions.
\( \implies 6 \left( \frac {x}{2} \right) + 6 \left( \frac {x}{3} \right) + 6(50000) = 6x \)
\( \implies 3x + 2x + 300000 = 6x \)
\( \implies 5x + 300000 = 6x \)
\( \implies 300000 = 6x - 5x \)
\( \implies x = 300000 \)
Therefore, Ramesh's total capital was Rs 3,00,000.
In simple words: Ramesh split his money. His wife got half, his son got one-third, and his daughter got Rs 50,000. By adding these parts, we find his total money was Rs 3,00,000.

🎯 Exam Tip: When working with fractions in an equation, multiplying by the least common multiple of the denominators is a great strategy to simplify the equation quickly.

Question 5. 5 times of any number is 48 more than its double. Find the number.
Answer: Let the required number be \( x \). Five times the number means \( 5x \). Twice the number (its double) means \( 2x \). The problem states that 5 times the number is 48 *more than* its double. This means if you take the double of the number and add 48, you get five times the number.
\( 5x = 2x + 48 \)
Now, we solve for \( x \).
\( \implies 5x - 2x = 48 \)
\( \implies 3x = 48 \)
\( \implies x = \frac {48}{3} \)
\( \implies x = 16 \)
So, the required number is 16.
In simple words: We found a number where if you multiply it by 5, the answer is 48 more than if you multiply it by 2. That number is 16.

🎯 Exam Tip: Translate word problems into algebraic equations carefully; phrases like "is more than" or "is less than" often indicate addition or subtraction in the equation.

Question 6. Distribute 45 in this way that one part is 7 less than three times of another part.
Answer: Let one part of the number 45 be \( x \). Since 45 is distributed into two parts, the second part will be \( 45 - x \). The problem says one part (\( x \)) is 7 less than three times the other part (\( 45 - x \)).
\( x = 3(45 - x) - 7 \)
Now, let's solve this equation for \( x \).
\( \implies x = 135 - 3x - 7 \) (First, multiply 3 by both terms inside the bracket).
\( \implies x = -3x + 128 \) (Combine the constant terms 135 and -7).
\( \implies x + 3x = 128 \) (Move the \( -3x \) to the left side by adding \( 3x \) to both sides).
\( \implies 4x = 128 \)
\( \implies x = \frac {128}{4} \)
\( \implies x = 32 \)
So, one part is 32. The other part is \( 45 - 32 = 13 \).
In simple words: We split the number 45 into two parts. One part is 32, and the other part is 13. The first part is 7 less than three times the second part.

🎯 Exam Tip: When splitting a total into two parts, if one part is \( x \), the other part is always (Total - \( x \)). This setup is crucial for these types of questions.

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.
Answer: Let Sujal's present age be \( x \) years. According to the problem, Ranu's present age is three times Sujal's age, so Ranu's age is \( 3x \) years. After 4 years: Sujal's age will be \( (x + 4) \) years. Ranu's age will be \( (3x + 4) \) years. The sum of their ages after 4 years will be 40 years.
\( (x + 4) + (3x + 4) = 40 \)
Now, let's combine like terms and solve for \( x \).
\( \implies x + 3x + 4 + 4 = 40 \)
\( \implies 4x + 8 = 40 \)
\( \implies 4x = 40 - 8 \)
\( \implies 4x = 32 \)
\( \implies x = \frac {32}{8} \) (This should be \( \frac{32}{4} \))
\( \implies x = 8 \)
Sujal's present age is 8 years. Ranu's present age is \( 3 \times 8 = 24 \) years.
In simple words: Sujal is 8 years old, and Ranu is 24 years old. In four years, if you add their ages together, it will be 40.

🎯 Exam Tip: Clearly define variables for present ages before calculating future or past ages. Always double-check calculations, especially divisions.

Question 8. Length of a rectangle exceeds its breadth by 6 meter. If its perimeter is 64 meter then find its length and breadth.
Answer: Let the breadth of the rectangle be \( x \) meters. The length of the rectangle exceeds its breadth by 6 meters, so the length is \( (x + 6) \) meters. The formula for the perimeter of a rectangle is \( 2 \times (\text{length} + \text{breadth}) \). We are given that the perimeter is 64 meters.
\( 2(\text{length} + \text{breadth}) = 64 \)
\( \implies 2(x + 6 + x) = 64 \) (Substitute the expressions for length and breadth).
\( \implies 2(2x + 6) = 64 \)
Now, we solve for \( x \).
\( \implies 4x + 12 = 64 \) (Multiply 2 by both terms inside the bracket).
\( \implies 4x = 64 - 12 \)
\( \implies 4x = 52 \)
\( \implies x = \frac {52}{4} \)
\( \implies x = 13 \)
So, the breadth of the rectangle is 13 meters. The length of the rectangle is \( x + 6 = 13 + 6 = 19 \) meters.
In simple words: The rectangle is 19 meters long and 13 meters wide. Its length is 6 meters more than its width, and its total boundary (perimeter) is 64 meters.

🎯 Exam Tip: Always write down the formulas for geometric shapes (like perimeter or area) correctly before substituting values, as this minimizes errors.

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.
Answer: Let the one's digit of the two-digit number be \( x \). The sum of the digits is 12, so the ten's digit must be \( (12 - x) \). The original number can be written as \( 10 \times (\text{ten's digit}) + (\text{one's digit}) \).
Original Number \( = 10(12 - x) + x \)
\( = 120 - 10x + x \)
\( = 120 - 9x \)
Now, if the digits are reversed, the new one's digit becomes \( (12 - x) \) and the new ten's digit becomes \( x \). The new number is:
New Number \( = 10(x) + (12 - x) \)
\( = 10x + 12 - x \)
\( = 9x + 12 \)
The problem states that the new number formed by reversing the digits is greater than the original number by 54.
New Number \( = \text{Original Number} + 54 \)
\( \implies 9x + 12 = (120 - 9x) + 54 \)
\( \implies 9x + 12 = 174 - 9x \) (Combine the constant terms on the right side: \( 120 + 54 \)).
\( \implies 9x + 9x = 174 - 12 \) (Move the \( -9x \) to the left and 12 to the right, changing signs).
\( \implies 18x = 162 \)
\( \implies x = \frac {162}{18} \)
\( \implies x = 9 \)
So, the one's digit is 9. The ten's digit is \( 12 - x = 12 - 9 = 3 \). Therefore, the original number is 39.
In simple words: We found a two-digit number, 39, where its digits (3 and 9) add up to 12. If you swap its digits to make 93, this new number is 54 more than the original number.

🎯 Exam Tip: For problems involving two-digit numbers, represent the number as \( 10 \times (\text{ten's digit}) + (\text{one's digit}) \). This algebraic representation is key to solving such questions.