CBSE Class 8 Mathematics Linear Equations Assignment Set 06

VIII - Mathematics Assignment - No. 06 - Linear Eqs in one variable

Question 1. Anil is 32 years older than his son Arun who is 12 years old. After how many years will the age of Anil be twice that of his son?
Answer: Let the number of years after which Anil's age will be twice of Arun's age be \(x\).
Present age of Arun = 12 years
Present age of Anil = 12 + 32 = 44 years
After \(x\) years:
Arun's age = \(12 + x\)
Anil's age = \(44 + x\)
According to the question:
\(44 + x = 2(12 + x)\)
\(44 + x = 24 + 2x\)
\(44 - 24 = 2x - x\)
\(x = 20\)
Therefore, after 20 years, Anil's age will be twice that of his son.
In simple words: We find Anil's present age, add \(x\) to both ages, set up the equation where Anil is twice as old, and solve for \(x\).

Exam Tip: Always add the unknown years \(x\) to both persons' present ages before setting up the equation.

Question 2. The ages of Ajay and Akhil are in the ratio 3 : 4. Five years ago, the ratio of their ages was 2 : 3. Find their present ages.
Answer: Let the present age of Ajay be \(3x\) and the present age of Akhil be \(4x\).
Five years ago:
Ajay's age = \(3x - 5\)
Akhil's age = \(4x - 5\)
According to the question, the ratio of their ages 5 years ago was 2 : 3:
\(\frac{3x - 5}{4x - 5} = \frac{2}{3}\)
Cross-multiplying:
\(3(3x - 5) = 2(4x - 5)\)
\(9x - 15 = 8x - 10\)
\(9x - 8x = 15 - 10\)
\(x = 5\)
Present age of Ajay = \(3 \times 5 = 15\) years
Present age of Akhil = \(4 \times 5 = 20\) years
In simple words: Use variables with a ratio like \(3x\) and \(4x\), subtract 5 from both for past ages, and cross-multiply to solve for \(x\).

Exam Tip: When ages are given in a ratio, always introduce a common variable \(x\) and express them as \(ax\) and \(bx\).

Question 3. The perimeter of a rectangle is 70 metres. If its width is three-fourths of its length, find its area.
Answer: Let the length of the rectangle be \(x\) metres.
Width of the rectangle = \(\frac{3}{4}x\) metres.
Perimeter of a rectangle = \(2(\text{length} + \text{width})\)
Given perimeter = 70 m
\(2 \left(x + \frac{3}{4}x\right) = 70\)
\(x + \frac{3}{4}x = \frac{70}{2}\)
\(\frac{7}{4}x = 35\)
\(7x = 35 \times 4\)
\(7x = 140\)
\(x = \frac{140}{7} = 20\) metres (Length)
Width = \(\frac{3}{4} \times 20 = 15\) metres
Area of the rectangle = \(\text{Length} \times \text{Width}\)
Area = \(20 \times 15 = 300\text{ m}^2\)
In simple words: Use the perimeter formula to find the length and width, then multiply them to get the area.

Exam Tip: Do not forget to include square units (like \(\text{m}^2\)) when writing the final answer for area.

Question 4. One-fifth of a number exceeds one-fifth of \((2x + 1)\) by \((-4)\). Find the number.
Answer: Let the number be \(x\).
According to the question, one-fifth of the number exceeds one-fifth of \((2x + 1)\) by \(-4\):
\(\frac{1}{5}x - \frac{1}{5}(2x + 1) = -4\)
Multiply the entire equation by 5 to clear the denominator:
\(x - (2x + 1) = -20\)
\(x - 2x - 1 = -20\)
\(-x - 1 = -20\)
\(-x = -20 + 1\)
\(-x = -19\)
\(x = 19\)
Therefore, the number is 19.
In simple words: Translate the sentence into an algebraic equation, multiply by 5 to remove fractions, and solve for \(x\).

Exam Tip: When subtracting an algebraic expression in parentheses like \(-(2x + 1)\), remember to change the signs of all terms inside.

Question 5. Two numbers are in the ratio 5 : 8. If the sum of the numbers is 182. Find the numbers.
Answer: Let the two numbers be \(5x\) and \(8x\).
According to the question, the sum of the numbers is 182:
\(5x + 8x = 182\)
\(13x = 182\)
\(x = \frac{182}{13}\)
\(x = 14\)
First number = \(5 \times 14 = 70\)
Second number = \(8 \times 14 = 112\)
Therefore, the numbers are 70 and 112.
In simple words: Represent the numbers as \(5x\) and \(8x\), add them to equal 182, find \(x\), and substitute it back.

Exam Tip: Verify your answer by adding the two resulting numbers to check if their sum equals 182 (\(70 + 112 = 182\)).

Question 6. Find three consecutive odd numbers whose sum is 159.
Answer: Let the three consecutive odd numbers be \(x\), \(x + 2\), and \(x + 4\).
According to the question, their sum is 159:
\(x + (x + 2) + (x + 4) = 159\)
\(3x + 6 = 159\)
\(3x = 159 - 6\)
\(3x = 153\)
\(x = \frac{153}{3} = 51\)
First number = 51
Second number = \(51 + 2 = 53\)
Third number = \(51 + 4 = 55\)
Therefore, the three consecutive odd numbers are 51, 53, and 55.
In simple words: Consecutive odd numbers differ by 2, so represent them as \(x\), \(x+2\), and \(x+4\), then add them to 159.

Exam Tip: Whether consecutive numbers are even or odd, their general forms are always represented by adding 2 successively (\(x, x+2, x+4\)).

Question 7. The sum of the digits of a two digit number is 15. If the number formed by reversing the digits is less than the original number by 27. Find the original number.
Answer: Let the digit at the units place be \(x\).
Since the sum of the digits is 15, the digit at the tens place is \(15 - x\).
Original number = \(10(15 - x) + x = 150 - 10x + x = 150 - 9x\)
Number formed by reversing the digits:
New units digit = \(15 - x\), New tens digit = \(x\)
Reversed number = \(10x + (15 - x) = 9x + 15\)
According to the question, the reversed number is less than the original number by 27:
\(\text{Original number} - \text{Reversed number} = 27\)
\((150 - 9x) - (9x + 15) = 27\)
\(150 - 9x - 9x - 15 = 27\)
\(135 - 18x = 27\)
\(-18x = 27 - 135\)
\(-18x = -108\)
\(x = \frac{-108}{-18} = 6\)
Units digit = 6
Tens digit = \(15 - 6 = 9\)
Original number = \(10(9) + 6 = 96\)
In simple words: Express digits as \(x\) and \(15-x\), write the number as \(10(\text{tens}) + \text{units}\), and set up the difference equation.

Exam Tip: A two-digit number with tens digit \(t\) and units digit \(u\) is always written algebraically as \(10t + u\).

Question 8. The digit in the tenth place of a two digit number is three times that in the unit place. If the digits in the number are reversed, the new number will be 36 less than the original number. Find the number.
Answer: Let the digit at the units place be \(x\).
The digit at the tens place is three times the units place = \(3x\).
Original number = \(10(3x) + x = 30x + x = 31x\)
When digits are reversed:
New units digit = \(3x\), New tens digit = \(x\)
Reversed number = \(10(x) + 3x = 10x + 3x = 13x\)
According to the question, the new number is 36 less than the original number:
\(\text{Original number} - \text{Reversed number} = 36\)
\(31x - 13x = 36\)
\(18x = 36\)
\(x = \frac{36}{18} = 2\)
Units digit = 2
Tens digit = \(3 \times 2 = 6\)
Original number = \(10(6) + 2 = 62\)
In simple words: Define the digits using \(x\) and \(3x\), write both the original and reversed numbers, and equate their difference to 36.

Exam Tip: Double-check the order of subtraction when a problem states "is less than the original number by 36" (Original - New = 36).

Question 9. The denominator of a rational number is greater than its numerator by 3. If 3 is subtracted from the numerator and 2 is added to its denominator, the new number becomes \(\frac{1}{5}\). Find the fraction.
Answer: Let the numerator of the rational number be \(x\).
Since the denominator is greater than the numerator by 3, the denominator = \(x + 3\).
Original fraction = \(\frac{x}{x + 3}\)
According to the question, if 3 is subtracted from the numerator and 2 is added to the denominator, the fraction becomes \(\frac{1}{5}\):
\(\frac{x - 3}{(x + 3) + 2} = \frac{1}{5}\)
\(\frac{x - 3}{x + 5} = \frac{1}{5}\)
Cross-multiplying:
\(5(x - 3) = 1(x + 5)\)
\(5x - 15 = x + 5\)
\(5x - x = 5 + 15\)
\(4x = 20\)
\(x = \frac{20}{4} = 5\)
Numerator = 5
Denominator = \(5 + 3 = 8\)
Therefore, the fraction is \(\frac{5}{8}\).
In simple words: Set numerator as \(x\) and denominator as \(x+3\), apply the given changes, cross-multiply, and solve for \(x\).

Exam Tip: Always write the final answer in the form of a fraction (\(\frac{\text{Numerator}}{\text{Denominator}}\)).

Question 10. A is now 9 years older than B. In 10 years, A will be twice as old as B was 10 years ago. Find their present ages.
Answer: Let the present age of B be \(x\) years.
Since A is 9 years older than B, the present age of A = \(x + 9\) years.
In 10 years:
A's age = \((x + 9) + 10 = x + 19\)
10 years ago:
B's age = \(x - 10\)
According to the question, in 10 years, A's age will be twice as old as B was 10 years ago:
\(x + 19 = 2(x - 10)\)
\(x + 19 = 2x - 20\)
\(19 + 20 = 2x - x\)
\(x = 39\)
Present age of B = 39 years
Present age of A = \(39 + 9 = 48\) years
Therefore, the present ages of A and B are 48 years and 39 years respectively.
In simple words: Define B's age as \(x\), write expressions for A in 10 years and B 10 years ago, then set up the equation and solve.

Exam Tip: Read age relationship sentences carefully to distinguish between future additions ("in 10 years") and past subtractions ("10 years ago").