CBSE Class 8 Mathematics Linear Equations Assignment Set 07

VIII - Mathematics Assignment No. 05 - Linear Equations in One Variable

Question 1. The sum of two positive numbers is 30. The greater number is \(\frac{3}{2}\) of the smaller. Find the larger number.
Answer: Let the smaller number be \(x\).
Then the greater number is \(\frac{3}{2}x\).
According to the question, the sum of the two numbers is 30:
\(x + \frac{3}{2}x = 30\)
\(\frac{5}{2}x = 30\)
\(5x = 30 \times 2\)
\(5x = 60\)
\(x = \frac{60}{5} = 12\)
Smaller number = 12
Larger number = \(\frac{3}{2} \times 12 = 3 \times 6 = 18\).
In simple words: Assume the smaller number as \(x\), find the larger one in terms of \(x\), add them to get 30, and solve for \(x\).

Exam Tip: Always define your variable clearly at the beginning of word problems and substitute back to find the specific number requested (the larger number).

Question 2. The sum of two positive numbers is 90 and their ratio is 7:8. Find the numbers.
Answer: Let the common ratio be \(x\).
The two numbers are \(7x\) and \(8x\).
According to the question, their sum is 90:
\(7x + 8x = 90\)
\(15x = 90\)
\(x = \frac{90}{15} = 6\)
First number = \(7 \times 6 = 42\)
Second number = \(8 \times 6 = 48\)
In simple words: When given a ratio like 7:8, write the numbers as \(7x\) and \(8x\), equate their sum to the given total, and solve for \(x\).

Exam Tip: Ratios can always be converted into algebraic expressions by introducing a common multiplier variable \(x\).

Question 3. The difference of two positive numbers is 56 and the quotient obtained on dividing one by the other is 5. Find the numbers.
Answer: Let the smaller number be \(x\).
Since the quotient of the two numbers is 5, the larger number is \(5x\).
According to the question, the difference between the two numbers is 56:
\(5x - x = 56\)
\(4x = 56\)
\(x = \frac{56}{4} = 14\)
Smaller number = 14
Larger number = \(5 \times 14 = 70\)
In simple words: A quotient of 5 means one number is 5 times the other. Subtract them to set up the equation for their difference.

Exam Tip: The phrase "quotient obtained on dividing one by the other is 5" directly translates to one variable being 5 times the other.

Question 4. The length of a rectangular field is greater than its width by 4 metres. If the perimeter of the field is 32 metres, find its area.
Answer: Let the width of the rectangular field be \(x\) metres.
Then the length of the field is \(x + 4\) metres.
Perimeter of a rectangle = \(2(\text{Length} + \text{Width})\)
Given perimeter = 32 metres:
\(2((x + 4) + x) = 32\)
\(2(2x + 4) = 32\)
\(2x + 4 = \frac{32}{2} = 16\)
\(2x = 16 - 4 = 12\)
\(x = \frac{12}{2} = 6\)
Width = 6 metres
Length = \(6 + 4 = 10\) metres
Area = \(\text{Length} \times \text{Width} = 10 \times 6 = 60 \text{ m}^2\).
In simple words: Find the length and width using the perimeter formula, then multiply them to get the area.

Exam Tip: Do not forget to include the correct square units (m\(^2\)) when calculating area.

Question 5. What number must be subtracted from the numerator and added to the denominator of \(\frac{13}{17}\) to give a rational number equal to \(\frac{1}{2}\)?
Answer: Let the number to be subtracted from the numerator and added to the denominator be \(x\).
The new numerator becomes \(13 - x\) and the new denominator becomes \(17 + x\).
According to the question, the new fraction is equal to \(\frac{1}{2}\):
\(\frac{13 - x}{17 + x} = \frac{1}{2}\)
Cross-multiplying gives:
\(2(13 - x) = 1(17 + x)\)
\(26 - 2x = 17 + x\)
\(26 - 17 = x + 2x\)
\(9 = 3x\)
\(x = \frac{9}{3} = 3\)
In simple words: Set up the modified fraction by subtracting \(x\) from the top and adding \(x\) to the bottom, then cross-multiply to solve for \(x\).

Exam Tip: Cross-multiplication is the fastest way to eliminate fractions when solving linear equations in one variable.

Question 6. The sum of the digits of a two-digit number is 9. The number formed by reversing the digits is greater than the original number by 9. Find the original number.
Answer: Let the units digit be \(x\).
Since the sum of the digits is 9, the tens digit is \(9 - x\).
Original number = \(10(\text{tens digit}) + \text{units digit} = 10(9 - x) + x = 90 - 10x + x = 90 - 9x\).
Number formed by reversing the digits:
New units digit = \(9 - x\), New tens digit = \(x\)
Reversed number = \(10x + (9 - x) = 9x + 9\).
According to the question, the reversed number is greater than the original number by 9:
\((9x + 9) - (90 - 9x) = 9\)
\(9x + 9 - 90 + 9x = 9\)
\(18x - 81 = 9\)
\(18x = 9 + 81 = 90\)
\(x = \frac{90}{18} = 5\)
Units digit = 5
Tens digit = \(9 - 5 = 4\)
Original number = 45.
In simple words: Express a two-digit number as \(10a + b\), set up the algebraic condition based on the reversed digits, and solve for the digits.

Exam Tip: For digit problems, always represent the number in expanded form: \(10 \times (\text{tens digit}) + (\text{units digit})\).

Question 7. The numerator of a rational number is less than the denominator by 3. If the numerator is multiplied by 3 and the denominator increased by 12, the new fraction becomes \(\frac{3}{4}\). Find the original fraction.
Answer: Let the denominator be \(x\).
Then the numerator is \(x - 3\).
Original fraction = \(\frac{x - 3}{x}\).
According to the question, if the numerator is multiplied by 3 and the denominator is increased by 12, the new fraction is \(\frac{3}{4}\):
\(\frac{3(x - 3)}{x + 12} = \frac{3}{4}\)
Divide both sides by 3:
\(\frac{x - 3}{x + 12} = \frac{1}{4}\)
Cross-multiplying gives:
\(4(x - 3) = 1(x + 12)\)
\(4x - 12 = x + 12\)
\(4x - x = 12 + 12\)
\(3x = 24\)
\(x = \frac{24}{3} = 8\)
Denominator = 8
Numerator = \(8 - 3 = 5\)
Original fraction = \(\frac{5}{8}\).
In simple words: Define denominator as \(x\), write numerator as \(x - 3\), apply the changes given in the problem, and cross-multiply.

Exam Tip: Simplifying common factors across the numerator and denominator before cross-multiplying makes calculations much simpler and less prone to error.

Question 8. What number must be subtracted from the numerator and added to the denominator of \(\frac{13}{17}\) to give a rational number equal to \(\frac{1}{2}\)?
Answer: Let the number to be subtracted from the numerator and added to the denominator be \(x\).
The modified fraction is \(\frac{13 - x}{17 + x}\).
Given that this fraction equals \(\frac{1}{2}\):
\(\frac{13 - x}{17 + x} = \frac{1}{2}\)
Cross-multiplying gives:
\(2(13 - x) = 1(17 + x)\)
\(26 - 2x = 17 + x\)
\(26 - 17 = 3x\)
\(9 = 3x\)
\(x = \frac{9}{3} = 3\)
In simple words: This is identical to Question 5; set up the equation with \(x\) and cross-multiply to find that 3 must be subtracted from the numerator and added to the denominator.

Exam Tip: Always verify your answer by substituting \(x = 3\) back into the fraction: \(\frac{13 - 3}{17 + 3} = \frac{10}{20} = \frac{1}{2}\).