CBSE Class 10 Mathematics Linear Equations Assignment Set 15

Class X Mathematics Assignment - Linear Equations

Q1. Solve for x: \(5x + \frac{7}{2} = \frac{3}{2}x - 14\) [Ans: \(x = -5\)]
Answer: Given equation is \(5x + \frac{7}{2} = \frac{3}{2}x - 14\)
Multiply the entire equation by 2 to clear the denominators:
\(2(5x) + 7 = 3x - 2(14)\)
\(10x + 7 = 3x - 28\)
Transpose terms containing \(x\) to the left side and constant terms to the right side:
\(10x - 3x = -28 - 7\)
\(7x = -35\)
\(x = \frac{-35}{7} = -5\)
In simple words: Clear the fractions by multiplying through by the denominator, then group all \(x\) terms together and numbers together to solve for \(x\).

Exam Tip: Always clear denominators first in linear equations with fractions to avoid calculation errors.

Q2. Solve for x: \(\frac{6x+1}{3} + 1 = \frac{(x-3)}{6}\) [Ans: \(x = -1\)]
Answer: Given equation is \(\frac{6x+1}{3} + 1 = \frac{x-3}{6}\)
Find the LCM of denominators (which is 6) and multiply the entire equation by 6:
\(6 \left(\frac{6x+1}{3}\right) + 6(1) = 6 \left(\frac{x-3}{6}\right)\)
\(2(6x + 1) + 6 = x - 3\)
Expand the brackets:
\(12x + 2 + 6 = x - 3\)
\(12x + 8 = x - 3\)
Transpose \(x\) to the left and constants to the right:
\(12x - x = -3 - 8\)
\(11x = -11\)
\(x = \frac{-11}{11} = -1\)
In simple words: Multiply every term by the LCM of denominators to remove fractions, then simplify and solve for \(x\).

Exam Tip: Do not forget to multiply the standalone constant terms (like 1) by the LCM as well.

Q3. Form the linear equation and solve it: "The present age of Anil's mother is three times the present age of Anil. After 5 years their ages will add to 66 years. Find their present ages" [Ans:- Anil's age = 14 yrs, Mother's age = 42 yrs]
Answer: Let the present age of Anil be \(x\) years.
Then, the present age of Anil's mother = \(3x\) years.
After 5 years, Anil's age will be \(x + 5\) years.
After 5 years, Mother's age will be \(3x + 5\) years.
According to the question, the sum of their ages after 5 years is 66:
\((x + 5) + (3x + 5) = 66\)
\(4x + 10 = 66\)
\(4x = 66 - 10\)
\(4x = 56\)
\(x = \frac{56}{4} = 14\)
Therefore, Anil's present age \(x = 14\) years.
Mother's present age \(3x = 3 \times 14 = 42\) years.
In simple words: Assign variables based on the given relation, add 5 to both ages for the future condition, and set up the sum equal to 66.

Exam Tip: Always clearly state what your variable represents at the beginning of word problems.

Q4. Make the linear equation and solve it, "Half of herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deers in the herd" [Ans:- 72]
Answer: Let the total number of deer in the herd be \(x\).
Number of deer grazing in the field = \(\frac{x}{2}\)
Remaining deer after grazing = \(x - \frac{x}{2} = \frac{x}{2}\)
Number of deer playing nearby = \(\frac{3}{4}\) of the remaining = \(\frac{3}{4} \times \frac{x}{2} = \frac{3x}{8}\)
Number of deer drinking water from the pond = 9
According to the problem, the sum of grazing deer, playing deer, and drinking deer equals the total herd \(x\):
\(\frac{x}{2} + \frac{3x}{8} + 9 = x\)
Find the LCM of denominators (8) and multiply through by 8:
\(4x + 3x + 72 = 8x\)
\(7x + 72 = 8x\)
\(72 = 8x - 7x\)
\(x = 72\)
Therefore, the total number of deer in the herd is 72.
In simple words: Break down the parts of the herd step-by-step, subtract from the remaining amount, and equate the sum of all parts to the total variable \(x\).

Exam Tip: Be careful with phrases like "three-fourths of the remaining"—always calculate the remainder first before applying fractions.

Q5. Solve for x: \(\frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{x+a+b} + \frac{1}{x}\) [Ans = \(-\left(\frac{a+b}{2}\right)\)]
Answer: Rearrange the terms by bringing \(\frac{1}{x}\) to the left side and \(\frac{1}{x+a}\) to the right side:
\(\frac{1}{x+a} - \frac{1}{x} = \frac{1}{x+a+b} - \frac{1}{x+b}\)
Take LCM on both sides:
Left side: \(\frac{x - (x+a)}{x(x+a)} = \frac{-a}{x(x+a)}\)
Right side: \(\frac{(x+b) - (x+a+b)}{(x+a+b)(x+b)} = \frac{-a}{(x+a+b)(x+b)}\)
Equating both sides:
\(\frac{-a}{x(x+a)} = \frac{-a}{(x+a+b)(x+b)}\)
Since \(a \neq 0\), divide both sides by \(-a\):
\(\frac{1}{x(x+a)} = \frac{1}{(x+a+b)(x+b)}\)
Cross-multiply:
\((x+a+b)(x+b) = x(x+a)\)
\(x^2 + bx + ax + ab + bx + b^2 = x^2 + ax\)
Cancel \(x^2 + ax\) from both sides:
\(2bx + b^2 + ab = 0\)
\(2bx = -b(a+b)\)
Since \(b \neq 0\), divide by \(b\):
\(2x = -(a+b)\)
\(x = -\frac{a+b}{2}\)
In simple words: Group terms with common parameters together before taking the LCM to simplify cross-multiplication significantly.

Exam Tip: Strategic grouping (like putting \(x+a\) with \(x\)) avoids expanding large quartic polynomials.

Q6. Solve for x: \(\left(\frac{x-1}{x-2}\right) - \left(\frac{x-2}{x-3}\right) = \left(\frac{x-3}{x-4}\right) - \left(\frac{x-4}{x-5}\right)\) [Ans \(x = \frac{7}{2}\)]
Answer: Convert each fraction into mixed fraction form by division:
\(\frac{x-1}{x-2} = \frac{(x-2)+1}{x-2} = 1 + \frac{1}{x-2}\)
\(\frac{x-2}{x-3} = \frac{(x-3)+1}{x-3} = 1 + \frac{1}{x-3}\)
\(\frac{x-3}{x-4} = \frac{(x-4)+1}{x-4} = 1 + \frac{1}{x-4}\)
\(\frac{x-4}{x-5} = \frac{(x-5)+1}{x-5} = 1 + \frac{1}{x-5}\)
Substitute these into the original equation:
\(\left(1 + \frac{1}{x-2}\right) - \left(1 + \frac{1}{x-3}\right) = \left(1 + \frac{1}{x-4}\right) - \left(1 + \frac{1}{x-5}\right)\)
Simplify by canceling the 1s:
\(\frac{1}{x-2} - \frac{1}{x-3} = \frac{1}{x-4} - \frac{1}{x-5}\)
Take LCM on both sides:
Left side: \(\frac{(x-3) - (x-2)}{(x-2)(x-3)} = \frac{-1}{(x-2)(x-3)}\)
Right side: \(\frac{(x-5) - (x-4)}{(x-4)(x-5)} = \frac{-1}{(x-4)(x-5)}\)
Equating both sides and canceling \(-1\):
\(\frac{1}{(x-2)(x-3)} = \frac{1}{(x-4)(x-5)}\)
Cross-multiply:
\((x-4)(x-5) = (x-2)(x-3)\)
\(x^2 - 9x + 20 = x^2 - 5x + 6\)
Cancel \(x^2\) from both sides:
\(-9x + 20 = -5x + 6\)
\(20 - 6 = 9x - 5x\)
\(14 = 4x\)
\(x = \frac{14}{4} = \frac{7}{2}\)
In simple words: Simplify complex algebraic fractions by splitting them into whole numbers and unit fractions before solving.

Exam Tip: Recognizing the algebraic identity pattern \(\frac{x-a}{x-b} = 1 + \frac{b-a}{x-b}\) saves immense calculation time.

Q7. The denominator of a fraction exceeds the numerator by 3 and if the numerator be increased by 7, the fraction is increased by unity. Find the fraction [Ans: \(\frac{4}{7}\)]
Answer: Let the numerator of the fraction be \(x\).
Then, the denominator of the fraction is \(x + 3\).
Original fraction = \(\frac{x}{x+3}\)
According to the second condition, if the numerator is increased by 7, the fraction increases by 1 (unity):
\(\frac{x + 7}{x + 3} = \frac{x}{x + 3} + 1\)
Rearrange the equation:
\(\frac{x + 7}{x + 3} - \frac{x}{x + 3} = 1\)
Combine terms with the same denominator:
\(\frac{x + 7 - x}{x + 3} = 1\)
\(\frac{7}{x + 3} = 1\)
Cross-multiply:
\(x + 3 = 7\)
\(x = 7 - 3 = 4\)
Numerator = 4
Denominator = \(4 + 3 = 7\)
Therefore, the required fraction is \(\frac{4}{7}\).
In simple words: Set numerator as \(x\) and denominator as \(x+3\), apply the given increase condition, and solve the simple linear equation.

Exam Tip: "Increased by unity" simply means adding 1 to the original value.

Q8. A's present age is to B's as 7:11 and 6 years hence their ages will be 9:13. Find the present age of each
Answer: Let the present age of A be \(7x\) years and the present age of B be \(11x\) years.
After 6 years, A's age will be \(7x + 6\) years.
After 6 years, B's age will be \(11x + 6\) years.
According to the question, the ratio of their ages after 6 years is \(9:13\):
\(\frac{7x + 6}{11x + 6} = \frac{9}{13}\)
Cross-multiply to solve for \(x\):
\(13(7x + 6) = 9(11x + 6)\)
\(91x + 78 = 99x + 54\)
Transpose terms:
\(78 - 54 = 99x - 91x\)
\(24 = 8x\)
\(x = \frac{24}{8} = 3\)
Therefore:
Present age of A = \(7x = 7 \times 3 = 21\) years.
Present age of B = \(11x = 11 \times 3 = 33\) years.
In simple words: Use a common ratio multiplier \(x\) to represent ages as \(7x\) and \(11x\), add 6 to both, and equate their ratio to \(\frac{9}{13}\).

Exam Tip: Whenever ages or quantities are given in a ratio like \(a:b\), always assume them as \(ax\) and \(bx\).