CBSE Class 11 Mathematics Linear Inequalities Assignment Set 03

EXTRA / HOT QUESTIONS

Question 1(i). Solve: \(\frac{5x - 2}{3} - \frac{7x - 3}{5} > \frac{x}{4}\)
Answer: Given inequality is \(\frac{5x - 2}{3} - \frac{7x - 3}{5} > \frac{x}{4}\)
Find the LCM of denominators 3, 5, and 4, which is 60.
Multiply both sides by 60:
\(20(5x - 2) - 12(7x - 3) > 15x\)
\(100x - 40 - 84x + 36 > 15x\)
\(16x - 4 > 15x\)
Subtract \(15x\) from both sides:
\(x - 4 > 0\)
\(x > 4\)
Therefore, the solution set is \((4, \infty)\).
In simple words: Simplify the fractions by finding a common denominator, then isolate x to find its range.

Exam Tip: Be careful with negative signs when expanding brackets, especially with terms like \(-12(7x - 3)\).

Question 1(ii). Solve: \(\frac{2x - 3}{4} + 8 \ge 2 + \frac{4x}{3}\)
Answer: Given inequality is \(\frac{2x - 3}{4} + 8 \ge 2 + \frac{4x}{3}\)
Subtract 2 from both sides and combine constants:
\(\frac{2x - 3}{4} + 6 \ge \frac{4x}{3}\)
Multiply the entire inequality by the LCM of denominators, which is 12:
\(3(2x - 3) + 72 \ge 16x\)
\(6x - 9 + 72 \ge 16x\)
\(6x + 63 \ge 16x\)
Subtract \(6x\) from both sides:
\(63 \ge 10x\)
\(10x \le 63\)
\(x \le \frac{63}{10}\)
Therefore, the solution set is \(\left(-\infty, \frac{63}{10}\right]\).
In simple words: Clear the fractions by multiplying by 12, group all x terms on one side, and solve for x.

Exam Tip: When reversing an inequality or writing intervals, note that a non-strict inequality (\(\le\)) uses a closed bracket.

Question 1(iii). Solve: \(2 - \frac{3x}{5} < 1 - \frac{x}{3} < 3 + \frac{4x}{2}\)
Answer: Split the compound inequality into two separate inequalities and solve simultaneously:
Part 1: \(2 - \frac{3x}{5} < 1 - \frac{x}{3}\)
Multiply by 15:
\(30 - 9x < 15 - 5x\)
\(30 - 15 < 9x - 5x\)
\(15 < 4x\)
\(x > \frac{15}{4}\)

Part 2: \(1 - \frac{x}{3} < 3 + 2x\)
Multiply by 3:
\(3 - x < 9 + 6x\)
\(3 - 9 < 6x + x\)
\(-6 < 7x\)
\(x > -\frac{6}{7}\)

Combining both conditions, \(x > \frac{15}{4}\) and \(x > -\frac{6}{7}\), the intersection is \(x > \frac{15}{4}\).
Therefore, the solution set is \(\left(\frac{15}{4}, \infty\right)\).
In simple words: Break the double inequality into two single inequalities, solve each separately, and find the overlapping solution.

Exam Tip: For compound inequalities, the final solution must satisfy all parts of the inequality at the same time.

Question 2(i). Solve graphically: \(2x + y \ge 3\), \(x - 2y \le -1\), \(x \ge 0\), \(y \ge 0\)
Answer: 1. Convert inequalities into equations to find boundary lines:
Line 1: \(2x + y = 3\) (Intercepts: \((1.5, 0)\) and \((0, 3)\))
Line 2: \(x - 2y = -1\) (Intercepts: \((-1, 0)\) and \((0, 0.5)\))
2. Test the origin \((0,0)\) for both inequalities:
For \(2x + y \ge 3\): \(0 + 0 \ge 3\) (False, so shade away from the origin).
For \(x - 2y \le -1\): \(0 - 0 \le -1\) (False, so shade away from the origin).
3. Incorporating \(x \ge 0\) and \(y \ge 0\), the common intersecting region in the first quadrant gives the required feasible region.
In simple words: Draw both boundary lines, test points to find the correct side to shade for each, and locate the overlapping region.

Exam Tip: Always shade lightly in pencil so you can clearly identify the final dark-bounded feasible region.

Question 2(ii). Solve graphically: \(x + 4y \le 4\), \(2x + 3y \le 6\), \(x \ge 0\), \(y \ge 0\)
Answer: 1. Consider the corresponding equations:
Line 1: \(x + 4y = 4\) (Intercepts: \((4, 0)\) and \((0, 1)\))
Line 2: \(2x + 3y = 6\) (Intercepts: \((3, 0)\) and \((0, 2)\))
2. Test origin \((0,0)\) for both:
\(0 + 0 \le 4\) (True)
\(0 + 0 \le 6\) (True)
3. Since both inequalities satisfy the origin and \(x, y \ge 0\), the feasible region is bounded by the axes and the lines containing the origin side.
In simple words: Plot the two line intercepts, test the origin, and shade the region below both lines in the first quadrant.

Exam Tip: Label all intersection points of the axes and boundary lines clearly on your graph.

Question 2(iii). Solve graphically: \(x + y \ge 1\), \(x \le 5\), \(y \le 4\), \(2x + 3y \le 12\), \(x \ge 0\), \(y \ge 0\)
Answer: 1. Consider boundary equations:
\(x + y = 1\) (Intercepts: \((1,0)\), \((0,1)\))
\(x = 5\) (Vertical line)
\(y = 4\) (Horizontal line)
\(2x + 3y = 12\) (Intercepts: \((6,0)\), \((0,4)\))
2. Test origin or substitute boundary conditions to determine the closed polygon region satisfying all constraints simultaneously in the first quadrant.
In simple words: Plot all four boundary lines and shade the region that satisfies all conditions at the same time.

Exam Tip: Check corner points by solving simultaneous linear equations where boundary lines intersect.

Question 3. The water acidity in a pool is considered normal when the average pH reading of 3 daily measurements is between 8.2 and 8.5. If the first two pH readings are 8.48 and 8.35, find the range of pH value for the third reading that will result in the acidity level being normal.
Answer: Let the third pH reading be \(x\).
The average of the three readings is given by:
\(\text{Average} = \frac{8.48 + 8.35 + x}{3}\)
According to the question, this average must be between 8.2 and 8.5:
\(8.2 < \frac{8.48 + 8.35 + x}{3} < 8.5\)
\(8.2 < \frac{16.83 + x}{3} < 8.5\)
Multiply all parts by 3:
\(24.6 < 16.83 + x < 25.5\)
Subtract 16.83 from all parts:
\(24.6 - 16.83 < x < 25.5 - 16.83\)
\(7.77 < x < 8.67\)
Therefore, the third reading must be between 7.77 and 8.67.
In simple words: Set up an inequality for the average of three numbers, simplify, and solve for the unknown third value.

Exam Tip: Read carefully whether inequalities are strict (\(<\)) or non-strict (\(\le\)) based on phrases like "between" versus "greater than or equal to".

Question 4. In the first 4 papers each of hundred marks Ravi got 90, 75, 73, 85 marks. If he wants an average of greater than or equal to 75 marks and less than 80 marks, find the range of marks she should score in the fifth paper.
Answer: Let the marks obtained in the fifth paper be \(x\).
Sum of marks in the first 4 papers = \(90 + 75 + 73 + 85 = 323\).
Total marks for 5 papers = \(323 + x\).
The required average is given by:
\(75 \le \frac{323 + x}{5} < 80\)
Multiply all parts by 5:
\(375 \le 323 + x < 400\)
Subtract 323 from all parts:
\(375 - 323 \le x < 400 - 323\)
\(52 \le x < 77\)
Therefore, Ravi should score more than or equal to 52 but less than 77 marks in the fifth paper.
In simple words: Calculate the total sum of the first four tests, set up an inequality for the 5-test average, and solve for the fifth test score.

Exam Tip: Ensure marks do not exceed the maximum possible score (100) when stating practical score ranges.

Question 5. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is atleast 61 cm find the minimum length of the shortest side.
Answer: Let the length of the shortest side be \(x\) cm.
Length of the longest side = \(3x\) cm.
Length of the third side = \(3x - 2\) cm.
The perimeter of a triangle is the sum of all its sides:
\(\text{Perimeter} = x + 3x + (3x - 2) = 7x - 2\)
Given that the perimeter is at least 61 cm:
\(7x - 2 \ge 61\)
Add 2 to both sides:
\(7x \ge 63\)
Divide by 7:
\(x \ge 9\)
Therefore, the minimum length of the shortest side is 9 cm.
In simple words: Express all sides in terms of a single variable x, add them to get the perimeter, and solve the inequality for the minimum value of x.

Exam Tip: "At least" translates mathematically to a greater-than-or-equal-to (\(\ge\)) sign.