ML Aggarwal Class 10 Maths Solutions Chapter 04 Linear Inequations

Question 1. Solve the inequation 3x - 11 < 3, where x ∈ {1, 2, 3, .........., 10}. Also represent its solution on number line.
Answer: We have 3x - 11 < 3. Adding 11 to both sides gives 3x < 14. Dividing by 3 yields x < 14/3, which equals x < 4.66. Since x must belong to {1, 2, 3, .........., 10}, the values that satisfy this are x ∈ {1, 2, 3, 4}. On the number line, this is shown using thick dots at positions 1, 2, 3, and 4.
In simple words: When you solve the inequality step by step, you find that x must be less than about 4.66. Looking at the given set of numbers, the answer is {1, 2, 3, 4}.

Exam Tip: Always check each candidate value from the given set by substituting it back into the original inequality to verify your solution set.

Question 2. Solve 2(x - 3) < 1, x ∈ {1, 2, 3, .........., 10}.
Answer: Expanding the left side gives 2x - 6 < 1. Adding 6 to both sides yields 2x < 7. Dividing by 2 produces x < 3.5. Since x belongs to {1, 2, 3, .........., 10}, the values satisfying this condition are x ∈ {1, 2, 3}.
In simple words: Solve by expanding and moving numbers across the inequality sign. The answer is the set {1, 2, 3}.

Exam Tip: Remember that when dividing an inequality by a positive number, the direction of the inequality symbol remains unchanged.

Question 3. Solve 5 - 4x > 2 - 3x, x ∈ W. Also represent its solution on number line.
Answer: Starting with 5 - 4x > 2 - 3x, we rearrange by moving all x terms to one side: 5 - 2 > -3x + 4x, which gives 3 > x, or equivalently x < 3. Since x belongs to the whole numbers W = {0, 1, 2, 3, ...}, the solution is x ∈ {0, 1, 2}. The number line shows thick dots at 0, 1, and 2.
In simple words: Collect the x terms and constants on opposite sides. Since x must be less than 3 and x must be a whole number, the answer is {0, 1, 2}.

Exam Tip: When variables are on both sides, move all variable terms to one side and constants to the other before simplifying.

Question 4. List the solution set of 30 – 4(2x – 1) < 30, given that x is a positive integer.
Answer: Expanding the expression gives 30 - 8x + 4 < 30, which simplifies to 34 - 8x < 30. Rearranging yields 34 - 30 < 8x, so 4 < 8x. Dividing by 8 produces x > 1/2. Since x must be a positive integer, the solution is x ∈ {1, 2, 3, 4, ...}.
In simple words: After simplifying, we find x is greater than 0.5. The positive integers greater than 0.5 are {1, 2, 3, 4, ...}.

Exam Tip: Always expand brackets carefully, and be mindful of the domain restriction (positive integer, whole number, natural number, etc.) when determining which values to include in the final solution.

Question 5. Solve: 2(x – 2) < 3x – 2, x ∈ {–3, –2, –1, 0, 1, 2, 3}.
Answer: Expanding gives 2x - 4 < 3x - 2. Moving x terms to the right and constants to the left yields 2x - 3x < -2 + 4, which simplifies to -x < 2. Multiplying by -1 (and reversing the inequality sign) gives x > -2. From the given set {–3, –2, –1, 0, 1, 2, 3}, the values greater than -2 are x ∈ {–1, 0, 1, 2, 3}.
In simple words: After solving, we need values of x that are larger than -2. From the provided set, these are {–1, 0, 1, 2, 3}.

Exam Tip: Be careful when multiplying or dividing by a negative number - always flip the inequality symbol. Verify a few values by direct substitution.

Question 6. If x is a negative integer, find the solution set of (2/3) + (1/3)(x + 1) > 0.
Answer: Expanding the left side gives (2/3) + (x/3) + (1/3) > 0. Combining like terms yields (x + 3)/3 > 0, which means x + 3 > 0, so x > -3. Since x must be a negative integer, the values satisfying this are x ∈ {-2, -1}.
In simple words: Simplify the fraction expression to find x > -3. The negative integers that are greater than -3 are only -2 and -1.

Exam Tip: When multiplying an inequality by a positive denominator, the direction of the inequality is preserved. List negative integers carefully - remember they include -1, -2, -3, and so on.

Question 7. Solve x – 3(2 + x) > 2(3x - 1), x ∈ {-3, -2, -1, 0, 1, 2}. Also represent its solution on the number line.
Answer: Expanding both sides gives x - 6 - 3x > 6x - 2. Simplifying the left side yields -2x - 6 > 6x - 2. Collecting x terms: -2x - 6x > -2 + 6, so -8x > 4. Dividing by -8 (and reversing the inequality) gives x < -1/2. From the given set {-3, -2, -1, 0, 1, 2}, the values less than -1/2 are x ∈ {-3, -2, -1}. These are shown as thick dots on the number line.
In simple words: Expand, collect x terms, and solve. We need numbers that are less than -0.5. From the set, that gives us {-3, -2, -1}.

Exam Tip: When dividing by a negative coefficient, remember to reverse the inequality sign. Check your work by testing a value inside and outside the solution set.

Question 8. Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solve x – 3 < 2x – 1.
Answer: Rearranging the inequality yields x - 2x < -1 + 3, which simplifies to -x < 2. Multiplying by -1 (and reversing) gives x > -2. Since all elements in the given set {1, 2, 3, 4, 5, 6, 7, 9} are greater than -2, the solution is the entire set x ∈ {1, 2, 3, 4, 5, 6, 7, 9}.
In simple words: After solving, we find x must be greater than -2. Every number in the given set satisfies this condition, so the answer is the complete set.

Exam Tip: Sometimes all members of a given set satisfy the inequality. Always check whether the domain restriction eliminates some solutions or includes all of them.

Question 9. List the solution set of the inequation (1/2) + 8x > 5x - (3/2), x ∈ ℤ.
Answer: Rearranging yields 8x - 5x > -(3/2) - (1/2), which simplifies to 3x > -2. Dividing by 3 gives x > -2/3. Since x must be an integer, the values satisfying this condition are x ∈ {0, 1, 2, 3, 4, ...}.
In simple words: After moving terms and simplifying, x must be greater than about -0.67. The integers greater than this are {0, 1, 2, 3, ...}.

Exam Tip: When working with fractions, combine them first before moving variables and constants. For integer solutions, identify which integers satisfy the inequality, starting from the smallest.

Question 10. List the solution set of: (11 - 2x)/5 ≥ (9 - 3x)/8 + 3/4, x ∈ ℕ.
Answer: First, rewrite the right side: (11 - 2x)/5 ≥ (9 - 3x + 6)/8 = (15 - 3x)/8. Multiplying both sides by 40 (the LCM of 5 and 8) gives 8(11 - 2x) ≥ 5(15 - 3x), which expands to 88 - 16x ≥ 75 - 15x. Rearranging yields -16x + 15x ≥ 75 - 88, so -x ≥ -13. Multiplying by -1 (and reversing) gives x ≤ 13. Since x is a natural number, the solution is x ∈ {1, 2, 3, 4, 5, ..., 13}.
In simple words: Clear the fractions by multiplying by a common factor. Solve to find x ≤ 13. The natural numbers up to and including 13 form the solution.

Exam Tip: When dealing with fractions, multiply through by the LCM to eliminate denominators. Remember that natural numbers start from 1, not 0.

Question 11. Find the values of x, which satisfy the inequation: -2 ≤ (1/2) - (2x/3) ≤ (5/6), x ∈ ℕ. Graph the solution set on the number line.
Answer: We solve this compound inequality in two parts. First, from -2 ≤ (1/2) - (2x/3), subtracting 1/2 from both sides gives -(5/2) ≤ -(2x/3). Multiplying by -6 (and reversing inequalities) yields 15 ≥ 4x, so x ≤ 15/4 = 3.75. Second, from (1/2) - (2x/3) ≤ 5/6, subtracting 1/2 gives -(2x/3) ≤ 1/3. Multiplying by -3/2 (and reversing) yields x ≥ -1/2. Since x is a natural number and must satisfy 3.75 ≥ x ≥ -0.5, the solution is x ∈ {1, 2, 3}. The number line shows thick dots at 1, 2, and 3.
In simple words: A compound inequality requires solving both parts. The natural numbers that satisfy both conditions are {1, 2, 3}.

Exam Tip: For compound inequalities, solve each part separately, then find the intersection of both solution sets. Graph all solutions that satisfy the domain restriction.

Question 12. If x ∈ W, find the solution set of (3x/5) - (2x - 1)/3 > 1. Also graph the solution set on the number line, if possible.
Answer: Finding a common denominator of 15, the inequality becomes (9x - 10x + 5)/15 > 1, which simplifies to (-x + 5)/15 > 1. Multiplying by 15 gives -x + 5 > 15, so -x > 10, which means x < -10. Since x must be a whole number (non-negative integer), there is no whole number less than -10. Therefore, the solution set is empty: x ∈ ∅. This cannot be represented on a number line.
In simple words: After solving, we find x < -10. But whole numbers are 0, 1, 2, ..., so there are no whole numbers less than -10. The solution set is empty.

Exam Tip: Not every inequality has a solution within the given domain. When the solution contradicts the domain restriction, the answer is the empty set, and it cannot be graphed.

Question 13(i). Solve: (x/2) + 5 ≤ (x/3) + 6 where x is a positive odd integer.
Answer: Rearranging gives (x/2) - (x/3) ≤ 6 - 5, which yields (3x - 2x)/6 ≤ 1, so x/6 ≤ 1. Multiplying by 6 produces x ≤ 6. Since x must be a positive odd integer, the values are x ∈ {1, 3, 5}.
In simple words: Collect variable terms on one side. Since x must be a positive odd number and x ≤ 6, the options are {1, 3, 5}.

Exam Tip: Identify the domain restriction clearly - in this case, positive odd integers - and apply it after solving the inequality.

Question 13(ii). Solve: (2x + 3)/3 ≥ (3x - 1)/4 where x is positive even integer.
Answer: Multiplying both sides by 12 gives 4(2x + 3) ≥ 3(3x - 1), which expands to 8x + 12 ≥ 9x - 3. Rearranging yields 8x - 9x ≥ -3 - 12, so -x ≥ -15. Multiplying by -1 (and reversing) gives x ≤ 15. Since x must be a positive even integer, the solution is x ∈ {2, 4, 6, 8, 10, 12, 14}.
In simple words: Clear the fractions and solve to find x ≤ 15. The positive even integers that do not exceed 15 are {2, 4, 6, 8, 10, 12, 14}.

Exam Tip: Multiply by a positive common multiple to clear fractions without changing the inequality direction. Then apply the domain restriction (even positive integer).

Question 14. Given that x ∈ I, solve the inequation and graph the solution on the number line: 3 ≥ (x - 4)/2 + x/3 ≥ 2.
Answer: Solving the left side: 3 ≥ (x - 4)/2 + x/3. Multiplying by 6 gives 18 ≥ 3(x - 4) + 2x = 3x - 12 + 2x = 5x - 12, so 5x ≤ 30 and x ≤ 6. Solving the right side: (x - 4)/2 + x/3 ≥ 2. Multiplying by 6 gives 3(x - 4) + 2x ≥ 12, so 5x - 12 ≥ 12, which yields 5x ≥ 24 and x ≥ 24/5 = 4.8. Combining both results: 4.8 ≤ x ≤ 6. Since x is an integer, the solution is x ∈ {5, 6}. The number line shows thick dots at 5 and 6.
In simple words: Solve both parts of the compound inequality separately. The integers that satisfy both conditions are {5, 6}.

Exam Tip: For compound inequalities, always solve both halves separately, then identify the overlap. Use the domain restriction (integers) to determine which values to include.

Question 15. Solve 1 ≥ 15 - 7x > 2x - 27, x ∈ ℕ.
Answer: From the left inequality 1 ≥ 15 - 7x, we get 7x ≥ 14, so x ≥ 2. From the right inequality 15 - 7x > 2x - 27, we get 15 + 27 > 2x + 7x, so 42 > 9x, which means x < 42/9 = 14/3 ≈ 4.67. Combining both: 2 ≤ x < 4.67. Since x is a natural number, the solution is x ∈ {2, 3, 4}.
In simple words: Break the compound inequality into two parts. Solve each to find the overlap. The natural numbers in this range are {2, 3, 4}.

Exam Tip: When an inequality contains multiple symbols in sequence (like a > b > c), split it into separate inequalities and solve both. The final answer is the intersection of both solution sets.

Question 16. If x ∈ ℤ, solve 2 + 4x < 2x – 5 ≤ 3x. Also represent its solution on the number line.
Answer: From 2 + 4x < 2x - 5, subtracting 2x gives 2 + 2x < -5, so 2x < -7 and x < -7/2 = -3.5. From 2x - 5 ≤ 3x, subtracting 2x gives -5 ≤ x, so x ≥ -5. Combining both inequalities: -5 ≤ x < -3.5. Since x is an integer, the solution is x ∈ {-5, -4}. The number line shows thick dots at -5 and -4.
In simple words: Solve both parts of the compound inequality separately. The integers that satisfy both are {-5, -4}.

Exam Tip: When dealing with a compound inequality, isolate x carefully in each part. Use the domain (integers, whole numbers, etc.) to filter out non-qualifying values.

Question 17. Solve 3x - 5 ≤ 6x + 4 < 11 + x, when (i) x ∈ W (ii) x ∈ ℤ. Represent the solution set on a real number in each case.
Answer: From 3x - 5 ≤ 6x + 4, we get -3x ≤ 9, so x ≥ -3. From 6x + 4 < 11 + x, we get 5x < 7, so x < 7/5 = 1.4. Combining both: -3 ≤ x < 1.4.

(i) If x ∈ W (whole numbers), then x ∈ {0, 1}.

(ii) If x ∈ ℤ (integers), then x ∈ {-3, -2, -1, 0, 1}.

For (i), the number line shows thick dots at 0 and 1. For (ii), the number line shows thick dots at -3, -2, -1, 0, and 1.
In simple words: Solve both parts to find -3 ≤ x < 1.4. Then apply each domain restriction: whole numbers give {0, 1}, while integers give {-3, -2, -1, 0, 1}.

Exam Tip: Always state the solution in terms of the domain. The same compound inequality yields different solution sets for different domains (W vs. ℤ). Graph each carefully.

Question 18. Solve: (5x - 7)/2 ≤ (4x - 10)/3, x ∈ ℝ and represent the solution set on the number line.
Answer: Multiplying both sides by 6 (the LCM of 2 and 3) gives 3(5x - 7) ≤ 2(4x - 10), which expands to 15x - 21 ≤ 8x - 20. Rearranging yields 15x - 8x ≤ -20 + 21, so 7x ≤ 1 and x ≤ 1/7. The solution set is {x : x ∈ ℝ, x ≤ 1/7}. On the number line, this is shown as a thick black line from 1/7 extending to the left, with a filled circle at 1/7 (indicating 1/7 is included).
In simple words: Clear the fractions by multiplying by a common denominator. Solve to get x ≤ 1/7. Every real number up to and including 1/7 is a solution.

Exam Tip: For real number domains, draw a line with an arrow indicating all values to the left (or right). Use a filled circle for ≤ or ≥, and an open circle for < or >.

Question 19. Solve and represent the solution set on the number line: (3x/5) - (2x - 1)/3 > 1, x ∈ ℝ.
Answer: Finding a common denominator of 15, the inequality becomes [3(3x) - 5(2x - 1)]/15 > 1, which simplifies to (9x - 10x + 5)/15 > 1. This gives (-x + 5)/15 > 1. Multiplying by 15 yields -x + 5 > 15, so -x > 10 and x < -10. The solution set is {x : x ∈ ℝ, x < -10}. On the number line, this is shown as a thick black line extending from -10 to the left, with an open circle at -10 (indicating -10 is not included).
In simple words: After clearing fractions and simplifying, we find x < -10. All real numbers less than -10 are solutions.

Exam Tip: Use an open circle when the boundary point is not included (< or >) and a filled circle when it is included (≤ or ≥). Extend the line with an arrow to show all values in that direction.

Question 20. Solve the following inequation and represent the solution on the number line: (3x/5) + 2 < x + 4 ≤ (x/2) + 5, x ∈ ℝ.
Answer: Solving the left part: (3x/5) + 2 < x + 4. Subtracting 2 from both sides gives (3x/5) < x + 2. Rearranging yields (3x/5) - x < 2, so (3x - 5x)/5 < 2, which means -2x/5 < 2. Multiplying by -5/2 (and reversing) gives x > -5. Solving the right part: x + 4 ≤ (x/2) + 5. Subtracting x from both sides gives 4 ≤ (x/2) - x + 5 = -(x/2) + 5. Rearranging yields 4 - 5 ≤ -(x/2), so -1 ≤ -(x/2). Multiplying by -2 (and reversing) gives 2 ≥ x, or x ≤ 2. Combining both: -5 < x ≤ 2. On the number line, this is shown as a thick line from just after -5 to 2, with an open circle at -5 and a filled circle at 2.
In simple words: Solve both inequalities separately. The solution is all numbers greater than -5 and less than or equal to 2.

Exam Tip: For compound inequalities with real numbers, find the intersection of both solution sets. Use open circles for strict inequalities (< or >) and filled circles for non-strict (≤ or ≥).

Question 20. Solve the inequation \( \frac{3x}{5} + 2 < x + 4 \leq \frac{x}{2} + 5 \), where x ∈ R. Write down the solution set and represent it on the real number line.
Answer: Working on the left side of the compound inequality, we have \( \frac{3x}{5} + 2 < x + 4 \). Multiplying by 5 and simplifying gives \( 3x + 10 < 5x + 20 \), which leads to \( -2x < 10 \) and thus \( x > -5 \). For the right side, \( x + 4 \leq \frac{x}{2} + 5 \) simplifies to \( 2x + 8 \leq x + 10 \), giving \( x \leq 2 \). Combining both results, the solution set is \( \{x : x \in \mathbb{R}, -5 < x \leq 2\} \).
In simple words: Break the compound inequality into two parts. Solve each separately, then combine the results to find where both conditions are true at the same time.

Exam Tip: Always solve compound inequalities by separating them into two parts and solving each independently. Check your final answer by substituting boundary values back into the original inequality.

Question 21. Solve the following inequation. Write down the solution set and represent it on the real number line. -5(x - 9) ≥ 17 - 9x > x + 2, x ∈ R.
Answer: For the left part, expanding -5(x - 9) ≥ 17 - 9x gives -5x + 45 ≥ 17 - 9x, which simplifies to 4x ≥ -28, so x ≥ -7. For the right part, 17 - 9x > x + 2 rearranges to -10x > -15, giving x < 1.5. Taking the intersection of both conditions, the solution set is \( \{x : -7 \leq x < 1.5\} \).
In simple words: Work through each inequality separately, then find the values of x that satisfy both at the same time.

Exam Tip: When the inequality has three parts connected by ≥ and >, you must satisfy both conditions simultaneously. Mark both boundary points clearly on the number line - one as a filled circle and one as an open circle.

Question 22. Solve the following inequation, write down the solution set and represent it on the real number line: -2 + 10x ≤ 13x + 10 < 24 + 10x, x ∈ Z.
Answer: Solving the left part: -2 + 10x ≤ 13x + 10 yields -3x ≤ 12, so x ≥ -4. For the right part: 13x + 10 < 24 + 10x gives 3x < 14, so x < \( \frac{14}{3} \) ≈ 4.67. Since x must be an integer, combining both conditions gives \( \{x : x \in \mathbb{Z}, -4 \leq x < \frac{14}{3}\} = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\} \).
In simple words: Separate the double inequality, solve both parts, then list only the integers that fall within the range.

Exam Tip: Always remember that when x ∈ Z (integers), your final answer must be a discrete set of whole numbers, not a continuous interval. Use filled dots to mark integer points on the number line.

Question 23. Solve the inequation 2x - 5 ≤ 5x + 4 < 11, where x ∈ I. Also represent the solution set on the number line.
Answer: Breaking down the compound inequality: 2x - 5 ≤ 5x + 4 leads to -3x ≤ 9, so x ≥ -3. And 5x + 4 < 11 gives 5x < 7, so x < 1.4. Since x must be an integer, the solution set becomes \( \{-3, -2, -1, 0, 1\} \).
In simple words: Solve both parts of the inequality separately. Then keep only the integers that satisfy both conditions at once.

Exam Tip: For integer solutions, convert decimal boundaries (like 1.4) to the nearest integer that actually satisfies the strict inequality - here, that means x = 1 is included but x = 2 is not.

Question 24. If x ∈ I, A is the solution set of 2(x - 1) < 3x - 1 and B is the solution set of 4x - 3 ≤ 8 + x, find A ∩ B.
Answer: For set A: solving 2(x - 1) < 3x - 1 gives 2x - 2 < 3x - 1, which simplifies to -x < 1, so x > -1. Thus A = {0, 1, 2, 3, ...}. For set B: solving 4x - 3 ≤ 8 + x gives 3x ≤ 11, so x ≤ \( \frac{11}{3} \) ≈ 3.67. Thus B = {..., -1, 0, 1, 2, 3}. The intersection A ∩ B contains integers that belong to both sets: \( \{0, 1, 2, 3\} \).
In simple words: Find all integers satisfying set A, find all integers satisfying set B, then pick only those appearing in both.

Exam Tip: For intersection problems, always identify the range for each set carefully. The intersection will be the narrower range where both conditions overlap.

Question 25. A = {x : 11x - 5 > 7x + 3, x ∈ R} and B = {x : 18x - 9 ≥ 15 + 12x, x ∈ R}. Find set A ∩ B and represent it on a number line.
Answer: For set A, the inequality 11x - 5 > 7x + 3 simplifies to 4x > 8, giving x > 2. For set B, the inequality 18x - 9 ≥ 15 + 12x simplifies to 6x ≥ 24, giving x ≥ 4. Since we need values satisfying both conditions, A ∩ B = {x : x ∈ R, x ≥ 4}, represented on the number line as a thick line starting at 4 (including 4) and extending to the right.
In simple words: Solve each inequality to get a range of values. The intersection is where both ranges overlap - in this case, all real numbers greater than or equal to 4.

Exam Tip: When finding intersections with real numbers, identify which condition is more restrictive. Always use a filled circle at the boundary where the condition includes equality (≥ or ≤).

Question 26. Given: P = {x : 5 < 2x - 1 ≤ 11, x ∈ R} and Q = {x : -1 ≤ 3 + 4x < 23, x ∈ I} where R = {real numbers}, I = {integers}. Represent P and Q on number line. Write down the elements of P ∩ Q.
Answer: For set P, solving the left inequality 5 < 2x - 1 gives 6 < 2x, so x > 3. Solving the right inequality 2x - 1 ≤ 11 gives 2x ≤ 12, so x ≤ 6. Therefore, P = {x : 3 < x ≤ 6}. For set Q, solving -1 ≤ 3 + 4x gives -4 ≤ 4x, so x ≥ -1. Solving 3 + 4x < 23 gives 4x < 20, so x < 5. Since x is an integer, Q = {-1, 0, 1, 2, 3, 4}. The intersection P ∩ Q contains integers from Q that also lie in P, which gives P ∩ Q = {1, 2, 3, 4}. (Note: The element 4 from Q satisfies 3 < 4 ≤ 6.)
In simple words: For P, find the continuous range of real numbers. For Q, list the integers in the given range. Then pick integers from Q that fall within P's range.

Exam Tip: Remember that P contains real numbers (shown as a thick line) and Q contains only integers (shown as dots). Their intersection will always be a finite set of integers.

Question 27. If x ∈ I, find the smallest value of x which satisfies the inequation \( 2x + \frac{5}{2} > \frac{5x}{3} + 2 \).
Answer: Rearranging the inequality: \( 2x - \frac{5x}{3} > 2 - \frac{5}{2} \). Multiplying through by 6 to clear denominators: 12x - 10x > 12 - 15, which gives 2x > -3, so x > -\( \frac{3}{2} \) = -1.5. Since x must be an integer and x > -1.5, the smallest integer satisfying this is x = -1.
In simple words: Work through the algebra to find the range of x. Then pick the smallest integer that falls within that range.

Exam Tip: When finding the smallest integer in a range like x > -1.5, remember that -1 is larger than -1.5, so -1 is indeed the smallest integer that works.

Question 28. Given 20 - 5x < 5(x + 8), find the smallest value of x, when (i) x ∈ I, (ii) x ∈ W, (iii) x ∈ N.
Answer: Simplifying the inequality: 20 - 5x < 5x + 40 leads to -10x < 20, so -x < 2, which gives x > -2. (i) When x ∈ I (integers), the smallest value is x = -1. (ii) When x ∈ W (whole numbers: 0, 1, 2, ...), the smallest value is x = 0. (iii) When x ∈ N (natural numbers: 1, 2, 3, ...), the smallest value is x = 1.
In simple words: First solve the inequality to get x > -2. Then, depending on which number set you're using, pick the smallest number from that set that is greater than -2.

Exam Tip: Remember the key differences: I includes all integers (negative, zero, positive), W includes non-negative integers starting from 0, and N includes only positive integers starting from 1.

Question 29. Solve the following inequation and represent the solution set on the number line: \( 4x - 19 < \frac{3x}{5} - 2 \leq -\frac{2}{5} + x \), x ∈ R.
Answer: Working on the left part: \( 4x - 19 < \frac{3x}{5} - 2 \). Multiplying by 5 and simplifying gives 20x - 95 < 3x - 10, which leads to 17x < 85, so x < 5. For the right part: \( \frac{3x}{5} - 2 \leq -\frac{2}{5} + x \). Multiplying by 5 gives 3x - 10 ≤ -2 + 5x, which simplifies to -2x ≤ 8, so x ≥ -4. Combining both, the solution set is \( \{x : x \in \mathbb{R}, -4 \leq x < 5\} \).
In simple words: Handle each inequality in the chain separately, then find where both are satisfied together.

Exam Tip: When multiplying or dividing by a negative number, always flip the inequality sign. Double-check your arithmetic when clearing fractions.

Question 30. Solve the given inequation and graph the solution on the number line: 2y - 3 < y + 1 ≤ 4y + 7; y ∈ R.
Answer: For the left side, 2y - 3 < y + 1 simplifies to y < 4. For the right side, y + 1 ≤ 4y + 7 simplifies to -3y ≤ 6, giving y ≥ -2. The solution set is \( \{y : y \in \mathbb{R}, -2 \leq y < 4\} \), represented by a thick line on the number line starting at -2 (filled circle) and ending at 4 (open circle).
In simple words: Solve each part of the chain. Then combine to find all values that work for both conditions.

Exam Tip: Use a filled dot for ≤ or ≥ (included boundary) and an open dot for < or > (excluded boundary) on your number line graph.

Question 31. Solve the following inequation, write down the solution set and represent it on the real number line. \( -3 + x \leq \frac{7x}{2} + 2 < 8 + 2x \), x ∈ I.
Answer: For the left part: \( -3 + x \leq \frac{7x}{2} + 2 \). Rearranging gives \( \frac{7x}{2} - x \geq -5 \), which simplifies to \( \frac{5x}{2} \geq -5 \), so x ≥ -2. For the right part: \( \frac{7x}{2} + 2 < 8 + 2x \). Rearranging gives \( \frac{7x}{2} - 2x < 6 \), which simplifies to \( \frac{3x}{2} < 6 \), so x < 4. Since x is an integer and -2 ≤ x < 4, the solution set is {-2, -1, 0, 1, 2, 3}.
In simple words: Solve each inequality separately. Combine them, then list the integers in the resulting range.

Exam Tip: When working with compound inequalities involving fractions, clear denominators early by multiplying through. Always check that your integer solution values actually satisfy the original inequality.

Question 32. Solve the following inequation, write the solution set and represent it on the real number line. \( 5x - 21 < \frac{5x}{7} - 6 \leq -3\frac{3}{7} + x \), x ∈ R.
Answer: For the left part: \( 5x - 21 < \frac{5x}{7} - 6 \). Multiplying by 7 gives 35x - 147 < 5x - 42, which simplifies to 30x < 105, so x < \( \frac{7}{2} \) = 3.5. For the right part: \( \frac{5x}{7} - 6 \leq -\frac{24}{7} + x \). Multiplying by 7 gives 5x - 42 ≤ -24 + 7x, which simplifies to 2x ≥ -18, so x ≥ -9. Combining both, the solution set is \( \{x : -9 \leq x < 3.5, x \in \mathbb{R}\} \).
In simple words: Clear fractions by multiplying, solve each part, then find the overlap.

Exam Tip: Convert mixed numbers to improper fractions before solving. Be careful when multiplying through by a denominator - apply it to every term in the inequality.

Question 33. Find the greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer.
Answer: Let the greatest integer be x. According to the condition, 2x + 7 > 3x simplifies to 7 > x, so x < 7. The greatest integer satisfying this inequality is x = 6.
In simple words: Set up an inequality from the problem statement, solve for the variable, then find the greatest whole number that fits.

Exam Tip: Read word problems carefully and translate each phrase into a mathematical expression. "Greater than" means >, and "the greatest integer" means you need the largest whole number in the solution range.

Question 34. One-third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water is greater than or equal to 3 metres. Find the length of the shortest pole.
Answer: Let the length of the pole be x metres. The part buried in mud is \( \frac{x}{3} \), and the part in water is \( \frac{x}{6} \). The part above water is \( x - \frac{x}{3} - \frac{x}{6} = x - \frac{2x}{6} - \frac{x}{6} = x - \frac{x}{2} = \frac{x}{2} \). Given that the part above water must be at least 3 metres, we have \( \frac{x}{2} \geq 3 \), which gives x ≥ 6. Therefore, the length of the shortest pole is 6 metres.
In simple words: Add up the fractions of the pole in different places. The remaining part must be at least 3 metres, so set up an inequality and solve for the pole's length.

Exam Tip: When dealing with parts of a whole, always make sure the fractions add up correctly. Express the part above water in terms of the total length, then use the given constraint to find the minimum length.

Question 1. If x ∈ {-3, -1, 0, 1, 3, 5}, then the solution set of the inequation 3x - 2 ≤ 8 is
(a) {-3, -1, 1, 3}
(b) {-3, -1, 0, 1, 3}
(c) {-3, -2, -1, 0, 1, 2, 3}
(d) {-3, -2, -1, 0, 1, 2}
Answer: (b) {-3, -1, 0, 1, 3}
In simple words: Work out the inequality by solving 3x - 2 ≤ 8, which gives x ≤ 10/3 or x ≤ 3.33. From the given set, pick all numbers that are less than or equal to 3.33.

Exam Tip: Always check each element from the given set by substituting it into the inequality - this confirms which values satisfy the condition.

Question 2. If x ∈ W, then the solution set of the inequation 3x + 11 ≥ x + 8 is
(a) {-2, -1, 0, 1, 2, ...}
(b) {-1, 0, 1, 2, ...}
(c) {0, 1, 2, 3, ...}
(d) {x : x ∈ R, x ≥ -3/2}
Answer: (c) {0, 1, 2, 3, ...}
In simple words: Solve the inequality to get x ≥ -3/2. Since x must be a whole number, the solution includes 0 and all positive whole numbers.

Exam Tip: Remember that W represents whole numbers, which start from 0. Even though the algebraic solution might include negative values, whole numbers cannot be negative.

Question 3. If x ∈ W, then the solution set of the inequation 5 - 4x ≥ 2 - 3x is
(a) {..., -2, -1, 0, 1, 2, 3}
(b) {1, 2, 3}
(c) {0, 1, 2, 3}
(d) {x : x ∈ R, x ≤ 3}
Answer: (c) {0, 1, 2, 3}
In simple words: Rearrange the inequality to find x ≤ 3. Since x belongs to whole numbers, the answer includes all whole numbers from 0 to 3.

Exam Tip: When the algebraic solution gives x ≤ a number, combine this with the domain constraint (W = whole numbers) to get the final answer set.

Question 4. If x ∈ I, then the solution set of the inequation 1 < 3x + 5 ≤ 11 is
(a) {-1, 0, 1, 2}
(b) {-2, -1, 0, 1}
(c) {-1, 0, 1}
(d) {x : x ∈ R, -4/3 < x ≤ 2}
Answer: (a) {-1, 0, 1, 2}
In simple words: This compound inequality has two parts. Solving the left side gives x > -4/3, and the right side gives x ≤ 2. Integers between these bounds are -1, 0, 1, and 2.

Exam Tip: For compound inequalities, split them into two separate parts and solve each one. Then find values that satisfy both conditions simultaneously.

Question 5. If x ∈ R, the solution set of 6 ≤ -3(2x - 4) < 12 is
(a) {x : x ∈ R, 0 < x ≤ 1}
(b) {x : x ∈ R, 0 ≤ x < 1}
(c) {0, 1}
(d) none of these
Answer: (a) {x : x ∈ R, 0 < x ≤ 1}
In simple words: Expand the expression -3(2x - 4) and solve both sides of the compound inequality separately to find the range of x values.

Exam Tip: When dealing with compound inequalities involving expressions, carefully expand and simplify each side. Track inequality direction changes, especially when multiplying or dividing by negative numbers.

Question 6. The solution set for the inequation 2x + 4 ≤ 14, x ∈ W is:
(a) {1, 2, 3, 4, 5}
(b) {0, 1, 2, 3, 4, 5}
(c) {1, 2, 3, 4}
(d) {0, 1, 2, 3, 4}
Answer: (b) {0, 1, 2, 3, 4, 5}
In simple words: Solve 2x + 4 ≤ 14 to get x ≤ 5. Since x is a whole number, the solution includes all whole numbers from 0 up to and including 5.

Exam Tip: Don't forget to include 0 when listing whole numbers - it's the smallest whole number and should be part of the solution set if the inequality permits it.

Question 7. Given, x + 2 ≤ x/3 + 3 and x is a prime number. The solution set for x is:
(a) ∅
(b) {0}
(c) {1}
(d) {0, 1}
Answer: (a) ∅
In simple words: Solving the inequality gives x ≤ 1.5. Prime numbers are natural numbers greater than 1 with no divisors other than 1 and themselves. No prime number exists that is less than or equal to 1.5, so the solution set is empty.

Exam Tip: Recall that prime numbers start from 2. When an inequality restricts x to a range that excludes all prime numbers, the solution set is the empty set.

Question. Assertion (A): If x ≤ -5, x ∈ W, the above inequation has no solution.
Reason (R): The whole numbers are always positive.
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (a) Assertion (A) is true, but Reason (R) is false
In simple words: The assertion is right because whole numbers begin at 0 and go upward, so none can be ≤ -5. The reason is wrong because whole numbers include 0, which is neither positive nor negative.

Exam Tip: For assertion-reason questions, evaluate each statement independently first. An assertion can be true while the reason supporting it is false - they must both be correct AND logically connected for option (c) to be right.

Question. Assertion (A): If -5 < x and x ≤ 6, x ∈ R, the above inequation has no solution.
Reason (R): Infinitely many real numbers lie between -5 and 6.
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, but Reason (R) is true
In simple words: The inequality definitely has solutions - countless real numbers fit between -5 and 6. So the assertion is wrong, but the reason explaining why is correct.

Exam Tip: Always verify that an assertion correctly matches reality. Here, the inequality clearly has many solutions, making the assertion false regardless of how reasonable the reason sounds.

Question. Assertion (A): If -3 ≤ x < 5 2/3, x ∈ Z, x has nine values.
Reason (R): x = 5 is included in the solution set.
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A)
In simple words: The range -3 ≤ x < 5.67 includes nine integers: -3, -2, -1, 0, 1, 2, 3, 4, 5. Both statements are factually correct, but the reason doesn't fully explain why there are exactly nine values - other integers matter too.

Exam Tip: When a reason is factually true but doesn't directly cause the assertion to be true, option (d) applies. The reason may be incomplete or tangential to the main claim.

Question. Assertion (A): Given below is the graphical representation of an inequality on number line. It represents real numbers lying between -5 and 3/2 but not -5 and 3/2.
Reason (R): The hole represents absence of the number -5.
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, but Reason (R) is true
In simple words: From the number line, the inequality is -5 < x ≤ 3/2 (not including -5, but including 3/2). The assertion claims it excludes both endpoints, which is incorrect. The reason correctly explains that the hole at -5 shows -5 is not included.

Exam Tip: On a number line, a hole (open circle) means that point is NOT included, while a filled dot means it IS included. Always read the assertion carefully to match it against what the number line shows.

Chapter Test

Question 1. Solve the inequation: 5x - 2 ≤ 3(3 - x) where x ∈ {-2, -1, 0, 1, 2, 3, 4}. Also represent its solution on the number line.
Answer: Working through the inequality, 5x - 2 ≤ 9 - 3x simplifies to 8x ≤ 11, giving x ≤ 11/8 or x ≤ 1.375. From the given set {-2, -1, 0, 1, 2, 3, 4}, the values satisfying this condition are x ∈ {-2, -1, 0, 1}. On the number line, the solution is marked with thick black dots at these four points.
In simple words: Rearrange the inequality to find the boundary value, then check which given numbers fall within the solution range.

Exam Tip: Always show both the algebraic solution and the graphical representation on a number line - this demonstrates complete understanding and typically earns full marks.

Question 2. Solve the inequations: 6x - 5 < 3x + 4, x ∈ I
Answer: Simplifying the inequality, 6x - 3x < 4 + 5 gives 3x < 9, so x < 3. Since x belongs to the set of integers, the solution set is {..., -2, -1, 0, 1, 2}, which includes all integers less than 3.
In simple words: Move all x terms to one side and numbers to the other, then divide to find x < 3. List all integers that are less than 3.

Exam Tip: When the domain is integers (I) or whole numbers (W), the algebraic solution is just the first step - you must then list or describe which specific values from that domain satisfy the inequality.

Question 3. Find the solution set of the inequation x + 5 ≤ 2x + 3; x ∈ R. Graph the solution set on the number line.
Answer: Rearranging, x - 2x ≤ 3 - 5 gives -x ≤ -2. Multiplying both sides by -1 and flipping the inequality sign yields x ≥ 2. The solution set is {x : x ∈ R, x ≥ 2}. On the number line, this is shown as a thick line starting from and including 2, extending to the right toward infinity.
In simple words: Collect variable terms on one side and constants on the other. When multiplying or dividing by a negative number, always flip the inequality sign.

Exam Tip: Use a filled circle at the boundary point (x = 2) to show it is included in the solution. The thick line extending right indicates all values greater than or equal to 2 are part of the solution.

Question 4. If x ∈ R (real numbers) and -1 < 3 - 2x ≤ 7, find solution set and represent it on a number line.
Answer: Splitting the compound inequality: from -1 < 3 - 2x, subtracting 3 from both sides and dividing by -2 (flipping the sign) gives x < 2. From 3 - 2x ≤ 7, subtracting 3 and dividing by -2 gives x ≥ -2. Combining both conditions: -2 ≤ x < 2. The solution set is {x : x ∈ R, -2 ≤ x < 2}. On the number line, a filled circle appears at -2 (included) and an open circle at 2 (not included), with a thick line connecting them.
In simple words: Break the compound inequality into two parts and solve each separately. Then combine the results to find the range that works for both.

Exam Tip: A filled circle shows inclusion of that endpoint; an open circle shows exclusion. Match your graphical representation carefully to whether endpoints are included or excluded in the solution set.

Question 5. Solve the inequation: (5x + 1)/7 - 4(x/7 + 2/5) ≤ 1 - 3/5 + (3x - 1)/7, x ∈ R.
Answer: Multiplying both sides by 35 to eliminate fractions: 25x + 5 - 4(5x + 14) ≤ 56 + 15x - 5. Expanding: 25x + 5 - 20x - 56 ≤ 56 + 15x - 5. Simplifying: 5x - 51 ≤ 51 + 15x. Collecting variable terms: -10x ≤ 102. Dividing by -10 and flipping the inequality sign: x ≥ -51/5. The solution set is {x : x ∈ R, x ≥ -51/5}.
In simple words: Find a common multiple of all denominators and multiply through to clear fractions. Then solve using standard inequality techniques, remembering to flip the sign when dividing by a negative number.

Exam Tip: When an inequality has multiple fractions, multiply by the least common multiple of denominators first - this simplifies the working significantly and reduces errors.

Question 6. Find the range of values of x, which satisfy 7 ≤ -4x + 2 < 12, x ∈ R. Graph these values of x on the real number line.
Answer: Splitting into two parts: from 7 ≤ -4x + 2, subtracting 2 gives 5 ≤ -4x, so 4x ≤ -5, and x ≤ -5/4. From -4x + 2 < 12, subtracting 2 gives -4x < 10, so 4x > -10, and x > -5/2. Combining both: -5/2 < x ≤ -5/4. The solution set is {x : x ∈ R, -5/2 < x ≤ -5/4}. On the number line, an open circle appears at -5/2 (not included) and a filled circle at -5/4 (included), with a thick line connecting them.
In simple words: Work through each side of the compound inequality separately, then combine the conditions to find the x values that satisfy both parts simultaneously.

Exam Tip: Pay close attention to which endpoints are included (≤ or ≥) versus excluded (< or >). These small symbols determine whether to use filled or open circles on your number line graph.

Question 7. If x ∈ R, solve 3 - 2x ≥ x + \( \frac{1-x}{3} \) > \( \frac{2x}{5} \). Also represent the solution on the number line.
Answer: Working with the left-hand side of the inequation:

\( 3 - 2x \geq x + \frac{1-x}{3} \)

\( 3 - 2x \geq \frac{3x + 1 - x}{3} \)

\( 3(3 - 2x) \geq 2x + 1 \)

\( 9 - 6x \geq 2x + 1 \)

\( 2x + 6x \leq 9 - 1 \)

\( 8x \leq 8 \)

\( x \leq 1 \) ............(1)

Now solving the right-hand side:

\( x + \frac{1-x}{3} > \frac{2x}{5} \)

\( \frac{3x + 1 - x}{3} > \frac{2x}{5} \)

\( 5(2x + 1) > 3 \times 2x \)

\( 10x + 5 > 6x \)

\( 10x - 6x > -5 \)

\( 4x > -5 \)

\( x > -\frac{5}{4} \) ............(2)

Combining equations (1) and (2):

Solution set = \( \left\{ x : -\frac{5}{4} < x \leq 1, x \in \mathbb{R} \right\} \)

The graph is shown as a thick black line starting from \( -\frac{5}{4} \) (excluding this point) and extending to 1 (including this point).
In simple words: Find the value of x that satisfies both parts of the combined inequality. Solve the left part to get one limit, solve the right part to get another limit, then combine them. The answer includes all numbers between these two limits that meet both conditions.

Exam Tip: Always work on each part of a compound inequation separately, then combine the results. Mark the endpoints correctly on the number line - use an open circle for excluded values and a filled circle for included values.

Question 8. Find positive integers which are such that if 6 is subtracted from five times the integer then the resulting number cannot be greater than four times the integer.
Answer: Let the positive integer = x

According to the problem:

\( 5x - 6 < 4x \)

\( 5x - 4x < 6 \)

\( x < 6 \)

Since x must be a positive integer, the solution set is \( \{1, 2, 3, 4, 5\} \).
In simple words: Set up the inequality from the words in the problem. Simplify to find what x must be. Then list all the positive whole numbers that work.

Exam Tip: Remember that "cannot be greater than" means you use the less-than symbol (<), not less-than-or-equal-to. For positive integers, always check that your final answer makes sense by testing a couple of values.

Question 9. Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is at least 3.
Answer: Let the first natural number = x

Then the second number = x + 1

And the third number = x + 2

Given condition:

\( \frac{1}{3}(x + 2) - \frac{1}{5}(x) \geq 3 \)

\( \frac{x}{3} - \frac{x}{5} \geq 3 - \frac{2}{3} \)

\( \frac{5x - 3x}{15} \geq \frac{9 - 2}{3} \)

\( \frac{2x}{15} \geq \frac{7}{3} \)

\( 2x \geq \frac{7 \times 15}{3} \)

\( 2x \geq 35 \)

\( x \geq \frac{35}{2} \)

\( x \geq 17\frac{1}{2} \)

Since x must be a natural number, the smallest value is x = 18.

Therefore:
\( x = 18 \)
\( x + 1 = 19 \)
\( x + 2 = 20 \)

The three smallest consecutive natural numbers are 18, 19, and 20.
In simple words: Write the three numbers using x. Use the given condition to create an inequality. Solve it to find the smallest starting value. Then list the three numbers in order.

Exam Tip: When working with "at least", use the greater-than-or-equal-to symbol (≥). After solving, always round up to the next whole number if needed, since the question asks for natural numbers.