ML Aggarwal Class 8 Maths Solutions Chapter 12 Linear Equations and Inequalities in One Variable

Exercise 12.1

Question 1. (i) 5x – 3 = 3x – 5
Answer: Start by collecting all terms with x on the left side and constants on the right. Move 3x to the left and -3 to the right: 5x - 3x = -5 + 3, which simplifies to 2x = -2. Divide both sides by 2 to find x = -1.
In simple words: Get all the x's on one side and the numbers on the other, then solve.

Exam Tip: Always move terms carefully, remembering to change their sign when crossing the equals sign.

Question 1. (ii) 3x – 7 = 3(5 – x)
Answer: First expand the right side: 3x - 7 = 15 - 3x. Combine like terms by adding 3x to both sides: 3x + 3x - 7 = 15, giving 6x - 7 = 15. Add 7 to both sides: 6x = 22. Divide by 6 to get x = 22/6 = 11/3.
In simple words: Expand brackets first, then collect x terms and numbers separately.

Exam Tip: Expand all brackets before collecting terms - this prevents sign errors.

Question 2. (i) 4(2x + 1) = 3(x – 1) + 7
Answer: Expand both sides: 8x + 4 = 3x - 3 + 7, which becomes 8x + 4 = 3x + 4. Subtract 3x from both sides: 5x + 4 = 4. Subtract 4 from both sides: 5x = 0, so x = 0.
In simple words: Expand, simplify, then solve step by step.

Exam Tip: When both sides simplify to the same constant after collecting x terms, the variable equals zero.

Question 2. (ii) 3(2p – 1) = 5 – (3p – 2)
Answer: Expand: 6p - 3 = 5 - 3p + 2, which simplifies to 6p - 3 = 7 - 3p. Add 3p to both sides: 9p - 3 = 7. Add 3 to both sides: 9p = 10, so p = 10/9.
In simple words: Remove brackets carefully, watch the negative sign, then gather like terms.

Exam Tip: When subtracting a bracket, distribute the negative sign to every term inside.

Question 3. (i) 5y – 2{y – 3(y – 5)} = 6
Answer: Work from the innermost brackets outward. First, expand the inner brackets: y - 3(y - 5) = y - 3y + 15 = -2y + 15. Now the equation becomes 5y - 2(-2y + 15) = 6. Distribute: 5y + 4y - 30 = 6. Combine: 9y = 36, giving y = 4.
In simple words: Remove inner brackets first, then work outward layer by layer.

Exam Tip: Nested brackets require careful step-by-step expansion - do not skip layers or mix the order.

Question 3. (ii) 0.3(6 – x) = 0.4(x + 8)
Answer: Expand both sides: 1.8 - 0.3x = 0.4x + 3.2. Collect x terms on one side: -0.3x - 0.4x = 3.2 - 1.8, which gives -0.7x = 1.4. Divide by -0.7: x = 1.4 / (-0.7) = -2.
In simple words: Expand, collect x's on one side, then solve by dividing.

Exam Tip: With decimals, move them carefully - dividing a negative by a negative gives a positive result.

Question 4. (i) (x – 1) / 3 = {(x + 2) / 6} + 3
Answer: Subtract (x + 2) / 6 from both sides: (x - 1) / 3 - (x + 2) / 6 = 3. Find a common denominator of 6: {2(x - 1) - (x + 2)} / 6 = 3. Simplify: (2x - 2 - x - 2) / 6 = 3, which becomes (x - 4) / 6 = 3. Multiply both sides by 6: x - 4 = 18, so x = 22.
In simple words: Use a common denominator, combine fractions, then multiply to clear the fraction.

Exam Tip: The LCM of the denominators is your common denominator - find it first before combining fractions.

Question 4. (ii) (x + 7) / 3 = 1 + {(3x – 2) / 5}
Answer: Rearrange: (x + 7) / 3 - (3x - 2) / 5 = 1. Use common denominator 15: {5(x + 7) - 3(3x - 2)} / 15 = 1. Expand: (5x + 35 - 9x + 6) / 15 = 1, giving (-4x + 41) / 15 = 1. Multiply by 15: -4x + 41 = 15. Solve: -4x = -26, so x = 13/2.
In simple words: Move terms to one side, find a common denominator, then clear fractions by multiplying.

Exam Tip: When you see multiple fractions with different denominators, use the LCM to combine them before solving.

Question 5. (i) {(y + 1) / 3} – {(y – 1) / 2} = (1 + 2y) / 3
Answer: Use common denominator 6 on the left side: {2(y + 1) - 3(y - 1)} / 6 = (1 + 2y) / 3. Simplify the numerator: (2y + 2 - 3y + 3) / 6 = (1 + 2y) / 3, giving (-y + 5) / 6 = (1 + 2y) / 3. Cross-multiply: 3(-y + 5) = 6(1 + 2y), which becomes -3y + 15 = 6 + 12y. Rearrange: -15y = -9, so y = 3/5.
In simple words: Combine the fractions on one side, then cross-multiply to clear all denominators.

Exam Tip: Cross-multiplication is fastest when you have one fraction on each side of the equals sign.

Question 5. (ii) (p / 3) + (p / 4) = 55 – {(p + 40) / 5}
Answer: Rearrange: (p / 3) + (p / 4) + (p + 40) / 5 = 55. Find LCM of 3, 4, 5 which is 60. Rewrite with common denominator: {20p + 15p + 12(p + 40)} / 60 = 55. Simplify numerator: (20p + 15p + 12p + 480) / 60 = 55, giving (47p + 480) / 60 = 55. Multiply by 60: 47p + 480 = 3300. Solve: 47p = 2820, so p = 60.
In simple words: Move all fractions to one side, use a common denominator, then multiply through and solve.

Exam Tip: For three or more fractions, finding the LCM prevents calculation errors when combining them.

Question 6. (i) n – {(n – 1) / 2} = 1 – {(n – 2) / 3}
Answer: Simplify the left side: (2n - n + 1) / 2 = (n + 1) / 2. Simplify the right side: (3 - n + 2) / 3 = (5 - n) / 3. Set them equal: (n + 1) / 2 = (5 - n) / 3. Cross-multiply: 3(n + 1) = 2(5 - n), giving 3n + 3 = 10 - 2n. Combine: 5n = 7, so n = 7/5.
In simple words: Rewrite each side as a single fraction, then cross-multiply to solve.

Exam Tip: When a whole number and a fraction are on the same side, combine them into one fraction first.

Question 6. (ii) {(3t – 2) / 3} + {(2t + 3) / 2} = t + (7 / 6)
Answer: Combine fractions on the left with common denominator 6: {2(3t - 2) + 3(2t + 3)} / 6 = (6t + 7) / 6. Expand numerator: (6t - 4 + 6t + 9) / 6 = (6t + 7) / 6. Simplify: (12t + 5) / 6 = (6t + 7) / 6. Multiply both sides by 6: 12t + 5 = 6t + 7. Solve: 6t = 2, giving t = 1/3.
In simple words: Get common denominators on both sides, multiply through to clear fractions, then solve.

Exam Tip: When denominators match on both sides after simplification, you can multiply through directly.

Question 7. (i) 4(3x + 2) – 5(6x – 1) = 2(x – 8) – 6(7x – 4)
Answer: Expand all brackets: 12x + 8 - 30x + 5 = 2x - 16 - 42x + 24. Combine terms on each side: -18x + 13 = -40x + 8. Add 40x to both sides: 22x + 13 = 8. Subtract 13: 22x = -5, so x = -5/22.
In simple words: Expand all brackets, combine like terms on each side, then solve.

Exam Tip: When expanding, be careful with negative signs in front of brackets - distribute them to each term inside.

Question 7. (ii) 3(5x + 7) + 5(2x – 11) = 3(8x – 5) – 15
Answer: Expand: 15x + 21 + 10x - 55 = 24x - 15 - 15. Combine on left: 25x - 34 = 24x - 30. Subtract 24x: x - 34 = -30. Add 34: x = 4.
In simple words: Expand, combine like terms, then isolate x.

Exam Tip: When x is on both sides, move all x terms to one side to isolate the variable.

Question 8. (i) (3 – 2x) / (2x + 5) = - (3 / 11)
Answer: Cross-multiply: 11(3 - 2x) = -3(2x + 5). Expand: 33 - 22x = -6x - 15. Collect x terms: -22x + 6x = -15 - 33, giving -16x = -48. Divide: x = 3.
In simple words: Cross-multiply to remove fractions, then expand and solve.

Exam Tip: Always cross-multiply first when you have a fraction equal to a fraction - it clears denominators in one step.

Question 8. (ii) (5p + 2) / (8 – 2p) = 7 / 6
Answer: Cross-multiply: 6(5p + 2) = 7(8 - 2p). Expand: 30p + 12 = 56 - 14p. Collect p terms: 30p + 14p = 56 - 12, giving 44p = 44. Divide: p = 1.
In simple words: Cross-multiply, expand, collect variables, then divide.

Exam Tip: Check your answer by substituting back into the original equation to verify.

Question 9. (i) 5 / x = 7 / (x – 4)
Answer: Cross-multiply: 5(x - 4) = 7x. Expand: 5x - 20 = 7x. Rearrange: -20 = 2x, so x = -10.
In simple words: Cross-multiply, expand, then solve the resulting simple equation.

Exam Tip: When x appears in a denominator, be alert to which value of x would make the denominator zero - that cannot be your answer.

Question 9. (ii) 4 / (2x + 3) = 5 / (x + 4)
Answer: Cross-multiply: 4(x + 4) = 5(2x + 3). Expand: 4x + 16 = 10x + 15. Rearrange: -6x = -1, so x = 1/6.
In simple words: Cross-multiply, expand both sides, collect x terms, then solve.

Exam Tip: Verify that your solution does not make any denominator zero in the original equation.

Question 10. (i) {(2x + 5) / 2} – {5x / (x – 1)} = x
Answer: Find a common denominator of 2(x - 1): {{(2x + 5)(x - 1) - 5x(2)} / {2(x - 1)}} = x. Expand numerator: {(2x² - 2x + 5x - 5 - 10x) / (2x - 2)} = x, which gives (2x² - 7x - 5) / (2x - 2) = x. Multiply both sides by (2x - 2): 2x² - 7x - 5 = x(2x - 2) = 2x² - 2x. Simplify: -7x - 5 = -2x, so -5x = 5, giving x = -1.
In simple words: Combine fractions with a common denominator, multiply through to clear fractions, then solve the resulting equation.

Exam Tip: When you have fractions with polynomial denominators, the common denominator is the product of all different denominators.

Question 10. (ii) 1 / 5 {(1 / 3x) – 5} = 1 / 3 {3 – (1 / x)}
Answer: Simplify each side. Left side: (1 / 5) × {(1 - 15x) / (3x)} = (1 - 15x) / (15x). Right side: (1 / 3) × {(3x - 1) / x} = (3x - 1) / (3x). Set equal: (1 - 15x) / (15x) = (3x - 1) / (3x). Cross-multiply: 3x(1 - 15x) = 15x(3x - 1). Divide both sides by 3x: 1 - 15x = 5(3x - 1) = 15x - 5. Rearrange: 1 + 5 = 15x + 15x, giving 6 = 30x, so x = 1/5.
In simple words: Simplify each side to a single fraction, cross-multiply, then solve.

Exam Tip: Nested fractions can be cleared by working methodically from the innermost denominator outward.

Question 11. (i) {(2x – 3) / (2x – 1)} = {(3x – 1) / (3x + 1)}
Answer: Cross-multiply: (2x - 3)(3x + 1) = (3x - 1)(2x - 1). Expand left side: 6x² + 2x - 9x - 3 = 6x² - 7x - 3. Expand right side: 6x² - 3x - 2x + 1 = 6x² - 5x + 1. Set equal: 6x² - 7x - 3 = 6x² - 5x + 1. Cancel 6x²: -7x - 3 = -5x + 1. Rearrange: -2x = 4, so x = -2.
In simple words: Cross-multiply, expand both sides, cancel matching terms, then solve.

Exam Tip: When squared terms appear on both sides, they often cancel - simplify before concluding you have a quadratic.

Question 11. (ii) {(2y + 3) / (3y + 2)} = {(4y + 5) / (6y + 7)}
Answer: Cross-multiply: (2y + 3)(6y + 7) = (4y + 5)(3y + 2). Expand left: 12y² + 14y + 18y + 21 = 12y² + 32y + 21. Expand right: 12y² + 8y + 15y + 10 = 12y² + 23y + 10. Set equal: 12y² + 32y + 21 = 12y² + 23y + 10. Cancel 12y²: 32y + 21 = 23y + 10. Rearrange: 9y = -11, so y = -11/9.
In simple words: Cross-multiply, expand, cancel like terms, then solve the remaining equation.

Exam Tip: Use FOIL or the distribution method carefully when expanding products of two binomials.

Question 12. If x = p + 1, find the value of p from the equation (1 / 2) (5x – 30) – (1/ 3) (1 + 7p) = 1 / 4
Answer: Substitute x = p + 1 into the given equation: (1 / 2) {5(p + 1) - 30} - (1 / 3) (1 + 7p) = 1 / 4. Simplify: (1 / 2)(5p + 5 - 30) - (1 / 3)(1 + 7p) = 1 / 4, which becomes (1 / 2)(5p - 25) - (1 / 3)(1 + 7p) = 1 / 4. Rewrite: (5p - 25) / 2 - (1 + 7p) / 3 = 1 / 4. Use common denominator 6: {3(5p - 25) - 2(1 + 7p)} / 6 = 1 / 4. Simplify numerator: (15p - 75 - 2 - 14p) / 6 = 1 / 4, giving (p - 77) / 6 = 1 / 4. Cross-multiply: 4(p - 77) = 6, so 4p - 308 = 6. Solve: 4p = 314, giving p = 157/2.
In simple words: Replace x with p + 1, combine fractions, then solve for p.

Exam Tip: When substituting, be careful to replace the variable completely and simplify before combining fractions.

Question 13. Solve {(x + 3) / 3} – {(x – 2) / 2} = 1, Hence find p if (1 / x) + P = 1
Answer: First, solve for x. Use common denominator 6: {2(x + 3) - 3(x - 2)} / 6 = 1. Expand: (2x + 6 - 3x + 6) / 6 = 1, which gives (-x + 12) / 6 = 1. Multiply by 6: -x + 12 = 6, so -x = -6, giving x = 6. Now use x = 6 in (1 / x) + P = 1: (1 / 6) + P = 1. Multiply by 6: 1 + 6P = 6, so 6P = 5, giving P = 5/6.
In simple words: First solve the equation to find x, then substitute that value into the second equation to find P.

Exam Tip: Two-part questions require you to complete the first part before using its answer in the second part.

Exercise 12.2

Question 1. Three more than twice a number is equal to four less than the number. Find the number.
Answer: Let the number be x. Twice the number is 2x, and three more than twice the number is 2x + 3. Four less than the number is x - 4. Set up the equation: 2x + 3 = x - 4. Move x to the left and constants to the right: 2x - x = -4 - 3, giving x = -7. Therefore, the number is -7.
In simple words: Write what you know into an equation, then solve for the unknown number.

Exam Tip: Always define your variable clearly and translate the words carefully into mathematical symbols.

Question 2. When four consecutive integers are added, the sum is 46. Find the integers.
Answer: Let the first integer be x. The next three consecutive integers are x + 1, x + 2, and x + 3. Their sum is x + (x + 1) + (x + 2) + (x + 3) = 46. Combine: 4x + 6 = 46, so 4x = 40, giving x = 10. The four consecutive integers are 10, 11, 12, and 13.
In simple words: Use x for the first number, then the rest are x + 1, x + 2, x + 3. Add them and solve.

Exam Tip: For consecutive integers, always add 1 to get the next one; for consecutive even or odd integers, add 2 instead.

Question 3. Manjula thinks a number and subtracts 7 / 3 from it. She multiplies the result by 6. The result now obtained is 2 less than twice the same number she thought of. What is the number?
Answer: Let the number be x. Subtracting 7/3 from it gives x - 7/3. Multiplying by 6 gives 6(x - 7/3) = 6x - 14. This equals 2 less than twice the number, which is 2x - 2. Set up: 6x - 14 = 2x - 2. Rearrange: 6x - 2x = -2 + 14, so 4x = 12, giving x = 3. The number is 3.
In simple words: Follow the steps she takes, write them as math, then solve.

Exam Tip: Break down multi-step word problems into individual operations and translate each into mathematical form.

Question 4. A positive number is 7 times another number. If 15 is added to both the numbers, then one of the new numbers becomes (5 / 2) times the other new number. What are the numbers?
Answer: Let one number be x. The other is x/7. When 15 is added to both, they become x + 15 and x/7 + 15. According to the condition, x + 15 = (5/2) × (x/7 + 15). Expand: 2(x + 15) = (5x/7) + 75, which becomes 2x + 30 = (5x/7) + 75. Rearrange: 2x - (5/7)x = 75 - 30, giving (9/7)x = 45. Solve: x = 35. The other number is 35/7 = 5. The numbers are 35 and 5.
In simple words: Define one number in terms of the other, add 15 to each, then use the new relationship to solve.

Exam Tip: When one number relates to another by multiplication or division, express both in terms of one variable.

Question 5. When three consecutive even integers are added, the sum is zero. Find the integers.
Answer: Let the first even integer be x. The next two are x + 2 and x + 4. Their sum is x + (x + 2) + (x + 4) = 0. Combine: 3x + 6 = 0, so 3x = -6, giving x = -2. The three consecutive even integers are -2, 0, and 2.
In simple words: For consecutive even integers, add 2 each time to move to the next one.

Exam Tip: Consecutive even integers differ by 2, and consecutive odd integers also differ by 2; only regular integers differ by 1.

Question 6. Find two consecutive odd integers such that two-fifth of the smaller exceeds two-ninth of the greater by 4.
Answer: Let the first odd integer be x. The next is x + 2. Two-fifth of the smaller is (2/5)x. Two-ninth of the greater is (2/9)(x + 2). Set up: (2/5)x = (2/9)(x + 2) + 4. Rearrange: (2/5)x - (2/9)(x + 2) = 4. Use common denominator 45: {18x - 10(x + 2)} / 45 = 4, which becomes (18x - 10x - 20) / 45 = 4, so (8x - 20) / 45 = 4. Multiply by 45: 8x - 20 = 180, giving 8x = 200, so x = 25. The two consecutive odd integers are 25 and 27.
In simple words: Write the relationship given, use fractions carefully, then solve.

Exam Tip: When comparing two quantities with fractions, align one on each side of the equals sign for clarity.

Question 7. The denominator of a fraction is 1 more than twice its numerator. If the numerator and denominator are both increased by 5, it becomes (3 / 5). Find the original fraction.
Answer: Let the numerator be x. Then the denominator is 2x + 1, so the original fraction is x / (2x + 1). When both are increased by 5, the new fraction is (x + 5) / (2x + 1 + 5) = (x + 5) / (2x + 6). This equals 3/5, so 5(x + 5) = 3(2x + 6). Expand: 5x + 25 = 6x + 18. Rearrange: -x = -7, so x = 7. The original fraction is 7 / (2(7) + 1) = 7/15.
In simple words: Define the fraction using a variable, apply the changes, set it equal to 3/5, then solve.

Exam Tip: When the relationship between numerator and denominator is given, use it to write the fraction in one variable.

Question 8. Find two positive numbers in the ratio 2: 5 such that their difference is 15.
Answer: Let the two numbers be 2x and 5x, which are in the ratio 2:5. Their difference is 5x - 2x = 15. Solve: 3x = 15, so x = 5. The numbers are 2(5) = 10 and 5(5) = 25. Therefore, the required numbers are 10 and 25.
In simple words: When numbers are in a given ratio, use that ratio to express both in terms of one variable.

Exam Tip: If numbers are in ratio a:b, write them as ax and bx; this makes finding their values straightforward.

Question 9. What number should be added to each of the numbers 12, 22, 42 and 72 so that the resulting numbers may be in proportion?
Answer: Let x be the number to be added. When added to each, the numbers become 12 + x, 22 + x, 42 + x, and 72 + x. For them to be in proportion: (12 + x) / (22 + x) = (42 + x) / (72 + x). Cross-multiply: (12 + x)(72 + x) = (42 + x)(22 + x). Expand left: 864 + 12x + 72x + x² = 864 + 84x + x². Expand right: 924 + 42x + 22x + x² = 924 + 64x + x². Set equal: 864 + 84x + x² = 924 + 64x + x². Cancel x²: 864 + 84x = 924 + 64x, so 20x = 60, giving x = 3. The number to be added is 3.
In simple words: Add x to each number, set the ratios equal (first to second as third to fourth), then solve.

Exam Tip: When four numbers are in proportion, the product of the first and fourth equals the product of the second and third (cross-multiply).

Question 10. The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What can be the original number?
Answer: Let the one's digit be x. Since the digits differ by 3, the ten's digit is x + 3. The original number is 10(x + 3) + x = 10x + 30 + x = 11x + 30. When the digits are interchanged, the one's digit becomes x + 3 and the ten's digit becomes x, so the new number is 10x + (x + 3) = 11x + 3. Their sum is (11x + 30) + (11x + 3) = 143, which simplifies to 22x + 33 = 143. Solve: 22x = 110, so x = 5. Wait, let me recalculate: 22x = 143 - 33 = 110, so x = 5. But check: if x = 5, then one's digit is 5, ten's digit is 8, original number is 85. Interchanged: 58. Sum: 85 + 58 = 143. ✓ The original number is 85. However, the problem states digits differ by 3, but doesn't specify which is larger; another possibility would exist if the one's digit were larger. But if one's digit = x and it exceeds ten's digit by 3, then ten's digit = x - 3. Original: 10(x - 3) + x = 11x - 30. Interchanged: 10x + (x - 3) = 11x - 3. Sum: (11x - 30) + (11x - 3) = 143 gives 22x - 33 = 143, so 22x = 176, x = 8. Original number would be 85 (same answer, different reasoning).
In simple words: Use a variable for one digit, express both digits and the numbers formed, then set up and solve the equation.

Exam Tip: For two-digit numbers, the value is 10 × (ten's digit) + (one's digit); remember this formula when working with digit problems.

Question 11. Sum of the digits of a two-digit number is 11. When we interchange the digits, it is found that the resulting new number is greater than the original number by 63. Find the two-digit number.
Answer: Let the unit's digit be x. Then the ten's digit is 11 - x. The original number can be written as x + 10(11 - x) = 110 - 9x. After swapping the digits, the new number becomes (11 - x) + 10x = 11 + 9x. According to the problem, the difference between the new and original numbers is 63, so (11 + 9x) - (110 - 9x) = 63. Simplifying: 11 + 9x - 110 + 9x = 63, which gives 18x = 162. Therefore, x = 9. The original number = 110 - 9(9) = 110 - 81 = 29.
In simple words: When you swap the digits of 29, you get 92. The difference is 92 - 29 = 63, which matches the given condition.

Exam Tip: Always express the two-digit number in the form (ten's digit × 10 + unit's digit) to set up the equation correctly; this avoids common errors in digit placement.

Question 12. Ritu is now four times as old as his brother Raju. In 4 years time, her age will be twice of Raju's age. What are their present ages?
Answer: Let Raju's current age be x years. Then Ritu's age is 4x years. In 4 years, Raju will be (x + 4) years old and Ritu will be (4x + 4) years old. Given that Ritu's age will be twice Raju's age at that time: 4x + 4 = 2(x + 4) = 2x + 8. Solving: 4x - 2x = 8 - 4, so 2x = 4, giving x = 2. Therefore, Raju is currently 2 years old and Ritu is 4 × 2 = 8 years old.
In simple words: Raju is 2 years old now, and Ritu is 8 years old - that's 4 times older. In 4 years, Raju will be 6 and Ritu will be 12, making Ritu twice as old.

Exam Tip: Always check your answer by substituting back into the original conditions to verify both relationships are satisfied.

Question 13. A father is 7 times as old as his son. Two years ago, the father was 13 times as old as his son. How old are they now?
Answer: Let the son's present age be x years. Then the father's age is 7x years. Two years ago, the son was (x - 2) years and the father was (7x - 2) years. According to the condition, the father's age two years back was 13 times the son's age then: 7x - 2 = 13(x - 2) = 13x - 26. Rearranging: 7x - 13x = -26 + 2, so -6x = -24, giving x = 4. Therefore, the son is currently 4 years old and the father is 7 × 4 = 28 years old.
In simple words: The son is 4 years old and the father is 28 years old now. Two years ago they were 2 and 26, and 26 is indeed 13 times 2.

Exam Tip: When dealing with past or future age problems, always express ages relative to that time period, then set up the equation based on the given relationship at that specific moment.

Question 14. The ages of Sona and Sonali are in the ratio 5: 3. Five years hence, the ratio of their ages will be 10: 7. Find their present ages.
Answer: Let Sona's present age be 5x years and Sonali's be 3x years. In 5 years, Sona will be (5x + 5) and Sonali will be (3x + 5). The ratio at that time will be 10:7, so (5x + 5)/(3x + 5) = 10/7. Cross-multiplying: 7(5x + 5) = 10(3x + 5), which gives 35x + 35 = 30x + 50. Simplifying: 5x = 15, so x = 3. Therefore, Sona is currently 5 × 3 = 15 years old and Sonali is 3 × 3 = 9 years old.
In simple words: Currently Sona is 15 and Sonali is 9, making the ratio 15:9 = 5:3. In 5 years they'll be 20 and 14, and 20:14 = 10:7.

Exam Tip: When a ratio is given, always express the quantities as multiples of a variable (like 5x and 3x); this preserves the ratio and simplifies solving the equation.

Question 15. An employee works in a company on a contract of 30 days on the condition that he will receive Rs 200 for each day he works and he will be fined Rs 20 for each day if he is absent. If he receives Rs 3800 in all, for how many days did he remain absent?
Answer: Let the employee be absent for x days. Then he works for (30 - x) days. His total earnings are calculated as: (days worked × Rs 200) - (days absent × Rs 20) = 3800. Substituting: (30 - x) × 200 - x × 20 = 3800, which gives 6000 - 200x - 20x = 3800. Simplifying: 6000 - 220x = 3800, so 220x = 2200, giving x = 10. Therefore, the employee was absent for 10 days.
In simple words: The employee worked 20 days and earned 20 × 200 = 4000. He was absent 10 days and lost 10 × 20 = 200. His net pay is 4000 - 200 = 3800.

Exam Tip: Earnings problems require careful attention to whether amounts are added or subtracted; always use positive earnings for work and negative amounts (or direct subtraction) for fines.

Question 16. I have a total of Rs 300 in coins of denomination Rs 1, Rs 2 and Rs 5. The number of coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Answer: Let the number of Rs 5 coins be x. Then the number of Rs 2 coins is 3x, and the number of Rs 1 coins is 160 - (x + 3x) = 160 - 4x. The total value equation is: 1 × (160 - 4x) + 2 × 3x + 5 × x = 300. Expanding: 160 - 4x + 6x + 5x = 300, which simplifies to 160 + 7x = 300. Solving: 7x = 140, so x = 20. Therefore, there are 20 coins of Rs 5, 60 coins of Rs 2 (since 3 × 20 = 60), and 80 coins of Rs 1 (since 160 - 20 - 60 = 80).
In simple words: You have 20 five-rupee coins worth 100 rupees, 60 two-rupee coins worth 120 rupees, and 80 one-rupee coins worth 80 rupees, totaling 300 rupees.

Exam Tip: In coin problems, always set up one equation for the total count and another for the total value; verify your answer by checking both conditions.

Question 17. A local bus is carrying 40 passengers, some with Rs 5 tickets and the remaining with Rs 7.50 tickets. If the total receipts from these passengers are Rs 230, find the number of passengers with Rs 5 tickets.
Answer: Let the number of passengers with Rs 5 tickets be x. Then the number with Rs 7.50 tickets is (40 - x). The total revenue equation is: 5x + 7.50(40 - x) = 230. Expanding: 5x + 300 - 7.5x = 230. Simplifying: -2.5x = -70, so x = 28. Therefore, 28 passengers have Rs 5 tickets.
In simple words: 28 passengers paid 5 rupees each (140 rupees total) and 12 passengers paid 7.50 rupees each (90 rupees total), giving 140 + 90 = 230 rupees.

Exam Tip: When dealing with decimal amounts in ticket problems, multiply through by 10 or 100 early on to eliminate decimals and reduce calculation errors.

Question 18. On a school picnic, a group of students agree to pay equally for the use of a full boat and pay Rs 10 each. If there had been 3 more students in the group, each would have paid Rs 2 less. How many students were there in the group?
Answer: Let the number of students in the group be x. The total boat cost is 10x rupees (since each pays Rs 10). If there were 3 more students, the group would have (x + 3) members. Each would then pay (10 - 2) = 8 rupees, so the total cost remains: 8(x + 3) = 10x. Expanding: 8x + 24 = 10x. Solving: 24 = 2x, so x = 12. Therefore, there were 12 students in the group.
In simple words: 12 students paying 10 rupees each pay 120 rupees total. If 15 students shared the same 120 rupees, each would pay 8 rupees, which is 2 rupees less.

Exam Tip: In problems where a fixed cost is shared among a changing number of people, equate the total cost (not the per-person rate) in both scenarios.

Question 19. Half of a 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 deer in the herd.
Answer: Let the total number of deer be x. Half are grazing: x/2. The remaining after grazing is x - x/2 = x/2. Three-fourths of the remaining are playing: (3/4) × (x/2) = 3x/8. The rest drinking water: x/2 - 3x/8 = 4x/8 - 3x/8 = x/8. We are told this equals 9: x/8 = 9. Solving: x = 72. Therefore, the herd has 72 deer.
In simple words: 36 deer are grazing, 27 are playing, and 9 are drinking water, making a total of 72 deer.

Exam Tip: When working with fractions of a total, carefully track what remains after each group is removed; using a clear step-by-step breakdown prevents errors.

Question 20. Sakshi takes some flowers in a basket and visits three temples one by one. At each temple, she offers one-half of the flowers from the basket. If she is left with 6 flowers at the end, find the number of flowers she had in the beginning.
Answer: Let the initial number of flowers be x. At the first temple, she offers x/2, leaving x/2. At the second temple, she offers (1/2) of x/2 = x/4, leaving x/2 - x/4 = x/4. At the third temple, she offers (1/2) of x/4 = x/8, leaving x/4 - x/8 = x/8. Given that x/8 = 6, we get x = 48. Therefore, she started with 48 flowers.
In simple words: She gave away half at the first temple (left with 24), half of that at the second temple (left with 12), and half again at the third temple (left with 6).

Exam Tip: In sequential fraction problems, track the amount remaining (not given away) after each step; working backwards from the final amount can also verify your answer quickly.

Question 21. Two supplementary angles differ by 50°. Find the measure of each angle.
Answer: Let one angle be x degrees. Since the angles are supplementary, the other is (180 - x) degrees. They differ by 50°, so x - (180 - x) = 50. Simplifying: x - 180 + x = 50, which gives 2x - 180 = 50. Solving: 2x = 230, so x = 115. The two angles are 115° and (180 - 115) = 65°.
In simple words: Two angles that add up to 180° are called supplementary. Here, 115° + 65° = 180°, and their difference is 115° - 65° = 50°.

Exam Tip: Always verify supplementary angle answers by checking that both the sum equals 180° and the difference matches the given value.

Question 22. If the angles of a triangle are in the ratio 5: 6: 7, find the angles.
Answer: Let the angles be 5x, 6x, and 7x degrees. Since the sum of angles in a triangle is 180°, we have 5x + 6x + 7x = 180. Simplifying: 18x = 180, so x = 10. The three angles are 5 × 10 = 50°, 6 × 10 = 60°, and 7 × 10 = 70°.
In simple words: The angles are 50°, 60°, and 70°. Their ratio is 50:60:70, which simplifies to 5:6:7, and they add up to 180°.

Exam Tip: When angles are given in a ratio, express them as multiples of a variable and use the fact that the angle sum in a triangle is always 180°.

Question 23. Two equal sides of an isosceles triangle are 3x - 1 and 2x + 2 units. The third side is 2x units. Find x and the perimeter of the triangle.
Answer: In an isosceles triangle, the two equal sides must be equal in length. Therefore: 3x - 1 = 2x + 2. Solving: 3x - 2x = 2 + 1, so x = 3. The two equal sides are each 3(3) - 1 = 8 units, and the third side is 2(3) = 6 units. The perimeter is 8 + 8 + 6 = 22 units.
In simple words: When x = 3, the two equal sides each measure 8 units and the base is 6 units, giving a perimeter of 22 units.

Exam Tip: In isosceles triangle problems, equate the expressions for the two equal sides first to find the variable; always verify the triangle inequality is satisfied before stating the perimeter.

Question 24. If each side of a triangle is increased by 4 cm, the ratio of the perimeters of the new triangle and the given triangle is 7: 5. Find the perimeter of the given triangle.
Answer: Let the perimeter of the original triangle be x cm. When each of the three sides is increased by 4 cm, the perimeter increases by 3 × 4 = 12 cm, making the new perimeter (x + 12) cm. The ratio of the new to original perimeter is 7:5, so (x + 12)/x = 7/5. Cross-multiplying: 5(x + 12) = 7x, which gives 5x + 60 = 7x. Solving: 60 = 2x, so x = 30. The perimeter of the given triangle is 30 cm.
In simple words: The original triangle has a perimeter of 30 cm. When all sides increase by 4 cm each, the perimeter becomes 30 + 12 = 42 cm, and the ratio 42:30 = 7:5.

Exam Tip: Remember that increasing each side by a fixed length increases the perimeter by (3 times that length) for a triangle; set up the ratio carefully using the new and original perimeters.

Question 25. The length of a rectangle is 5 cm less than twice its breadth. If the length is decreased by 3 cm and breadth increased by 2 cm, the perimeter of the resulting rectangle is 72 cm. Find the area of the original rectangle.
Answer: Let the breadth of the original rectangle be x cm. Then the length is (2x - 5) cm. When the length is decreased by 3 cm, the new length becomes (2x - 5 - 3) = (2x - 8) cm. When the breadth is increased by 2 cm, the new breadth becomes (x + 2) cm. The perimeter of the new rectangle is 2[(2x - 8) + (x + 2)] = 2(3x - 6) = 6x - 12 cm. Setting this equal to 72: 6x - 12 = 72, so 6x = 84, giving x = 14. The original rectangle has breadth 14 cm and length 2(14) - 5 = 23 cm. The area is 23 × 14 = 322 cm².
In simple words: The original rectangle is 23 cm long and 14 cm wide, with an area of 322 square cm. After changes, it becomes 20 cm by 16 cm with a perimeter of 72 cm.

Exam Tip: Always express length and breadth in terms of the same variable using the given relationship; then apply the perimeter formula to the modified rectangle, not the original.

Question 26. A rectangle is 10 cm long and 8 cm wide. When each side of the rectangle is increased by x cm, its perimeter is doubled. Find the equation in x and hence find the area of the new rectangle.
Answer: The original perimeter is 2(10 + 8) = 36 cm. When each side is increased by x cm, the new length is (10 + x) cm and the new breadth is (8 + x) cm. The new perimeter is 2[(10 + x) + (8 + x)] = 2(18 + 2x) = 36 + 4x cm. Since the perimeter doubles: 36 + 4x = 2 × 36 = 72. Solving: 4x = 36, so x = 9. The new rectangle has length 10 + 9 = 19 cm and breadth 8 + 9 = 17 cm. The area of the new rectangle is 19 × 17 = 323 cm².
In simple words: The original rectangle (10 by 8 cm) has a perimeter of 36 cm. Adding 9 cm to each side creates a 19 by 17 cm rectangle with a perimeter of 72 cm and an area of 323 square cm.

Exam Tip: When each dimension changes by the same amount, express the new dimensions and new perimeter in terms of that change variable; doubling the perimeter means multiplying the original perimeter by 2.

Question 27. A steamer travels 90 km downstream in the same time as it takes to travel 60 km upstream. If the speed of the stream is 5 km/hr, find the speed of the steamer in still water.
Answer: Let the steamer's speed in still water be x km/h. The downstream speed is (x + 5) km/h and the upstream speed is (x - 5) km/h. Since time is the same in both cases: 90/(x + 5) = 60/(x - 5). Cross-multiplying: 90(x - 5) = 60(x + 5), which gives 90x - 450 = 60x + 300. Solving: 30x = 750, so x = 25. The steamer's speed in still water is 25 km/h.
In simple words: At 25 km/h in still water, the steamer goes 30 km/h downstream and 20 km/h upstream. It takes 3 hours to cover 90 km downstream or 3 hours to cover 60 km upstream.

Exam Tip: In boat/stream problems, always use time = distance/speed and equate the times when they are equal; downstream speed = steamer speed + stream speed, upstream speed = steamer speed - stream speed.

Question 28. A steamer goes downstream and covers the distance between two ports in 5 hours while it covers the same distance upstream in 6 hours. If the speed of the stream is 1 km/h, find the speed of the steamer in still water and the distance between two ports.
Answer: Let the steamer's speed in still water be x km/h. The downstream speed is (x + 1) km/h and the upstream speed is (x - 1) km/h. The distance is the same in both directions, so: 5(x + 1) = 6(x - 1). Expanding: 5x + 5 = 6x - 6. Solving: x = 11. The steamer's speed in still water is 11 km/h. The distance is 5(11 + 1) = 5 × 12 = 60 km.
In simple words: The steamer moves at 11 km/h in still water, 12 km/h downstream, and 10 km/h upstream. It covers 60 km in 5 hours going downstream or 6 hours going upstream.

Exam Tip: When the same distance is covered in different times with different speeds, equate the distances (speed × time in each direction) to solve for the unknown speed.

Question 29. Distance between two places A and B is 350 km. Two cars start simultaneously from A and B towards each other and the distance between them after 4 hours is 62 km. If the speed of one car is 8 km/h less than the speed of other cars, find the speed of each car.
Answer: Let the speed of car C₁ be x km/h and the speed of car C₂ be (x - 8) km/h. In 4 hours, C₁ travels 4x km and C₂ travels 4(x - 8) km. Together they cover 4x + 4(x - 8) = 8x - 32 km. Since 62 km remains between them: 8x - 32 = 350 - 62 = 288. Solving: 8x = 320, so x = 40. Car C₁ has a speed of 40 km/h and car C₂ has a speed of 40 - 8 = 32 km/h.
In simple words: The faster car travels at 40 km/h and covers 160 km in 4 hours. The slower car travels at 32 km/h and covers 128 km in 4 hours. Together they cover 288 km, leaving 62 km between them.

Exam Tip: In problems where two objects move toward each other, the total distance covered by both in time t equals (combined speed) × t; set this equal to (original distance - remaining distance).

Exercise 12.3

Question 1. If the replacement set = {-7, -5, -3, -1, 1, 3}, find the solution set of: (i) x > - 2 (ii) x < - 2 (iii) x > 2 (iv) -5 < x ≤ 5 (v) -8 < x < 1 (vi) 0 ≤ x ≤ 4
Answer:
(i) The solution set of x > - 2 is {-1, 1, 3}
(ii) The solution set of x < - 2 is {-7, -5, -3}
(iii) The solution set of x > 2 is {3}
(iv) The solution set of -5 < x ≤ 5 is {-3, -1, 1, 3}
(v) The solution set of -8 < x < 1 is {-7, -5, -3, -1}
(vi) The solution set of 0 ≤ x ≤ 4 is {1, 3}
In simple words: For each inequality, choose only the values from the replacement set that make the inequality true.

Exam Tip: When finding a solution set from a given replacement set, always check each value individually against the inequality condition; never assume continuity.

Question 2. Represent the solution of the following inequalities graphically: (i) x ≤ 4, x ε N (ii) x < 5, x ε W (iii) -3 ≤ x < 3, x ε I
Answer:
(i) For x ≤ 4, x ε N, the solution set is {1, 2, 3, 4}. These are shown as filled dots on a number line.


(ii) For x < 5, x ε W, the solution set is {0, 1, 2, 3, 4}. These are shown as filled dots on a number line.


(iii) For -3 ≤ x < 3, x ε I, the solution set is {-3, -2, -1, 0, 1, 2}. These are shown as filled dots on a number line.


In simple words: Draw a number line and mark the values that satisfy the inequality with filled circles; the filled circles show which numbers are part of the solution.

Exam Tip: When representing inequalities graphically, use filled circles for values included in the solution (≤ or ≥) and open circles for values excluded (< or >).

Question 3. If the replacement set is {-6, -4, -2, 0, 2, 4, 6}; then represent the solution set of the inequality -4 ≤ x < 4 graphically.
Answer: For the inequality -4 ≤ x < 4, we need values from the replacement set that satisfy this condition. These are {-4, -2, 0, 2}. These are shown as filled circles on a number line.


In simple words: The solution set includes -4 (since x can equal -4) but excludes 4 (since x must be less than 4). The values -4, -2, 0, and 2 are marked on the number line.

Exam Tip: Pay close attention to whether inequality symbols include equality (≤ or ≥) or not (< or >); this affects which boundary values are included in the solution.

Question 4. Find the solution set of the inequality x < 4 if the replacement set is (i) {1, 2, 3, ....., 10} (ii) {-1, 0, 1, 2, 5, 8} (iii) {-5, 10} (iv) {5, 6, 7, 8, 9, 10}
Answer:
(i) For replacement set {1, 2, 3, ....., 10}, the solution set of x < 4 is {1, 2, 3}
(ii) For replacement set {-1, 0, 1, 2, 5, 8}, the solution set of x < 4 is {-1, 0, 1, 2}
(iii) For replacement set {-5, 10}, the solution set of x < 4 is {-5}
(iv) For replacement set {5, 6, 7, 8, 9, 10}, the solution set of x < 4 is { } (empty set)
In simple words: For each replacement set, choose only those elements that are less than 4; if none exist, the solution set is empty.

Exam Tip: The solution set depends entirely on the given replacement set; always check each element individually and only include those that satisfy the inequality condition.

Question 5. If the replacement set = {-6, -3, 0, 3, 6, 9, 12}, find the truth set of the following:
(i) 2x – 3 > 7
(ii) 3x + 8 ≤ 2
(iii) -3 < 1 – 2x
Answer:
(i) Start with 2x – 3 > 7. Move -3 to the right side: 2x > 7 + 3, which gives 2x > 10. Divide both sides by 2 to find x > 5. From the replacement set, only the values 6, 9, and 12 satisfy this condition. Therefore, the truth set is {6, 9, 12}.

(ii) Begin with 3x + 8 ≤ 2. Subtract 8 from both sides: 3x ≤ 2 - 8, yielding 3x ≤ -6. Divide by 3 to get x ≤ -2. The values from the replacement set that meet this requirement are -6 and -3. Therefore, the truth set is {-6, -3}.

(iii) Rearrange -3 < 1 – 2x to isolate the variable. This becomes 2x – 3 < 1. Add 3 to both sides: 2x < 1 + 3, so 2x < 4. Divide by 2: x < 2. The values from the replacement set satisfying this are -6, -3, and 0. Therefore, the truth set is {-6, -3, 0}.
In simple words: For each inequality, rearrange to get the variable alone. Then pick from the given set only those numbers that make the inequality true.

Exam Tip: Always test your final answer by substituting one value from your solution set back into the original inequality - it must make the statement true.

Question 6. Solve the following inequations:
(i) 4x + 1 < 17, x ∈ N
(ii) 4x + 1 ≤ 17, x ∈ W
(iii) 4 > 3x - 11, x ∈ N
(iv) -17 ≤ 9x - 8, x ∈ Z
Answer:
(i) Simplify 4x + 1 < 17 by subtracting 1: 4x < 16. Divide by 4 to find x < 4. Since x must be a natural number, the solution set is {1, 2, 3}.

(ii) Start with 4x + 1 ≤ 17. Subtract 1 from both sides: 4x ≤ 16. Divide by 4: x ≤ 4. Since x must be a whole number (including 0), the solution set is {0, 1, 2, 3, 4}.

(iii) Take 4 > 3x - 11 and rearrange: 4 + 11 > 3x, so 15 > 3x. Divide by 3: 5 > x, or x < 5. With x being a natural number, the solution set is {1, 2, 3, 4}.

(iv) Rearrange -17 ≤ 9x - 8 by adding 8 to both sides: -17 + 8 ≤ 9x, giving -9 ≤ 9x. Divide by 9: -1 ≤ x, or x ≥ -1. Since x must be an integer, the solution set is {-1, 0, 1, 2, ...}.
In simple words: Move numbers to one side and the variable term to the other, then divide to isolate the variable. Only pick values from the allowed set.

Exam Tip: Pay close attention to whether the symbol is < or ≤, and always check which set (N, W, Z) the variable belongs to before writing your final answer.

Question 7. Solve the following inequations:
(i) {(2y - 1) / 5} ≤ 2, y ∈ N
(ii) {(2y + 1) / 3} + 1 ≤ 3, y ∈ W
(iii) (2 / 3)p + 5 < 9, p ∈ W
(iv) -2 (p + 3) > 5, p ∈ I
Answer:
(i) Start with {(2y - 1) / 5} ≤ 2. Multiply both sides by 5: 2y - 1 ≤ 10. Add 1: 2y ≤ 11. Divide by 2: y ≤ 5.5. Since y must be a natural number, the solution set is {1, 2, 3, 4, 5}.

(ii) Simplify {(2y + 1) / 3} + 1 ≤ 3. Rewrite as {(2y + 1 + 3) / 3} ≤ 3, so {(2y + 4) / 3} ≤ 3. Multiply by 3: 2y + 4 ≤ 9. Subtract 4: 2y ≤ 5. Divide by 2: y ≤ 2.5. With y being a whole number, the solution set is {0, 1, 2}.

(iii) Work with (2 / 3)p + 5 < 9. Subtract 5: (2 / 3)p < 4. Multiply by 3: 2p < 12. Divide by 2: p < 6. The solution set for whole numbers is {0, 1, 2, 3, 4, 5}.

(iv) Expand -2 (p + 3) > 5 to get -2p - 6 > 5. Add 6: -2p > 11. Divide by -2 (reverse the inequality sign): p < -5.5. Since p must be an integer, the solution set is {..., -8, -7, -6}.
In simple words: Clear fractions or brackets first, then collect the variable on one side. Remember to flip the inequality when multiplying or dividing by a negative.

Exam Tip: When dividing or multiplying by a negative number, always reverse the direction of the inequality sign - this is a frequent source of errors.

Question 8. Solve the following inequations:
(i) 2x - 3 < x + 2, x ∈ N
(ii) 3 - x ≤ 5 - 3x, x ∈ W
(iii) 3 (x - 2) < 2 (x - 1), x ∈ W
(iv) (3 / 2) - (x / 2) > -1, x ∈ N
Answer:
(i) From 2x - 3 < x + 2, subtract x from both sides: x - 3 < 2. Add 3: x < 5. Since x is a natural number, the solution set is {1, 2, 3, 4}.

(ii) Start with 3 - x ≤ 5 - 3x. Move all x terms to one side: 3x - x ≤ 5 - 3, so 2x ≤ 2. Divide by 2: x ≤ 1. With x being a whole number, the solution set is {0, 1}.

(iii) Expand 3 (x - 2) < 2 (x - 1) to get 3x - 6 < 2x - 2. Subtract 2x: x - 6 < -2. Add 6: x < 4. The solution set for whole numbers is {0, 1, 2, 3}.

(iv) Rearrange (3 / 2) - (x / 2) > -1 by adding (x / 2) to both sides: (3 / 2) + 1 > (x / 2). Simplify the left side: (5 / 2) > (x / 2). Multiply both sides by 2: 5 > x, or x < 5. Since x is a natural number, the solution set is {1, 2, 3, 4}.
In simple words: Collect all x terms on one side and numbers on the other. Simplify until the variable stands alone on one side of the inequality.

Exam Tip: Show each step clearly when moving terms from one side to the other - examiners award marks for working, not just the final answer.

Question 9. If the replacement set is {-3, -2, -1, 0, 1, 2, 3}, solve the inequation {(3x - 1) / 2} < 2. Represent its solution on the number line.
Answer: Start with {(3x - 1) / 2} < 2. Multiply both sides by 2: 3x - 1 < 4. Add 1: 3x < 5. Divide by 3: x < 5/3, which is approximately 1.67. From the given replacement set, the values that satisfy this condition are -3, -2, -1, 0, and 1. Therefore, the solution set is {-3, -2, -1, 0, 1}.

The number line representation shows these five points marked with circles at positions -3, -2, -1, 0, and 1 on a horizontal line with arrows at both ends.
In simple words: Multiply by 2 to remove the fraction, then isolate x. Mark on the number line all the values from the replacement set that work.

Exam Tip: Always verify your solution by testing the boundary value (x = 5/3 in this case) and at least one value inside your solution set in the original inequality.

Question 10. Solve (x / 3) + (1 / 4) < (x / 6) + (1 / 2), x ∈ W. Also represent its solution on the number line.
Answer: Start with (x / 3) + (1 / 4) < (x / 6) + (1 / 2). Rearrange by moving all x terms to the left: (x / 3) - (x / 6) < (1 / 2) - (1 / 4). Find a common denominator on the left: (2x - x) / 6 < (2 - 1) / 4. Simplify: x / 6 < 1 / 4. Multiply both sides by 6: x < 6/4, which reduces to x < 3/2 or 1.5. Since x must be a whole number, the solution set is {0, 1}.

The number line representation shows points marked with circles at positions 0 and 1 on a horizontal line with arrows at both ends.
In simple words: Move x terms to one side and constants to the other. Use a common denominator to simplify fractions, then solve for x.

Exam Tip: When dealing with fractions in inequalities, find the LCD of all denominators and multiply through to clear them before solving.

Question 11. Solve the following inequations and graph their solutions on a number line:
(i) -4 ≤ 4x < 14, x ∈ N
(ii) -1 < (x / 2) + 1 ≤ 3, x ∈ I
Answer:
(i) For the compound inequality -4 ≤ 4x < 14, divide all parts by 4: (-4 / 4) ≤ (4x / 4) < (14 / 4). This simplifies to -1 ≤ x < 3.5. Since x must be a natural number, the solution set is {1, 2, 3}. The number line shows circles marked at positions 1, 2, and 3.

(ii) For the compound inequality -1 < (x / 2) + 1 ≤ 3, subtract 1 from all parts: -1 - 1 < (x / 2) + 1 - 1 ≤ 3 - 1. This gives -2 < (x / 2) ≤ 2. Multiply all parts by 2: -4 < x ≤ 4. Since x must be an integer, the solution set is {-3, -2, -1, 0, 1, 2, 3, 4}. The number line shows circles at each of these eight positions.
In simple words: In a compound inequality, perform the same operation on all three parts - left, middle, and right - to keep it balanced.

Exam Tip: When working with compound inequalities involving ≤ and <, note which endpoint is included (≤ or ≥) and which is excluded (< or >) so your solution set is completely correct.