ML Aggarwal Class 7 Maths Solutions Chapter 09 Linear Equations and Inequalities

Question 1. Solve: (i) 2(3 - 2x) = 13 (ii) \( \frac{3}{5}y - 2 = \frac{10}{7} \)
Answer:
(i) Expanding the bracket: \( \Rightarrow 6 - 4x = 13 \)
\( \Rightarrow -4x = 13 - 6 \)
\( \Rightarrow -4x = 7 \)
\( \Rightarrow x = -\frac{7}{4} \)
Therefore, \( x = -\frac{7}{4} \)
(ii) Rearranging: \( \Rightarrow \frac{3}{5}y = \frac{10}{7} + 2 \)
\( \Rightarrow \frac{3}{5}y = \frac{10 + 20}{7} \)
\( \Rightarrow \frac{3}{5}y = \frac{27}{7} \)
\( \Rightarrow y = \frac{27}{7} \times \frac{5}{3} \)
\( \Rightarrow y = \frac{9}{2} \)
\( \Rightarrow y = 4\frac{1}{2} \)
Therefore, \( y = 4\frac{1}{2} \)
In simple words: To solve equations, expand any brackets first. Then collect all terms with the unknown variable on one side and numbers on the other. Divide both sides to find the final answer.

Exam Tip: Always write out every step clearly. Check your answer by substituting it back into the original equation.

Question 2. Solve: (i) \( \frac{x}{2} = 5 + \frac{x}{3} \) (ii) \( 2\left(x - \frac{3}{2}\right) = 11 \)
Answer:
(i) Find the LCM of 2 and 3, which is 6. Multiply both sides by 6:
\( \Rightarrow \frac{x}{2} \times 6 = 5 \times 6 + \frac{x}{3} \times 6 \)
\( \Rightarrow 3x = 30 + 2x \)
\( \Rightarrow 3x - 2x = 30 \)
\( \Rightarrow x = 30 \)
Therefore, \( x = 30 \)
(ii) Expanding the bracket: \( \Rightarrow 2x - 3 = 11 \)
\( \Rightarrow 2x = 11 + 3 \)
\( \Rightarrow 2x = 14 \)
\( \Rightarrow x = 7 \)
Therefore, \( x = 7 \)
In simple words: When fractions appear, find the LCM of the denominators and multiply the whole equation by it. This clears the fractions, making the equation easier to solve.

Exam Tip: Remember to multiply every term on both sides by the LCM, not just the fractions. Verify your answer by plugging it back into the original equation.

Question 3. Solve: (i) 7(x - 2) = 2(2x - 4) (ii) 21 - 3(x - 7) = x + 20
Answer:
(i) Expanding both sides: \( \Rightarrow 7x - 14 = 4x - 8 \)
\( \Rightarrow 7x - 4x = -8 + 14 \)
\( \Rightarrow 3x = 6 \)
\( \Rightarrow x = 2 \)
Therefore, \( x = 2 \)
(ii) Expanding the bracket: \( \Rightarrow 21 - 3x + 21 = x + 20 \)
\( \Rightarrow 42 - 3x = x + 20 \)
\( \Rightarrow 42 - 20 = x + 3x \)
\( \Rightarrow 22 = 4x \)
\( \Rightarrow x = \frac{22}{4} = \frac{11}{2} = 5\frac{1}{2} \)
Therefore, \( x = 5\frac{1}{2} \)
In simple words: Expand brackets on both sides of the equation. Move variable terms to one side and constant terms to the other. Then divide to find the answer.

Exam Tip: Be careful with signs when expanding brackets, especially with a negative sign in front. Double-check your arithmetic when collecting like terms.

Question 4. Solve: (i) \( 3x - \frac{1}{3} = 2\left(x - \frac{1}{2}\right) + 5 \) (ii) \( \frac{2m}{3} - \frac{m}{5} = 7 \)
Answer:
(i) Expanding the right side: \( \Rightarrow 3x - \frac{1}{3} = 2x - 1 + 5 \)
\( \Rightarrow 3x - \frac{1}{3} = 2x + 4 \)
\( \Rightarrow 3x - 2x = 4 + \frac{1}{3} \)
\( \Rightarrow x = \frac{12 + 1}{3} = \frac{13}{3} = 4\frac{1}{3} \)
Therefore, \( x = 4\frac{1}{3} \)
(ii) Find the LCM of 3 and 5, which is 15. Multiply both sides by 15:
\( \Rightarrow \frac{2m}{3} \times 15 - \frac{m}{5} \times 15 = 7 \times 15 \)
\( \Rightarrow 10m - 3m = 105 \)
\( \Rightarrow 7m = 105 \)
\( \Rightarrow m = 15 \)
Therefore, \( m = 15 \)
In simple words: Remove fractions by finding the LCM and multiplying through. Simplify step by step, keeping the variable on one side and numbers on the other.

Exam Tip: When combining fractions, convert everything to a common denominator first. Always show the LCM calculation in your working.

Question 5. Solve: (i) \( \frac{x + 1}{5} - \frac{x - 7}{2} = 1 \) (ii) \( \frac{3p - 2}{7} - \frac{p - 2}{4} = 2 \)
Answer:
(i) Find the LCM of 5 and 2, which is 10. Multiply both sides by 10:
\( \Rightarrow (x + 1) \times 2 - (x - 7) \times 5 = 1 \times 10 \)
\( \Rightarrow 2(x + 1) - 5(x - 7) = 10 \)
\( \Rightarrow 2x + 2 - 5x + 35 = 10 \)
\( \Rightarrow -3x + 37 = 10 \)
\( \Rightarrow -3x = 10 - 37 \)
\( \Rightarrow -3x = -27 \)
\( \Rightarrow x = 9 \)
Therefore, \( x = 9 \)
(ii) Find the LCM of 7 and 4, which is 28. Multiply both sides by 28:
\( \Rightarrow (3p - 2) \times 4 - (p - 2) \times 7 = 2 \times 28 \)
\( \Rightarrow 4(3p - 2) - 7(p - 2) = 56 \)
\( \Rightarrow 12p - 8 - 7p + 14 = 56 \)
\( \Rightarrow 5p + 6 = 56 \)
\( \Rightarrow 5p = 50 \)
\( \Rightarrow p = 10 \)
Therefore, \( p = 10 \)
In simple words: When you have fractions being subtracted, first clear all fractions by multiplying by the LCM. Then expand the brackets and solve as usual.

Exam Tip: Pay careful attention to signs when subtracting fractions. Expand each bracket completely before combining like terms.

Question 6. Solve: (i) \( \frac{1}{2}(x + 5) - \frac{1}{3}(x - 2) = 4 \) (ii) \( \frac{2x - 3}{6} - \frac{x - 5}{2} = \frac{x}{6} \)
Answer:
(i) Find the LCM of 2 and 3, which is 6. Multiply both sides by 6:
\( \Rightarrow \frac{1}{2}(x + 5) \times 6 - \frac{1}{3}(x - 2) \times 6 = 4 \times 6 \)
\( \Rightarrow 3(x + 5) - 2(x - 2) = 24 \)
\( \Rightarrow 3x + 15 - 2x + 4 = 24 \)
\( \Rightarrow x + 19 = 24 \)
\( \Rightarrow x = 5 \)
Therefore, \( x = 5 \)
(ii) Find the LCM of 2 and 6, which is 6. Multiply both sides by 6:
\( \Rightarrow (2x - 3) - (x - 5) \times 3 = x \)
\( \Rightarrow 2x - 3 - 3x + 15 = x \)
\( \Rightarrow -x + 12 = x \)
\( \Rightarrow 12 = 2x \)
\( \Rightarrow x = 6 \)
Therefore, \( x = 6 \)
In simple words: Identify the smallest common multiple of all denominators. Multiply every term by this number to remove fractions. Then solve the resulting simple equation.

Exam Tip: After multiplying by the LCM, make sure each term is correctly simplified. Always simplify fractions before final multiplication.

Question 7. Solve: (i) \( \frac{x - 4}{7} - \frac{x + 4}{5} = \frac{x + 3}{7} \) (ii) \( \frac{x - 1}{5} + \frac{x - 2}{2} = \frac{x}{3} + 1 \)
Answer:
(i) Find the LCM of 5 and 7, which is 35. Multiply both sides by 35:
\( \Rightarrow (x - 4) \times 5 - (x + 4) \times 7 = (x + 3) \times 5 \)
\( \Rightarrow 5x - 20 - 7x - 28 = 5x + 15 \)
\( \Rightarrow -2x - 48 = 5x + 15 \)
\( \Rightarrow -48 - 15 = 5x + 2x \)
\( \Rightarrow -63 = 7x \)
\( \Rightarrow x = -9 \)
Therefore, \( x = -9 \)
(ii) Find the LCM of 5, 2, and 3, which is 30. Multiply both sides by 30:
\( \Rightarrow (x - 1) \times 6 + (x - 2) \times 15 = x \times 10 + 1 \times 30 \)
\( \Rightarrow 6(x - 1) + 15(x - 2) = 10x + 30 \)
\( \Rightarrow 6x - 6 + 15x - 30 = 10x + 30 \)
\( \Rightarrow 21x - 36 = 10x + 30 \)
\( \Rightarrow 21x - 10x = 30 + 36 \)
\( \Rightarrow 11x = 66 \)
\( \Rightarrow x = 6 \)
Therefore, \( x = 6 \)
In simple words: When there are many fractions with different denominators, find the LCM of all of them. Multiply the entire equation by the LCM to clear all fractions at once.

Exam Tip: For equations with three or more different denominators, write out the LCM calculation clearly. This shows the examiner your working method.

Question 8. Solve: (i) y + 1.2y = 4.4 (ii) 15% of x = 21
Answer:
(i) Combining like terms: \( \Rightarrow 2.2y = 4.4 \)
\( \Rightarrow y = \frac{4.4}{2.2} \)
\( \Rightarrow y = 2 \)
Therefore, \( y = 2 \)
(ii) Converting percentage to fraction: \( \Rightarrow \frac{15}{100} \times x = 21 \)
\( \Rightarrow x = 21 \times \frac{100}{15} \)
\( \Rightarrow x = \frac{2100}{15} \)
\( \Rightarrow x = 140 \)
Therefore, \( x = 140 \)
In simple words: When combining decimal terms, add the coefficients first. For percentage equations, rewrite the percentage as a fraction, then isolate the variable.

Exam Tip: With decimal coefficients, combine them first to simplify the equation. For percentage problems, convert to a fraction out of 100 before solving.

Question 9. Solve: (i) 2p + 20% of (2p - 1) = 7 (ii) 3(2x - 1) + 25% of x = 97
Answer:
(i) Converting the percentage: \( \Rightarrow 2p + \frac{20}{100}(2p - 1) = 7 \)
\( \Rightarrow 2p + \frac{1}{5}(2p - 1) = 7 \)
Multiplying both sides by 5:
\( \Rightarrow 5\left(2p + \frac{1}{5}(2p - 1)\right) = 7 \times 5 \)
\( \Rightarrow 10p + (2p - 1) = 35 \)
\( \Rightarrow 10p + 2p - 1 = 35 \)
\( \Rightarrow 12p = 36 \)
\( \Rightarrow p = 3 \)
Therefore, \( p = 3 \)
(ii) Converting the percentage: \( \Rightarrow 3(2x - 1) + \frac{25}{100}x = 97 \)
\( \Rightarrow 6x - 3 + \frac{1}{4}x = 97 \)
Multiplying both sides by 4:
\( \Rightarrow 6x \times 4 - 3 \times 4 + \frac{1}{4}x \times 4 = 97 \times 4 \)
\( \Rightarrow 24x - 12 + x = 388 \)
\( \Rightarrow 25x - 12 = 388 \)
\( \Rightarrow 25x = 400 \)
\( \Rightarrow x = 16 \)
Therefore, \( x = 16 \)
In simple words: Change each percentage to a fraction (like 20% becomes 1/5). Multiply the whole equation by the denominator to clear the fractions, then solve.

Exam Tip: Always simplify percentages to their lowest fraction form. When you have both whole terms and fractional terms, use the LCM of all denominators to multiply through.

Question 10. Find the value of p if the value of x⁴ - 3x³ - px - 5 is equal to 23 when x = -2
Answer: We are told that x⁴ - 3x³ - px - 5 equals 23 when x = -2. Substituting x = -2 into the expression:

\( (-2)^4 - 3(-2)^3 - p(-2) - 5 = 23 \)

\( 16 - 3(-8) + 2p - 5 = 23 \)

\( 16 + 24 + 2p - 5 = 23 \)

\( 35 + 2p = 23 \)

\( 2p = 23 - 35 \)

\( 2p = -12 \)

\( p = -6 \)
In simple words: Plug in x = -2 and work through the arithmetic step by step. You'll get an equation with p that you can solve to find p = -6.

Exam Tip: Substitute the given value carefully and track the signs - the (-2)³ term becomes positive 8 when you apply the negative sign. Verify your answer by substituting p back into the original expression.

Exercise 9.2

Question 1. If 7 is added to five times a number, the result is 57. Find the number.
Answer: Let the unknown number be x. Five times this number gives 5x. When we add 7 to this, we get 5x + 7, which equals 57 according to the problem.

\( 5x + 7 = 57 \)

\( 5x = 57 - 7 \)

\( 5x = 50 \)

\( x = \frac{50}{5} \)

\( x = 10 \)

The required number is 10.
In simple words: Call the mystery number x. Five times it, plus 7, makes 57. Subtract 7 first, then divide by 5 to get x = 10.

Exam Tip: Always define your variable at the start and set up the equation matching the words of the problem exactly. Check: 5(10) + 7 = 57 ✓

Question 2. Find a number, such that one-fourth of the number is 3 more than 7.
Answer: Let x be the number we are looking for. One-fourth of this number is \( \frac{x}{4} \). The phrase "3 more than 7" means 7 + 3 = 10. According to the problem:

\( \frac{x}{4} = 10 \)

\( x = 10 \times 4 \)

\( x = 40 \)

The required number is 40.
In simple words: One-quarter of the number equals 10. To find the whole number, multiply 10 by 4 to get 40.

Exam Tip: Read carefully - "3 more than 7" is a specific phrase meaning addition, not a separate condition. Verify: 40 ÷ 4 = 10 ✓

Question 3. A number is as much greater than 15 as it is less than 51. Find the number.
Answer: Let x be the number. The phrase "as much greater than 15" means the distance above 15, which is (x - 15). The phrase "as it is less than 51" means the distance below 51, which is (51 - x). Since these distances are equal:

\( x - 15 = 51 - x \)

\( x + x = 51 + 15 \)

\( 2x = 66 \)

\( x = 33 \)

The required number is 33.
In simple words: The number is equally far above 15 as it is below 51. This means it sits right in the middle between 15 and 51. The middle is 33.

Exam Tip: This is an "equidistant" problem - set the two distance expressions equal. Verify: 33 - 15 = 18 and 51 - 33 = 18 ✓

Question 4. If \( \frac{1}{2} \) is subtracted from a number and the difference is multiplied by 4, the result is 5. What is the number?
Answer: Let x be the number. Subtracting \( \frac{1}{2} \) from it gives \( x - \frac{1}{2} \). When this difference is multiplied by 4, we get \( 4\left(x - \frac{1}{2}\right) = 5 \).

\( 4x - 2 = 5 \)

\( 4x = 5 + 2 \)

\( 4x = 7 \)

\( x = \frac{7}{4} \)

The required number is \( \frac{7}{4} \).
In simple words: Take away one-half from the number, then multiply the result by 4 to get 5. Working backwards, divide by 4 first, then add one-half to find the original number.

Exam Tip: Expand the bracket first: 4 times (x - 1/2) becomes 4x - 2. This step prevents sign errors. Verify: \( 4 \times \frac{7}{4} - 2 = 7 - 2 = 5 \) ✓

Question 5. The sum of two numbers is 80 and the greater number exceeds twice the smaller by 11. Find the numbers.
Answer: Let the smaller number be x. Since the sum of the two numbers is 80, the greater number is 80 - x. We are told the greater number is 11 more than twice the smaller:

\( 80 - x = 2x + 11 \)

\( 80 - 11 = 2x + x \)

\( 69 = 3x \)

\( x = 23 \)

The smaller number is 23. The greater number is 80 - 23 = 57.

The two numbers are 57 and 23.
In simple words: One number is 23, the other is 57. Add them: 23 + 57 = 80. Also, 57 is 11 more than twice 23 (which is 46), so 57 = 46 + 11.

Exam Tip: Always assign the variable to the unknown. Here, calling the smaller number x makes setting up the second equation (about "exceeds") cleaner. Check both conditions in your final answer.

Question 6. Find three consecutive odd natural numbers whose sum is 87.
Answer: Let x be the smallest of the three consecutive odd numbers. Since consecutive odd numbers differ by 2, the next two are x + 2 and x + 4. Their sum is 87:

\( x + (x + 2) + (x + 4) = 87 \)

\( 3x + 6 = 87 \)

\( 3x = 87 - 6 \)

\( 3x = 81 \)

\( x = 27 \)

The three consecutive odd numbers are 27, 27 + 2 = 29, and 27 + 4 = 31.
In simple words: Three odd numbers in a row, each separated by 2. They add to 87. Divide 87 by 3 to find the middle one (approximately), then adjust to get 27, 29, and 31.

Exam Tip: Remember consecutive odd numbers increase by 2, not 1. Verify: 27 + 29 + 31 = 87 ✓

Question 7. In a class of 35 students, the number of girls is two-fifth of the number of boys. Find the number of girls in the class.
Answer: Let x be the number of boys. Then the number of girls is \( \frac{2}{5}x \). Since the total is 35 students:

\( x + \frac{2}{5}x = 35 \)

\( \frac{5x + 2x}{5} = 35 \)

\( \frac{7x}{5} = 35 \)

\( 7x = 175 \)

\( x = 25 \)

The number of boys is 25. The number of girls is \( \frac{2}{5} \times 25 = 10 \).

The number of girls in the class is 10.
In simple words: There are 25 boys and 10 girls in the class. Two-fifths of 25 is indeed 10. Together they make 35 students.

Exam Tip: Combine fractions before solving: rewrite \( x + \frac{2x}{5} \) as \( \frac{7x}{5} \) to avoid working with two separate terms. Always verify both parts of your answer against the original conditions.

Question 8. A chair costs Rs.250 and a table costs Rs.400. If Ananya purchased a certain number of chairs and two tables for Rs.2800, find the number of chairs she purchased.
Answer: Let x be the number of chairs. The cost of x chairs is Rs.250x. Two tables cost Rs.(2 × 400) = Rs.800. The total amount spent is Rs.2800:

\( 250x + 800 = 2800 \)

\( 250x = 2800 - 800 \)

\( 250x = 2000 \)

\( x = 8 \)

The number of chairs purchased is 8.
In simple words: Subtract the cost of the two tables (Rs.800) from the total (Rs.2800) to get Rs.2000 for chairs. Divide by Rs.250 per chair to get 8 chairs.

Exam Tip: Calculate the fixed costs (tables) first, then solve for the variable quantity (chairs). Verify: 8 × 250 + 2 × 400 = 2000 + 800 = 2800 ✓

Question 9. Aparna got Rs.27,840 as her monthly salary and over-time. Her salary exceeds the over-time by Rs.16,560. What is her monthly salary?
Answer: Let Aparna's monthly salary be Rs.x. Since her salary exceeds the over-time payment by Rs.16,560, the over-time payment is Rs.(x - 16,560). The sum of these two amounts is Rs.27,840:

\( x + (x - 16,560) = 27,840 \)

\( 2x - 16,560 = 27,840 \)

\( 2x = 27,840 + 16,560 \)

\( 2x = 44,400 \)

\( x = 22,200 \)

Aparna's monthly salary is Rs.22,200.
In simple words: Her salary is Rs.16,560 more than what she earned as over-time. Together, they add to Rs.27,840. So salary + over-time = 27,840, and salary - over-time = 16,560. Solving these gives salary = Rs.22,200.

Exam Tip: When a quantity "exceeds" another by a fixed amount, use subtraction: the smaller = the larger minus the difference. Always define which part is which before writing the equation.

Question 10. Heena has only Rs.2 and Rs.5 coins in her purse. If in all she has 80 coins in her purse amounting to Rs.232, find the number of Rs.5 coins.
Answer: Let x be the number of Rs.5 coins. Since there are 80 coins total, the number of Rs.2 coins is 80 - x. The total value is Rs.232:

\( 5x + 2(80 - x) = 232 \)

\( 5x + 160 - 2x = 232 \)

\( 3x + 160 = 232 \)

\( 3x = 232 - 160 \)

\( 3x = 72 \)

\( x = 24 \)

The number of Rs.5 coins is 24.
In simple words: 24 five-rupee coins and 56 two-rupee coins. Five-rupee coins are worth 24 × 5 = 120 rupees. Two-rupee coins are worth 56 × 2 = 112 rupees. Total is 120 + 112 = 232 rupees.

Exam Tip: This is a "coin/note" problem - always set up two equations: one for the count (80 coins) and one for the value (Rs.232). Expand brackets carefully and combine like terms.

Question 11. A purse contains Rs.550 in notes of denominations of Rs.10 and Rs.50. If the number of Rs.50 notes is one less than that of Rs.10 notes, then find the number of Rs.50 notes.
Answer: Let x be the number of Rs.10 notes. The number of Rs.50 notes is x - 1. The value of the Rs.10 notes is Rs.10x, and the value of the Rs.50 notes is Rs.50(x - 1). The total value is Rs.550:

\( 10x + 50(x - 1) = 550 \)

\( 10x + 50x - 50 = 550 \)

\( 60x - 50 = 550 \)

\( 60x = 600 \)

\( x = 10 \)

The number of Rs.10 notes is 10. The number of Rs.50 notes is 10 - 1 = 9.
In simple words: There are 10 notes of Rs.10 (worth Rs.100) and 9 notes of Rs.50 (worth Rs.450). Together: 100 + 450 = Rs.550.

Exam Tip: The condition "one less than" is a relative relationship - always express one quantity in terms of the other. Double-check by computing the total value at the end.

Question 12. After 12 years, I shall be 3 times as old as I was 4 years ago. Find my present age.
Answer: Let x be my present age in years. After 12 years, my age will be (x + 12) years. Four years ago, my age was (x - 4) years. According to the problem:

\( x + 12 = 3(x - 4) \)

\( x + 12 = 3x - 12 \)

\( 12 + 12 = 3x - x \)

\( 24 = 2x \)

\( x = 12 \)

My present age is 12 years.
In simple words: In 12 years, I will be 24 years old. That is 3 times my age from 4 years ago, which was 8 years old. Check: 3 × 8 = 24 ✓

Exam Tip: Set up ages carefully: "after 12 years" means add 12 to present age, and "4 years ago" means subtract 4. The word "shall be" indicates a future condition - use the future age on one side and relate it to a past age on the other.

Question 13. Two equal sides of an isosceles triangle are (3x - 1) units 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 have the same length. Therefore:

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

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

\( x = 3 \)

Each equal side measures 3(3) - 1 = 9 - 1 = 8 units. The third side measures 2(3) = 6 units. The perimeter is 8 + 8 + 6 = 22 units.
In simple words: Two sides of the triangle are called equal, so we set them equal and solve for x. Once x = 3, we calculate all three side lengths and add them to find the perimeter.

Exam Tip: In an isosceles triangle, the two equal sides MUST be equal in length - use this fact to form your equation. After finding x, substitute it into ALL three side expressions and verify they make sense (all positive, triangle inequality holds).

Question 14. The length of a rectangular plot is 6 m less than thrice its breadth. Find the dimensions of the plot if its perimeter is 148 m.
Answer: Let the breadth of the rectangular plot be x metres. The length is 6 m less than three times the breadth, so length = (3x - 6) metres. The perimeter of a rectangle is 2(length + breadth):

\( 2[(3x - 6) + x] = 148 \)

\( 2(4x - 6) = 148 \)

\( 4x - 6 = 74 \)

\( 4x = 80 \)

\( x = 20 \)

The breadth is 20 m and the length is 3(20) - 6 = 60 - 6 = 54 m.

The dimensions of the plot are 54 m and 20 m.
In simple words: Call the breadth 20 m. The length is 3 times 20 minus 6, which is 54 m. Going around: 54 + 20 + 54 + 20 = 148 m.

Exam Tip: Always define your variable clearly - here, breadth is simpler than length because length depends on breadth. The perimeter formula is 2(l + b), not 2l + b. After solving, verify both that x is positive and that the dimensions satisfy the given condition ("6 m less than thrice").

Question 15. Two complementary angles differ by 20°. Find the measure of each angle.
Answer: Let one angle be x degrees. Since the two angles differ by 20°, the other angle is (x - 20) degrees. Complementary angles sum to 90°:

\( x + (x - 20) = 90 \)

\( 2x - 20 = 90 \)

\( 2x = 110 \)

\( x = 55 \)

One angle is 55° and the other is 55 - 20 = 35°.

The two angles are 35° and 55°.
In simple words: Two angles add to 90 degrees and differ by 20 degrees. The larger is 55° and the smaller is 35°. Check: 55 + 35 = 90 and 55 - 35 = 20.

Exam Tip: Complementary angles always sum to exactly 90°. Use this fact directly as your main equation, and express one angle in terms of the other using the "differ by" condition. Verify both the sum and the difference in your final answer.

Exercise 9.3

Question 1. If the replacement set is {-5, -3, -1, 0, 1, 3, 4}, find the solution set of:
(i) x < -2
(ii) x > 1
(iii) x ≥ -1
(iv) -5 < x < 3
(v) -3 ≤ x < 4
(vi) 0 ≤ x < 7
Answer: The replacement set is {-5, -3, -1, 0, 1, 3, 4}.

(i) For x < -2, we select all values in the replacement set that are less than -2. These are -5 and -3.
Hence, the solution set is {-5, -3}.

(ii) For x > 1, we select all values in the replacement set that are greater than 1. These are 3 and 4.
Hence, the solution set is {3, 4}.

(iii) For x ≥ -1, we select all values in the replacement set that are greater than or equal to -1. These are -1, 0, 1, 3, and 4.
Hence, the solution set is {-1, 0, 1, 3, 4}.

(iv) For -5 < x < 3, we select all values in the replacement set that are strictly between -5 and 3. These are -3, -1, 0, and 1.
Hence, the solution set is {-3, -1, 0, 1}.

(v) For -3 ≤ x < 4, we select all values in the replacement set that are greater than or equal to -3 and less than 4. These are -3, -1, 0, 1, and 3.
Hence, the solution set is {-3, -1, 0, 1, 3}.

(vi) For 0 ≤ x < 7, we select all values in the replacement set that are greater than or equal to 0 and less than 7. These are 0, 1, 3, and 4.
Hence, the solution set is {0, 1, 3, 4}.
In simple words: For each condition, go through the given set and pick the numbers that satisfy it. A "<" sign means strictly less (not equal), while "≤" means you can include the boundary number too.

Exam Tip: Pay close attention to whether a boundary is included (≤, ≥) or excluded (<, >). Even one misplaced number in your solution set will cost marks. Always list the solution set in ascending order.

Question 1. (iii) For x ≥ -1, find the solution set from the given replacement set.
Answer: From the replacement set, the values that satisfy x ≥ -1 (meaning they are -1 or larger) are -1, 0, 1, 3, and 4. Therefore, the solution set is {-1, 0, 1, 3, 4}.
In simple words: Pick all numbers from the list that are -1 or bigger. Those numbers make up your answer set.

Exam Tip: When you see the symbol ≥, it means "greater than or equal to" - so the boundary number itself is always included in the solution.

Question 1. (iv) For -5 < x < 3, find the solution set from the given replacement set.
Answer: From the replacement set, the values that lie between -5 and 3 (not including -5 and 3 themselves) are -3, -1, 0, and 1. Therefore, the solution set is {-3, -1, 0, 1}.
In simple words: Pick all numbers strictly between -5 and 3 - they must be bigger than -5 and smaller than 3, but not equal to either boundary.

Exam Tip: The symbols < and > never include the boundary numbers, while ≤ and ≥ always do - this is the key difference to remember.

Question 1. (v) For -3 ≤ x < 4, find the solution set from the given replacement set.
Answer: From the replacement set, the values that are greater than or equal to -3 and less than 4 are -3, -1, 0, 1, and 3. Therefore, the solution set is {-3, -1, 0, 1, 3}.
In simple words: Pick numbers that are -3 or larger, but smaller than 4 (not including 4 itself).

Exam Tip: When one inequality uses ≤ and the other uses <, be careful to include only the boundary with ≤ and exclude the one with <.

Question 1. (vi) For 0 ≤ x < 7, find the solution set from the given replacement set.
Answer: From the replacement set, the values that are greater than or equal to 0 and less than 7 are 0, 1, 3, and 4. Therefore, the solution set is {0, 1, 3, 4}.
In simple words: Pick all numbers that are 0 or bigger, but smaller than 7.

Exam Tip: Always double-check your boundary points - include 0 here since the inequality says ≤, but exclude 7 since it says <.

Question 2. Represent the solution set of the following inequalities graphically:
(i) x ≤ 3, x ∈ N
(ii) x < 4, x ∈ W
(iii) -2 ≤ x < 4, x ∈ I
(iv) -3 ≤ x ≤ 2, x ∈ I
Answer:
(i) For x ≤ 3 where x is a natural number, the solution set is {1, 2, 3}. On a number line, mark these three values with thick dots to show they satisfy the condition.

(ii) For x < 4 where x is a whole number, the solution set is {0, 1, 2, 3}. On a number line, mark these four values with thick dots.

(iii) For -2 ≤ x < 4 where x is an integer, the solution set is {-2, -1, 0, 1, 2, 3}. On a number line, mark these six values with thick dots.

(iv) For -3 ≤ x ≤ 2 where x is an integer, the solution set is {-3, -2, -1, 0, 1, 2}. On a number line, mark these six values with thick dots.
In simple words: Draw a number line and put heavy dots on only those numbers that make the inequality true. Each dot shows one answer.

Exam Tip: Always respect the domain (N, W, or I) - natural numbers start at 1, whole numbers start at 0, and integers include negatives. Mark only values within that domain.

Question 3. Solve the following inequalities:
(i) 4 - x > -2, x ∈ N
(ii) 3x + 1 ≤ 8, x ∈ W
Also represent their solutions on the number line.
Answer:
(i) Solving 4 - x > -2 where x is a natural number:

\( \Rightarrow 4 - x > -2 \)
\( \Rightarrow -x > -6 \)
\( \Rightarrow x < 6 \) (Dividing by -1 and reversing the symbol)

Since x must be a natural number and x < 6, the solution set is {1, 2, 3, 4, 5}. On the number line, mark these five values with thick dots.

(ii) Solving 3x + 1 ≤ 8 where x is a whole number:

\( \Rightarrow 3x + 1 \leq 8 \)
\( \Rightarrow 3x \leq 7 \)
\( \Rightarrow x \leq \frac{7}{3} \)
\( \Rightarrow x \leq 2\frac{1}{3} \)

Since x must be a whole number and x ≤ 2.33..., the solution set is {0, 1, 2}. On the number line, mark these three values with thick dots.
In simple words: Solve the inequality like you solve an equation - do the same thing to both sides. If you divide or multiply by a negative number, flip the symbol. Then check which values from your allowed set (N or W) work.

Exam Tip: Remember to reverse the inequality symbol when multiplying or dividing by a negative number - this is the most common mistake. Always verify your answer by testing one value from your solution set.

Question 4. Solve 3 - 4x < x - 12, x ∈ {-1, 0, 1, 2, 3, 4, 5, 6, 7}
Answer: Solving the inequality:

\( \Rightarrow 3 - 4x < x - 12 \)
\( \Rightarrow 3 + 12 < x + 4x \)
\( \Rightarrow 15 < 5x \)
\( \Rightarrow 3 < x \)
\( \Rightarrow x > 3 \)

From the replacement set {-1, 0, 1, 2, 3, 4, 5, 6, 7}, the values greater than 3 are 4, 5, 6, and 7. Therefore, the solution set is {4, 5, 6, 7}.
In simple words: Rearrange the inequality to get x by itself on one side. Collect all x terms on the right and numbers on the left. Then pick only the values from your given set that satisfy the final inequality.

Exam Tip: Always move variable terms to one side and constant terms to the other. When you divide by 5 in the last step, the symbol stays the same because you are dividing by a positive number.

Question 5. Solve -7 < 4x + 1 ≤ 23, x ∈ I
Answer: Solving the compound inequality:

\( \Rightarrow -7 < 4x + 1 \leq 23 \)
\( \Rightarrow -7 - 1 < 4x \leq 23 - 1 \)
\( \Rightarrow -8 < 4x \leq 22 \)
\( \Rightarrow \frac{-8}{4} < x \leq \frac{22}{4} \)
\( \Rightarrow -2 < x \leq \frac{11}{2} \)
\( \Rightarrow -2 < x \leq 5.5 \)

Since x must be an integer, the integers greater than -2 and less than or equal to 5.5 are -1, 0, 1, 2, 3, 4, and 5. Therefore, the solution set is {-1, 0, 1, 2, 3, 4, 5}.
In simple words: When you have a compound inequality, treat it like two inequalities joined together - subtract 1 from all parts, then divide all parts by 4. Find which integers fall in the range you get.

Exam Tip: In compound inequalities, whatever you do to one part must be done to all parts equally. Since 5.5 is not an integer, the largest integer that satisfies x ≤ 5.5 is 5.

Question 1. Fill in the blanks:
(i) A linear equation in one variable cannot have more than .... solution.
(ii) If five times a number is 50, then the number is .....
(iii) The number 4 is the .... of the equation 2y - 5 = 3
(iv) The equation for the statement '5 less than thrice a number x is 7' is ......
(v) ...... is a solution of the equation 4x + 9 = 5
(vi) If 3x + 7 = 1, then the value of 5x + 13 is .......
(vii) In natural numbers, 4x + 5 = -7 has ....... solution.
(viii) In integers, 3x - 1 = 4 has ....... solution.
(ix) 5x + ....... = 13 has the solution -3
(x) If a number is increased by 15, it becomes 50. Then the number is .......
(xi) If 63 exceed another number by 21, then the other number is .......
(xii) If x ∈ W, then the solution set of x < 2 is ......
Answer:
(i) A linear equation in one variable cannot have more than one solution.

(ii) If five times a number is 50, then the number is 10 (since 50 ÷ 5 = 10).

(iii) The number 4 is the solution (or root) of the equation 2y - 5 = 3, because 2(4) - 5 = 8 - 5 = 3.

(iv) The equation for the statement '5 less than thrice a number x is 7' is 3x - 5 = 7.

(v) Solving 4x + 9 = 5 gives 4x = -4, so x = -1. Therefore -1 is a solution of the equation 4x + 9 = 5.

(vi) If 3x + 7 = 1, then 3x = -6, so x = -2. Therefore 5x + 13 = 5(-2) + 13 = -10 + 13 = 3.

(vii) Solving 4x + 5 = -7 gives 4x = -12, so x = -3. Since -3 is not a natural number, in natural numbers 4x + 5 = -7 has no solution.

(viii) Solving 3x - 1 = 4 gives 3x = 5, so x = 5/3. Since 5/3 is not an integer, in integers 3x - 1 = 4 has no solution.

(ix) Putting x = -3 in 5x + ___ = 13 gives 5(-3) + ___ = 13, so -15 + ___ = 13. Therefore ___ = 28.

(x) If a number is increased by 15, it becomes 50. Then the number is 50 - 15 = 35.

(xi) If 63 exceed another number by 21, then the other number is 63 - 21 = 42.

(xii) If x ∈ W, then the solution set of x < 2 is {0, 1}.
In simple words: For fill-in questions, solve each one like a small puzzle - substitute values, undo operations step by step, or use the given numbers to find the blank.

Exam Tip: Watch out for domain restrictions - natural numbers don't include 0 or negatives, while whole numbers do include 0. Always check whether a fractional answer counts as a valid solution in the given number system.

Question 2. State whether the following statements are true (T) or false (F):
(i) We can add (or subtract) the same number or expression to both sides of an equation.
(ii) We can divide both sides of an equation by the same non-zero number.
(iii) 3x - 5 = 2(x + 3) + 7 is a linear equation in one variable.
(iv) The solution of the equation 3(x - 4) = 30 is x = 6
(v) The solution of the equation 3x - 5 = 2 is x = 7/3
(vi) The solution of a linear equation in one variable is always an integer.
(vii) 4x + 5 < 65 is not an equation.
(viii) 2x + 1 = 7 and 3x - 5 = 4 have the same solution.
(ix) 9/4 is a solution of the equation 5x - 1 = 8
(x) If 5 is a solution of variable x in the equation (5x - 7)/2 = y, then the value of y is 18
(xi) One-fourth of a number added to itself gives 10, can be represented as x/4 + 10 = x
Answer:
(i) True. We can add or subtract the same number or expression to both sides of an equation - this preserves equality.

(ii) True. We can divide both sides of an equation by the same non-zero number and the equation remains balanced.

(iii) True. It contains only one variable x with the highest power of 1, making it a linear equation in one variable.

(iv) False. Solving 3(x - 4) = 30 gives 3x - 12 = 30, so 3x = 42, hence x = 14 (not 6).

(v) True. Solving 3x - 5 = 2 gives 3x = 7, so x = 7/3.

(vi) False. The solution of a linear equation in one variable can be a fraction - for example, x = 7/3 is a valid fractional solution.

(vii) True. 4x + 5 < 65 is an inequality, not an equation (it uses < instead of =).

(viii) True. Solving 2x + 1 = 7 gives x = 3, and solving 3x - 5 = 4 also gives x = 3, so both equations have the same solution.

(ix) False. Putting x = 9/4 in 5x - 1 gives 5(9/4) - 1 = 45/4 - 1 = 45/4 - 4/4 = 41/4, which is not equal to 8.

(x) False. Putting x = 5 gives y = (5(5) - 7)/2 = (25 - 7)/2 = 18/2 = 9 (not 18).

(xi) False. The correct representation is x/4 + x = 10, not x/4 + 10 = x. The phrase means "one-fourth of a number added to the number itself equals 10".
In simple words: For true/false, always work through the math yourself - don't just trust what the statement says. Check by substituting values or solving it your own way.

Exam Tip: Many false statements contain one small error - like saying x = 6 when it's actually 14, or leaving off a negative sign. Read each statement carefully and verify by calculation, not just by reading.

Question 3. Which of the following is not a linear equation in one variable?
(1) 3x - 1 = 7
(2) 5y - 2 = 3(y + 2)
(3) 2x - 3 = 7/2
(4) 7p + q = 3
Answer: (4) 7p + q = 3
A linear equation in one variable must contain only one variable with highest power 1. Option (4) contains two variables, p and q, so it is not a linear equation in one variable. Options (1), (2), and (3) each contain only one variable with power 1, making them linear equations in one variable.
In simple words: Count the different letters (variables) in each equation. If there is more than one letter, it is not a linear equation in one variable.

Exam Tip: Always look for the word "one variable" - it means the equation must have only one letter, and that letter cannot be squared, cubed, or have any power higher than 1.

Question 4. The solution of the equation (1/3)(2y - 1) = 3 is
(1) 5
(2) 3
(3) 2
(4) 1
Answer: (1) 5
Solving the equation:

\( \Rightarrow \frac{1}{3}(2y - 1) = 3 \)
\( \Rightarrow 2y - 1 = 9 \)
\( \Rightarrow 2y = 10 \)
\( \Rightarrow y = 5 \)

Therefore, y = 5 is the correct solution.
In simple words: Multiply both sides by 3 to remove the fraction. Then add 1 to both sides, and divide by 2.

Exam Tip: When you see a fraction multiplying the variable expression, multiply both sides by the denominator first to clear the fraction - this makes the rest of the work cleaner.

Question 5. x = -1 is a solution of the equation
(1) x - 5 = 6
(2) 2x + 5 = 7
(3) 2(x - 2) + 6 = 0
(4) 3x + 5 = 4
Answer: (3) 2(x - 2) + 6 = 0
To find which equation has x = -1 as a solution, substitute -1 for x in each option:

(1) (-1) - 5 = -6 ≠ 6
(2) 2(-1) + 5 = 3 ≠ 7
(3) 2((-1) - 2) + 6 = 2(-3) + 6 = -6 + 6 = 0 ✓
(4) 3(-1) + 5 = 2 ≠ 4

Only option (3) gives 0 when x = -1, so x = -1 satisfies 2(x - 2) + 6 = 0.
In simple words: Plug in x = -1 into each equation and see which one becomes a true statement (both sides equal).

Exam Tip: Substitution is the fastest way to check if a value is a solution - you don't have to solve the whole equation, just plug in and verify.

Question 6. If 3(3n - 10) = 2n + 5, then the value of n is
(1) 12
(2) 5
(3) 3
(4) -5
Answer: (2) 5
Solving the equation:

\( \Rightarrow 3(3n - 10) = 2n + 5 \)
\( \Rightarrow 9n - 30 = 2n + 5 \)
\( \Rightarrow 9n - 2n = 5 + 30 \)
\( \Rightarrow 7n = 35 \)
\( \Rightarrow n = 5 \)

Therefore, n = 5 is the correct answer.
In simple words: Expand the bracket first, then move all n terms to the left and numbers to the right, then divide to find n.

Exam Tip: Always expand brackets before collecting like terms - skipping this step often leads to sign errors.

Question 7. -1 is not a solution of the equation
(1) x + 1 = 0
(2) 3x + 4 = 1
(3) 5x + 7 = 2
(4) x - 1 = 2
Answer: (4) x - 1 = 2
To find which equation does not have x = -1 as a solution, test -1 in each option:

(1) (-1) + 1 = 0 ✓
(2) 3(-1) + 4 = 1 ✓
(3) 5(-1) + 7 = 2 ✓
(4) (-1) - 1 = -2 ≠ 2

Option (4) is not satisfied by x = -1, so -1 is not a solution of x - 1 = 2.
In simple words: Substitute -1 into each equation. Three will work out to true statements, but one won't. That's your answer.

Exam Tip: The question asks for the equation that -1 does NOT solve, so look for the false statement, not the true ones.

Question 8. The value of p for which the expressions p - 13 and 2p + 1 become equal is
(1) 0
(2) 14
(3) -14
(4) 5
Answer: (3) -14
For the expressions to be equal, set them equal and solve:

\( \Rightarrow p - 13 = 2p + 1 \)
\( \Rightarrow -13 - 1 = 2p - p \)
\( \Rightarrow -14 = p \)
\( \Rightarrow p = -14 \)

Therefore, p = -14 is the correct answer.
In simple words: Write "expression 1 = expression 2" and solve for p like a normal equation.

Exam Tip: When moving terms across the equals sign, flip their signs - this is what causes confusion. Moving p to the right turns it into -p; moving -13 to the right turns it into +13.

Question 9. The equation which cannot be solved in integers is
(1) 5x - 3 = -18
(2) 3y - 5 = y - 1
(3) 3p + 8 = 3 + p
(4) 9z + 8 = 4z - 7
Answer: (3) 3p + 8 = 3 + p
Solving 3p + 8 = 3 + p:

\( \Rightarrow 3p + 8 = 3 + p \)
\( \Rightarrow 3p - p = 3 - 8 \)
\( \Rightarrow 2p = -5 \)
\( \Rightarrow p = -\frac{5}{2} \)

Since -5/2 is not an integer, this equation cannot be solved in integers. All other equations have integer solutions.
In simple words: Solve each equation. If the answer is a fraction (like -5/2), then that equation cannot be solved in integers. If the answer is a whole number, it can be.

Exam Tip: "Cannot be solved in integers" means the solution is not a whole number - even if it's a valid number, it doesn't count as an integer solution.

Question 10. The solution of which of the following equations is neither an integer nor a fraction?
(1) 2x + 5 = 1
(2) 3x - 7 = 0
(3) 5x - 7 = x + 1
(4) 4x + 7 = x + 2
Answer: Working through option 4:

\( 4x + 7 = x + 2 \)

\( 4x - x = 2 - 7 \)

\( 3x = -5 \)

\( x = -\frac{5}{3} \)

The value \( -\frac{5}{3} \) is a negative rational number. It is neither a whole number (integer) nor a positive fraction.
In simple words: When you solve this equation, you get a negative fraction. That is not an integer and not a regular positive fraction.

Exam Tip: Check each option by solving - an integer is a whole number (positive, negative, or zero), and a standard fraction is positive. Any negative fraction or decimal fails both conditions.

Question 11. If the sum of two consecutive even numbers is 54, then the smaller number is
(1) 25
(2) 26
(3) 27
(4) 28
Answer: Let the smaller even number be x. The next even number is x + 2.

\( x + (x + 2) = 54 \)

\( 2x + 2 = 54 \)

\( 2x = 52 \)

\( x = 26 \)

The smaller even number is 26.
In simple words: Two even numbers that come next to each other always differ by 2. Add them together to get 54, then solve for the smaller one.

Exam Tip: Always set up consecutive even/odd numbers as x and x+2 (they differ by 2, not 1). This is a common setup in these problems.

Question 12. If the sum of two consecutive odd numbers is 28, then the bigger number is
(1) 19
(2) 17
(3) 15
(4) 13
Answer: Let the smaller odd number be x. The larger odd number is x + 2.

\( x + (x + 2) = 28 \)

\( 2x + 2 = 28 \)

\( 2x = 26 \)

\( x = 13 \)

The larger odd number = x + 2 = 13 + 2 = 15.
In simple words: Consecutive odd numbers also differ by 2. The smaller one is 13, so the bigger one is 15.

Exam Tip: Remember that both even and odd consecutive numbers use the formula x and x+2. Make sure to find what the question asks - here it asks for the bigger number, not the smaller one.

Question 13. If 5 added to thrice an integer is -7, then the integer is
(1) -6
(2) -5
(3) -4
(4) 4
Answer: Let the integer be x.

\( 3x + 5 = -7 \)

\( 3x = -7 - 5 \)

\( 3x = -12 \)

\( x = -4 \)

The integer is -4.
In simple words: Triple the unknown number and add 5. This gives -7. Work backwards to find the number is -4.

Exam Tip: "5 added to thrice" means 3x + 5, not 5 + 3 added separately. Set up the equation carefully before solving.

Question 14. If the length of a rectangle is twice its breadth and its perimeter is 120 m, then its length is
(1) 20 m
(2) 40 m
(3) 60 m
(4) 30 m
Answer: Let the breadth be x m. Then the length is 2x m.

Perimeter of rectangle = \( 2(length + breadth) \)

\( 2(2x + x) = 120 \)

\( 2(3x) = 120 \)

\( 6x = 120 \)

\( x = 20 \)

Length = 2x = 2 × 20 = 40 m.
In simple words: The breadth is 20 m and the length is twice that, which is 40 m. Together they give a perimeter of 120 m.

Exam Tip: Always use the perimeter formula correctly: 2(l+b). Set up your relationship between length and breadth first before substituting into the formula.

Question 15. If the difference of two complementary angles is 10°, then the smaller angle is
(1) 40°
(2) 50°
(3) 45°
(4) 85°
Answer: Complementary angles add up to 90°. Let the smaller angle be x°. The larger angle is (x + 10)°.

\( x + (x + 10) = 90 \)

\( 2x + 10 = 90 \)

\( 2x = 80 \)

\( x = 40° \)

The smaller angle is 40°.
In simple words: Two angles that add to 90° are complementary. If one is 10° bigger than the other, the smaller one must be 40°.

Exam Tip: Know the key definitions: complementary angles sum to 90°, supplementary angles sum to 180°. Use these as your constraint equation.

Question 16. If the difference of two supplementary angles is 30°, then the larger angle is
(1) 60°
(2) 75°
(3) 90°
(4) 105°
Answer: Supplementary angles add up to 180°. Let the larger angle be x°. The smaller angle is (x - 30)°.

\( x + (x - 30) = 180 \)

\( 2x - 30 = 180 \)

\( 2x = 210 \)

\( x = 105° \)

The larger angle is 105°.
In simple words: Two angles that add to 180° are supplementary. One is 30° bigger than the other. The bigger one is 105°.

Exam Tip: For supplementary angles, the sum is always 180° (not 90° as with complementary). Double-check your angle definition before setting up the equation.

Question 17. If x ∈ W, the solution set of the inequality -2 ≤ x < 3 is
(1) {-2, -1, 0, 1, 2}
(2) {-1, 0, 1, 2, 3}
(3) {0, 1, 2, 3}
(4) {0, 1, 2}
Answer: The symbol W stands for whole numbers: {0, 1, 2, 3, ...}. The inequality -2 ≤ x < 3 includes all x values from -2 up to (but not including) 3. Since x must be a whole number, only 0, 1, and 2 satisfy both conditions.

Solution set: {0, 1, 2}.
In simple words: Whole numbers start at 0. The range goes from -2 to just under 3. The only whole numbers in that range are 0, 1, and 2.

Exam Tip: Remember that W (whole numbers) does NOT include negative numbers - it starts at 0. The symbol < means "not equal to" at that boundary, so x = 3 is excluded.

Statement I-II Type Questions

Question 18. Statement I: A boy's age is half the age of his father. 5 years ago, their ages were in the ratio 3 : 7. Based on this, we can say that the boy is 20 years old. Statement II: We can add or subtract the same algebraic expression to both sides of an equation.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer:

Checking Statement I: Let the boy's age be x years. Then father's age = 2x years. Five years back, the boy was (x - 5) years old and the father was (2x - 5) years old.

\( \frac{x - 5}{2x - 5} = \frac{3}{7} \)

\( 7(x - 5) = 3(2x - 5) \)

\( 7x - 35 = 6x - 15 \)

\( x = 20 \)

So the boy is 20 years old. Statement I is true.

Checking Statement II: Adding or subtracting the same algebraic term to both sides of an equation keeps the equation balanced and valid. This is a fundamental property of equations. Statement II is true.

Both statements are true. The answer is option (3).
In simple words: The boy is indeed 20 years old based on the ratio given. You can also move terms across an equation by adding or subtracting the same thing on both sides - both ideas are correct.

Exam Tip: For Statement I type questions, always verify the calculation with the given conditions. For properties like Statement II, recall the basic rules that make equations work (addition/subtraction property of equality).

Question 19. Statement I: -x > -1 is the same as x > 1. Statement II: We can multiply or divide both sides of a linear equality with any non-zero integer.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer:

Checking Statement I: When you multiply or divide an inequality by a negative number, the inequality symbol reverses. Starting with -x > -1, multiply both sides by -1:

\( (-1) \times (-x) < (-1) \times (-1) \)

\( x < 1 \)

This is NOT the same as x > 1. Statement I is false.

Checking Statement II: For inequalities, multiplying or dividing by a NEGATIVE integer reverses the inequality symbol. Therefore, you cannot multiply or divide both sides by ANY non-zero integer while keeping the symbol unchanged. Statement II is false.

Both statements are false. The answer is option (4).
In simple words: When you multiply an inequality by a negative number, the > or < flips the other way. So -x > -1 actually becomes x < 1, not x > 1. And you have to be careful with negative multipliers in inequalities.

Exam Tip: A key rule for inequalities: multiplying or dividing by a negative number REVERSES the symbol. This is different from equations. Always check whether you are working with an equation or an inequality.

Question 20. Statement I: Given 1 - 2x ≥ 5, where x ∈ I. We can say that x ≤ -2. Statement II: Solution set of an inequality can be represented on a number line.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer:

Checking Statement I:

\( 1 - 2x \geq 5 \)

\( -2x \geq 5 - 1 \)

\( -2x \geq 4 \)

Divide both sides by -2 (and reverse the inequality symbol):

\( x \leq -2 \)

Statement I is true.

Checking Statement II: The solution set of any inequality (like x ≤ -2) can be shown on a number line using dots and arrows. Statement II is true.

Both statements are true. The answer is option (3).
In simple words: When you divide by a negative number in an inequality, flip the symbol. The solution x ≤ -2 can be drawn on a number line with a closed dot at -2 and an arrow pointing left.

Exam Tip: Always remember to reverse the inequality symbol when dividing or multiplying by a negative. For visualizing solutions, a closed dot means "equal to included" (≤ or ≥), and an open dot means "not equal to" (< or >).

Check Your Progress

Question 1. Solve the following equations:
(i) 2(x - 5) + 3(x - 2) = 8 + 7(x - 4)
(ii) \( \frac{3(2x - 5)}{4} - \frac{5(7 - 5x)}{6} = \frac{7x}{3} \)
Answer:

(i)

\( 2(x - 5) + 3(x - 2) = 8 + 7(x - 4) \)

\( 2x - 10 + 3x - 6 = 8 + 7x - 28 \)

\( 5x - 16 = 7x - 20 \)

\( -16 + 20 = 7x - 5x \)

\( 4 = 2x \)

\( x = 2 \)

(ii) First find the LCM of 4, 6, and 3. LCM = 12.

Multiply all terms by 12:

\( 3 \times 3(2x - 5) - 2 \times 5(7 - 5x) = 4 \times 7x \)

\( 9(2x - 5) - 10(7 - 5x) = 28x \)

\( 18x - 45 - 70 + 50x = 28x \)

\( 68x - 115 = 28x \)

\( 40x = 115 \)

\( x = \frac{115}{40} = \frac{23}{8} = 2\frac{7}{8} \)

In simple words: For (i), expand the brackets, collect x terms on one side and numbers on the other, then divide. For (ii), clear the fractions first by multiplying by the LCM, then solve as normal.

Exam Tip: When fractions are in an equation, always find the LCM of all denominators and multiply the entire equation by it. This eliminates fractions and makes the working cleaner.

Question 2. A number exceeds its three-fifth by 22. Find the number.
Answer: Let the number be x. Its three-fifth = \( \frac{3x}{5} \).

According to the problem:

\( x - \frac{3x}{5} = 22 \)

\( \frac{5x - 3x}{5} = 22 \)

\( \frac{2x}{5} = 22 \)

\( 2x = 110 \)

\( x = 55 \)

The number is 55.
In simple words: The number minus its three-fifths equals 22. Simplify the left side to get two-fifths of the number equals 22. Solve to find the number.

Exam Tip: Translate "A exceeds B by C" into the equation A - B = C. This phrasing means A is larger than B by the amount C.

Question 3. When 9 is added to twice a number, the result is 3 more than thrice the number. Find the number.
Answer: Let the number be x. Twice the number = 2x, and thrice the number = 3x.

According to the problem:

\( 2x + 9 = 3x + 3 \)

\( 9 - 3 = 3x - 2x \)

\( 6 = x \)

The number is 6.
In simple words: Double the number and add 9. This equals triple the number plus 3. Moving all x terms to one side and constants to the other gives you the answer.

Exam Tip: For word problems, set up the equation by translating each phrase carefully. "Twice a number is 3 more than thrice" means the first expression equals the second expression plus 3.

Question 4. The ten's digit of a two digit number is twice the unit's digit. The sum of the number and its unit's digit is 66. Find the number.
Answer: Let the unit's digit be x. Then the ten's digit = 2x.

A two-digit number with ten's digit = 2x and unit's digit = x has value = 10(2x) + x = 20x + x = 21x.

According to the problem:

\( 21x + x = 66 \)

\( 22x = 66 \)

\( x = 3 \)

Unit's digit = 3, ten's digit = 2 × 3 = 6, and the number = 21 × 3 = 63.

The required number is 63.
In simple words: The unit digit is 3 and the ten digit is 6 (which is twice 3). The number is 63. Adding 63 and 3 gives 66.

Exam Tip: For a two-digit number with ten's digit = a and unit's digit = b, the value is 10a + b. Always write the number in this form before setting up your equation.

Question 5. A student bought some pens at Rs.8 each and some pencils at Rs.1.50 each. If the total number of pens and pencils purchased is 16 and their total cost is Rs.50, how many pens did he buy?
Answer: Let the number of pens bought be x. Then the number of pencils = 16 - x.

Cost of x pens at Rs.8 each = 8x.

Cost of (16 - x) pencils at Rs.1.50 each = 1.50(16 - x).

Total cost:

\( 8x + 1.50(16 - x) = 50 \)

\( 8x + 24 - 1.50x = 50 \)

\( 6.50x = 50 - 24 \)

\( 6.50x = 26 \)

\( x = 4 \)

The student bought 4 pens.
In simple words: If you buy 4 pens at Rs.8 each, that costs Rs.32. You also buy 12 pencils at Rs.1.50 each, costing Rs.18. Together that is Rs.50 for 16 items.

Exam Tip: In two-item cost problems, always express one quantity in terms of the other (if total is 16, then pencils = 16 - pens). Set up one equation with one unknown, then solve.

Question 6. Arvind is eight years older than his sister. In three years, he will be twice as old as his sister. How old are they now?
Answer: Let the sister's current age be x years. Since Arvind is eight years older, his present age is (x + 8) years. After three years, the sister will be (x + 3) years old and Arvind will be (x + 11) years old. According to the given condition, Arvind's age after three years will be twice his sister's age:
\( x + 11 = 2(x + 3) \)
\( x + 11 = 2x + 6 \)
\( 11 - 6 = 2x - x \)
\( 5 = x \)
\( x = 5 \)
Therefore, the sister is currently 5 years old and Arvind is currently 5 + 8 = 13 years old.
In simple words: Set up an equation using their ages three years from now. The sister will be x + 3 and Arvind will be x + 11. Since Arvind will be twice as old, solve to find x = 5.

Exam Tip: Always define your variable clearly at the start. Write equations based on the future condition (after 3 years), not the present, since the problem states what happens in the future.

Question 7. The angles of a triangle are in the ratio 1 : 2 : 3. Find their measure in degrees.
Answer: Let the three angles be x°, 2x°, and 3x°. Since the sum of all angles in a triangle must equal 180°:
\( x + 2x + 3x = 180 \)
\( 6x = 180 \)
\( x = \frac{180}{6} \)
\( x = 30 \)
The three angles are: x° = 30°, 2x° = 2 × 30° = 60°, and 3x° = 3 × 30° = 90°.
In simple words: Write each angle as a multiple of x using the given ratio. Add them together, set equal to 180°, and solve for x. Then multiply by each ratio number to find each angle.

Exam Tip: Always verify your answer by adding the three angles - they must sum to exactly 180°. This check takes just a few seconds and prevents careless errors.

Question 8. Solve the following inequations and represent their solution on a number line:
(i) \( \frac{2x - 1}{3} \leq 2\frac{1}{2}, x \in W \)
(ii) \( -1 < \frac{2x}{3} + 1 \leq 4, x \in I \)
Answer:
(i) Solving the first inequation:
\( \frac{2x - 1}{3} \leq 2\frac{1}{2} \)
\( \frac{2x - 1}{3} \leq \frac{5}{2} \)
\( 2(2x - 1) \leq 3 \times 5 \)
\( 4x - 2 \leq 15 \)
\( 4x \leq 17 \)
\( x \leq \frac{17}{4} \)
Since x ∈ W (whole numbers), the solution set is {0, 1, 2, 3, 4}. These values are shown as thick dots on the number line.

(ii) Solving the second inequation:
\( -1 < \frac{2x}{3} + 1 \leq 4 \)
\( -1 - 1 < \frac{2x}{3} \leq 4 - 1 \)
\( -2 < \frac{2x}{3} \leq 3 \)
\( -6 < 2x \leq 9 \)
\( -3 < x \leq \frac{9}{2} \)
Since x ∈ I (integers) and \( \frac{9}{2} = 4.5 \), the integers greater than -3 and less than or equal to 4.5 are: -2, -1, 0, 1, 2, 3, 4. The solution set is {-2, -1, 0, 1, 2, 3, 4}. These values are shown as thick dots on the number line.

In simple words: For part (i), simplify the fraction inequality step-by-step until you isolate x. Choose only whole numbers that satisfy the result. For part (ii), handle the compound inequality (two inequality signs) by performing the same operation on all three parts.

Exam Tip: When solving compound inequations with two inequality signs, perform every operation on all three parts simultaneously - this keeps all parts balanced and reduces errors. Always list all integers in the final set, counting carefully from the lower to upper bound.

Question 9. If \( \frac{7m + 2}{3} = -11 \), then find the value of \( 2m^3 + 10m^2 + 4m - 3 \).
Answer: First, find the value of m from the given equation:
\( \frac{7m + 2}{3} = -11 \)
\( 7m + 2 = -33 \)
\( 7m = -33 - 2 \)
\( 7m = -35 \)
\( m = -5 \)
Now substitute m = -5 into \( 2m^3 + 10m^2 + 4m - 3 \):
\( 2(-5)^3 + 10(-5)^2 + 4(-5) - 3 \)
\( = 2(-125) + 10(25) - 20 - 3 \)
\( = -250 + 250 - 20 - 3 \)
\( = -23 \)
Therefore, \( 2m^3 + 10m^2 + 4m - 3 = -23 \).
In simple words: First solve for m by clearing the fraction. Then plug that value into the expression and work through the powers carefully, keeping track of negative signs.

Exam Tip: When substituting negative numbers into powers, enclose them in parentheses - this prevents sign errors. Always calculate powers before multiplying by coefficients.

Question 10. Two persons start moving from two points A and B in opposite directions towards each other. One person start moving from A at the speed of 4 km/h and meets the other person coming from B after 6 hours. If the distance between A and B is 42 km, find the speed of the other person.
Answer: Let the speed of the person from B be x km/h. In 6 hours, the person starting from A covers a distance of 4 × 6 = 24 km. In the same 6 hours, the person from B covers a distance of 6x km. Since they move towards each other and meet, the sum of the distances they cover equals the total distance between A and B:
\( 24 + 6x = 42 \)
\( 6x = 42 - 24 \)
\( 6x = 18 \)
\( x = \frac{18}{6} \)
\( x = 3 \)
Therefore, the speed of the other person is 3 km/h.
In simple words: When two people move towards each other and meet, the total distance traveled by both equals the distance between their starting points. Find how far the first person goes, then use this to find how far the second person goes.

Exam Tip: In relative motion problems where two objects move towards each other, always add their distances - never subtract. This is the key principle for these "meeting" problems.

Question 11. There are some benches in a classroom. If 4 students sit on each bench then 3 benches remains empty and if 3 students sit on each bench then 3 students remain standing. Find the number of students in the class.
Answer: Let the number of benches be x. In the first scenario, if 4 students sit on each bench and 3 benches stay empty, then only (x - 3) benches are occupied. The number of students is 4(x - 3). In the second scenario, if 3 students sit on each bench, all x benches are used and 3 students remain standing. The total number of students is 3x + 3. Since the number of students is the same in both cases:
\( 4(x - 3) = 3x + 3 \)
\( 4x - 12 = 3x + 3 \)
\( 4x - 3x = 3 + 12 \)
\( x = 15 \)
The number of benches is 15. The total number of students is 3x + 3 = 3(15) + 3 = 45 + 3 = 48.
In simple words: In both situations, the student count stays the same. Write two different expressions for the number of students and set them equal. Solve for the number of benches first, then find the number of students.

Exam Tip: These bench-and-student problems require you to write two separate expressions for the same quantity. Carefully track which benches are occupied versus empty in each scenario - this is where most mistakes occur.