GSEB Class 11 Maths Solutions Chapter 6 Linear Inequalities Exercise 6.1

Question 1. Solve \( 24x < 100 \), when
(i) x is a natural number.
(ii) x is an integer.
Answer: We need to solve the inequality \( 24x < 100 \).
Divide both sides by 24:
\( x < \frac{100}{24} \)
\( x < \frac{25}{6} \)
\( x < 4.16... \)
(i) When x is a natural number:
Natural numbers that are less than 4.16... are 1, 2, 3, 4.
So, the solution for natural numbers is \( \{1, 2, 3, 4\} \).
(ii) When x is an integer:
Integers that are less than 4.16... are ..., -4, -3, -2, -1, 0, 1, 2, 3, 4.
So, the solution for integers is \( \{..., -4, -3, -2, -1, 0, 1, 2, 3, 4\} \).
In simple words: First, simplify the inequality by dividing to find the value of x. Then, identify which natural numbers and which integers fit into that range. Remember that natural numbers start from 1, and integers include negative numbers, zero, and positive numbers.

Exam Tip: When dividing an inequality, if you divide by a negative number, always remember to reverse the inequality sign. Also, carefully distinguish between natural numbers, integers, and real numbers for the solutions.

Question 2. Solve \( -12x > 30 \), when
(i) x is a natural number,
(ii) x is an integer.
Answer: We need to solve the inequality \( -12x > 30 \).
Divide both sides by -12. Since we are dividing by a negative number, reverse the inequality sign:
\( x < \frac{30}{-12} \)
\( x < \frac{-5}{2} \)
\( x < -2.5 \)
(i) When x is a natural number:
There are no natural numbers (1, 2, 3, ...) that are less than -2.5.
So, there is no solution for natural numbers.
(ii) When x is an integer:
Integers that are less than -2.5 are ..., -5, -4, -3.
So, the solution for integers is \( \{..., -5, -4, -3\} \).
In simple words: First, solve the inequality. When you divide by a negative number, flip the sign. Then, look for natural numbers (whole numbers from 1 up) and integers (all whole numbers, positive, negative, and zero) that fit the answer.

Exam Tip: Be very careful when dividing or multiplying an inequality by a negative number; always remember to flip the direction of the inequality sign. This is a common mistake.

Question 3. Solve \( 5x - 3 < 7 \), when
(i) x is a natural number
(ii) x is a real number.
Answer: We need to solve the inequality \( 5x - 3 < 7 \).
Add 3 to both sides:
\( 5x < 7 + 3 \)
\( 5x < 10 \)
Divide both sides by 5:
\( x < \frac{10}{5} \)
\( x < 2 \)
(i) When x is a natural number:
The only natural number less than 2 is 1.
So, the solution is \( x = 1 \).
(ii) When x is a real number:
All real numbers less than 2 satisfy this inequality. This can be written in interval notation as \( (-\infty, 2) \).
The number line illustration is given below:

In simple words: First, solve for x. You will find that x must be smaller than 2. For natural numbers, only 1 works. For real numbers, any number less than 2 is a solution. On a number line, this means an open circle at 2 and a thick line going towards negative infinity.

Exam Tip: Pay close attention to whether the variable is restricted to natural numbers, integers, or real numbers, as this changes how you write the solution set. An open circle on a number line means the endpoint is not included.

Question 4. Solve \( 3x + 8 > 2 \), when
(i) x is an integer
(ii) x is a real number.
Answer: We need to solve the inequality \( 3x + 8 > 2 \).
Subtract 8 from both sides:
\( 3x > 2 - 8 \)
\( 3x > -6 \)
Divide both sides by 3:
\( x > \frac{-6}{3} \)
\( x > -2 \)
(i) When x is an integer:
Integers that are greater than -2 are -1, 0, 1, 2, 3, ...
So, the solution for integers is \( \{-1, 0, 1, 2, 3, ...\} \).
(ii) When x is a real number:
All real numbers greater than -2 satisfy this inequality. This can be written in interval notation as \( (-2, \infty) \).
The number line illustration is given below:

In simple words: First, solve for x, which means x must be larger than -2. For integers, this includes -1, 0, 1, and so on. For real numbers, it includes all numbers from just after -2 up to infinity. The number line will have an open circle at -2 and a thick line stretching to the right.

Exam Tip: Remember to express integer solutions as a set of discrete values and real number solutions as an interval. An open circle on the number line indicates that the boundary value is not part of the solution.

Solve the following inequalities in questions 5 to 16 for real x:

Question 5. The inequality is \( 4x + 3 < 6x + 7 \).
Answer: We need to solve the inequality \( 4x + 3 < 6x + 7 \).
Transpose 6x to the left-hand side and 3 to the right-hand side:
\( 4x - 6x < 7 - 3 \)
\( -2x < 4 \)
Divide both sides by -2. Since we are dividing by a negative number, reverse the inequality sign:
\( x > \frac{4}{-2} \)
\( x > -2 \)
The solution for real x is the interval \( (-2, \infty) \).
The number line illustration is given below:

In simple words: To solve this, move the x terms to one side and numbers to the other. Remember to flip the inequality sign if you divide by a negative number. The result will be x is greater than -2, meaning all real numbers from just above -2 to infinity.

Exam Tip: Always double-check your arithmetic, especially when rearranging terms. A common error is forgetting to change the inequality direction when multiplying or dividing by a negative number.

Question 6. The inequality is \( 3x - 7 > 5x - 1 \).
Answer: We need to solve the inequality \( 3x - 7 > 5x - 1 \).
Transpose 5x to the left-hand side and -7 to the right-hand side:
\( 3x - 5x > -1 + 7 \)
\( -2x > 6 \)
Divide both sides by -2. Since we are dividing by a negative number, reverse the inequality sign:
\( x < \frac{6}{-2} \)
\( x < -3 \)
The solution for real x is the interval \( (-\infty, -3) \).
The number line illustration is given below:

In simple words: Move the terms to simplify the inequality. When you divide by a negative, change the direction of the sign. The result means x must be smaller than -3, covering all real numbers from negative infinity up to, but not including, -3.

Exam Tip: Ensure that you correctly transpose terms across the inequality sign, changing their operation (addition becomes subtraction, multiplication becomes division). Always flip the inequality sign when multiplying or dividing by a negative value.

Question 7. The inequality is \( 3(x - 1) \le 2(x - 3) \).
Answer: We need to solve the inequality \( 3(x - 1) \le 2(x - 3) \).
Simplify both sides by distributing:
\( 3x - 3 \le 2x - 6 \)
Transpose 2x to the left-hand side and -3 to the right-hand side:
\( 3x - 2x \le -6 + 3 \)
\( x \le -3 \)
The solution for real x is the interval \( (-\infty, -3] \).
The number line illustration is given below:

In simple words: First, expand both sides of the inequality. Then, collect all x terms on one side and constant terms on the other. This gives x is less than or equal to -3. This means all real numbers from negative infinity up to and including -3 are solutions.

Exam Tip: When an inequality includes "equal to" (\( \le \) or \( \ge \)), the endpoint on the number line should be a filled circle, and the interval notation should use a square bracket (e.g., `]`).

Question 8. The inequality is \( 3(2 - x) \ge 2(1 - x) \).
Answer: We need to solve the inequality \( 3(2 - x) \ge 2(1 - x) \).
Simplify both sides by distributing:
\( 6 - 3x \ge 2 - 2x \)
Transpose -2x to the left-hand side and 6 to the right-hand side:
\( -3x + 2x \ge 2 - 6 \)
\( -x \ge -4 \)
Multiply both sides by -1. Since we are multiplying by a negative number, reverse the inequality sign:
\( x \le 4 \)
The solution for real x is the interval \( (-\infty, 4] \).
The number line illustration is given below:

In simple words: First, expand the inequality. Then, move all x terms to one side and numbers to the other. Be sure to flip the inequality sign when multiplying by -1. The solution means x is 4 or any smaller real number, so the line will go left from a filled circle at 4.

Exam Tip: Always perform distribution correctly first. A common mistake is not changing the inequality direction when multiplying or dividing by a negative value, especially at the last step.

Question 9. The inequality is \( x + \frac{x}{2} + \frac{x}{3} < 11 \).
Answer: We need to solve the inequality \( x + \frac{x}{2} + \frac{x}{3} < 11 \).
Find a common denominator for the fractions, which is 6:
\( \frac{6x}{6} + \frac{3x}{6} + \frac{2x}{6} < 11 \)
Combine the terms on the left-hand side:
\( \frac{6x + 3x + 2x}{6} < 11 \)
\( \frac{11x}{6} < 11 \)
Multiply both sides by 6:
\( 11x < 11 \times 6 \)
\( 11x < 66 \)
Divide both sides by 11:
\( x < \frac{66}{11} \)
\( x < 6 \)
The solution for real x is the interval \( (-\infty, 6) \).
The number line illustration is given below:

In simple words: To solve this, first add the fractions with x by finding a common bottom number. Then, multiply and divide to isolate x. The answer means x must be smaller than 6, so it includes all real numbers from negative infinity up to, but not including, 6.

Exam Tip: When dealing with fractional inequalities, multiply all terms by the least common multiple (LCM) of the denominators to eliminate the fractions, making the equation simpler to solve.

Question 10. The inequality is \( \frac{x}{3} > \frac{x}{2} + 1 \).
Answer: We need to solve the inequality \( \frac{x}{3} > \frac{x}{2} + 1 \).
Transpose \( \frac{x}{2} \) to the left-hand side:
\( \frac{x}{3} - \frac{x}{2} > 1 \)
Find a common denominator for the fractions, which is 6:
\( \frac{2x}{6} - \frac{3x}{6} > 1 \)
Combine the terms on the left-hand side:
\( \frac{2x - 3x}{6} > 1 \)
\( \frac{-x}{6} > 1 \)
Multiply both sides by 6:
\( -x > 6 \)
Multiply both sides by -1. Since we are multiplying by a negative number, reverse the inequality sign:
\( x < -6 \)
The solution for real x is the interval \( (-\infty, -6) \).
The number line illustration is given below:

In simple words: First, move the x terms to one side, combine the fractions using a common denominator. Then, multiply to remove the denominator. Finally, multiply by -1 to solve for x, remembering to flip the inequality sign. The answer shows that x is less than -6.

Exam Tip: Always collect like terms, especially when dealing with fractions. Using the LCM to clear denominators can prevent fractional arithmetic errors. Don't forget to flip the sign if you multiply or divide by a negative number.

Question 11. The inequality is \( \frac{3(x - 2)}{5} \le \frac{5(2 - x)}{3} \).
Answer: We need to solve the inequality \( \frac{3(x - 2)}{5} \le \frac{5(2 - x)}{3} \).
Multiply both sides by the Least Common Multiple (LCM) of 5 and 3, which is 15:
\( 15 \times \frac{3(x - 2)}{5} \le 15 \times \frac{5(2 - x)}{3} \)
\( 3 \times 3(x - 2) \le 5 \times 5(2 - x) \)
\( 9(x - 2) \le 25(2 - x) \)
Simplify both sides by distributing:
\( 9x - 18 \le 50 - 25x \)
Transpose -25x to the left-hand side and -18 to the right-hand side:
\( 9x + 25x \le 50 + 18 \)
\( 34x \le 68 \)
Divide both sides by 34:
\( x \le \frac{68}{34} \)
\( x \le 2 \)
The solution for real x is the interval \( (-\infty, 2] \).
The number line illustration is given below:

In simple words: Get rid of the fractions by multiplying everything by the smallest number both denominators can divide into. Then, expand the brackets, move x terms to one side, and numbers to the other. Finally, divide to find x, which should be less than or equal to 2.

Exam Tip: Clearing denominators by multiplying by the LCM is crucial for solving fractional inequalities. Be careful with signs when distributing and transposing terms.

Question 12. The inequality is \( \frac{1}{2}\left(\frac{3x}{5} + 4\right) \ge \frac{1}{3}(x - 6) \).
Answer: We need to solve the inequality \( \frac{1}{2}\left(\frac{3x}{5} + 4\right) \ge \frac{1}{3}(x - 6) \).
First, simplify the left-hand side:
\( \frac{1}{2}\left(\frac{3x + 20}{5}\right) \ge \frac{1}{3}(x - 6) \)
\( \frac{3x + 20}{10} \ge \frac{x - 6}{3} \)
Multiply both sides by the LCM of 10 and 3, which is 30:
\( 30 \times \frac{3x + 20}{10} \ge 30 \times \frac{x - 6}{3} \)
\( 3(3x + 20) \ge 10(x - 6) \)
Simplify both sides by distributing:
\( 9x + 60 \ge 10x - 60 \)
Transpose 10x to the left-hand side and 60 to the right-hand side:
\( 9x - 10x \ge -60 - 60 \)
\( -x \ge -120 \)
Multiply both sides by -1. Since we are multiplying by a negative number, reverse the inequality sign:
\( x \le 120 \)
The solution for real x is the interval \( (-\infty, 120] \).
The number line illustration is given below:

In simple words: First, simplify each side of the inequality, especially the left side by combining terms in the bracket. Then, get rid of all fractions by multiplying by the smallest common multiple of the denominators. Distribute and move terms, and finally, divide by -1, remembering to reverse the inequality sign. This means x is 120 or smaller.

Exam Tip: When simplifying complex inequalities, break it down into steps: clear parentheses, combine like terms, eliminate fractions, then isolate the variable. This systematic approach reduces errors.

Question 13. The inequality is \( 2(2x + 3) - 10 < 6(x - 2) \).
Answer: We need to solve the inequality \( 2(2x + 3) - 10 < 6(x - 2) \).
Simplify both sides by distributing:
\( 4x + 6 - 10 < 6x - 12 \)
\( 4x - 4 < 6x - 12 \)
Transpose 6x to the left-hand side and -4 to the right-hand side:
\( 4x - 6x < -12 + 4 \)
\( -2x < -8 \)
Divide both sides by -2. Since we are dividing by a negative number, reverse the inequality sign:
\( x > \frac{-8}{-2} \)
\( x > 4 \)
The solution for real x is the interval \( (4, \infty) \).
The number line illustration is given below:

In simple words: Expand the brackets on both sides and simplify the numbers. Move all terms with x to one side and plain numbers to the other. When you divide by a negative number at the end, remember to flip the direction of the inequality sign. This means x is larger than 4.

Exam Tip: Be careful with signs when performing arithmetic, especially subtraction of negative numbers. Always distribute the numbers outside the parentheses to all terms inside.

Question 14. The inequality is \( 37 - (3x + 5) \ge 9x - 8(x - 3) \).
Answer: We need to solve the inequality \( 37 - (3x + 5) \ge 9x - 8(x - 3) \).
Simplify both sides by distributing:
\( 37 - 3x - 5 \ge 9x - 8x + 24 \)
\( 32 - 3x \ge x + 24 \)
Transpose x to the left-hand side and 32 to the right-hand side:
\( -3x - x \ge 24 - 32 \)
\( -4x \ge -8 \)
Divide both sides by -4. Since we are dividing by a negative number, reverse the inequality sign:
\( x \le \frac{-8}{-4} \)
\( x \le 2 \)
The solution for real x is the interval \( (-\infty, 2] \).
The number line illustration is given below:

In simple words: First, simplify both sides by distributing the numbers outside the brackets and combining similar terms. Then, gather all x terms on one side and constant numbers on the other. Finally, divide by -4, remembering to flip the inequality sign, which means x is 2 or smaller.

Exam Tip: Be very careful when removing parentheses, especially when a minus sign precedes them; remember to change the sign of every term inside the parentheses. This is a frequent source of errors.

Question 15. The inequality is \( \frac{x}{4} < \frac{5x - 2}{3} - \frac{7x - 3}{5} \).
Answer: We need to solve the inequality \( \frac{x}{4} < \frac{5x - 2}{3} - \frac{7x - 3}{5} \).
Multiply each term by the LCM of 4, 3, and 5, which is 60:
\( 60 \times \frac{x}{4} < 60 \times \frac{5x - 2}{3} - 60 \times \frac{7x - 3}{5} \)
\( 15x < 20(5x - 2) - 12(7x - 3) \)
Simplify both sides by distributing:
\( 15x < 100x - 40 - 84x + 36 \)
\( 15x < (100x - 84x) + (-40 + 36) \)
\( 15x < 16x - 4 \)
Transpose 16x to the left-hand side:
\( 15x - 16x < -4 \)
\( -x < -4 \)
Multiply both sides by -1. Since we are multiplying by a negative number, reverse the inequality sign:
\( x > 4 \)
The solution for real x is the interval \( (4, \infty) \).
The number line illustration is given below:

In simple words: To clear the fractions, multiply every part of the inequality by the smallest common multiple of the bottom numbers. Then, open the brackets, combine similar terms, and move x terms to one side. Finally, divide by a negative number, so remember to flip the inequality sign. The answer means x is larger than 4.

Exam Tip: Be meticulous with calculations, especially when dealing with multiple fractions and negative signs. Distribute correctly and combine like terms before isolating the variable. Remember the rule about reversing the inequality sign.

Question 16. The inequality is \( \frac{2x - 1}{3} \ge \frac{3x - 2}{4} - \frac{2 - x}{5} \).
Answer: We need to solve the inequality \( \frac{2x - 1}{3} \ge \frac{3x - 2}{4} - \frac{2 - x}{5} \).
Multiply each term by the LCM of 3, 4, and 5, which is 60:
\( 60 \times \frac{2x - 1}{3} \ge 60 \times \frac{3x - 2}{4} - 60 \times \frac{2 - x}{5} \)
\( 20(2x - 1) \ge 15(3x - 2) - 12(2 - x) \)
Simplify both sides by distributing:
\( 40x - 20 \ge 45x - 30 - 24 + 12x \)
\( 40x - 20 \ge (45x + 12x) + (-30 - 24) \)
\( 40x - 20 \ge 57x - 54 \)
Transpose 57x to the left-hand side and -20 to the right-hand side:
\( 40x - 57x \ge -54 + 20 \)
\( -17x \ge -34 \)
Divide both sides by -17. Since we are dividing by a negative number, reverse the inequality sign:
\( x \le \frac{-34}{-17} \)
\( x \le 2 \)
The solution for real x is the interval \( (-\infty, 2] \).
The number line illustration is given below:

In simple words: Clear the fractions by multiplying by the smallest common multiple of 3, 4, and 5. Then, expand all brackets, collect x terms on one side and numbers on the other. Finally, divide by -17, remembering to flip the inequality sign. This indicates x is 2 or smaller.

Exam Tip: Handling complex inequalities requires meticulous care with distribution and combining like terms. Pay extra attention to the minus signs, as they often lead to errors. Always reverse the inequality sign when multiplying or dividing by a negative number.

Question 17. The inequality is \( 3x - 2 < 2x + 1 \).
Answer: We need to solve the inequality \( 3x - 2 < 2x + 1 \).
Transpose 2x to the left-hand side and -2 to the right-hand side:
\( 3x - 2x < 1 + 2 \)
\( x < 3 \)
The solution for real x is the interval \( (-\infty, 3) \).
The number line illustration is given below:

In simple words: Move the x terms to one side and the constant numbers to the other. Simplify both sides. The solution indicates that x is smaller than 3, meaning all real numbers from negative infinity up to, but not including, 3.

Exam Tip: Basic linear inequalities are like equations, but remember to maintain the inequality sign. Always gather x terms and constants separately for clear simplification.

Question 18. The inequality is \( 5x - 3 \ge 3x - 5 \).
Answer: We need to solve the inequality \( 5x - 3 \ge 3x - 5 \).
Transpose 3x to the left-hand side and -3 to the right-hand side:
\( 5x - 3x \ge -5 + 3 \)
\( 2x \ge -2 \)
Divide both sides by 2:
\( x \ge \frac{-2}{2} \)
\( x \ge -1 \)
The solution for real x is the interval \( [-1, \infty) \).
The number line illustration is given below:

In simple words: Move the x terms to one side and the numbers to the other, then simplify. The answer means x is -1 or any larger real number, including -1 itself.

Exam Tip: A filled circle on the number line and a square bracket in interval notation indicate that the endpoint is included in the solution set.

Question 19. The inequality is \( 3(1 - x) < 2(x + 4) \).
Answer: We need to solve the inequality \( 3(1 - x) < 2(x + 4) \).
Simplify both sides by distributing:
\( 3 - 3x < 2x + 8 \)
Transpose 2x to the left-hand side and 3 to the right-hand side:
\( -3x - 2x < 8 - 3 \)
\( -5x < 5 \)
Divide both sides by -5. Since we are dividing by a negative number, reverse the inequality sign:
\( x > \frac{5}{-5} \)
\( x > -1 \)
The solution for real x is the interval \( (-1, \infty) \).
The number line illustration is given below:

In simple words: First, expand the expressions on both sides. Then, move all x terms to one side and numbers to the other. When you divide by a negative number, be sure to flip the inequality sign. The answer means x is larger than -1.

Exam Tip: Be careful with signs, especially when distributing a negative number. Always verify that the inequality direction is correctly reversed when dividing or multiplying by a negative value.

Question 20. The inequality is \( \frac{x}{2} < \frac{5x - 2}{3} - \frac{7x - 3}{5} \).
Answer: We need to solve the inequality \( \frac{x}{2} < \frac{5x - 2}{3} - \frac{7x - 3}{5} \).
Multiply each term by the LCM of 2, 3, and 5, which is 30:
\( 30 \times \frac{x}{2} < 30 \times \frac{5x - 2}{3} - 30 \times \frac{7x - 3}{5} \)
\( 15x < 10(5x - 2) - 6(7x - 3) \)
Simplify both sides by distributing:
\( 15x < 50x - 20 - 42x + 18 \)
\( 15x < (50x - 42x) + (-20 + 18) \)
\( 15x < 8x - 2 \)
Transpose 8x to the left-hand side:
\( 15x - 8x < -2 \)
\( 7x < -2 \)
Divide both sides by 7:
\( x < \frac{-2}{7} \)
The solution for real x is the interval \( \left(-\infty, -\frac{2}{7}\right) \).
The number line illustration is given below:

In simple words: To solve this, first clear all fractions by multiplying by the smallest common multiple of the denominators. Then, expand the brackets, combine similar terms, and move the x terms to one side. Finally, divide to find x, which means x is smaller than \( -\frac{2}{7} \).

Exam Tip: Be meticulous with calculations involving multiple fractions and terms. Ensure proper distribution, especially when a negative sign precedes a fraction. Always simplify the expression as much as possible before isolating the variable.

Question 21. Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks?
Answer: Let Ravi get x marks in the third unit test.
The marks obtained in the first two tests are 70 and 75.
The average marks obtained by Ravi across three tests will be \( \frac{70 + 75 + x}{3} \).
According to the problem, the average must be at least 60 marks:
\( \frac{70 + 75 + x}{3} \ge 60 \)
\( \frac{145 + x}{3} \ge 60 \)
Multiply both sides by 3:
\( 145 + x \ge 60 \times 3 \)
\( 145 + x \ge 180 \)
Subtract 145 from both sides:
\( x \ge 180 - 145 \)
\( x \ge 35 \)
Therefore, Ravi should get at least 35 marks in the third unit test to achieve an average of at least 60 marks.
In simple words: To find the lowest score Ravi needs, set up an inequality where his total marks (including x for the third test) divided by three is 60 or more. Solve this to find that x must be 35 or higher.

Exam Tip: When calculating averages, always sum all the values and divide by the number of values. For "at least" problems, use the \( \ge \) inequality sign.

Question 22. To receive 'grade A' in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita's marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade 'A' in the course?
Answer: Let Sunita obtain x marks in the fifth examination.
The marks in the first four examinations are 87, 92, 94, and 95.
The average of marks for 5 examinations will be \( \frac{87 + 92 + 94 + 95 + x}{5} \).
Summing the given marks:
\( 87 + 92 + 94 + 95 = 368 \)
So, the average is \( \frac{368 + x}{5} \).
For Sunita to receive a 'grade A', her average must be at least 90 marks:
\( \frac{368 + x}{5} \ge 90 \)
Multiply both sides by 5:
\( 368 + x \ge 90 \times 5 \)
\( 368 + x \ge 450 \)
Subtract 368 from both sides:
\( x \ge 450 - 368 \)
\( x \ge 82 \)
Therefore, Sunita should obtain at least 82 marks in the fifth examination to get a 'grade A'.
In simple words: To get an A, Sunita needs an average of 90 or more across five tests. Add up her first four scores, then add x for the fifth test, divide by five, and set it greater than or equal to 90. Solving this shows she needs at least 82 marks in the last test.

Exam Tip: For problems involving minimum scores, clearly define the unknown variable (e.g., x for the missing score), set up the average inequality, and solve for x. Ensure your final answer is phrased in context.

Question 23. Find all pairs of consecutive odd positive integers both of which are smaller than 10, such that their sum is more than 11?
Answer: Let x be the smaller of the two consecutive odd positive integers. Then the other odd integer is \( x + 2 \).
We are given two conditions:
1. Both integers are smaller than 10.
\( x < 10 \) and \( x + 2 < 10 \)
Since x is a positive odd integer, \( x \) must be less than 10. The possible odd positive integers less than 10 are 1, 3, 5, 7, 9.
2. Their sum is more than 11.
\( x + (x + 2) > 11 \)
\( 2x + 2 > 11 \)
\( 2x > 11 - 2 \)
\( 2x > 9 \)
\( x > \frac{9}{2} \)
\( x > 4.5 \)
Now, we combine the conditions for x:
x must be an odd positive integer.
x must be less than 10.
x must be greater than 4.5.
The odd positive integers that satisfy \( x < 10 \) and \( x > 4.5 \) are 5, 7, 9.
Let's check the pairs:
If \( x = 5 \), the consecutive odd integer is \( x + 2 = 7 \).
Both 5 and 7 are smaller than 10. (True)
Their sum is \( 5 + 7 = 12 \), which is more than 11. (True)
So, (5, 7) is a valid pair.
If \( x = 7 \), the consecutive odd integer is \( x + 2 = 9 \).
Both 7 and 9 are smaller than 10. (True)
Their sum is \( 7 + 9 = 16 \), which is more than 11. (True)
So, (7, 9) is a valid pair.
If \( x = 9 \), the consecutive odd integer is \( x + 2 = 11 \).
11 is not smaller than 10. (False)
So, (9, 11) is not a valid pair.
Thus, the possible pairs of consecutive odd positive integers are (5, 7) and (7, 9).
In simple words: First, set up x as the smaller odd integer and x+2 as the next. Make sure both are under 10. Then, write an inequality that their sum is over 11. Solve for x and check which odd numbers fit all the rules to find the pairs.

Exam Tip: Clearly define your variables (e.g., x and x+2 for consecutive odd/even integers). Systematically list all conditions as inequalities, solve them, and then check each potential solution against all original conditions.

Question 24. Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23?
Answer: Let x be the smaller of the two consecutive even positive integers. Then the other even integer is \( x + 2 \).
We are given two conditions:
1. Both integers are larger than 5.
\( x > 5 \) and \( x + 2 > 5 \)
Since x is an even positive integer, \( x \) must be greater than 5. Possible even positive integers greater than 5 are 6, 8, 10, 12, ...
2. Their sum is less than 23.
\( x + (x + 2) < 23 \)
\( 2x + 2 < 23 \)
\( 2x < 23 - 2 \)
\( 2x < 21 \)
\( x < \frac{21}{2} \)
\( x < 10.5 \)
Now, we combine the conditions for x:
x must be an even positive integer.
x must be greater than 5.
x must be less than 10.5.
The even positive integers that satisfy \( x > 5 \) and \( x < 10.5 \) are 6, 8, 10.
Let's check the pairs:
If \( x = 6 \), the consecutive even integer is \( x + 2 = 8 \).
Both 6 and 8 are larger than 5. (True)
Their sum is \( 6 + 8 = 14 \), which is less than 23. (True)
So, (6, 8) is a valid pair.
If \( x = 8 \), the consecutive even integer is \( x + 2 = 10 \).
Both 8 and 10 are larger than 5. (True)
Their sum is \( 8 + 10 = 18 \), which is less than 23. (True)
So, (8, 10) is a valid pair.
If \( x = 10 \), the consecutive even integer is \( x + 2 = 12 \).
Both 10 and 12 are larger than 5. (True)
Their sum is \( 10 + 12 = 22 \), which is less than 23. (True)
So, (10, 12) is a valid pair.
If \( x = 12 \), the consecutive even integer is \( x + 2 = 14 \).
Their sum is \( 12 + 14 = 26 \), which is not less than 23. (False)
Thus, the possible pairs of consecutive even positive integers are (6, 8), (8, 10), and (10, 12).
In simple words: Define x as the first even integer and x+2 as the second. Set up two conditions: both integers must be greater than 5, and their sum must be less than 23. Solve these inequalities for x. Then, find the even numbers that fit both rules and list the valid pairs.

Exam Tip: Remember that consecutive even (or odd) integers differ by 2. Always set up and solve all given conditions as separate inequalities, then find the intersection of their solution sets.

Question 25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side?
Answer: Let the length of the shortest side be x cm.
According to the problem:
The longest side is 3 times the shortest side, so its length is \( 3x \) cm.
The third side is 2 cm shorter than the longest side, so its length is \( (3x - 2) \) cm.
The perimeter of the triangle is the sum of the lengths of all three sides:
Perimeter = \( x + 3x + (3x - 2) \)
Perimeter = \( 7x - 2 \)
We are given that the perimeter of the triangle is at least 61 cm:
\( 7x - 2 \ge 61 \)
Add 2 to both sides:
\( 7x \ge 61 + 2 \)
\( 7x \ge 63 \)
Divide both sides by 7:
\( x \ge \frac{63}{7} \)
\( x \ge 9 \)
Thus, the minimum length of the shortest side is 9 cm.
We can also find the lengths of the other sides for this minimum value:
Shortest side = 9 cm
Longest side = \( 3 \times 9 = 27 \) cm
Third side = \( 3 \times 9 - 2 = 27 - 2 = 25 \) cm
In simple words: First, use 'x' for the shortest side, then write expressions for the other two sides based on 'x'. Add all three side lengths to get the perimeter. Set this perimeter to be 61 or more, then solve for 'x'. The smallest value of 'x' you find is the minimum length for the shortest side.

Exam Tip: Clearly define variables for each side of the triangle. Translate word problems into mathematical inequalities step-by-step. Remember that "at least" translates to \( \ge \).

Question 26. A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths for the shortest board, if the third piece is to be at least 5 cm longer than the second?
Answer: Let x be the length of the shortest board in cm.
According to the problem:
The second length is 3 cm longer than the shortest: \( (x + 3) \) cm.
The third length is twice as long as the shortest: \( 2x \) cm.
The total length of the board is 91 cm. The sum of the three pieces must be less than or equal to 91 cm:
\( x + (x + 3) + 2x \le 91 \)
\( 4x + 3 \le 91 \)
Subtract 3 from both sides:
\( 4x \le 91 - 3 \)
\( 4x \le 88 \)
Divide by 4:
\( x \le \frac{88}{4} \)
\( x \le 22 \)
Now, we also have an additional condition: the third piece must be at least 5 cm longer than the second piece.
Third piece length: \( 2x \)
Second piece length: \( x + 3 \)
So, \( 2x \ge (x + 3) + 5 \)
\( 2x \ge x + 8 \)
Subtract x from both sides:
\( x \ge 8 \)
Combining both conditions, \( x \ge 8 \) and \( x \le 22 \), we get the possible lengths for the shortest board:
\( 8 \le x \le 22 \)
Thus, the shortest board should be at least 8 cm long but not more than 22 cm long. The possible lengths for the shortest board are between 8 cm and 22 cm, inclusive.
In simple words: Assign x to the shortest board. Express the other two board lengths using x. First, set up an inequality that the sum of all three lengths is 91 cm or less. Second, set up an inequality that the third length is at least 5 cm longer than the second. Solve both inequalities and find the range of x that satisfies both conditions.

Exam Tip: In problems with multiple conditions, ensure you translate each condition into a separate inequality. Solve each inequality independently, and then find the intersection of all solution sets to get the final answer range.