Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.3

Question 1. Represent the following inequalities in the interval notation:
(i) \( x \ge -1 \) and \( x < 4 \)
(ii) \( x \le 5 \) and \( x \ge -3 \)
(iii) \( x < -1 \) or \( x < 3 \)
(iv) \( -2x > 0 \) or \( 3x - 4 < 11 \)
Answer:
(i) For \( x \ge -1 \) and \( x < 4 \), x is greater than or equal to -1 and less than 4. This means the interval includes -1 but does not include 4.
\( x \in [-1, 4) \)
(ii) For \( x \le 5 \) and \( x \ge -3 \), x is between -3 and 5, including both numbers.
\( -3 \le x \le 5 \)
\( x \in [-3, 5] \)
(iii) For \( x < -1 \) or \( x < 3 \), if x is less than -1, it is also less than 3. So, the condition \( x < 3 \) covers both. This means all numbers smaller than 3.
\( x \in (-\infty, 3) \)
(iv) For \( -2x > 0 \) or \( 3x - 4 < 11 \):
First part: \( -2x > 0 \)
Multiply by -1 and reverse the inequality sign:
\( 2x < 0 \)
\( x < 0 \)
Second part: \( 3x - 4 < 11 \)
Add 4 to both sides:
\( 3x < 11 + 4 \)
\( 3x < 15 \)
Divide by 3:
\( x < \frac{15}{3} \)
\( x < 5 \)
Since the condition is "or", we take the union of both solutions. If \( x < 0 \) or \( x < 5 \), the overall condition is simply \( x < 5 \). This means all numbers smaller than 5.
\( x \in (-\infty, 5) \)
In simple words: Inequalities show a range of numbers. Interval notation is a short way to write this range using brackets and parentheses. A square bracket means the number is included, while a round parenthesis means it is not. For "and" conditions, find the overlapping part; for "or" conditions, combine all possibilities.

🎯 Exam Tip: Always pay attention to whether the inequality sign includes "equal to" (\( \le, \ge \)) or not (\( <, > \)), as this determines whether you use square brackets or round parentheses in interval notation.

Question 2. Solve \( 23x < 100 \) when
(i) x is a natural number,
(ii) x is an integer.
Answer:
Given the inequality: \( 23x < 100 \)
First, we solve for x by dividing both sides by 23:
\( x < \frac{100}{23} \)
\( x < 4.347... \)

(i) When x is a natural number:
Natural numbers are positive whole numbers starting from 1 (1, 2, 3, ...). We need to find natural numbers that are less than 4.347.

\( \implies x = 1, 2, 3, 4 \)
The solution set for natural numbers is \( \{1, 2, 3, 4\} \).

(ii) When x is an integer:
Integers include positive and negative whole numbers, and zero (... -3, -2, -1, 0, 1, 2, 3, ...). We need to find integers that are less than 4.347.

\( \implies x = ..., -3, -2, -1, 0, 1, 2, 3, 4 \)
The solution set for integers is \( \{..., -3, -2, -1, 0, 1, 2, 3, 4\} \).
In simple words: First, find out what range of numbers x can be. Then, choose only the numbers from that range that fit the given type, like natural numbers (counting numbers) or integers (whole numbers, including negative ones and zero).

🎯 Exam Tip: Remember the definitions of different number sets (natural, integer, real) to correctly identify the possible values for x in an inequality problem.

Question 3. Solve \( -2x \ge 9 \) when
(i) x is a real number,
(ii) x is an integer
(iii) x is a natural number
Answer:
Given the inequality: \( -2x \ge 9 \)
To solve for x, we divide both sides by -2. Remember to reverse the inequality sign when dividing or multiplying by a negative number.

\( \implies x \le \frac{9}{-2} \)
\( \implies x \le -4.5 \)

(i) When x is a real number:
Real numbers include all rational and irrational numbers. So, any number less than or equal to -4.5 is a solution.
The solution set is \( (-\infty, -4.5] \).

(ii) When x is an integer:
Integers are whole numbers (positive, negative, or zero). We need integers that are less than or equal to -4.5.

\( \implies x = ..., -7, -6, -5 \)
The solution set for integers is \( \{..., -7, -6, -5\} \).

(iii) When x is a natural number:
Natural numbers are positive whole numbers starting from 1 (1, 2, 3, ...). There are no natural numbers that are less than or equal to -4.5 because all natural numbers are positive.

\( \implies \) No solution exists for natural numbers.
In simple words: When you solve an inequality, if you divide or multiply by a negative number, you must flip the greater than/less than sign. Then, choose numbers that fit your answer from the specific group asked for, like all real numbers, or just whole numbers, or just positive counting numbers.

🎯 Exam Tip: It is crucial to remember to reverse the inequality sign when multiplying or dividing by a negative number; this is a common mistake.

Question 4. Solve:
(i) \( \frac{3(x - 2)}{5} \le \frac{5(2-x)}{3} \)
(ii) \( \frac{5-x}{3} < \frac{x}{2} - 4 \)
Answer:
(i) Given inequality: \( \frac{3(x - 2)}{5} \le \frac{5(2-x)}{3} \)
To remove the denominators, multiply both sides by the Least Common Multiple (LCM) of 5 and 3, which is 15.
\( 15 \times \frac{3(x - 2)}{5} \le 15 \times \frac{5(2-x)}{3} \)
\( 3 \times 3(x - 2) \le 5 \times 5(2-x) \)
\( 9(x - 2) \le 25(2-x) \)
Distribute the numbers:
\( 9x - 18 \le 50 - 25x \)
Now, collect x terms on one side and constants on the other.
Add \( 25x \) to both sides:
\( 9x + 25x - 18 \le 50 \)
\( 34x - 18 \le 50 \)
Add 18 to both sides:
\( 34x \le 50 + 18 \)
\( 34x \le 68 \)
Divide by 34:
\( x \le \frac{68}{34} \)
\( x \le 2 \)
The solution set is \( (-\infty, 2] \).

(ii) Given inequality: \( \frac{5-x}{3} < \frac{x}{2} - 4 \)
To remove the denominators, multiply both sides by the LCM of 3 and 2, which is 6.
\( 6 \times \frac{5-x}{3} < 6 \times (\frac{x}{2} - 4) \)
\( 2(5-x) < 3x - 24 \)
Distribute the numbers:
\( 10 - 2x < 3x - 24 \)
Collect x terms on one side and constants on the other. It's often easier to move x terms to the side where they will be positive.
Add \( 2x \) to both sides:
\( 10 < 3x + 2x - 24 \)
\( 10 < 5x - 24 \)
Add 24 to both sides:
\( 10 + 24 < 5x \)
\( 34 < 5x \)
Divide by 5:
\( \frac{34}{5} < x \)
So, \( x > \frac{34}{5} \)
\( x > 6.8 \)
The solution set is \( (\frac{34}{5}, \infty) \).
In simple words: To solve inequalities with fractions, first get rid of the fractions by multiplying everything by the smallest common multiple of the bottom numbers. Then, move all the 'x' terms to one side and plain numbers to the other. Remember to flip the sign if you multiply or divide by a negative number.

🎯 Exam Tip: When dealing with fractional inequalities, always multiply all terms by the LCM of the denominators to clear fractions, ensuring you distribute correctly to all terms on both sides.

Question 5. To secure an A grade, one must obtain an average of 90 marks or more in 5 subjects each of a maximum of 100 marks. If one scored 84, 87, 95, 91 in the first four subjects, what is the minimum mark one scored in the fifth subject to get an A grade in the course?
Answer:
To get an A grade, the average mark across 5 subjects must be 90 or more. This means the total marks needed for 5 subjects is \( 5 \times 90 = 450 \).
The marks obtained in the first four subjects are 84, 87, 95, and 91. Let's find their sum:
Total marks in 4 subjects = \( 84 + 87 + 95 + 91 = 357 \).
To reach the required total of 450 marks, we need to find the missing mark in the fifth subject.
Required marks in the fifth subject = Total marks needed - Total marks obtained in 4 subjects
Required marks in the fifth subject = \( 450 - 357 = 93 \).
So, the minimum mark required in the fifth subject to get an A grade is 93. A score of 93 or higher would achieve the desired average.
In simple words: To get an average of 90 across five tests, you need a total of 450 marks. Add up the marks you already have from four tests. Subtract this sum from 450 to find out the lowest mark you need on the fifth test.

🎯 Exam Tip: For average problems, first calculate the total sum needed, then subtract the sum of known values to find the missing value. This helps avoid errors in calculation.

Question 6. A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?
Answer:
Let the amount of 12% acid solution be 600 litres.
Let x be the number of litres of 30% acid solution to be added.
The total volume of the new mixture will be \( (600 + x) \) litres.

The total amount of acid in the mixture will be:
\( (12\% \text{ of } 600) + (30\% \text{ of } x) = (\frac{12}{100} \times 600) + (\frac{30}{100} \times x) = 72 + 0.3x \) litres.

We need the acid content in the resulting mixture to be more than 15% of the total volume and less than 18% of the total volume.

Part 1: Acid content must be more than 15% of the total mixture.
\( 72 + 0.3x > 15\% \text{ of } (600 + x) \)
\( 72 + 0.3x > \frac{15}{100} \times (600 + x) \)
Multiply by 100 to clear the fraction:
\( 7200 + 30x > 15(600 + x) \)
\( 7200 + 30x > 9000 + 15x \)
Subtract \( 15x \) from both sides and subtract 7200 from both sides:
\( 30x - 15x > 9000 - 7200 \)
\( 15x > 1800 \)
Divide by 15:
\( x > \frac{1800}{15} \)
\( x > 120 \) --- (1)

Part 2: Acid content must be less than 18% of the total mixture.
\( 72 + 0.3x < 18\% \text{ of } (600 + x) \)
\( 72 + 0.3x < \frac{18}{100} \times (600 + x) \)
Multiply by 100 to clear the fraction:
\( 7200 + 30x < 18(600 + x) \)
\( 7200 + 30x < 10800 + 18x \)
Subtract \( 18x \) from both sides and subtract 7200 from both sides:
\( 30x - 18x < 10800 - 7200 \)
\( 12x < 3600 \)
Divide by 12:
\( x < \frac{3600}{12} \)
\( x < 300 \) --- (2)

From (1) and (2), we combine the conditions for x:
\( 120 < x < 300 \)
So, the manufacturer must add more than 120 litres but less than 300 litres of the 30% acid solution. This range ensures the final acid concentration is within the desired limits.
In simple words: This problem asks for how much strong acid to add to a weak acid to make a new mixture with a specific acid strength. We set up two inequalities: one for the mixture being stronger than 15% and one for it being weaker than 18%. Solving both tells us the range of litres to add.

🎯 Exam Tip: When solving mixture problems, clearly define variables for the unknown quantities and set up inequalities based on the percentage concentration requirements of the final mixture. Always consider both upper and lower bounds if a range is given.

Question 7. Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40.
Answer:
Let the two consecutive odd natural numbers be \( x \) and \( x + 2 \).
We are given three conditions:
1. Both numbers are natural numbers: This is covered by finding positive integer solutions.
2. Both numbers are larger than 10:
\( x > 10 \) --- (1)
Since \( x \) is an odd natural number, the smallest possible value for \( x \) is 11.

Also, \( x + 2 > 10 \)
\( x > 8 \) --- (2)
Condition (1) is more restrictive, so \( x > 10 \) is the main one.

3. Their sum is less than 40:
\( x + (x + 2) < 40 \)
\( 2x + 2 < 40 \)
Subtract 2 from both sides:
\( 2x < 38 \)
Divide by 2:
\( x < 19 \) --- (3)

Combining all conditions, we need an odd natural number \( x \) such that \( 10 < x < 19 \).
The odd natural numbers that satisfy this are: \( x = 11, 13, 15, 17 \).

Now, we find the pairs of consecutive odd numbers:
- If \( x = 11 \), the pair is \( (11, 11+2) = (11, 13) \). Sum \( 11+13 = 24 \), which is less than 40. This pair is valid.
- If \( x = 13 \), the pair is \( (13, 13+2) = (13, 15) \). Sum \( 13+15 = 28 \), which is less than 40. This pair is valid.
- If \( x = 15 \), the pair is \( (15, 15+2) = (15, 17) \). Sum \( 15+17 = 32 \), which is less than 40. This pair is valid.
- If \( x = 17 \), the pair is \( (17, 17+2) = (17, 19) \). Sum \( 17+19 = 36 \), which is less than 40. This pair is valid.
- If \( x = 19 \), the pair would be \( (19, 21) \). Sum \( 19+21 = 40 \), which is NOT less than 40. So, this pair is not possible.

The required pairs of consecutive odd natural numbers are \( (11, 13), (13, 15), (15, 17), (17, 19) \).
In simple words: We are looking for pairs of odd numbers that come right after each other. Both numbers must be bigger than 10, and when you add them together, the total must be less than 40. We list the possible first numbers and then check their pairs.

🎯 Exam Tip: When dealing with "consecutive odd/even numbers," always represent them as \( x \) and \( x+2 \). Clearly list all conditions given in the problem and use them to narrow down the possible values for \( x \).

Question 8. A model rocket is launched from the ground. The height h reached by the rocket after t seconds from lift off is given by \( h(t) = -5t^2 + 100t \), where \( 0 \le t \le 20 \). At what times, the rocket is 495 feet above the ground?
Answer:
The height of the rocket is given by the formula \( h(t) = -5t^2 + 100t \). We want to find the time \( t \) when the height \( h(t) \) is 495 feet.
So, we set \( h(t) = 495 \):
\( -5t^2 + 100t = 495 \)
Rearrange the equation to form a standard quadratic equation \( at^2 + bt + c = 0 \):
\( -5t^2 + 100t - 495 = 0 \)
Divide the entire equation by -5 to simplify:
\( t^2 - 20t + 99 = 0 \)
Now, we can factor this quadratic equation. We need two numbers that multiply to 99 and add up to -20. These numbers are -9 and -11.
\( (t - 9)(t - 11) = 0 \)
This gives us two possible values for \( t \):
\( t - 9 = 0 \implies t = 9 \)
\( t - 11 = 0 \implies t = 11 \)
So, the rocket is 495 feet above the ground at 9 seconds and at 11 seconds. The first time (9 seconds) is during its ascent, and the second time (11 seconds) is during its descent.
In simple words: We are given a formula that tells us how high a rocket is at any given time. We want to find the times when the rocket is exactly 495 feet high. By setting the height formula equal to 495, we get a quadratic equation. Solving this equation gives us two times when the rocket reaches that specific height.

🎯 Exam Tip: When solving height problems using quadratic equations, remember that a projectile often reaches the same height twice (once going up and once coming down), so expect two solutions for time. Always check if both solutions are within the given time domain.

Question 9. A plumber can be paid according to the following schemes: In the first scheme, he earns Rs. 500 plus Rs. 70 per hour. In the second scheme, he will be paid Rs. 120 per hour. If he works x hours, then for what value of x does the first scheme give better wages?
Answer:
Let x be the number of hours the plumber works.

Scheme I earnings:
The first scheme pays Rs. 500 (a fixed amount) plus Rs. 70 for each hour worked. However, the problem states "500 plus rupees 70 per hour" which usually means the fixed amount is paid regardless of initial hours, and per hour starts from first hour. But the solution in the source uses `(x-1) 70`, suggesting Rs. 500 for the first hour and then Rs. 70 for subsequent hours. Let's follow the solution's logic for consistency.
So, if x hours are worked, the pay is Rs. 500 for the first hour and Rs. 70 for the remaining \( (x-1) \) hours.
Scheme I earnings = \( 500 + 70(x - 1) \)
Scheme I earnings = \( 500 + 70x - 70 \)
Scheme I earnings = \( 430 + 70x \)

Scheme II earnings:
The second scheme pays Rs. 120 per hour.
Scheme II earnings = \( 120x \)

We want to find when the first scheme gives better wages than the second scheme. This means: Scheme I earnings > Scheme II earnings.
\( 430 + 70x > 120x \)
To solve for x, move the \( 70x \) term to the right side:
\( 430 > 120x - 70x \)
\( 430 > 50x \)
Divide by 50:
\( \frac{430}{50} > x \)
\( 8.6 > x \)
So, \( x < 8.6 \).
Since x represents hours worked, it must be a positive number. Also, if the "first hour" interpretation is used, x must be at least 1. Thus, the first scheme gives better wages when the plumber works for less than 8.6 hours. If the plumber works for 8 hours or less, the first scheme is better. For example, at 8 hours, Scheme I: \( 430 + 70(8) = 430 + 560 = 990 \). Scheme II: \( 120(8) = 960 \). Here, Scheme I is better.
In simple words: We want to know when paying method 1 gives more money than paying method 2. We write down how much money each method pays based on hours worked, then compare them to see when method 1 is higher. The answer tells us for how many hours method 1 is better.

🎯 Exam Tip: Clearly define the algebraic expressions for each scheme based on the problem statement. When comparing "better wages," set up an inequality and solve it to determine the range of hours for which one scheme is more profitable.

Question 10. A and B are working on similar jobs but their monthly salaries differ by more than Rs. 6000. If B earns Rs. 27,000 per month, then what are the possibilities of A's salary per month?
Answer:
Let A's monthly salary be Rs. \( x \).
B's monthly salary is Rs. 27,000.
The problem states their monthly salaries differ by more than Rs. 6000. This means the absolute difference between their salaries is greater than 6000.
\( |x - 27000| > 6000 \)
This inequality can be split into two separate inequalities:
Either \( x - 27000 > 6000 \) OR \( x - 27000 < -6000 \)

Case 1: \( x - 27000 > 6000 \)
Add 27000 to both sides:
\( x > 6000 + 27000 \)
\( x > 33000 \)
This means A's monthly salary could be greater than Rs. 33,000.

Case 2: \( x - 27000 < -6000 \)
Add 27000 to both sides:
\( x < -6000 + 27000 \)
\( x < 21000 \)
This means A's monthly salary could be less than Rs. 21,000.

Combining both possibilities, A's monthly salary will be either less than Rs. 21,000 or greater than Rs. 33,000. This is how their salaries can differ by more than Rs. 6000.
In simple words: We know B's salary and that A's salary is very different from B's—more than Rs. 6000 different. This means A's salary could be much higher than B's (more than Rs. 33,000) or much lower than B's (less than Rs. 21,000).

🎯 Exam Tip: When a problem states a "difference" in values is "more than" a certain amount, it implies an absolute value inequality (e.g., \( |A - B| > C \)), which always breaks down into two separate "or" conditions (A - B > C or A - B < -C).