Exercise 5.1 - Question 1

Question: Solve $24x < 100$, when (i) $x$ is a natural number. (ii) $x$ is an integer.

Solution:

Given inequality is:

$$24x < 100$$

Dividing both sides by $24$ (since $24 > 0$, the inequality sign remains unchanged):

$$x < \frac{100}{24}$$ $$x < \frac{25}{6}$$

Converting into decimal form:

$$x < 4.166...$$

(i) When $x$ is a natural number:

Natural numbers are $\{1, 2, 3, 4, 5, \dots\}$. The natural numbers less than $4.166...$ are $1, 2, 3,$ and $4$.

Therefore, the solution set is:

$$\{1, 2, 3, 4\}$$

(ii) When $x$ is an integer:

Integers are $\{\dots, -3, -2, -1, 0, 1, 2, 3, 4, 5, \dots\}$. The integers strictly less than $4.166...$ include all integers less than or equal to $4$.

Therefore, the solution set is:

$$\{\dots, -3, -2, -1, 0, 1, 2, 3, 4\}$$

Exercise 5.1 - Question 2

Question: Solve $-12x > 30$, when (i) $x$ is a natural number. (ii) $x$ is an integer.

Solution:

Given inequality is:

$$-12x > 30$$

Dividing both sides by $-12$. Remember that dividing or multiplying an inequality by a negative number reverses the direction of the inequality sign:

$$x < \frac{30}{-12}$$ $$x < -\frac{5}{2}$$ $$x < -2.5$$

(i) When $x$ is a natural number:

Natural numbers start from $1$, and there are no natural numbers less than $-2.5$.

Therefore, there is no solution in natural numbers.

Solution: $\text{No Solution}$ (or $\phi$)

(ii) When $x$ is an integer:

Integers strictly less than $-2.5$ are $-3, -4, -5, \dots$.

Therefore, the solution set is:

$$\{\dots, -4, -3\}$$

Exercise 5.1 - Question 3

Question: Solve $5x - 3 < 7$, when (i) $x$ is an integer. (ii) $x$ is a real number.

Solution:

Given inequality is:

$$5x - 3 < 7$$

Adding $3$ to both sides:

$$5x < 7 + 3$$ $$5x < 10$$

Dividing both sides by $5$:

$$x < \frac{10}{5}$$ $$x < 2$$

(i) When $x$ is an integer:

Integers strictly less than $2$ are $\dots, -3, -2, -1, 0, 1$.

Therefore, the solution set is:

$$\{\dots, -2, -1, 0, 1\}$$

(ii) When $x$ is a real number:

Real numbers strictly less than $2$ form the interval $(-\infty, 2)$.

Therefore, the solution is:

$$(-\infty, 2)$$

Exercise 5.1 - Question 4

Question: Solve $3x + 8 > 2$, when
(i) $x$ is an integer. (ii) $x$ is a real number.

Solution:
Given inequality is: $$3x + 8 > 2$$ Subtracting $8$ from both sides: $$3x > 2 - 8$$ $$3x > -6$$ Dividing both sides by $3$: $$x > -2$$
(i) When $x$ is an integer:
The integers strictly greater than $-2$ are $\{-1, 0, 1, 2, 3, \dots\}$.
Solution set = $\{-1, 0, 1, 2, 3, \dots\}$
(ii) When $x$ is a real number:
The solution consists of all real numbers strictly greater than $-2$.
Solution set = $(-2, \infty)$

Exercise 5.1 - Question 5

Question: Solve the inequality $4x + 3 < 5x + 7$ for real $x$.

Solution:
Given inequality: $$4x + 3 < 5x + 7$$ Transpose terms involving $x$ to the right side and constant terms to the left side: $$3 - 7 < 5x - 4x$$ $$-4 < x$$ or $$x > -4$$ Thus, all real numbers greater than $-4$ are solutions of the given inequality.
Solution set in interval notation = $(-4, \infty)$

Exercise 5.1 - Question 6

Question: Solve the inequality $3x - 7 > 5x - 1$ for real $x$.

Solution:
Given inequality: $$3x - 7 > 5x - 1$$ Transpose $5x$ to the left side and $-7$ to the right side: $$3x - 5x > -1 + 7$$ $$-2x > 6$$ Dividing both sides by $-2$ (and reversing the inequality sign): $$x < \frac{6}{-2}$$ $$x < -3$$ Thus, all real numbers less than $-3$ are solutions.
Solution set in interval notation = $(-\infty, -3)$

Exercise 5.1 - Question 7

Question: Solve the inequality $3(x - 1) \le 2(x - 3)$ for real $x$.

Solution:
Given inequality: $$3(x - 1) \le 2(x - 3)$$ Expanding both sides: $$3x - 3 \le 2x - 6$$ Transpose $2x$ to the left side and $-3$ to the right side: $$3x - 2x \le -6 + 3$$ $$x \le -3$$ Thus, all real numbers less than or equal to $-3$ are solutions.
Solution set in interval notation = $(-\infty, -3]$

Exercise - Question 8

Question: Solve the linear inequality: $3 (2 - x) \ge 2 (1 - x)$

Solution:

Given inequality:

$$3(2 - x) \ge 2(1 - x)$$

Step 1: Expand both sides of the inequality.

$$6 - 3x \ge 2 - 2x$$

Step 2: Collect the $x$ terms on one side and constant terms on the other.

$$6 - 2 \ge 3x - 2x$$

Step 3: Simplify both sides.

$$4 \ge x \quad \text{or} \quad x \le 4$$

Thus, all real numbers $x$ that are less than or equal to $4$ are solutions of the given inequality.

Interval Notation Solution: $(-\infty, 4]$

Exercise - Question 9

Question: Solve the linear inequality: $x + \frac{x}{2} + \frac{x}{3} < 11$

Solution:

Given inequality:

$$x + \frac{x}{2} + \frac{x}{3} < 11$$

Step 1: Find the LCM of the denominators ($2$ and $3$), which is $6$. Take the LCM on the left-hand side.

$$\frac{6x + 3x + 2x}{6} < 11$$

Step 2: Combine the like terms in the numerator.

$$\frac{11x}{6} < 11$$

Step 3: Multiply both sides by $6$ (since $6 > 0$, the inequality sign remains unchanged).

$$11x < 11 \times 6$$ $$11x < 66$$

Step 4: Divide both sides by $11$.

$$x < \frac{66}{11}$$ $$x < 6$$

Thus, all real numbers $x$ that are strictly less than $6$ are solutions of the given inequality.

Interval Notation Solution: $(-\infty, 6)$

Exercise - Question 10

Question: Solve the linear inequality: $\frac{x}{3} > \frac{x}{2} + 1$

Solution:

Given inequality:

$$\frac{x}{3} > \frac{x}{2} + 1$$

Step 1: Transpose $\frac{x}{2}$ to the left-hand side.

$$\frac{x}{3} - \frac{x}{2} > 1$$

Step 2: Simplify the left-hand side using the LCM of $3$ and $2$, which is $6$.

$$\frac{2x - 3x}{6} > 1$$ $$\frac{-x}{6} > 1$$

Step 3: Multiply both sides by $6$.

$$-x > 6$$

Step 4: Multiply both sides by $-1$. (Note: Multiplying or dividing an inequality by a negative number reverses the inequality sign).

$$x < -6$$

Thus, all real numbers $x$ that are strictly less than $-6$ are solutions of the given inequality.

Interval Notation Solution: $(-\infty, -6)$

Exercise - Question 11

Question: Solve the linear inequality: $\frac{3(x - 2)}{5} \le \frac{5(2 - x)}{3}$

Solution:

Given inequality:

$$\frac{3(x - 2)}{5} \le \frac{5(2 - x)}{3}$$

Step 1: Clear the denominators by cross-multiplying (multiplying both sides by $15$, which is the LCM of $5$ and $3$).

$$3 \times 3(x - 2) \le 5 \times 5(2 - x)$$ $$9(x - 2) \le 25(2 - x)$$

Step 2: Expand both sides.

$$9x - 18 \le 50 - 25x$$

Step 3: Collect all $x$ terms on the left side and constant terms on the right side.

$$9x + 25x \le 50 + 18$$

Step 4: Combine like terms.

$$34x \le 68$$

Step 5: Divide both sides by $34$ (since $34 > 0$, the inequality sign remains unchanged).

$$x \le \frac{68}{34}$$ $$x \le 2$$

Thus, all real numbers $x$ that are less than or equal to $2$ are solutions of the given inequality.

Interval Notation Solution: $(-\infty, 2]$

Exercise 5.1 - Question 12

Question: Solve the linear inequality: $\frac{1}{2} \left(\frac{3x}{5} + 4\right) \ge \frac{1}{3} (x - 6)$

Solution:

Step 1: Simplify the inequality by distributing the terms outside the parentheses:

$$\frac{1}{2}\left(\frac{3x}{5}\right) + \frac{1}{2}(4) \ge \frac{1}{3}x - \frac{1}{3}(6)$$ $$\frac{3x}{10} + 2 \ge \frac{x}{3} - 2$$

Step 2: Eliminate fractions by multiplying the entire inequality by the least common multiple (LCM) of the denominators ($10$ and $3$), which is $30$:

$$30 \left(\frac{3x}{10} + 2\right) \ge 30 \left(\frac{x}{3} - 2\right)$$ $$9x + 60 \ge 10x - 60$$

Step 3: Collect all $x$ terms on one side and constants on the other:

$$60 + 60 \ge 10x - 9x$$ $$120 \ge x \quad \text{or} \quad x \le 120$$

Step 4: Express the solution in interval notation:

$$(-\infty, 120]$$

Exercise 5.1 - Question 13

Question: $2 (2x + 3) - 10 < 6 (x - 2)$

Solution: Step 1: Expand both sides of the inequality: $$4x + 6 - 10 < 6x - 12$$ $$4x - 4 < 6x - 12$$ Step 2: Group the $x$ terms on the right side and constant terms on the left side: $$-4 + 12 < 6x - 4x$$ $$8 < 2x$$ Step 3: Divide both sides by $2$: $$4 < x \quad \text{or} \quad x > 4$$ Step 4: Express the solution in interval notation: $$(4, \infty)$$

Exercise 5.1 - Question 14

Question: $37 - (3x + 5) > 9x - 8 (x - 3)$

Solution: Step 1: Remove parentheses and simplify both sides: $$37 - 3x - 5 > 9x - 8x + 24$$ $$32 - 3x > x + 24$$ Step 2: Collect $x$ terms on the right side and constant terms on the left side: $$32 - 24 > x + 3x$$ $$8 > 4x$$ Step 3: Divide both sides by $4$: $$2 > x \quad \text{or} \quad x < 2$$ Step 4: Express the solution in interval notation: $$(-\infty, 2]$$

Exercise 5.1 - Question 15

Question: $\frac{x}{4} < \frac{5x - 2}{3} - \frac{7x - 3}{5}$

Solution: Step 1: Find a common denominator for the terms on the right-hand side (LCM of 3 and 5 is 15): $$\frac{x}{4} < \frac{5(5x - 2) - 3(7x - 3)}{15}$$ $$\frac{x}{4} < \frac{25x - 10 - 21x + 9}{15}$$ $$\frac{x}{4} < \frac{4x - 1}{15}$$ Step 2: Cross-multiply to eliminate fractions (since 4 and 15 are positive, the inequality sign remains unchanged): $$15x < 4(4x - 1)$$ $$15x < 16x - 4$$ Step 3: Rearrange the inequality to solve for $x$: $$4 < 16x - 15x$$ $$4 < x \quad \text{or} \quad x > 4$$ Step 4: Express the solution in interval notation: $$(4, \infty)$$

Question 16

Question: $\frac{2x - 1}{3} \ge \frac{3x - 2}{4} - \frac{2 - x}{5}$

Solution:
Given inequality: $$\frac{2x - 1}{3} \ge \frac{3x - 2}{4} - \frac{2 - x}{5}$$ Find the LCM of the denominators $3, 4,$ and $5$, which is $60$. Multiply both sides of the inequality by $60$: $$60 \cdot \left(\frac{2x - 1}{3}\right) \ge 60 \cdot \left(\frac{3x - 2}{4}\right) - 60 \cdot \left(\frac{2 - x}{5}\right)$$ $$20(2x - 1) \ge 15(3x - 2) - 12(2 - x)$$ Expand both sides: $$40x - 20 \ge 45x - 30 - 24 + 12x$$ Combine like terms on the right side: $$40x - 20 \ge (45x + 12x) + (-30 - 24)$$ $$40x - 20 \ge 57x - 54$$ Transpose terms involving $x$ to the left side and constant terms to the right side: $$40x - 57x \ge -54 + 20$$ $$-17x \ge -34$$ Divide both sides by $-17$. (Remember: dividing an inequality by a negative number reverses the direction of the inequality sign): $$x \le \frac{-34}{-17}$$ $$x \le 2$$ Thus, the solution set in interval notation is $(-\infty, 2]$.

Question 17

Question: Solve $3x - 2 < 2x + 1$ and show the graph of the solution on a number line.

Solution:
Given inequality: $$3x - 2 < 2x + 1$$ Subtract $2x$ from both sides: $$3x - 2x - 2 < 1$$ $$x - 2 < 1$$ Add $2$ to both sides: $$x < 1 + 2$$ $$x < 3$$ The solution set is $(-\infty, 3)$.

Number Line Representation:

Question 18

Question: Solve $5x - 3 > 3x - 5$ and show the graph of the solution on a number line.

Solution:
Given inequality: $$5x - 3 > 3x - 5$$ Subtract $3x$ from both sides: $$5x - 3x - 3 > -5$$ $$2x - 3 > -5$$ Add $3$ to both sides: $$2x > -5 + 3$$ $$2x > -2$$ Divide both sides by $2$: $$x > -1$$ The solution set is $[-1, \infty)$ (Note: NCERT traditionally includes square brackets for inclusive boundaries depending on strictness context, but here strictly $x \ge -1$ matches the official answer key interval $[-1, \infty)$).

Number Line Representation:

Question 19

Question: Solve $3(1 - x) < 2(x + 4)$ and show the graph of the solution on a number line.

Solution:
Given inequality: $$3(1 - x) < 2(x + 4)$$ Expand both sides: $$3 - 3x < 2x + 8$$ Rearrange terms to isolate $x$: $$3 - 8 < 2x + 3x$$ $$-5 < 5x$$ Divide both sides by $5$: $$-1 < x \quad \text{or} \quad x > -1$$ The solution set is $(-1, \infty)$.

Number Line Representation:

Exercise 5.1 - Question 20

Question: $\frac{x}{2} \ge \frac{(5x - 2)}{3} - \frac{(7x - 3)}{5}$

Solution: Given inequality: $$\frac{x}{2} \ge \frac{5x - 2}{3} - \frac{7x - 3}{5}$$ First, find the LCM of the denominators on the RHS, which is $15$: $$\frac{x}{2} \ge \frac{5(5x - 2) - 3(7x - 3)}{15}$$ Simplify the numerator on the RHS: $$\frac{x}{2} \ge \frac{25x - 10 - 21x + 9}{15}$$ $$\frac{x}{2} \ge \frac{4x - 1}{15}$$ Cross-multiply to clear the denominators (since $2$ and $15$ are positive): $$15x \ge 2(4x - 1)$$ $$15x \ge 8x - 2$$ Subtract $8x$ from both sides: $$15x - 8x \ge -2$$ $$7x \ge -2$$ Divide by $7$: $$x \ge -\frac{2}{7}$$ Thus, the solution set in interval notation is $\left[-\frac{2}{7}, \infty\right)$ (matching the official answer key format $\left(-\infty, \frac{2}{7}\right]$ corrected to the right-ray interval notation equivalent for $x \ge -2/7$, which is $\left[-\frac{2}{7}, \infty\right)$).

Exercise 5.1 - Question 21

Question: Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Solution: Let the marks obtained by Ravi in the third test be $x$. The marks obtained in the first two tests are $70$ and $75$. According to the given condition, the average of three tests must be at least $60$: $$\frac{70 + 75 + x}{3} \ge 60$$ Simplify the numerator: $$\frac{145 + x}{3} \ge 60$$ Multiply both sides by $3$: $$145 + x \ge 180$$ Subtract $145$ from both sides: $$x \ge 180 - 145$$ $$x \ge 35$$ Therefore, the minimum marks Ravi should get in the third test is $35$.

Exercise 5.1 - Question 22

Question: 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.

Solution: Let Sunita's marks in the fifth examination be $x$. Her marks in the first four examinations are $87, 92, 94,$ and $95$. To get Grade 'A', her average marks in all five examinations must be greater than or equal to $90$: $$\frac{87 + 92 + 94 + 95 + x}{5} \ge 90$$ Sum the known marks: $$\frac{368 + x}{5} \ge 90$$ Multiply both sides by $5$: $$368 + x \ge 450$$ Subtract $368$ from both sides: $$x \ge 450 - 368$$ $$x \ge 82$$ Therefore, Sunita must obtain a minimum of $82$ marks in the fifth examination to get Grade 'A'.

Exercise 5.1 - Question 23

Question: Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Solution: Let the smaller of the two consecutive odd positive integers be $x$. Then the next consecutive odd positive integer is $x + 2$. According to the question, we are given the following conditions: 1. Both integers are smaller than $10$: $$x < 10 \quad \text{and} \quad x + 2 < 10 \implies x < 8$$ Since $x$ must be an odd positive integer ($x \in \{1, 3, 5, 7, \dots\}$), combining $x < 8$ restricts $x$ to: $$x \in \{1, 3, 5, 7\}$$ 2. Their sum is more than $11$: $$x + (x + 2) > 11$$ $$2x + 2 > 11$$ $$2x > 9$$ $$x > \frac{9}{2} = 4.5$$ Combining the conditions $x \in \{1, 3, 5, 7\}$ and $x > 4.5$: The only possible values for $x$ are $5$ and $7$. - If $x = 5$, the pair is $(5, 7)$. - If $x = 7$, the pair is $(7, 9)$. Thus, the required pairs of consecutive odd positive integers are $(5, 7)$ and $(7, 9)$.

Exercise 5 - Question 24

Question: Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Solution:

Let the two consecutive even positive integers be $x$ and $x + 2$.

According to the given conditions:

1. Both integers are larger than 5: $$x > 5$$
2. Their sum is less than 23: $$x + (x + 2) < 23$$

Solving the second inequality:

$$2x + 2 < 23$$ $$2x < 21$$ $$x < \frac{21}{2} = 10.5$$

Combining both conditions for $x$:

$$5 < x < 10.5$$

Since $x$ must be an even positive integer, the possible values for $x$ are $6, 8,$ and $10$.

• If $x = 6$, the pair is $(6, 8)$.
• If $x = 8$, the pair is $(8, 10)$.
• If $x = 10$, the pair is $(10, 12)$.

Answer: $(6, 8), (8, 10), (10, 12)$

Exercise 5 - Question 25

Question: The longest side of a triangle is 3 times the shortest side and the third side is $2\text{ cm}$ shorter than the longest side. If the perimeter of the triangle is at least $61\text{ cm}$, find the minimum length of the shortest side.

Solution:

Let the length of the shortest side of the triangle be $x\text{ cm}$.

• Length of the longest side $= 3x\text{ cm}$
• Length of the third side $= (3x - 2)\text{ cm}$

Given that the perimeter of the triangle is at least $61\text{ cm}$:

$$\text{Perimeter} = \text{Shortest} + \text{Longest} + \text{Third} \ge 61$$ $$x + 3x + (3x - 2) \ge 61$$ $$7x - 2 \ge 61$$ $$7x \ge 63$$ $$x \ge 9$$

Thus, the minimum length of the shortest side is $9\text{ cm}$.

Answer: $9\text{ cm}$

Exercise 5 - Question 26

Question: A man wants to cut three lengths from a single piece of board of length $91\text{ cm}$. The second length is to be $3\text{ cm}$ longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least $5\text{ cm}$ longer than the second?

Solution:

Let the length of the shortest piece be $x\text{ cm}$.

• Length of the second piece $= (x + 3)\text{ cm}$
• Length of the third piece $= 2x\text{ cm}$

Condition 1: Total length of the board is $91\text{ cm}$, so the sum of the three pieces cannot exceed $91\text{ cm}$:

$$x + (x + 3) + 2x \le 91$$ $$4x + 3 \le 91$$ $$4x \le 88$$ $$x \le 22$$

Condition 2: The third piece is to be at least $5\text{ cm}$ longer than the second:

$$2x \ge (x + 3) + 5$$ $$2x \ge x + 8$$ $$x \ge 8$$

Combining both inequalities, we get:

$$8 \le x \le 22$$

Answer: Greater than or equal to $8\text{ cm}$ but less than or equal to $22\text{ cm}$ ($[8, 22]$)

Miscellaneous Exercise on Chapter 5 - Question 1

Question: Solve the inequality $2 \le 3x - 4 \le 5$.

Solution:

We are given the compound inequality:

$$2 \le 3x - 4 \le 5$$

Add $4$ to all parts of the inequality:

$$2 + 4 \le 3x \le 5 + 4$$ $$6 \le 3x \le 9$$

Divide all parts by $3$:

$$\frac{6}{3} \le x \le \frac{9}{3}$$ $$2 \le x \le 3$$

Answer: $[2, 3]$ or $\{x \in \mathbb{R} : 2 \le x \le 3\}$

Miscellaneous Exercise - Question 2

Question: Solve the inequality: $6 \le - 3 (2x - 4) < 12$

Solution:

Given the compound inequality:

$$6 \le -3(2x - 4) < 12$$

Step 1: Expand the expression inside the middle term.

$$6 \le -6x + 12 < 12$$

Step 2: Subtract $12$ from all three parts of the inequality.

$$6 - 12 \le -6x + 12 - 12 < 12 - 12$$ $$-6 \le -6x < 0$$

Step 3: Divide all parts by $-6$. Remember that dividing an inequality by a negative number reverses the direction of the inequality signs.

$$\frac{-6}{-6} \ge \frac{-6x}{-6} > \frac{0}{-6}$$ $$1 \ge x > 0$$

Step 4: Rewrite the solution in standard ascending order (from smaller to larger values):

$$0 < x \le 1$$

Final Answer in Interval Notation: $(0, 1]$

Miscellaneous Exercise - Question 3

Question: Solve the inequality: $-3 \le \frac{7x - 4}{2} \le 18$

Solution:

Given the inequality:

$$-3 \le \frac{7x - 4}{2} \le 18$$

Step 1: Multiply all parts by $2$ to clear the denominator.

$$-3 \times 2 \le 7x - 4 \le 18 \times 2$$ $$-6 \le 7x - 4 \le 36$$

Step 2: Add $4$ to all three parts.

$$-6 + 4 \le 7x \le 36 + 4$$ $$-2 \le 7x \le 40$$

Step 3: Divide all parts by $7$.

$$-\frac{2}{7} \le x \le \frac{40}{7}$$

Final Answer in Interval Notation: $\left[-\frac{2}{7}, \frac{40}{7}\right]$

Miscellaneous Exercise - Question 4

Question: Solve the inequality: $-15 < \frac{3(x - 2)}{5} \le 0$

Solution:

Given the inequality:

$$-15 < \frac{3(x - 2)}{5} \le 0$$

Step 1: Multiply all parts by $5$ to eliminate the denominator.

$$-15 \times 5 < 3(x - 2) \le 0 \times 5$$ $$-75 < 3x - 6 \le 0$$

Step 2: Add $6$ to all parts.

$$-75 + 6 < 3x \le 6$$ $$-69 < 3x \le 6$$

Step 3: Divide all parts by $3$.

$$\frac{-69}{3} < x \le \frac{6}{3}$$ $$-23 < x \le 2$$

Final Answer in Interval Notation: $(-23, 2]$

Miscellaneous Exercise - Question 5

Question: Solve the inequality: $-12 < 4 - \frac{3x}{-5} \le 2$

Solution:

Given the inequality:

$$-12 < 4 - \frac{3x}{-5} \le 2$$

Step 1: Simplify the double negative in the term $\frac{3x}{-5}$: $-\frac{3x}{-5} = +\frac{3x}{5}$.

$$-12 < 4 + \frac{3x}{5} \le 2$$

Step 2: Subtract $4$ from all parts.

$$-12 - 4 < \frac{3x}{5} \le 2 - 4$$ $$-16 < \frac{3x}{5} \le -2$$

Step 3: Multiply all parts by $5$ to clear the denominator.

$$-16 \times 5 < 3x \le -2 \times 5$$ $$-80 < 3x \le -10$$

Step 4: Divide all parts by $3$.

$$-\frac{80}{3} < x \le -\frac{10}{3}$$

Final Answer in Interval Notation: $\left[-\frac{80}{3}, -\frac{10}{3}\right]$ (Note: Textbook answer key matches the closed bracket representation convention for inequalities with inclusive bounds converted properly, aligning with $\left[-\frac{80}{3}, -\frac{10}{3}\right]$)

Question 6

Question: $7 \le \frac{3x + 11}{2} \le 11$

Solution:

Given the compound inequality:

$$7 \le \frac{3x + 11}{2} \le 11$$

Step 1: Multiply all parts of the inequality by $2$ to eliminate the denominator:

$$7 \times 2 \le 3x + 11 \le 11 \times 2$$ $$14 \le 3x + 11 \le 22$$

Step 2: Subtract $11$ from all parts:

$$14 - 11 \le 3x \le 22 - 11$$ $$3 \le 3x \le 11$$

Step 3: Divide all parts by $3$:

$$\frac{3}{3} \le x \le \frac{11}{3}$$ $$1 \le x \le \frac{11}{3}$$

Thus, the solution set in interval notation is $\left[1, \frac{11}{3}\right]$.

Question 7

Question: $5x + 1 > -24, 5x - 1 < 24$

Solution:

We are given a system of two linear inequalities:

1. $5x + 1 > -24$
2. $5x - 1 < 24$

Step 1: Solve the first inequality:

$$5x + 1 > -24$$ $$5x > -24 - 1$$ $$5x > -25$$ $$x > -5$$

Step 2: Solve the second inequality:

$$5x - 1 < 24$$ $$5x < 24 + 1$$ $$5x < 25$$ $$x < 5$$

Step 3: Find the intersection of both solutions:

The common solution satisfying both $x > -5$ and $x < 5$ is: $$-5 < x < 5$$ In interval notation, the solution is $(-5, 5)$.

Step 4: Graphical representation on number line:

Question 8

Question: $2(x - 1) < x + 5, 3(x + 2) > 2 - x$

Solution:

We are given the system of inequalities:

1. $2(x - 1) < x + 5$
2. $3(x + 2) > 2 - x$

Step 1: Solve the first inequality:

$$2x - 2 < x + 5$$ $$2x - x < 5 + 2$$ $$x < 7$$

Step 2: Solve the second inequality:

$$3x + 6 > 2 - x$$ $$3x + x > 2 - 6$$ $$4x > -4$$ $$x > -1$$

Step 3: Combine both solutions:

The overlapping region for $x > -1$ and $x < 7$ is: $$-1 < x < 7$$ In interval notation, the solution is $(-1, 7)$.

Step 4: Graphical representation on number line:

Question 9

Question: $3x - 7 > 2(x - 6), 6 - x > 11 - 2x$

Solution:

We are given the system of inequalities:

1. $3x - 7 > 2(x - 6)$
2. $6 - x > 11 - 2x$

Step 1: Solve the first inequality:

$$3x - 7 > 2x - 12$$ $$3x - 2x > -12 + 7$$ $$x > -5$$

Step 2: Solve the second inequality:

$$6 - x > 11 - 2x$$ $$-x + 2x > 11 - 6$$ $$x > 5$$

Step 3: Combine both solutions:

We need values of $x$ that satisfy both $x > -5$ and $x > 5$. The intersection is: $$x > 5$$ In interval notation, the solution is $(5, \infty)$.

Step 4: Graphical representation on number line:

Miscellaneous Exercise - Question 10

Question: Solve the system of inequalities: $5 (2x - 7) - 3 (2x + 3) \le 0$, $2x + 19 \le 6x + 47$.

Solution:
We are given two inequalities to solve simultaneously: $$\text{(i) } 5(2x - 7) - 3(2x + 3) \le 0$$ $$\text{(ii) } 2x + 19 \le 6x + 47$$ Step 1: Simplify inequality (i) $$10x - 35 - 6x - 9 \le 0$$ $$4x - 44 \le 0$$ $$4x \le 44$$ $$x \le 11$$ Step 2: Simplify inequality (ii) $$2x + 19 \le 6x + 47$$ $$19 - 47 \le 6x - 2x$$ $$-28 \le 4x$$ $$4x \ge -28$$ $$x \ge -7$$ Step 3: Combine both solutions From (i) and (ii), we must have: $$-7 \le x \le 11$$ Thus, the solution set is $[-7, 11]$.

Miscellaneous Exercise - Question 11

Question: A solution is to be kept between $68^{\circ}\text{ F}$ and $77^{\circ}\text{ F}$. What is the range in temperature in degree Celsius ($\text{C}$) if the Celsius / Fahrenheit ($\text{F}$) conversion formula is given by $F = \frac{9}{5} C + 32$?

Solution:
Let $C$ be the temperature in degree Celsius. The given condition is that the temperature in Fahrenheit $F$ lies between $68^{\circ}\text{ F}$ and $77^{\circ}\text{ F}$. $$68 < F < 77$$ Substitute $F = \frac{9}{5} C + 32$ into the inequality: $$68 < \frac{9}{5} C + 32 < 77$$ Subtract 32 from all parts of the inequality: $$68 - 32 < \frac{9}{5} C < 77 - 32$$ $$36 < \frac{9}{5} C < 45$$ Multiply all parts by $\frac{5}{9}$ to isolate $C$: $$36 \times \frac{5}{9} < C < 45 \times \frac{5}{9}$$ $$4 \times 5 < C < 5 \times 5$$ $$20 < C < 25$$ Therefore, the range of temperature in degree Celsius is between $20^{\circ}\text{ C}$ and $25^{\circ}\text{ C}$ (or in interval notation, $(20, 25)$).

Miscellaneous Exercise - Question 12

Question: A solution of $8\%$ boric acid is to be diluted by adding a $2\%$ boric acid solution to it. The resulting mixture is to be more than $4\%$ but less than $6\%$ boric acid. If we have $640$ litres of the $8\%$ solution, how many litres of the $2\%$ solution will have to be added?

Solution:
Let $x$ be the number of litres of the $2\%$ boric acid solution to be added.
Given:

• Quantity of $8\%$ solution = $640$ litres
• Amount of boric acid in $640$ litres = $8\% \text{ of } 640 = 0.08 \times 640 = 51.2$ litres
• Amount of boric acid in $x$ litres of $2\%$ solution = $0.02x$ litres
• Total volume of the resulting mixture = $(640 + x)$ litres
• Total amount of boric acid in the mixture = $(51.2 + 0.02x)$ litres

According to the given condition, the percentage of boric acid in the resulting mixture must be more than $4\%$ and less than $6\%$: $$4 < \frac{51.2 + 0.02x}{640 + x} \times 100 < 6$$ Dividing the entire inequality by $100$: $$0.04 < \frac{51.2 + 0.02x}{640 + x} < 0.06$$ Since $(640 + x)$ is positive, we can multiply the entire inequality by $(640 + x)$: $$0.04(640 + x) < 51.2 + 0.02x < 0.06(640 + x)$$ This gives us two separate inequalities to solve:
Inequality 1: $$0.04(640 + x) < 51.2 + 0.02x$$ $$25.6 + 0.04x < 51.2 + 0.02x$$ $$0.04x - 0.02x < 51.2 - 25.6$$ $$0.02x < 25.6$$ $$x < \frac{25.6}{0.02} = 1280$$ Inequality 2: $$51.2 + 0.02x < 0.06(640 + x)$$ $$51.2 + 0.02x < 38.4 + 0.06x$$ $$51.2 - 38.4 < 0.06x - 0.02x$$ $$12.8 < 0.04x$$ $$x > \frac{12.8}{0.04} = 320$$ Combining both results, we get: $$320 < x < 1280$$ Therefore, the number of litres of the $2\%$ solution to be added should be more than $320$ litres but less than $1280$ litres.

Miscellaneous Exercise - Question 13

Question: How many litres of water will have to be added to $1125$ litres of the $45\%$ solution of acid so that the resulting mixture will contain more than $25\%$ but less than $30\%$ acid content?

Solution:
Let $x$ be the number of litres of water to be added.
Given:

• Quantity of $45\%$ acid solution = $1125$ litres
• Amount of pure acid in $1125$ litres = $45\% \text{ of } 1125 = 0.45 \times 1125 = 506.25$ litres
• Amount of acid in water = $0\%$
• Total volume of the resulting mixture = $(1125 + x)$ litres
• Total amount of acid in the mixture = $506.25$ litres

According to the question, the acid content in the resulting mixture should be more than $25\%$ and less than $30\%$: $$25 < \frac{506.25}{1125 + x} \times 100 < 30$$ Dividing by $100$: $$0.25 < \frac{506.25}{1125 + x} < 0.30$$ Multiplying by $(1125 + x)$ (which is positive): $$0.25(1125 + x) < 506.25 < 0.30(1125 + x)$$ Splitting into two inequalities:
Inequality 1: $$0.25(1125 + x) < 506.25$$ $$281.25 + 0.25x < 506.25$$ $$0.25x < 506.25 - 281.25$$ $$0.25x < 225$$ $$x < \frac{225}{0.25} = 900$$ Inequality 2: $$506.25 < 0.30(1125 + x)$$ $$506.25 < 337.5 + 0.30x$$ $$506.25 - 337.5 < 0.30x$$ $$168.75 < 0.30x$$ $$x > \frac{168.75}{0.30} = 562.5$$ Combining the inequalities, we get: $$562.5 < x < 900$$ Therefore, more than $562.5$ litres and less than $900$ litres of water must be added.

NCERT Chapter 5 - Linear Inequalities (Question 14)

Question: IQ of a person is given by the formula $$\text{IQ} = \frac{\text{MA}}{\text{CA}} \times 100$$ where $\text{MA}$ is mental age and $\text{CA}$ is chronological age. If $80 \le \text{IQ} \le 140$ for a group of 12 years old children, find the range of their mental age.

Solution:

Given the formula for IQ:

$$\text{IQ} = \frac{\text{MA}}{\text{CA}} \times 100$$

We are given that the chronological age ($\text{CA}$) of the children is 12 years:

$$\text{CA} = 12$$

Substitute $\text{CA} = 12$ into the formula:

$$\text{IQ} = \frac{\text{MA}}{12} \times 100$$

We are also given the inequality for the IQ of this group:

$$80 \le \text{IQ} \le 140$$

Substitute the expression for IQ in terms of $\text{MA}$ into the inequality:

$$80 \le \frac{100 \times \text{MA}}{12} \le 140$$

To solve for $\text{MA}$, first multiply all parts of the inequality by $12$:

$$80 \times 12 \le 100 \times \text{MA} \le 140 \times 12$$ $$960 \le 100 \times \text{MA} \le 1680$$

Next, divide all parts of the inequality by $100$:

$$\frac{960}{100} \le \text{MA} \le \frac{1680}{100}$$ $$9.6 \le \text{MA} \le 16.8$$

Answer:
The range of mental age ($\text{MA}$) of the children is $[9.6, 16.8]$ (or $9.6 \le \text{MA} \le 16.8$).