Linear Equations and Linear Inequalities

(i) Solving,

⇒ 2(3 - 2x) = 13

⇒ 6 - 4x = 13

⇒ -4x = 13 - 6

⇒ -4x = 7

⇒ $x = -\frac{7}{4}$

Hence, x = $-\frac{7}{4}$.

(ii) Solving,

$$\begin{aligned}\Rightarrow \frac{3}{5}y - 2 = \frac{7}{10}\\[1em] \Rightarrow \frac{3}{5}y = \frac{7}{10} + 2\\[1em] \Rightarrow \frac{3}{5}y = \frac{7 + 20}{10}\\[1em] \Rightarrow \frac{3}{5}y = \frac{27}{10}\\[1em] \Rightarrow y = \frac{27}{10} \times \frac{5}{3}\\[1em] \Rightarrow y = \frac{9}{2}\\[1em] \Rightarrow y = 4\frac{1}{2}\end{aligned}$$

Hence, y = $4\frac{1}{2}$.

(i) Solving,

LCM of 2 and 3:

$$\begin{array}{l|r} 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 2 and 3 = 2 x 3 = 6

Multiplying both sides by 6.

⇒ $\frac{x}{2}\times 6 = 5 \times 6 + \frac{x}{3}\times 6$

⇒ 3x = 30 + 2x

⇒ 3x - 2x = 30

⇒ x = 30

Hence, x = 30.

(ii) Solving,

⇒ $2\left(x - \frac{3}{2}\right) = 11$

⇒ 2x - 3 = 11

⇒ 2x = 11 + 3

⇒ 2x = 14

⇒ $x = \frac{14}{2}$

⇒ x = 7

Hence, x = 7.

(i) Solving,

⇒ 7(x - 2) = 2(2x - 4)

⇒ 7x - 14 = 4x - 8

⇒ 7x - 4x = -8 + 14

⇒ 3x = 6

⇒ $x = \frac{6}{3}$

⇒ x = 2

Hence, x = 2.

(ii) Solving,

⇒ 21 - 3(x - 7) = x + 20

⇒ 21 - 3x + 21 = x + 20

⇒ 42 - 3x = x + 20

⇒ 42 - 20 = x + 3x

⇒ 22 = 4x

$$\begin{aligned}\Rightarrow x = \frac{22}{4}\\[1em] \Rightarrow x = \frac{11}{2}\\[1em] \Rightarrow x = 5\frac{1}{2}\end{aligned}$$

Hence, x = $5\frac{1}{2}$.

(i) Solving,

$$\begin{aligned}\Rightarrow 3x - \frac{1}{3} = 2\left(x - \frac{1}{2}\right) + 5\\[1em] \Rightarrow 3x - \frac{1}{3} = 2x - 1 + 5\\[1em] \Rightarrow 3x - \frac{1}{3} = 2x + 4\\[1em] \Rightarrow 3x - 2x = 4 + \frac{1}{3}\\[1em] \Rightarrow x = \frac{12 + 1}{3}\\[1em] \Rightarrow x = \frac{13}{3}\\[1em] \Rightarrow x = 4\frac{1}{3}\end{aligned}$$

Hence, x = $4\frac{1}{3}$.

(ii) Solving,

LCM of 3 and 5:

$$\begin{array}{l|r} 3 & 3, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 3 and 5 = 3 x 5 = 15

Multiplying both sides by 15,

⇒ $\frac{2m}{3}\times 15 - \frac{m}{5}\times 15 = 7 \times 15$

⇒ 10m - 3m = 105

⇒ 7m = 105

⇒ $m = \frac{105}{7}$

⇒ m = 15

Hence, m = 15.

(i) Solving,

LCM of 5 and 2:

$$\begin{array}{l|r} 2 & 5, 2 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 5 and 2 = 5 x 2 = 10

Multiplying both sides by 10,

⇒ $\Big(\frac{x+1}{5}\Big)\times 10 - \Big(\frac{x-7}{2}\Big) \times 10 = 1 \times 10$

⇒ 2(x + 1) - 5(x - 7) = 10

⇒ 2x + 2 - 5x + 35 = 10

⇒ -3x + 37 = 10

⇒ -3x = 10 - 37

⇒ -3x = -27

⇒ $x = \frac{-27}{-3}$

⇒ x = 9

Hence, x = 9.

(ii) Solving,

LCM of 7 and 4:

$$\begin{array}{l|r} 2 & 7, 4 \\ \hline 2 & 7, 2 \\ \hline 7 & 7, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 7 and 4 = 2 x 2 x 7 = 28

Multiplying both sides by 28,

⇒ $\Big(\frac{3p-2}{7}\Big) \times 28 - \Big(\frac{p-2}{4}\Big)\times 28 = 2 \times 28$

⇒ 4(3p - 2) - 7(p - 2) = 56

⇒ 12p - 8 - 7p + 14 = 56

⇒ 5p + 6 = 56

⇒ 5p = 56 - 6

⇒ 5p = 50

⇒ $p = \frac{50}{5}$

⇒ p = 10

Hence, p = 10.

(i) Solving,

LCM of 2 and 3:

$$\begin{array}{l|r} 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 2 and 3 = 2 x 3 = 6

Given, equation :

$\frac{1}{2}(x + 5) - \frac{1}{3}(x - 2) = 4$

Multiplying both sides by 6,

⇒ $\frac{1}{2}(x + 5)\times 6- \frac{1}{3}(x - 2)\times 6 = 4 \times 6$

⇒ 3(x + 5) - 2(x - 2) = 24

⇒ 3x + 15 - 2x + 4 = 24

⇒ x + 19 = 24

⇒ x = 24 - 19

⇒ x = 5

Hence, x = 5.

(ii) Solving,

LCM of 2 and 6:

$$\begin{array}{l|r} 2 & 2, 6 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

LCM of 2 and 6 = 2 x 3 = 6

Given, equation :

$\frac{2x-3}{6} - \frac{x-5}{2} = \frac{x}{6}$

Multiplying both sides by 6,

⇒ $\Big(\frac{2x-3}{6}\Big) \times 6 - \Big(\frac{x-5}{2}\Big) \times 6 = \frac{x}{6} \times 6$

⇒ (2x - 3) - 3(x - 5) = x

⇒ 2x - 3 - 3x + 15 = x

⇒ -x + 12 = x

⇒ 12 = x + x

⇒ 12 = 2x

⇒ $x = \frac{12}{2}$

⇒ x = 6

Hence, x = 6.

(i) Solving,

LCM of 5 and 7:

$$\begin{array}{l|r} 5 & 7, 5 \\ \hline 7 & 7, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 5 and 7 = 5 x 7 = 35

Given,

$\Rightarrow \frac{x-4}{7} - \frac{x+4}{5} = \frac{x+3}{7}$

Multiplying both sides by 35,

$\Rightarrow \Big(\frac{x-4}{7}\Big) \times 35 - \Big(\frac{x+4}{5}\Big) \times 35 = \Big(\frac{x+3}{7}\Big) \times 35$

⇒ 5(x - 4) - 7(x + 4) = 5(x + 3)

⇒ 5x - 20 - 7x - 28 = 5x + 15

⇒ -2x - 48 = 5x + 15

⇒ -48 - 15 = 5x + 2x

⇒ -63 = 7x

⇒ $x = \frac{-63}{7}$

⇒ x = -9

Hence, x = -9.

(ii) Solving,

LCM of 5, 2 and 3:

$$\begin{array}{l|r} 2 & 5, 2, 3 \\ \hline 3 & 5, 1, 3 \\ \hline 5 & 5, 1, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 5, 2 and 3 = 5 x 2 x 3 = 30

Given,

$\frac{x-1}{5} + \frac{x-2}{2} = \frac{x}{3} + 1$

Multiplying both sides by 30,

⇒ $\Big(\frac{x-1}{5}\Big) \times 30 + \Big(\frac{x-2}{2}\Big) \times 30 = \Big(\frac{x}{3} + 1 \Big) \times 30$

⇒ 6(x - 1) + 15(x - 2) = 10x + 30

⇒ 6x - 6 + 15x - 30 = 10x + 30

⇒ 21x - 36 = 10x + 30

⇒ 21x - 10x = 30 + 36

⇒ 11x = 66

⇒ $x = \frac{66}{11}$

⇒ x = 6

Hence, x = 6.

(i) Solving,

⇒ y + 1.2y = 4.4

⇒ 2.2y = 4.4

⇒ $y = \frac{4.4}{2.2}$

⇒ y = 2

Hence, y = 2.

(ii) Solving,

$$\begin{aligned}\Rightarrow 15% \text{ of } x = 21\\[1em] \Rightarrow\frac{15}{100} \times x = 21\\[1em] \Rightarrow x = 21 \times \frac{100}{15}\\[1em] \Rightarrow x = \frac{2100}{15}\\[1em] \Rightarrow x = 140\end{aligned}$$

Hence, x = 140.

(i) Solving,

$$\begin{aligned}\Rightarrow 2p + 20% \text{ of } (2p - 1) = 7\\[1em] \Rightarrow 2p + \frac{20}{100}(2p - 1) = 7\\[1em] \Rightarrow 2p + \frac{1}{5}(2p - 1) = 7\\[1em]\end{aligned}$$

Multiplying both sides by 5,

$\Rightarrow 5\Big(2p + \frac{1}{5}(2p - 1)\Big) = 7 \times 5$

⇒ 10p + (2p - 1) = 35

⇒ 10p + 2p - 1 = 35

⇒ 12p = 35 + 1

⇒ 12p = 36

⇒ $p = \frac{36}{12}$

⇒ p = 3

Hence, p = 3.

(ii) Solving,

$$\begin{aligned}\Rightarrow 3(2x - 1) + 25% \text{ of } x = 97\\[1em] \Rightarrow 6x - 3 + \frac{25}{100}x = 97\\[1em] \Rightarrow 6x - 3 + \frac{1}{4}x = 97\end{aligned}$$

Multiplying both sides by 4,

$\Rightarrow 6x \times 4 - 3 \times 4 + \frac{1}{4}x \times 4 = 97 \times 4$

⇒ 24x - 12 + x = 388

⇒ 25x - 12 = 388

⇒ 25x = 388 + 12

⇒ 25x = 400

⇒ $x = \frac{400}{25}$

⇒ x = 16

Hence, x = 16.

Given, the value of x⁴ - 3x³ - px - 5 is 23 when x = -2.

Substituting x = -2,

⇒ (-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

⇒ $p = \frac{-12}{2}$

⇒ p = -6

Hence, p = -6.

Let the required number be x.

Five times the number = 5x.

According to the problem,

⇒ 5x + 7 = 57

⇒ 5x = 57 - 7

⇒ 5x = 50

⇒ $x = \frac{50}{5}$

⇒ x = 10

Hence, the required number is 10.

Let the required number be x.

One-fourth of the number = $\frac{x}{4}$.

3 more than 7 = 7 + 3 = 10.

According to the problem,

⇒ $\frac{x}{4} = 10$

⇒ $x = 10 \times 4$

⇒ x = 40

Hence, the required number is 40.

Let the required number be x.

As the number is as much greater than 15 as it is less than 51,

⇒ x - 15 = 51 - x

⇒ x + x = 51 + 15

⇒ 2x = 66

⇒ $x = \frac{66}{2}$

⇒ x = 33

Hence, the required number is 33.

Let the required number be x.

On subtracting $\frac{1}{2}$ from the number, the difference = $x - \frac{1}{2}$.

According to the problem,

⇒ $4\left(x - \frac{1}{2}\right) = 5$

⇒ 4x - 2 = 5

⇒ 4x = 5 + 2

⇒ 4x = 7

⇒ $x = \frac{7}{4}$

Hence, the required number is $\frac{7}{4}$.

Let the smaller number be x.

As the sum of the two numbers is 80, the greater number = 80 - x.

As the greater number exceeds twice the smaller by 11,

⇒ 80 - x = 2x + 11

⇒ 80 - 11 = 2x + x

⇒ 69 = 3x

⇒ $x = \frac{69}{3}$

⇒ x = 23

∴ The greater number = 80 - 23 = 57.

Hence, the two numbers are 57 and 23.

Let the smallest of the three consecutive odd natural numbers be x.

Then the next two odd natural numbers are x + 2 and x + 4.

According to the problem,

⇒ x + (x + 2) + (x + 4) = 87

⇒ 3x + 6 = 87

⇒ 3x = 87 - 6

⇒ 3x = 81

⇒ $x = \frac{81}{3}$

⇒ x = 27

∴ x + 2 = 27 + 2 = 29 and x + 4 = 27 + 4 = 31.

Hence, the three consecutive odd natural numbers are 27, 29 and 31.

Let the number of boys be x.

Then the number of girls = $\frac{2}{5}x$.

As there are 35 students in all,

$$\begin{aligned}\Rightarrow x + \frac{2}{5}x = 35\\[1em] \Rightarrow \frac{5x + 2x}{5} = 35\\[1em] \Rightarrow \frac{7x}{5} = 35\\[1em] \Rightarrow 7x = 175\\[1em] \Rightarrow x = \frac{175}{7}\end{aligned}$$

⇒ x = 25

∴ The number of girls = $\frac{2}{5} \times 25 = 10$.

Hence, the number of girls in the class is 10.

Let the number of chairs purchased be x.

Cost of x chairs = ₹250x.

Cost of two tables = ₹(2 × 400) = ₹800.

According to the problem,

⇒ 250x + 800 = 2800

⇒ 250x = 2800 - 800

⇒ 250x = 2000

⇒ $x = \frac{2000}{250}$

⇒ x = 8

Hence, the number of chairs purchased is 8.

Let Aparna's monthly salary be ₹x.

As her salary exceeds the over-time by ₹16,560, the over-time payment = ₹(x - 16,560).

According to the problem,

⇒ x + (x - 16,560) = 27,840

⇒ 2x - 16,560 = 27,840

⇒ 2x = 27,840 + 16,560

⇒ 2x = 44,400

⇒ $x = \frac{44,400}{2}$

⇒ x = 22,200

Hence, Aparna's monthly salary is ₹22,200.

Let the number of ₹5 coins be x.

As there are 80 coins in all, the number of ₹2 coins = 80 - x.

Value of ₹5 coins = ₹5x and value of ₹2 coins = ₹2(80 - x).

According to the problem,

⇒ 5x + 2(80 - x) = 232

⇒ 5x + 160 - 2x = 232

⇒ 3x + 160 = 232

⇒ 3x = 232 - 160

⇒ 3x = 72

⇒ $x = \frac{72}{3}$

⇒ x = 24

Hence, the number of ₹5 coins is 24.

Let the number of ₹10 notes be x.

As the number of ₹50 notes is one less than that of ₹10 notes, the number of ₹50 notes = x - 1.

Value of ₹10 notes = ₹10x and value of ₹50 notes = ₹50(x - 1).

According to the problem,

⇒ 10x + 50(x - 1) = 550

⇒ 10x + 50x - 50 = 550

⇒ 60x - 50 = 550

⇒ 60x = 550 + 50

⇒ 60x = 600

⇒ $x = \frac{600}{60}$

⇒ x = 10

∴ The number of ₹50 notes = x - 1 = 10 - 1 = 9.

Hence, the number of ₹50 notes is 9.

Let my present age be x years.

After 12 years, my age = (x + 12) years.

4 years ago, my age = (x - 4) years.

According to the problem,

⇒ x + 12 = 3(x - 4)

⇒ x + 12 = 3x - 12

⇒ 12 + 12 = 3x - x

⇒ 24 = 2x

⇒ $x = \frac{24}{2}$

⇒ x = 12

Hence, my present age is 12 years.

As the two equal sides of the isosceles triangle are equal in length,

⇒ 3x - 1 = 2x + 2

⇒ 3x - 2x = 2 + 1

⇒ x = 3

Now, the length of each equal side = (3x - 1) = (3 × 3 - 1) = 8 units.

The length of the third side = 2x = 2 × 3 = 6 units.

∴ Perimeter of the triangle = 8 + 8 + 6 = 22 units.

Hence, x = 3 and the perimeter of the triangle is 22 units.

Let the breadth of the rectangular plot be x metres.

As the length is 6 m less than thrice the breadth, the length = (3x - 6) metres.

According to the problem,

⇒ 2[(3x - 6) + x] = 148

⇒ 2(4x - 6) = 148

⇒ $4x - 6 = \frac{148}{2}$

⇒ 4x - 6 = 74

⇒ 4x = 74 + 6

⇒ 4x = 80

⇒ $x = \frac{80}{4}$

⇒ x = 20

∴ The length = (3 × 20 - 6) = 54 m and the breadth = 20 m.

Hence, the dimensions of the plot are 54 m and 20 m.

As the two angles differ by 20°, let the angles be x° and (x - 20)°.

As the two angles are complementary, their sum is 90°.

⇒ x + (x - 20) = 90

⇒ 2x - 20 = 90

⇒ 2x = 90 + 20

⇒ 2x = 110

⇒ $x = \frac{110}{2}$

⇒ x = 55

∴ One angle = 55° and the other angle = (55 - 20)° = 35°.

Hence, the two angles are 35° and 55°.

The replacement set is {-5, -3, -1, 0, 1, 3, 4}.

(i) For x < -2, the values from the replacement set that are less than -2 are -5 and -3.

Hence, the solution set is {-5, -3}.

(ii) For x > 1, the values from the replacement set that are greater than 1 are 3 and 4.

Hence, the solution set is {3, 4}.

(iii) For x ≥ -1, the values from the replacement set that are greater than or equal to -1 are -1, 0, 1, 3 and 4.

Hence, the solution set is {-1, 0, 1, 3, 4}.

(iv) For -5 < x < 3, the values from the replacement set that lie between -5 and 3 (excluding -5 and 3) are -3, -1, 0 and 1.

Hence, the solution set is {-3, -1, 0, 1}.

(v) For -3 ≤ x < 4, the values from the replacement set that are greater than or equal to -3 and less than 4 are -3, -1, 0, 1 and 3.

Hence, the solution set is {-3, -1, 0, 1, 3}.

(vi) For 0 ≤ x < 7, the values from the replacement set that are greater than or equal to 0 and less than 7 are 0, 1, 3 and 4.

Hence, the solution set is {0, 1, 3, 4}.

(i) For x ≤ 3, x ∈ N, the solution set is {1, 2, 3}.

The solution set is shown by thick dots on the number line.

(ii) For x < 4, x ∈ W, the solution set is {0, 1, 2, 3}.

The solution set is shown by thick dots on the number line.

(iii) For -2 ≤ x < 4, x ∈ I, the solution set is {-2, -1, 0, 1, 2, 3}.

The solution set is shown by thick dots on the number line.

(iv) For -3 ≤ x ≤ 2, x ∈ I, the solution set is {-3, -2, -1, 0, 1, 2}.

The solution set is shown by thick dots on the number line.

(i) Solving 4 - x > -2, x ∈ N,

⇒ 4 - x > -2

⇒ -x > -2 - 4

⇒ -x > -6

⇒ x< 6 (Dividing by -1 and reversing the symbol)

As x ∈ N, the solution set is {1, 2, 3, 4, 5}.

The solution set is shown by thick dots on the number line.

(ii) Solving 3x + 1 ≤ 8, x ∈ W,

⇒ 3x + 1 ≤ 8

⇒ 3x ≤ 8 - 1

⇒ 3x ≤ 7

⇒ $x ≤ \frac{7}{3}$

⇒ $x ≤ 2\frac{1}{3}$

As x ∈ W, the solution set is {0, 1, 2}.

The solution set is shown by thick dots on the number line.

Solving,

⇒ 3 - 4x < x - 12

⇒ 3 + 12 < x + 4x

⇒ 15 < 5x

⇒ 3 < x

⇒ x > 3

As x ∈ {-1, 0, 1, 2, 3, 4, 5, 6, 7}, the values greater than 3 are 4, 5, 6 and 7.

Hence, the solution set is {4, 5, 6, 7}.

Solving -7 < 4x + 1 ≤ 23,

$$\begin{aligned}\Rightarrow -7 \lt 4x + 1 ≤ 23\\[1em] \Rightarrow -7 - 1 \lt 4x ≤ 23 - 1\\[1em] \Rightarrow -8 \lt 4x ≤ 22\\[1em] \Rightarrow \frac{-8}{4} \lt x ≤ \frac{22}{4}\\[1em] \Rightarrow -2 \lt x ≤ \frac{11}{2}\end{aligned}$$

As x ∈ I, the integers greater than -2 and less than or equal to $\frac{11}{2}$ (= 5.5) are -1, 0, 1, 2, 3, 4 and 5.

Hence, the solution set is {-1, 0, 1, 2, 3, 4, 5}.

(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 $\frac{50}{5}$ = 10.

(iii) The number 4 is the solution (root) of the equation 2y - 5 = 3, since 2(4) - 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. ∴ -1 is a solution of the equation 4x + 9 = 5.

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

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

(viii) Solving 3x - 1 = 4 gives 3x = 5, so x = $\frac{5}{3}$, which 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, hence ___ = 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}.

(i) True. We can add or subtract the same number or expression to both sides of an equation.

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

(iii) True. It contains only one variable x with highest power 1, so it is 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 = $\frac{7}{3}$.

(vi) False. The solution of a linear equation in one variable may be a fraction, e.g. x = $\frac{7}{3}$.

(vii) True. 4x + 5 < 65 is an inequality, not an equation.

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

(ix) False. Putting x = $\frac{9}{4}$ in 5x - 1 gives $5 \times \frac{9}{4} - 1 = \frac{45}{4} - 1 = \frac{41}{4} \neq 8$.

(x) False. Putting x = 5 gives y = $\frac{5(5) - 7}{2} = \frac{18}{2} = 9$ (not 18).

(xi) False. The correct representation is $\frac{x}{4}$ + x = 10.

A linear equation in one variable contains only one variable with highest power 1.

The equation 7p + q = 3 contains two variables p and q, so it is not a linear equation in one variable.

Hence, Option 4 is the correct answer.

Solving the equation,

⇒ $\frac{1}{3}(2y - 1) = 3$

⇒ 2y - 1 = 9

⇒ 2y = 10

⇒ y = 5

Hence, Option 1 is the correct answer.

Putting x = -1 in 2(x - 2) + 6, we get 2(-1 - 2) + 6 = 2(-3) + 6 = -6 + 6 = 0.

So x = -1 satisfies the equation 2(x - 2) + 6 = 0.

Hence, Option 3 is the correct answer.

Solving the equation,

⇒ 3(3n - 10) = 2n + 5

⇒ 9n - 30 = 2n + 5

⇒ 9n - 2n = 5 + 30

⇒ 7n = 35

⇒ n = 5

Hence, Option 2 is the correct answer.

Putting x = -1 in x - 1, we get -1 - 1 = -2, which is not equal to 2.

So x = -1 does not satisfy the equation x - 1 = 2.

Hence, Option 4 is the correct answer.

For the expressions to be equal,

⇒ p - 13 = 2p + 1

⇒ -13 - 1 = 2p - p

⇒ -14 = p

⇒ p = -14

Hence, Option 3 is the correct answer.

Solving 3p + 8 = 3 + p,

⇒ 3p + 8 = 3 + p

⇒ 3p - p = 3 - 8

⇒ 2p = -5

⇒ $p = -\frac{5}{2}$

As $-\frac{5}{2}$ is not an integer, this equation cannot be solved in integers.

Hence, Option 3 is the correct answer.

Solving 4x + 7 = x + 2,

⇒ 4x + 7 = x + 2

⇒ 4x - x = 2 - 7

⇒ 3x = -5

⇒ $x = -\frac{5}{3}$

The solution $-\frac{5}{3}$ is a negative rational number, which is neither a (whole number) integer nor a (positive) fraction.

Hence, Option 4 is the correct answer.

Let the smaller even number be x, then the next even number is x + 2.

⇒ x + (x + 2) = 54

⇒ 2x + 2 = 54

⇒ 2x = 52

⇒ x = 26

Hence, Option 2 is the correct answer.

Let the smaller odd number be x, then the bigger odd number is x + 2.

⇒ x + (x + 2) = 28

⇒ 2x + 2 = 28

⇒ 2x = 26

⇒ x = 13

∴ The bigger number = x + 2 = 13 + 2 = 15.

Hence, Option 3 is the correct answer.

Let the integer be x.

⇒ 3x + 5 = -7

⇒ 3x = -7 - 5

⇒ 3x = -12

⇒ x = -4

Hence, Option 3 is the correct answer.

Let the breadth be x m, then the length is 2x m.

⇒ 2(2x + x) = 120

⇒ 2(3x) = 120

⇒ 6x = 120

⇒ x = 20

∴ The length = 2x = 2 × 20 = 40 m.

Hence, Option 2 is the correct answer.

Let the smaller angle be x°, then the larger angle is (x + 10)°.

As the angles are complementary, their sum is 90°.

⇒ x + (x + 10) = 90

⇒ 2x + 10 = 90

⇒ 2x = 80

⇒ x = 40°

Hence, Option 1 is the correct answer.

Let the larger angle be x°, then the smaller angle is (x - 30)°.

As the angles are supplementary, their sum is 180°.

⇒ x + (x - 30) = 180

⇒ 2x - 30 = 180

⇒ 2x = 210

⇒ x = 105°

Hence, Option 4 is the correct answer.

As x ∈ W (whole numbers), x can take values 0, 1, 2, 3, ...

For -2 ≤ x < 3, the whole numbers satisfying this are 0, 1 and 2.

∴ The solution set is {0, 1, 2}.

Hence, Option 4 is the correct answer.

Checking Statement I:

Let the boy's age be x years, then the father's age = 2x years.

5 years ago, the boy's age = (x - 5) years and the father's age = (2x - 5) years.

⇒ $\frac{x - 5}{2x - 5} = \frac{3}{7}$

⇒ 7(x - 5) = 3(2x - 5)

⇒ 7x - 35 = 6x - 15

⇒ 7x - 6x = -15 + 35

⇒ x = 20

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

Checking Statement II: We can add or subtract the same algebraic expression to both sides of an equation. This is a valid rule, so Statement II is true.

Hence, Option 3 is the correct answer.

Checking Statement I: Multiplying both sides of -x > -1 by -1 reverses the inequality symbol, giving x < 1, not x > 1. So Statement I is false.

Checking Statement II: For an inequality, multiplying or dividing both sides by a negative integer reverses the inequality symbol. So we cannot multiply or divide both sides by any non-zero integer while keeping the symbol unchanged. So Statement II is false.

Hence, Option 4 is the correct answer.

Checking Statement I:

⇒ 1 - 2x ≥ 5

⇒ -2x ≥ 5 - 1

⇒ -2x ≥ 4

⇒ $x ≤ -2 \quad \text{(Dividing by -2 and reversing the symbol)}$

So x ≤ -2. Statement I is true.

Checking Statement II: The solution set of an inequality can be represented on a number line. So Statement II is true.

Hence, Option 3 is the correct answer.

(i) Solving,

⇒ 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

⇒ $x = \frac{4}{2}$

⇒ x = 2

Hence, x = 2.

(ii) Solving,

LCM of 4, 6 and 3:

$$\begin{array}{l|r} 2 & 4, 6, 3 \\ \hline 2 & 2, 3, 3 \\ \hline 3 & 1, 3, 3 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 4, 6 and 3 = 2 x 2 x 3 = 12

Given Equation,

$\frac{3(2x-5)}{4} - \frac{5(7-5x)}{6} = \frac{7x}{3}$

Multiplying both sides by 12,

$$\begin{aligned}\Rightarrow \frac{3(2x-5)}{4} \times 12 - \frac{5(7-5x)}{6} \times 12 = \frac{7x}{3} \times 12\\[1em] \Rightarrow 3 \times 3(2x - 5) - 2 \times 5(7 - 5x) = 4 \times 7x\\[1em] \Rightarrow 9(2x - 5) - 10(7 - 5x) = 28x\\[1em] \Rightarrow 18x - 45 - 70 + 50x = 28x\\[1em] \Rightarrow 68x - 115 = 28x\\[1em] \Rightarrow 68x - 28x = 115\\[1em] \Rightarrow 40x = 115\\[1em] \Rightarrow x = \frac{115}{40}\\[1em] \Rightarrow x = \frac{23}{8}\\[1em] \Rightarrow x = 2\frac{7}{8}\\[1em]\end{aligned}$$

Hence, x = $2\frac{7}{8}$.

Let the required number be x.

Its three-fifth = $\frac{3}{5}x$.

According to the problem,

$$\begin{aligned}\Rightarrow x - \frac{3}{5}x = 22\\[1em] \Rightarrow \frac{5x - 3x}{5} = 22\\[1em] \Rightarrow \frac{2x}{5} = 22\\[1em] \Rightarrow 2x = 110\\[1em] \Rightarrow x = \frac{110}{2}\\[1em] \Rightarrow x = 55\end{aligned}$$

Hence, the required number is 55.

Let the required 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

⇒ x = 6

Hence, the required number is 6.

Let the unit's digit be x.

Then the ten's digit = 2x.

∴ The number = 10 × (2x) + x = 20x + x = 21x.

According to the problem,

⇒ 21x + x = 66

⇒ 22x = 66

⇒ $x = \frac{66}{22}$

⇒ x = 3

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

Hence, the required number is 63.

Let the number of pens bought be x.

As the total number of pens and pencils is 16, the number of pencils = 16 - x.

Cost of pens = ₹8x and cost of pencils = ₹1.50(16 - x).

According to the problem,

⇒ 8x + 1.50(16 - x) = 50

⇒ 8x + 24 - 1.5x = 50

⇒ 6.5x + 24 = 50

⇒ 6.5x = 50 - 24

⇒ 6.5x = 26

⇒ $x = \frac{26}{6.5}$

⇒ x = 4

Hence, the student bought 4 pens.

Let the present age of the sister be x years.

As Arvind is eight years older, Arvind's present age = (x + 8) years.

In three years, the sister's age = (x + 3) years and Arvind's age = (x + 8 + 3) = (x + 11) years.

According to the problem,

⇒ x + 11 = 2(x + 3)

⇒ x + 11 = 2x + 6

⇒ 11 - 6 = 2x - x

⇒ 5 = x

⇒ x = 5

∴ The sister's present age = 5 years and Arvind's present age = 5 + 8 = 13 years.

Hence, Arvind is 13 years old and his sister is 5 years old.

Let the angles of the triangle be x°, 2x° and 3x°.

As the sum of the angles of a triangle is 180°,

⇒ x + 2x + 3x = 180

⇒ 6x = 180

⇒ $x = \frac{180}{6}$

⇒ x = 30

∴ The angles are 30°, 2 × 30° = 60° and 3 × 30° = 90°.

Hence, the angles of the triangle are 30°, 60° and 90°.

(i) Solving,

$$\begin{aligned}\Rightarrow \frac{2x-1}{3} ≤ 2\frac{1}{2}, x ∈ W,\\[1em] \Rightarrow \frac{2x-1}{3} ≤ \frac{5}{2}\\[1em] \Rightarrow 2(2x - 1) ≤ 3 \times 5\\[1em] \Rightarrow 4x - 2 ≤ 15\\[1em] \Rightarrow 4x ≤ 17\\[1em] \Rightarrow x ≤ \frac{17}{4}\end{aligned}$$

As x ∈ W, the solution set is {0, 1, 2, 3, 4}.

The solution set is shown by thick dots on the number line.

(ii) Solving, -1 < $\frac{2x}{3}$ + 1 ≤ 4, x ∈ I,

$$\begin{aligned}\Rightarrow -1 \lt \frac{2x}{3} + 1 ≤ 4\\[1em] \Rightarrow -1 - 1 \lt \frac{2x}{3} ≤ 4 - 1\\[1em] \Rightarrow -2 \lt \frac{2x}{3} ≤ 3\\[1em] \Rightarrow -6 \lt 2x ≤ 9\\[1em] \Rightarrow -3 \lt x ≤ \frac{9}{2}\end{aligned}$$

As x ∈ I, the integers greater than -3 and less than or equal to $\frac{9}{2}$ (= 4.5) are -2, -1, 0, 1, 2, 3 and 4.

∴ The solution set is {-2, -1, 0, 1, 2, 3, 4}.

The solution set is shown by thick dots on the number line.

First, finding the value of m,

⇒ $\frac{7m+2}{3} = -11$

⇒ 7m + 2 = -33

⇒ 7m = -33 - 2

⇒ 7m = -35

⇒ m = -5

Now, substituting m = -5 in 2m³ + 10m² + 4m - 3,

2(-5)³ + 10(-5)² + 4(-5) - 3 = 2(-125) + 10(25) - 20 - 3 = -250 + 250 - 20 - 3 = -23

Hence, 2m³ + 10m² + 4m - 3 = -23.

Let the speed of the other person be x km/h.

In 6 hours, the person from A covers = 4 × 6 = 24 km.

In 6 hours, the person from B covers = 6x km.

As they move towards each other and meet, the sum of the distances covered equals the distance between A and B.

⇒ 24 + 6x = 42

⇒ 6x = 42 - 24

⇒ 6x = 18

⇒ $x = \frac{18}{6}$

⇒ x = 3

Hence, the speed of the other person is 3 km/h.

Let the number of benches be x.

If 4 students sit on each bench and 3 benches remain empty, the number of students = 4(x - 3).

If 3 students sit on each bench and 3 students remain standing, the number of students = 3x + 3.

As 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 students = 3x + 3 = 3 × 15 + 3 = 45 + 3 = 48.

Hence, the number of students in the class is 48.