Linear Equations and Linear Inequalities

Solve :

(i) 2(3 - 2x) = 13

(ii) $\dfrac{3}{5} y - 2 = \dfrac{7}{10}$

Answer

(i) Solving,

⇒ 2(3 - 2x) = 13

⇒ 6 - 4x = 13

⇒ -4x = 13 - 6

⇒ -4x = 7

⇒ $x = -\dfrac{7}{4}$

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

(ii) Solving,

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

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

Solve :

(i) $\dfrac{x}{2} = 5 + \dfrac{x}{3}$

(ii) $2\left(x - \dfrac{3}{2}\right)$ = 11

Answer

(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.

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

⇒ 3x = 30 + 2x

⇒ 3x - 2x = 30

⇒ x = 30

Hence, x = 30.

(ii) Solving,

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

⇒ 2x - 3 = 11

⇒ 2x = 11 + 3

⇒ 2x = 14

⇒ $x = \dfrac{14}{2}$

⇒ x = 7

Hence, x = 7.

Solve :

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

(ii) 21 - 3(x - 7) = x + 20

Answer

(i) Solving,

⇒ 7(x - 2) = 2(2x - 4)

⇒ 7x - 14 = 4x - 8

⇒ 7x - 4x = -8 + 14

⇒ 3x = 6

⇒ $x = \dfrac{6}{3}$

⇒ 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

$\Rightarrow x = \dfrac{22}{4}\\[1em] \Rightarrow x = \dfrac{11}{2}\\[1em] \Rightarrow x = 5\dfrac{1}{2}$

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

Solve :

(i) 3x - $\dfrac{1}{3} = 2\left(x - \dfrac{1}{2}\right)$ + 5

(ii) $\dfrac{2m}{3} - \dfrac{m}{5}$ = 7

Answer

(i) Solving,

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

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

(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,

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

⇒ 10m - 3m = 105

⇒ 7m = 105

⇒ $m = \dfrac{105}{7}$

⇒ m = 15

Hence, m = 15.

Solve :

(i) $\dfrac{x+1}{5} - \dfrac{x-7}{2}$ = 1

(ii) $\dfrac{3p-2}{7} - \dfrac{p-2}{4}$ = 2

Answer

(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,

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

⇒ 2(x + 1) - 5(x - 7) = 10

⇒ 2x + 2 - 5x + 35 = 10

⇒ -3x + 37 = 10

⇒ -3x = 10 - 37

⇒ -3x = -27

⇒ $x = \dfrac{-27}{-3}$

⇒ 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,

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

⇒ 4(3p - 2) - 7(p - 2) = 56

⇒ 12p - 8 - 7p + 14 = 56

⇒ 5p + 6 = 56

⇒ 5p = 56 - 6

⇒ 5p = 50

⇒ $p = \dfrac{50}{5}$

⇒ p = 10

Hence, p = 10.

Solve :

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

(ii) $\dfrac{2x-3}{6} - \dfrac{x-5}{2} = \dfrac{x}{6}$

Answer

(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 :

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

Multiplying both sides by 6,

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

⇒ 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 :

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

Multiplying both sides by 6,

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

⇒ (2x - 3) - 3(x - 5) = x

⇒ 2x - 3 - 3x + 15 = x

⇒ -x + 12 = x

⇒ 12 = x + x

⇒ 12 = 2x

⇒ $x = \dfrac{12}{2}$

⇒ x = 6

Hence, x = 6.

Solve :

(i) $\dfrac{x-4}{7} - \dfrac{x+4}{5} = \dfrac{x+3}{7}$

(ii) $\dfrac{x-1}{5} + \dfrac{x-2}{2} = \dfrac{x}{3} + 1$

Answer

(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,

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

Multiplying both sides by 35,

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

⇒ 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

⇒ $x = \dfrac{-63}{7}$

⇒ 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,

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

Multiplying both sides by 30,

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

⇒ 6(x - 1) + 15(x - 2) = 10x + 30

⇒ 6x - 6 + 15x - 30 = 10x + 30

⇒ 21x - 36 = 10x + 30

⇒ 21x - 10x = 30 + 36

⇒ 11x = 66

⇒ $x = \dfrac{66}{11}$

⇒ x = 6

Hence, x = 6.

Solve :

(i) y + 1.2y = 4.4

(ii) 15% of x = 21

Answer

(i) Solving,

⇒ y + 1.2y = 4.4

⇒ 2.2y = 4.4

⇒ $y = \dfrac{4.4}{2.2}$

⇒ y = 2

Hence, y = 2.

(ii) Solving,

$\Rightarrow 15$

Hence, x = 140.

Solve :

(i) 2p + 20% of (2p - 1) = 7

(ii) 3(2x - 1) + 25% of x = 97

Answer

(i) Solving,

$\Rightarrow 2p + 20$

Multiplying both sides by 5,

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

⇒ 10p + (2p - 1) = 35

⇒ 10p + 2p - 1 = 35

⇒ 12p = 35 + 1

⇒ 12p = 36

⇒ $p = \dfrac{36}{12}$

⇒ p = 3

Hence, p = 3.

(ii) Solving,

$\Rightarrow 3(2x - 1) + 25$

Multiplying both sides by 4,

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

⇒ 24x - 12 + x = 388

⇒ 25x - 12 = 388

⇒ 25x = 388 + 12

⇒ 25x = 400

⇒ $x = \dfrac{400}{25}$

⇒ x = 16

Hence, x = 16.

Find the value of p if the value of x⁴ - 3x³ - px - 5 is equal to 23 when x = -2

Answer

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

⇒ $p = \dfrac{-12}{2}$

⇒ p = -6

Hence, p = -6.

If 7 is added to five times a number, the result is 57. Find the number.

Answer

Let the required number be x.

Five times the number = 5x.

According to the problem,

⇒ 5x + 7 = 57

⇒ 5x = 57 - 7

⇒ 5x = 50

⇒ $x = \dfrac{50}{5}$

⇒ x = 10

Hence, the required number is 10.

Find a number, such that one-fourth of the number is 3 more than 7.

Answer

Let the required number be x.

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

3 more than 7 = 7 + 3 = 10.

According to the problem,

⇒ $\dfrac{x}{4} = 10$

⇒ $x = 10 \times 4$

⇒ x = 40

Hence, the required number is 40.

A number is as much greater than 15 as it is less than 51. Find the number.

Answer

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

⇒ $x = \dfrac{66}{2}$

⇒ x = 33

Hence, the required number is 33.

If $\dfrac{1}{2}$ is subtracted from a number and the difference is multiplied by 4, the result is 5. What is the number?

Answer

Let the required number be x.

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

According to the problem,

⇒ $4\left(x - \dfrac{1}{2}\right) = 5$

⇒ 4x - 2 = 5

⇒ 4x = 5 + 2

⇒ 4x = 7

⇒ $x = \dfrac{7}{4}$

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

The sum of two numbers is 80 and the greater number exceeds twice the smaller by 11. Find the numbers.

Answer

Let the smaller number be x.

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

⇒ $x = \dfrac{69}{3}$

⇒ x = 23

∴ The greater number = 80 - 23 = 57.

Hence, the two numbers are 57 and 23.

Find three consecutive odd natural numbers whose sum is 87.

Answer

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

⇒ $x = \dfrac{81}{3}$

⇒ 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.

In a class of 35 students, the number of girls is two-fifth of the number of boys. Find the number of girls in the class.

Answer

Let the number of boys be x.

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

As there are 35 students in all,

$\Rightarrow x + \dfrac{2}{5}x = 35\\[1em] \Rightarrow \dfrac{5x + 2x}{5} = 35\\[1em] \Rightarrow \dfrac{7x}{5} = 35\\[1em] \Rightarrow 7x = 175\\[1em] \Rightarrow x = \dfrac{175}{7}$

⇒ x = 25

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

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

A chair costs ₹250 and a table costs ₹400. If Ananya purchased a certain number of chairs and two tables for ₹2800, find the number of chairs she purchased.

Answer

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

⇒ $x = \dfrac{2000}{250}$

⇒ x = 8

Hence, the number of chairs purchased is 8.

Aparna got ₹27,840 as her monthly salary and over-time. Her salary exceeds the over-time by ₹16,560. What is her monthly salary?

Answer

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

⇒ $x = \dfrac{44,400}{2}$

⇒ x = 22,200

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

Heena has only ₹2 and ₹5 coins in her purse. If in all she has 80 coins in her purse amounting to ₹232, find the number of ₹5 coins.

Answer

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

⇒ $x = \dfrac{72}{3}$

⇒ x = 24

Hence, the number of ₹5 coins is 24.

A purse contains ₹550 in notes of denominations of ₹10 and ₹50. If the number of ₹50 notes is one less than that of ₹10 notes, then find the number of ₹50 notes.

Answer

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

⇒ $x = \dfrac{600}{60}$

⇒ x = 10

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

Hence, the number of ₹50 notes is 9.

After 12 years, I shall be 3 times as old as I was 4 years ago. Find my present age.

Answer

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

⇒ $x = \dfrac{24}{2}$

⇒ x = 12

Hence, my present age is 12 years.

Two equal sides of an isosceles triangle are (3x - 1) units and (2x + 2) units. The third side is 2x units. Find x and the perimeter of the triangle.

Answer

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.

The length of a rectangular plot is 6 m less than thrice its breadth. Find the dimensions of the plot if its perimeter is 148 m.

Answer

Let the breadth of the rectangular plot be x metres.

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

⇒ $4x - 6 = \dfrac{148}{2}$

⇒ 4x - 6 = 74

⇒ 4x = 74 + 6

⇒ 4x = 80

⇒ $x = \dfrac{80}{4}$

⇒ 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.

Two complementary angles differ by 20°. Find the measure of each angle.

Answer

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

⇒ $x = \dfrac{110}{2}$

⇒ x = 55

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

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

If the replacement set is {-5, -3, -1, 0, 1, 3, 4}, find the solution set of:

(i) x < -2

(ii) x > 1

(iii) x ≥ -1

(iv) -5 < x < 3

(v) -3 ≤ x < 4

(vi) 0 ≤ x < 7

Answer

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

(i) For x < -2, 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}.

Represent the solution set of the following inequalities graphically:

(i) x ≤ 3, x ∈ N

(ii) x < 4, x ∈ W

(iii) -2 ≤ x < 4, x ∈ I

(iv) -3 ≤ x ≤ 2, x ∈ I

Answer

(i) For x ≤ 3, 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.

Solve the following inequalities:

(i) 4 - x > -2, x ∈ N

(ii) 3x + 1 ≤ 8, x ∈ W

Also represent their solutions on the number line.

Answer

(i) Solving 4 - x > -2, 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

⇒ $x ≤ \dfrac{7}{3}$

⇒ $x ≤ 2\dfrac{1}{3}$

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

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

Solve 3 - 4x < x - 12, x ∈ {-1, 0, 1, 2, 3, 4, 5, 6, 7}

Answer

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}.

Solve -7 < 4x + 1 ≤ 23, x ∈ I

Answer

Solving -7 < 4x + 1 ≤ 23,

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

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

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

Fill in the blanks:

(i) A linear equation in one variable cannot have more than .... solution.

(ii) If five times a number is 50, then the number is .....

(iii) The number 4 is the .... of the equation 2y - 5 = 3

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

(v) ...... is a solution of the equation 4x + 9 = 5

(vi) If 3x + 7 = 1, then the value of 5x + 13 is .......

(vii) In natural numbers, 4x + 5 = -7 has ....... solution.

(viii) In integers, 3x - 1 = 4 has ....... solution.

(ix) 5x + ....... = 13 has the solution -3

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

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

(xii) If x ∈ W, then the solution set of x < 2 is ......

Answer

(i) A linear equation in one variable cannot have more than one solution.

(ii) If five times a number is 50, then the number is $\dfrac{50}{5}$ = 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 = $\dfrac{5}{3}$, 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}.

State whether the following statements are true (T) or false (F):

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

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

(iii) 3x - 5 = 2(x + 3) + 7 is a linear equation in one variable.

(iv) The solution of the equation 3(x - 4) = 30 is x = 6

(v) The solution of the equation 3x - 5 = 2 is x = $\dfrac{7}{3}$

(vi) The solution of a linear equation in one variable is always an integer.

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

(viii) 2x + 1 = 7 and 3x - 5 = 4 have the same solution.

(ix) $\dfrac{9}{4}$ is a solution of the equation 5x - 1 = 8

(x) If 5 is a solution of variable x in the equation $\dfrac{5x-7}{2}$ = y, then the value of y is 18

(xi) One-fourth of a number added to itself gives 10, can be represented as $\dfrac{x}{4}$ + 10 = x

Answer

(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 = $\dfrac{7}{3}$.

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

(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 = $\dfrac{9}{4}$ in 5x - 1 gives $5 \times \dfrac{9}{4} - 1 = \dfrac{45}{4} - 1 = \dfrac{41}{4} \neq 8$.

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

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

Which of the following is not a linear equation in one variable?

1. 3x - 1 = 7

2. 5y - 2 = 3(y + 2)

3. 2x - 3 = $\dfrac{7}{2}$

4. 7p + q = 3

Answer

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.

The solution of the equation $\dfrac{1}{3}$(2y - 1) = 3 is

1. 5

2. 3

3. 2

4. 1

Answer

Solving the equation,

⇒ $\dfrac{1}{3}(2y - 1) = 3$

⇒ 2y - 1 = 9

⇒ 2y = 10

⇒ y = 5

Hence, Option 1 is the correct answer.

x = -1 is a solution of the equation

1. x - 5 = 6

2. 2x + 5 = 7

3. 2(x - 2) + 6 = 0

4. 3x + 5 = 4

Answer

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.

If 3(3n - 10) = 2n + 5, then the value of n is

1. 12

2. 5

3. 3

4. -5

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.

-1 is not a solution of the equation

1. x + 1 = 0

2. 3x + 4 = 1

3. 5x + 7 = 2

4. x - 1 = 2

Answer

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.

The value of p for which the expressions p - 13 and 2p + 1 become equal is

1. 0

2. 14

3. -14

4. 5

Answer

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.

The equation which cannot be solved in integers is

1. 5x - 3 = -18

2. 3y - 5 = y - 1

3. 3p + 8 = 3 + p

4. 9z + 8 = 4z - 7

Answer

Solving 3p + 8 = 3 + p,

⇒ 3p + 8 = 3 + p

⇒ 3p - p = 3 - 8

⇒ 2p = -5

⇒ $p = -\dfrac{5}{2}$

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

Hence, Option 3 is the correct answer.

The solution of which of the following equations is neither an integer nor a fraction?

1. 2x + 5 = 1

2. 3x - 7 = 0

3. 5x - 7 = x + 1

4. 4x + 7 = x + 2

Answer

Solving 4x + 7 = x + 2,

⇒ 4x + 7 = x + 2

⇒ 4x - x = 2 - 7

⇒ 3x = -5

⇒ $x = -\dfrac{5}{3}$

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

Hence, Option 4 is the correct answer.

If the sum of two consecutive even numbers is 54, then the smaller number is

1. 25

2. 26

3. 27

4. 28

Answer

Let the smaller even number be x, 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.

If the sum of two consecutive odd numbers is 28, then the bigger number is

1. 19

2. 17

3. 15

4. 13

Answer

Let the smaller odd number be x, 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.

If 5 added to thrice an integer is -7, then the integer is

1. -6

2. -5

3. -4

4. 4

Answer

Let the integer be x.

⇒ 3x + 5 = -7

⇒ 3x = -7 - 5

⇒ 3x = -12

⇒ x = -4

Hence, Option 3 is the correct answer.

If the length of a rectangle is twice its breadth and its perimeter is 120 m, then its length is

1. 20 m

2. 40 m

3. 60 m

4. 30 m

Answer

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

⇒ 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.

If the difference of two complementary angles is 10°, then the smaller angle is

1. 40°

2. 50°

3. 45°

4. 85°

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.

If the difference of two supplementary angles is 30°, then the larger angle is

1. 60°

2. 75°

3. 90°

4. 105°

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.

If x ∈ W, the solution set of the in equation -2 ≤ x < 3 is

1. {-2, -1, 0, 1, 2}

2. {-1, 0, 1, 2, 3}

3. {0, 1, 2, 3}

4. {0, 1, 2}

Answer

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.

Statement I: A boy's age is half the age of his father. 5 years ago, their ages were in the ratio 3 : 7. Based on this, we can say that the boy is 20 years old.

Statement II: We can add or subtract the same algebraic expression to both sides of an equation.

1. Statement I is true but statement II is false.

2. Statement I is false but statement II is true.

3. Both Statement I and statement II are true.

4. Both Statement I and statement II are false.

Answer

Checking Statement I:

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

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

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

⇒ 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.

Statement I: -x > -1 is the same as x > 1

Statement II: We can multiply or divide both sides of a linear equality with any non-zero integer.

1. Statement I is true but statement II is false.

2. Statement I is false but statement II is true.

3. Both Statement I and statement II are true.

4. Both Statement I and statement II are false.

Answer

Checking Statement I: 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.

Statement I: Given 1 - 2x ≥ 5, where x ∈ I. We can say that x ≤ -2

Statement II: Solution set of an inequality can be represented on a number line.

1. Statement I is true but statement II is false.

2. Statement I is false but statement II is true.

3. Both Statement I and statement II are true.

4. Both Statement I and statement II are false.

Answer

Checking Statement I:

⇒ 1 - 2x ≥ 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.

Solve the following equations:

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

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

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

⇒ $x = \dfrac{4}{2}$

⇒ 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,

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

Multiplying both sides by 12,

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

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

A number exceeds its three-fifth by 22. Find the number.

Answer

Let the required number be x.

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

According to the problem,

$\Rightarrow x - \dfrac{3}{5}x = 22\\[1em] \Rightarrow \dfrac{5x - 3x}{5} = 22\\[1em] \Rightarrow \dfrac{2x}{5} = 22\\[1em] \Rightarrow 2x = 110\\[1em] \Rightarrow x = \dfrac{110}{2}\\[1em] \Rightarrow x = 55$

Hence, the required number is 55.

When 9 is added to twice a number, the result is 3 more than thrice the number. Find the number.

Answer

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.

The ten's digit of a two digit number is twice the unit's digit. The sum of the number and its unit's digit is 66. Find the number.

Answer

Let the unit's digit be x.

Then the ten's digit = 2x.

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

According to the problem,

⇒ 21x + x = 66

⇒ 22x = 66

⇒ $x = \dfrac{66}{22}$

⇒ 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.

A student bought some pens at ₹8 each and some pencils at ₹1.50 each. If the total number of pens and pencils purchased is 16 and their total cost is ₹50, how many pens did he buy?

Answer

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

⇒ $x = \dfrac{26}{6.5}$

⇒ x = 4

Hence, the student bought 4 pens.

Arvind is eight years older than his sister. In three years, he will be twice as old as his sister. How old are they now?

Answer

Let the 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.

The angles of a triangle are in the ratio 1 : 2 : 3. Find their measure in degrees.

Answer

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

⇒ $x = \dfrac{180}{6}$

⇒ x = 30

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

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

Solve the following inequations and represent their solution on a number line:

(i) $\dfrac{2x-1}{3} ≤ 2\dfrac{1}{2}$, x ∈ W

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

Answer

(i) Solving,

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

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 < $\dfrac{2x}{3}$ + 1 ≤ 4, x ∈ I,

$\Rightarrow -1 \lt \dfrac{2x}{3} + 1 ≤ 4\\[1em] \Rightarrow -1 - 1 \lt \dfrac{2x}{3} ≤ 4 - 1\\[1em] \Rightarrow -2 \lt \dfrac{2x}{3} ≤ 3\\[1em] \Rightarrow -6 \lt 2x ≤ 9\\[1em] \Rightarrow -3 \lt x ≤ \dfrac{9}{2}$

As x ∈ I, the integers greater than -3 and less than or equal to $\dfrac{9}{2}$ (= 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.

If $\dfrac{7m+2}{3}$ = -11, then find the value of 2m³ + 10m² + 4m - 3

Answer

First, finding the value of m,

⇒ $\dfrac{7m+2}{3} = -11$

⇒ 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.

Two persons start moving from two points A and B in opposite directions towards each other. One person start moving from A at the speed of 4 km/h and meets the other person coming from B after 6 hours. If the distance between A and B is 42 km, find the speed of the other person.

Answer

Let the speed of the 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

⇒ $x = \dfrac{18}{6}$

⇒ x = 3

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

There are some benches in a classroom. If 4 students sit on each bench then 3 benches remains empty and if 3 students sit on each bench then 3 students remain standing. Find the number of students in the class.

Answer

Let the number of benches be x.

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.