# NCERT Solutions for Class 11 Maths Chapter 5 - Linear Inequalities

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Exercise 6.1

1. Solve 24x < 100, when:

(i) x is a natural number.

(ii) x is an integer.

Sol. Given, 24x < 100

On dividing both sides by 24, we get

$$x\lt\frac{100}{24}\\\Rarr\space x\lt\frac{50}{12}\\\Rarr\space x\lt\frac{25}{6}\\\text{Convert this into mixed fraction.}\\\Rarr\space x\lt4\frac{1}{6}$$

(i) When x is a natural number.

It is clear, natural number should less than $$4\frac{1}{6}.$$

∴ The solution set is {1, 2, 3, 4}.

(ii) When x is an integer.

It is clear, integer number should be less than

$$4\frac{1}{6}.$$

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

2. Solve – 12x > 30, when:

(i) x is a natural number.

(ii) x is an integer.

Sol. Given, – 12x > 30

On dividing both sides by – 12, we get

$$\Rarr\space x\lt\frac{30}{-12}\\\Rarr\space x\lt\frac{-5}{2}$$

(i) When x is a natural number then there is no solution of a given inequality because

$$\frac{-5}{2}$$

is a negative number and natural numbers are positive numbers.

(ii) When x is an integer.

It is clear, integer number should less than

$$\frac{-5}{2}$$

∴ The solution set is {…, – 4, – 3}.

3. Solve 5x – 3 < 7, when:

(i) x is an integer.

(ii) x is a real number.

Sol. Given, 5x – 3 < 7

On adding both sides by 3, we get

5x < 10

On dividing both sides by 5, we get

x < 2

(i) When x is an integer.

It is clear, integer number should less than 2.

∴ The solution set is {…, – 1, 0, 1}.

(ii) When x is a real number.

The solution of given inequality is given by x < 2 i.e., all real numbers x should less than 2.

∴ The solution set is (– ∞, 2).

4. Solve 3x + 8 > 2, when:

(i) x is an integer.

(ii) x is a real number.

Sol. Given, 3x + 8 > 2

On adding both sides by – 8, we get

3x > – 6

On dividing both sides by 3, we get

x > – 2

(i) When x is an integer. It is clear, the integer number should greater than – 2. Hence the solution set is {– 1, 0, 1, 2, …}.

(ii) When x is a real number, the solution set is (– 2, ∞).

Because the solution of given inequality is given by x > – 2 i.e., all real numbers should greater than – 2.

Solve the inequalities in Exercises 5 to 16 for real x:

5. 4x + 3 < 5x + 7.

Sol. Given, 4x + 3 < 5x + 7

On adding both sides by – 4x, we get

3 < x + 7

On adding both sides by – 7, we get

– 4 < x

It is clear, x should be greater than – 4.

∴ The solution set is (– 4, ∞).

6. 3x – 7 > 5x – 1.

Sol. Given, 3x – 7 > 5x – 1

On adding both sides by – 3x, we get

– 7 > 2x – 1

On adding both sides by 1, we get

– 6 > 2x

On dividing both sides by 2, we get

– 3 > x

It is clear, x should less than – 3.

∴ The solution set is (– ∞, – 3).

7. 3(x – 1) ≤ 2(x – 3).

Sol. Given, 3(x – 1) ≤ 2(x – 3)

⇒ 3x – 3 ≤ 2x – 6

On adding both sides by – 2x, we get

⇒ x – 3 ≤ – 6

Adding 3 on both sides, we get

x ≤ – 3

It is clear, x should less than or equal to – 3.

∴ the solution set is (– ∞, – 3].

8. 3(2 – x) ≥ 2(1 – x).

Sol. Given, 3(2 – x) ≥ 2(1 – x)

⇒ 6 – 3x ≥ 2 – 2x

Adding 3x on both sides, we get

⇒ 6 ≥ 2 + x

Adding – 2 on both sides, we get

⇒ 4 ≥ x

It is clear, x should greater than or equal to 4.

∴ The solution set is (– ∞, 4].

$$\textbf{9.}\space \textbf{x +}\frac{\textbf{x}}{\textbf{2}}+\frac{\textbf{x}}{\textbf{2}}\lt\textbf{11.}$$

$$\textbf{Sol.}\space\text{Given}\space x+\frac{x}{2}+\frac{x}{3}\lt11\\\text{Taking LCM of 2, 3 is 6.}\\\Rarr\space\frac{6x+3x+2x}{6}\lt 11\\\Rarr\space\frac{11x}{6}\lt11$$

$$\text{Multiplying}\space\frac{6}{11}\space\text{on both sides, we get}\\\Rarr\space x\lt6$$

It is clear, x should less than 6.

∴ The solution set is (– ∞, 6).

$$\textbf{10.}\space\frac{\textbf{x}}{\textbf{3}}\gt\frac{\textbf{x}}{\textbf{2}}\textbf{+1}\\\textbf{Sol.}\space\text{Given}\space\frac{x}{3}\gt\frac{x}{2}+1\\\text{Transferring the term}\space\frac{x}{2}\text{to LHS,}\\\frac{x}{3}-\frac{x}{2}\gt1$$

Taking LCM of 2, 3 is 6,

$$\frac{2x-3x}{6}\gt 1\\\frac{-x}{6}\gt1$$

Mutiplyhing – 6 on both sides, we get

x < – 6

It is clear, x should less than – 6.

∴ The solution set is (– ∞, – 6).

$$\textbf{11.}\space\frac{\textbf{3(x-2)}}{\textbf{5}}\leq\frac{\textbf{5(2-x)}}{\textbf{3}}\textbf{.}\\\textbf{Sol.}\space\text{Given,}\frac{3(x-2)}{5}\leq\frac{5(2-x)}{3}\\\Rarr\space\frac{3x-6}{5}\leq\frac{10-5x}{3}$$

Multiplying 15 on both sides, we get

⇒ 3(3x – 6) ≤ 5(10 – 5x)

⇒ 9x – 18 ≤ 50 – 25x

Now, adding 25x on both sides, we get

⇒ 34x – 18 ≤ 50

Adding 18 on both sides, we get

⇒ 34x ≤ 68

Dividing 34 on both sides, we get

$$x\leq\frac{68}{34}\\\Rarr\space x\leq 2$$

It is clear, x should less than or equal to 2.

∴ The solution set is (– ∞, 2].

$$\textbf{12.}\space\frac{\textbf{1}}{\textbf{2}}\bigg(\frac{\textbf{3x}}{\textbf{5}}\textbf{+ 4}\bigg)\geq\frac{\textbf{1}}{\textbf{3}}\textbf{(x-6).}\\\textbf{Sol.\space}\text{Given}\space\frac{1}{2}\bigg(\frac{3x}{5}+4\bigg)\geq\frac{1}{3}(x-6)\\\Rarr\space\frac{3}{10}x+2\geq\frac{x}{3}-2\\\Rarr\space\frac{3x+20}{10}\geq\frac{x-6}{3}$$

Multiplying 30 on both sides, we get

⇒ 3(3x + 20) ≥ 10(x – 6)

⇒ 9x + 60 ≥ 10x – 60

Subtracting 9x on both sides, we get

60 ≥ x – 60

Adding 60 on both sides, we get

120 ≥ x

It is clear, x should less than or equal to 120.

∴ the solution set = (– ∞, 120].

13. 2(2x + 3) – 10 < 6(x – 2).

Sol. Given, 2(2x + 3) – 10 < 6(x – 2)

4x + 6 – 10 < 6x – 12

4x – 4 < 6x – 12

Substracting 4x on both sides, we get

– 4 < 2x – 12

Adding 12 on both sides, we get

8 < 2x

Dividing 2 on both sides, we get

4 < x

It is clear, x should greater than 4.

∴ The solution set is (4, ∞).

14. 37 – (3x + 5) ≥ 9x – 8(x – 3).

Sol. Given, 37 – (3x + 5) ≥ 9x – 8(x – 3)

⇒ 37 – 3x – 5 ≥ 9x – 8x + 24

⇒ 32 – 3x ≥ x + 24

Adding 3x on both sides, we get

⇒ 32 ≥ 4x + 24

Subtracting 24 on both sides, we get

⇒ + 8 ≥ 4x

Dividing 4 on both sides, we get

⇒ 2 ≥ x

It is clear x, should less than or equal to 2.

∴ The solution set is (– ∞, 2].

$$\textbf{15.}\space\frac{\textbf{x}}{\textbf{4}}\lt\frac{\textbf{(5x-2)}}{\textbf{3}}-\frac{\textbf{(7x-3)}}{\textbf{5}}\textbf{.}\\\textbf{Sol.}\space\text{Given},\space\frac{x}{4}\lt\frac{(5x-2)}{3}-\frac{(7x-3)}{5}\\\text{Taking LCM of 3, 5 is 15 on RHS, we get}\\\frac{x}{4}\lt\frac{5(5x-2)-3(7x-3)}{15}\\\Rarr\space\frac{x}{4}\lt\frac{25x-10-21x+9}{15}\\\Rarr\space\frac{x}{4}\lt\frac{4x-1}{15}$$

Multiplying 60 on both sides, we get

⇒ 15x < 4(4x – 1)

⇒ 15x < 16x – 4

Adding 4 on both sides, we get

15x + 4 < 16x

Subtracting 15x on both sides, we get

⇒ 4 < x

It is clear, x should greater than 4.

∴ The solution set is (4, ∞).

$$\textbf{16.\space}\frac{\textbf{2x-1}}{\textbf{3}}\geq\frac{\textbf{3x-2}}{\textbf{4}}-\bigg(\frac{\textbf{2-x}}{\textbf{5}}\bigg)\textbf{.}\\\textbf{Sol.}\space\text{Given,}\frac{2x-1}{3}\geq\frac{3x-2}{4}-\bigg(\frac{2-x}{5}\bigg)$$

Taking LCM of 4, 5 is 20.

$$\frac{2x-1}{3}\geq\frac{5(3x-2)-4(2-x)}{20}\\\Rarr\space\frac{2x-1}{3}\geq\frac{15x-10-8+4x}{20}\\\Rarr\space\frac{2x-1}{3}\geq\frac{19x-18}{20}$$

Multiplying by 60 on both sides, we get

⇒ 20(2x – 1) ≥ 3(19x – 18)

⇒ 40x – 20 ≥ 57x – 54

Subtracting 40x on both sides, we get

⇒ – 20 ≥ 17x – 54

Adding 54 on both sides, we get

⇒ 34 ≥ 17x

Dividing by 17 on both sides, we get

2 ≥ x

It is clear, x should less than or equal to 2.

∴ The solution set is (– ∞, 2].

17. 3x – 2 < 2x + 1.

Sol. Given, 3x – 2 < 2x + 1

Adding 2 on both sides, we get

3x < 2x + 3

Subtracting 2x on both sides, we get

x < 3

It is clear, x should less than 3.

∴ The solution set is (– ∞, 3).

The graphical representation of the solution of given inequality is as follows:

18. 5x – 3 ≥ 3x – 5.

Sol. 5x – 3 ≥ 3x – 5

Adding 3 on both sides, we get

5x ≥ 3x – 2

Subtracting 3x on both sides, we get

2x ≥ – 2

Dividing by 2 on both sides, we get

x ≥ – 1

It is clear, x should greater than or equal to – 1.

∴ The solution set is [– 1, ∞).

The graphical representation of the solution of the given inequality is as follows:

19. 3(1 – x) < 2(x + 4).

Sol. Given 3(1 – x) < 2(x + 4)

⇒ 3 – 3x < 2x + 8

Adding 3x on both sides, we get

3 < 5x + 8

Subtracting 8 on both sides, we get

– 5 < 5x

Dividing by 5 on both sides, we get

– 1 < x

It is clear, x should greater than – 1.

∴ The solution set is (– 1, ∞).

The graphical representation of the solution of the given inequality is as follows:

$$\textbf{20.}\space\frac{\textbf{x}}{\textbf{2}}\lt\frac{\textbf{5x-2}}{\textbf{3}}-\frac{\textbf{7x-3}}{\textbf{5}}.\\\textbf{Sol.}\text{Given,}\space\frac{x}{2}\lt\frac{5x-2}{3}-\frac{7x-3}{5}\\\text{Taking LCM 3 and 5 is 15,}\\\frac{x}{2}\lt\frac{5(5x-2)-3(7x-3)}{15}\\\Rarr\frac{x}{2}\lt\frac{25x-10-21x+19}{15}\\\frac{x}{2}\lt\frac{4x-1}{15}$$

Multiplying 30 on both sides, we get

15x < 2(4x – 1)

15x < 8x – 2

Subtracting 8x on both sides, we get

7x < – 2

Dividing 7 on both sides, we get

$$x\lt\frac{-2}{7}\\\text{It is clear, x should less than}\space\frac{-2}{7}.$$

$$\therefore\space\text{The solution set is}\bigg(-\infty,\frac{-2}{7}\bigg)$$

The graphical representation of the solution of the given inequality is as follow:

21. Ravi obtained 70 and 75 marks in first two unit tests. Find the number if minimum marks he should get in the third test to have an average of atleast 60 marks.

Sol. Let Ravi got x marks in third unit test.

Since, the student should have an average at least 60 marks.

$$\therefore\space\leq\space\frac{\text{Sum of all test marks}}{\text{Number of tests}}\\\leq\frac{70+75+x}{3}\\\leq\frac{145+x}{3}\\\text{Therefore, we can write as}\\\frac{145+x}{3}\geq60$$

Multiplying by 3 on both sides, we get

145 + x ≥ 180

Subtracting 145 on both sides, we get

x ≥ 35

Hence, Ravi should get greater than or equal to 35 marks in 3rd unit test.

22. To receive grade A in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade A in the course.

Sol. Let Sunita got x marks in her fifth exam.

$$\therefore\space\text{Average =}\space\frac{\text{sum of marks in all exams}}{\text{number of exams}}$$

$$=\frac{87+92+94+95+x}{5}\\=\frac{368+x}{5}$$

Since, average should be greater than or equal to 90.

$$\Rarr\space\frac{368+x}{5}\geq90$$

Multiplying by 5 on both sides, we get

368 + x ≥ 450

Subtracting 368 on both sides, we get

x ≥ 82

Hence, Sunita should get greater than or equal to 82 marks in fifth exam.

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

Sol. Let x be the smaller of the two consecutive odd positive integers.

And other integer is x + 2.

∵ Both the integers are smaller than 10 as giv
en,

⇒ x + 2 < 10

x < 8 …(i)

Also, it is given that sum of two integers is mroe than 11.

∴ x + (x + 2) > 11

2x + 2 > 11

$$x\gt\frac{9}{2}$$

x > 4.5 …(ii)

Thus, from equation (i) and (ii), we have x is an odd integer and it can take values 5 and 7 as both are odd.

Hence, possible pairs are (5, 7) and (7, 9).

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

Sol. Let the numbers are 2x and 2x + 2.

According to the question,

2x > 5 and 2x + 2 > 5

$$x\gt\frac{5}{2}\space\text{and}\space2x\gt3\\x\gt\frac{3}{2}$$

Also, 2x + 2x + 2 < 23

4x + 2 < 23

4x < 21

$$x\lt\frac{21}{4}\\\therefore\space x\epsilon\bigg(\frac{5}{2},\frac{21}{4}\bigg)$$

Hence the integers are 3, 4, 5.

When x = 3

2x = 6

and 2x + 2 = 8

When x = 4

2x = 8

and 2x + 2 = 10

When x = 5

2x = 10

and 2x + 2 = 12

Hence the required numbers are

(6, 8), (8, 10), (10, 12).

25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Sol. Let x cm be the shortest side.

According to the question,

longest side = 3x cm

and third side = (3x – 2) cm

Since perimeter of triangle ≥ 61 cm

⇒ 3x + x + 3x – 2 ≥ 61

⇒ 7x – 2 ≥ 61

⇒ 7x ≥ 63

⇒ x ≥ 9

∴ The shortest side must be greater than and equal to 9 cm.

26. A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and 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 cm longer than the second?

[Hint: If x is the length of the shortest board, then x, (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5]

Sol. Let x cm be the shortest side.

According to the question,

Second length = (x + 3) cm

and third length = 2x cm

Also, total length = 91 cm

⇒ x + x + 3 + 2x ≤ 91 cm

⇒ 4x + 3 ≤ 91

⇒ 4x ≤ 88

⇒ x ≤ 22

Again, given, third piece of at least 5 cm longer than the second piece.

2x ≥ (x + 3) + 5

2x ≥ x + 8

x ≥ 8

∴ x ∈ (8, 22)

⇒ 8 ≤ x ≤ 22

Hence, shortest board should be greater than or equal to 8 but less than or equal to 22 cm.

Exercise 6.2

Solve the following inequalities graphically in two-dimensional plane.

1. x + y < 5.

Sol. Given, x + y < 5   …(i)

Let the inequations as a strict equation i.e.,

Check (0, 0) satisfy the in equation (i)

0 + 0 < 5

0 < 5 (true)

So, the shaded region will be origin side.

∴ The graph of the inequations is

2. 2x + y ≥ 6.

Sol. Given, 2x + y ≥ 6 …(i)

Let the inequation as a strict equation.

i.e., 2x + y = 6

Now

 x 3 0 y 0 6

Check (0, 0) satisfy the inequation (i)

2 × 0 + 0 ≥ 6

0 ≥ 6 (false)

so, the shaded region will be non-origin side.

∴ The graph of the inequation is

3. 3x + 4y ≤ 12.

Sol. Given 3x + 4y ≤ 12 …(i)

Let the inequation as a strict equation i.e.,

3x + 4y = 12

Now,

 x 4 0 y 0 3

Check (0, 0) satisfy the inequation (i)

3 × 0 + 4 × 0 ≤ 12

0 ≤ 12 (true)

So, the shaded region will be origin side.

∴ The graph of the inequation is

4. y + 8 ≥ 2x.

Sol. Given, y + 8 ≥ 2x …(i)

Let the inequation as a strict equation

i.e., y + 8 = 2x

⇒ 2x – y = 8

Now,

 x 0 4 y -8 0

Check (0, 0) satisfy the inequation (i)

+ 8 ≥ 2 × 0

⇒ 8 ≥ 0 (true)

So, the shaded region will be origin side.

∴ The graph of the inequation is

5. x – y ≤ 2.

Sol. Given, x – y ≤ 2 …(i)

Let the inequation as a strict equation i.e.,

x – y = 2

Now,

 x 2 0 y 0 -2

Check (0, 0) satisfy the inequation (i),

0 – 0 ≤ 2 (true)

So, the shaded region will be origin side.

∴ The graph of the inequation is

6. 2x – 3y > 6.

Sol. Given, 2x – 3y > 6 …(i)

Let the inequation as a strict equation.

i.e., 2x – 3y = 6

Now,

 x 0 3 y -2 0

Check (0, 0) satisfy the inequation (i)

2 × 0 – 3 × 0 > 6

0 > 6 (false)

∴ The graph of the inequation is

7. – 3x + 2y ≥ – 6.

Sol. Given, – 3x + 2y ≥ – 6 …(i)

Let the inequation as a strict equation.

i.e., – 3x + 2y = – 6

⇒ 3x – 2y = 6

Now,

 x 0 2 y -3 0

Check (0, 0) satisfy the inequation (i)

– 3 × 0 + 2 × 0 ≥ – 6

0 ≥ – 6 (true)

∴ The graph of the inequation is

8. 3y – 5x < 30.

Sol. Given, 3x – 5x < 30 …(i)

Let the inequation as a strict inequation.

i.e., 3y – 5x = 30

Now,

 x 0 -6 y 10 0

Check (0, 0) satisfy the inequation (i)

3 × 0 – 5 × 0 < 30

⇒ 0 < 30 (true)

So, the shaded region will be origin side.

∴ The graph of the inequation is

9. y < – 2.

Sol. Given y < – 2 …(i)

Let the inequation as a strict equation i.e.,

y = – 2

∴ y = – 2 is parallel to the x-axis.

Check (0, 0) satisfy the equation (i).

⇒ 0 < – 2 (false)

So, the shaded region will be non-origin side.

Thus, the graph of the inequation is

10. x > – 3.

Sol. Given, x > – 3 …(i)

Let the inequation as a strict equation.

i.e., x = – 3

∴ The equation x = – 3 is parallel to y-axis.

Check (0, 0) satisfy the inequation (i).

0 > – 3 (true)

Thus the graph of the inequation.

Exercise 6.3

Solve the following system of inequalities graphically:

1. x ≥ 3, y ≥ 2.

Sol. Given, inequalities

x ≥ 3 …(i)

and

y ≥ 2 …(ii)

Let the inequations as a strict equations.

⇒ x = 3 and y = 2

∴ x = 3, y = 2 are parallel to y-axis and x-axis respectively.

Now, check (0, 0) satisfy the inequations (i) and (ii),

0 ≥ 3 (false)

and

0 ≥ 2 (false)

Thus, the inequations does not lies on origin.

∴ The region to be included will be on the right side of the two equalities drawn on the graph.

The region which is covered by two inequalities will give the required answer.

2. 3x + 2y ≤ 12, x ≥ 1, y ≥ 2.

Sol. Given inequalities

3x + 2y ≤ 12 …(i)

x ≥ 1 …(ii)

y ≥ 2 …(iii)

Let the inequations as a straight equation.

⇒ 3x + 2y = 12

x = 1

y = 2

Now,

3x + 2y = 12

 x 0 4 y 6 0

and x = 1, y = 2 are parallel to y, x axes respectively.

Chekc (0, 0) satisfy the inequations (i), (ii) and (iii).

⇒ 3 × 0 + 2 × 0 ≤ 12

0 ≤ 12 (true)

⇒ 0 ≥ 1 (false)

and 0 ≥ 2 (false)

∴ Only inequality (i) towards the origin.

The region to be included in the solution would be towards the left of the equality y ≥ 2. The shaded region in the graph will give the answer to the required inequalities as it is the region which is covered by all the given three inequalities at the same time satisfying all the given conditions.

Thus, the graphically representation is

3. 2x + y ≥ 6, 3x + 4y ≤ 12,

Sol. Given inequalities

2x + y ≥ 6 …(i)

3x + 4y ≤ 12 …(ii)

Let the above inequations as a strict equation.

⇒ 2x + y = 6

3x + 4y = 12

Now, the coordinates are

2x + y = 6

 x 0 3 y 6 0

and 3x + 4y = 12

 x 0 4 y 3 0

Check the origin satisfy the inequation (i) and (ii)

⇒ 2 × 0 + 0 ≥ 6

0 ≥ 6 (false)

and  3 × 0 + 4 × 0 ≤ 12

0 ≤ 12 (true)

The region on the right of the equation is the region required. The solution is the region which is common to the graphs of both the inequalities. The shaded region is the required region.

∴ The graphical representation is

4. x + y ≥ 4, 2x – y > 0.

Sol. Given inequalities

x + y ≥ 4 …(i)

and

2x – y > 0 …(ii)

Let the inequations as a strict equation.

x + y = 4, 2x – y = 0

Now, the coordinates are

x + y = 4

 x 0 4 y 4 0

and  2x – y = 0

 x 0 1 y 0 2

Check the origin satisfy the inequation (i).

⇒ 0 + 0 ≥ 4

0 ≥ 4 (false)

The required region would be on the right of line’s graph.

and (1, 0) in equation (ii),

2 × 1 – 0 > 0

2 > 0 (true)

The required region would be on the left of line’s graph.

The region to be included in the solution would be on the left side of the 2x – y < 0. The shaded region is the required solution of the inequalities.

∴ The graphical representation is

5. 2x – y > 1, x – 2y < – 1.

Sol. Given inequalities,

2x – y > 1 …(i)

and

x – 2y < – 1 …(ii)

Let the above equations as a strict equation.

⇒ 2x – y = 1

and

x – 2y = – 1

Now, the co-ordinate are

2x – y = 1

 x 0 1/2 y -1 0

and  x – 2y = – 1

 x 0 -1 y 1/2 0

Check origin satisfies both the inequations (i) and (ii),

2 × 0 – 0 > 1

0 > 1 (false)

The origin does not lie in the solution region.

The required region would be on the right of the line’s graph.

and   0 – 2 × 0 < – 1

0 < – 1 (false)

The origin does not lie in the solution area, the required area would be on the left side of the line’s graph.

Hence, the shaded area is the required solution of the given inequalities.

∴ The graphical representation is

6. x + y ≤ 6, x + y ≥ 4.

Sol. Given inequalities are

x + y ≤ 6 …(i)

and

x + y ≥ 4 …(ii)

Let the above inequations as a strict equations.

⇒ x + y = 6 and x + y = 4

Now, the co-ordinates are

x + y = 6

 x 6 0 y 0 6

and x + y = 4

 x 0 4 y 4 0

Check the origin satisfies the inequations (i) and (ii),

0 + 0 ≤ 6

0 ≤ 6 (true)

The origin would be included in the area of the line’s graph. So the required solution of the equation would be on the left side of the line graph which will be including and origin.

0 + 0 ≥ 4

0 ≥ 4 (false)

The origin would not be included in the required area. The solution area will be above the line graph or the area on the right of line graph.

Hence, the shaded region in the graph is the required graph area.

∴ The graphical representation is

7. 2x + y ≥ 8, x + 2y ≥ 10.

Sol. Given inequalities are

2x + y ≥ 8 …(i)

and

x + 2y ≥ 10 …(ii)

Let the above equations as a strict equations.

2x + y = 8

and

x + 2y = 10

Now, the co-ordinates are

2x + y = 8

 x 0 4 y 8 0

and x + 2y = 10

 x 0 10 y 5 0

Check the origin satisfies the inequalities (i) and (ii)

2 × 0 + 0 ≥ 8

0 ≥ 8 (false)

The origin is not included in the solution area and the requires area would be the area to the right of line’s graph.

and

0 + 2 × 0 ≥ 10

0 ≥ 10 (false)

The origin would not lie in the required solution area. The required area would be the left of the line graph.

The shaded area in the graph is the required solution of the given inequalities.

∴ The graphical representation is

8. x + y ≤ 9, y > x, x ≥ 0.

Sol. Given, inequalities are

x + y ≤ 9 …(i)

y > x …(ii)

x ≥ 0 …(iii)

Let the inequations as a strict equations.

x + y = 9

y = x and x = 0

Now, the co-ordinates are

x + y = 9

 x 0 9 y 9 0

x = y

 x 1 2 y 1 2

and x = 0 is parallel to y-axis.

Check origin satisfies the inequations.

0 + 0 ≤ 9

0 ≤ 9 (true)

The reqired area would be including the origin and hence will lie on the left side of the line’s graph.

(0, 1) in equation (ii)

1 > 0 (true)

The required area would be including the origin and hence will lie on the left side of the line’s graph.

and (1, 0) in equation (iii)

1 ≥ 0 (true)

The area of the required line’s graph would be on the right side of the line’s graph.

Therefore, the shaded area is the required solution of the given inequalities.

∴ The graphical representation is

9. 5x + 4y ≤ 20, x ≥ 1, y ≥ 2.

Sol. Given, inequalities

5x + 4y ≤ 20 …(i)

x ≥ 1 …(ii)

y ≥ 2 …(iii)

Let the above inequalities as a strict equations.

⇒ 5x + 4y = 20

x = 1

and

y = 2

So, the coordinates of 5x + 4y = 20 is

 x 0 4 y 5 0

and x = 1 is parallel to y-axis.

y = 2 is parallel to x-axis.

Check the origin satisfies the inequations (i), (ii) and (iii),

⇒ 5 × 0 + 4 × 0 ≤ 20

0 ≤ 20 (true)

The origin would lie in the solution area. The required area of the line’s graph is on the left side of the graph.

⇒ 0 ≥ 1 (false)

The origin would not lie in the required area. The required area on the graph will be on the right side of the line’s graph.

and 0 ≥ 2 (false)

Similarly, origin would not lie in the required area. The graph will be on left side.

Hence, the required area would be on the right side of the line’s graph.

The shaded area on the graph shows the required solution of the given inequalities.

∴ The graphical representation is

10. 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0.

Sol. Given, inequalities are

3x + 4y ≤ 60 …(i)

x – 3y ≤ 30 …(ii)

x ≥ 0 …(iii)

y ≥ 0 …(iv)

Let the above inequations as a strict equations.

⇒ 3x + 4y = 60

x + 3y = 30

x = 0

y = 0

Now, the co-ordinates are

3x + 4y = 60

 x 0 20 y 15 0

and x + 3y = 30

 x 0 30 y 10 0

Check (0, 0) satisfies the equation (i), (ii)

3 × 0 + 4 × 0 ≤ 60

0 ≤ 60 (True)

The origin would lie in the solution area of the line’s graph. The required solution area would be on the left of the line’s graph.

0 + 3 × 0 ≤ 30

0 ≤ 30 (true)

Similarly, origin lie in the solution and area would be on the left of the line’s graph.

and (1, 0) in equation (iii)

1 ≥ 0 (true)

Similarly, origin lie in the solution and area would be on the left of the line’s graph.

(0, 1) in equation (iv)

1 ≥ 0 (true)

Similarly, origin lie in the solution and area

would be on the left of the line’s graph.

The given inequalities imply the solution lies in the first quadrant only.

Hence, the solution of the inequalities is given by region in the graph.

∴ The graphical representation

11. 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6.

Sol. Given inequalities are

2x + y ≥ 4 …(i)

x + y ≤ 3 …(ii)

2x – 3y ≤ 6 …(iii)

Let the above equations as a strict equations.

2x + y = 4

x + y = 3

2x – 3y = 6

Now, the co-ordinates are

2x + y = 4

 x 0 2 y 4 0

x + y = 3

 x 0 3 y 3 0

and 2x – 3y = 6

 x 0 3 y -2 0

Check the origin satisfies the inequations (i), (ii) and (iii), we get

2 × 0 + 0 ≥ 4

0 ≥ 4 (false)

The origin doesn’t lies in the solution area of the line’s graph. The solution area would be given by the right side of the line’s graph.

and  0 + 0 ≤ 3

0 ≤ 3 (true)

The solution area would include the origin and

hence would be on the left side of the line’s graph.

and 2 × 0 – 3 × 0 ≤ 6

0 ≤ 6 (true)

The origin lies in the solution area and area would be on the left of the line’s graph.

Hence, the shaded area in the graph is the reqired solution area for the given inequalities.

∴ The graphical representation is

12. x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1.

Sol. Given inequalities are

x – 2y ≤ 3 …(i)

3x + 4y ≥ 12 …(ii)

x ≥ 0 …(iii)

y ≥ 1 …(iv)

Let the above inequations as a strict equations

x – 2y = 3

3x + 4y = 12

x = 0

y = 1

Now, the co-ordinates are

x – 2y = 3

 x 0 3 y -3/2 0

and 3x + 4y = 12

 x 0 4 y 3 0

Check origin satisfies the inequations (i) and (ii),
0 – 2 × 2 ≤ 3

0 ≤ 3 (true)

The solution area would be on the left of the line’s graph.

3 × 0 + 4 × 0 ≥ 12

0 ≥ 12 (false)

The solution area would include the origin and the required soluton area would be on the right side of the line’s graph.

and (1, 0) in equation (iii) and (iv)

⇒ 1 ≥ 0 (true)

The region would be the region above the x-axis on the graph.

and  0 ≥ 1 (false)

The solution area would be on the left side of the line’s graph.

∴ The graphical representation is

13. 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0.

Sol. Given inequalities are

4x + 3y ≤ 60 …(i)

y ≥ 2x …(ii)

x ≥ 3 …(iii)

x, y ≥ 0 …(iv)

Let the above inequations as a strict equations,

4x + 3y = 60

y = 2x

x = 3

x, y = 0

Now, the co-ordinates are

4x + 3y = 60

 x 0 15 y 20 0

and y = 2x

Check the origin satisfies inequations (i) and (iii), we get

⇒ 4 × 0 + 3 × 0 ≤ 60

0 ≤ 60 (true)

The origin would lie in the solution area. The required area would include be on the left of the line’s graph.

⇒ 0 ≥ 3 (false)

The origin doesn’t lie in the solution area. Hence the required solution area would be the right of the line’s graph.

and (1, 1) inequations (ii)

1 ≥ 2 (false)

Since, given both x and y are greater than 0.

The solution area would be in the first quadrant only.

The shaded region in the graph shows the solution area for the given inequalities.

∴ The graphical representation is

14. 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0.

Sol. Given inequalities

3x + 2y ≤ 150 …(i)

x + 4y ≤ 80 …(ii)

x ≤ 15 …(iii)

x, y ≥ 0 …(iv)

Let the above inequations as a straight equations.

⇒ 3x + 2y = 150

x + 4y = 80

x = 15

x, y = 0

Now, the co-ordinates are

3x + 2y = 150

 x 0 50 y 75 0

x + 4y = 80

 x 0 80 y 20 0

Check the origin satisfies the equations (i), (ii) and (iii), we get

3 × 0 + 2 × 0 ≤ 150

0 ≤ 150 (true)

The solution area for the line would be on the left side of the line’s graph which would be including the origin too.

and

0 + 4 × 0 ≤ 80

The origin would be included in the solution area. The required solution area would be towards the left of line’s graph.

0 ≤ 80 (true)

and 0 ≤ 15 (true)

The origin would be included in the solution area. The required solution area would be towards the left of line’s graph.

and (1, 0) in equation (iv), we get

1 ≥ 0 (true)

The origin would be included in the solution area. The required solution area would be towards the left of line’s graph.
and (0, 1) in equation (v), we get

1 ≥ 0 (true)

The origin would be included in the solution area. The required solution area would be towards the left of line’s graph.

Since x and y are greater than 0, the solution would lie in the Ist quadrant. The shaded area in the graph satisfies all the given inequalities and hence it is the solution area for given inequalities.

∴ The graphical representation is

15. x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0.

Sol. Given inequalities are

x + 2y ≤ 10 …(i)

x + y ≥ 1 …(ii)

x – y ≤ 0 …(iii)

x, y ≥ 0 …(iv)

Let the above in equations as a strict equations

x + 2y = 10

x + y = 1

x – y = 0

x, y = 0

Now, the co-ordinates are

x + 2y = 10

 x 0 10 y 5 0

and x + y = 1

 x 0 1 y 1 0

and x – y = 0

 x 1 2 y 1 2

Check, origin satisfies the equations (i) and (ii),

we get

0 + 2 × 0 ≤ 10

0 ≤ 10 (true)

The solution area would be towards origin including the same. The solution area would be towards the left of the line’s graph.

and 0 + 0 ≥ 1

0 ≥ 1 (false)

The origin would not be including into the solution area. The require solution area would be towards right of the line’s graph.

and (0, 1) in equation (iii)

0 – 1 ≤ 0

– 1 ≤ 0 (true)

The solution area would be towards origin including the same. The solution area would be towards the left of the line’s graph.

Since, both x and y are greater than 0.

The solution area would be in the Ist quadrant. Hence, the solution area for the given inequalities would be the shaded area of the graph satisfying all the given inequalities.

∴ The graphical representation is

Miscellaneous Exercise

Direction (Q. Nos. 1 to 6): Solve the inequalities:

1. 2 ≤ 3x – 4 ≤ 5.

Sol. Given, 2 ≤ 3x – 4 ≤ 5

6 ≤ 3x ≤ 9

On dividing by 3, we get

2 ≤ x ≤ 3

All real numbers x greater than or equal to 2 but less than or equal to 3 are solution of given equality.

∴ The solution set is [2, 3].

2. 6 ≤ – 3(2x – 4) < 12.

Sol. Given, 6 ≤ – 3(2x – 4) < 12

⇒ 6 ≤ – 6x + 12 < 12

Subtracting 12 on both sides

⇒ – 6 ≤ – 6x < 0

On multiplying by (– 1), we get

⇒ 0 < 6x ≤ 6

On dividing by 6, we get

⇒ 0 < x ≤ 1

All real numbers x greater than 0 but less than or equal to 1 are solution of given equality.

∴ The solution set (0, 1].

$$\textbf{3.}\space \textbf{-3}\leq\textbf{4}-\frac{\textbf{7x}}{\textbf{2}}\leq\textbf{18.}\\\textbf{Sol.}\space\text{Given,}\space -3\leq4-\frac{7x}{2}\leq18$$

Taking LCM of 1, 2 in 2, then

$$-3\leq\frac{8-7x}{2}\leq18$$

On multiplying by 2 throughout

– 6 ≤ 8 – 7x ≤ 36

Subtracting 8 on both sides

– 14 ≤ – 7x ≤ 28

On dividing by – 7, we get

– 4 ≤ x ≤ + 2

All real numbers x greater than or equal to – 4 but less than or equal to 2 are solution of given equality.

∴ The solution set is [– 4, 2].

$$\textbf{4.\space-15}\lt\frac{\textbf{3(x-2)}}{\textbf{5}}\leq\textbf{0.}\\\textbf{Sol.}\space\text{Given,}\space-15\lt\frac{3(x-2)}{5}\leq0$$

On multiplying by 5 throughout

– 75 ≤ 3x – 6 ≤ 0

– 69 ≤ 3x ≤ 6

On dividing by 3, we get

– 23 < x ≤ 2

All real numbers x greater than – 23 but less than or equal to 2 are solution of given equality.

∴ The solution set is (– 23, 2].

$$\textbf{5. -12}\lt\textbf{4}-\frac{\textbf{3x}}{\textbf{-5}}\leq\textbf{2.}\\\textbf{Sol.}\space\text{Given}\space-12\lt4-\frac{3x}{(-5)}\leq2\\\Rarr\space-12\lt4+\frac{3x}{5}\leq2$$

Taking LCM of 1, 5 is 5.

$$\Rarr\space-12\lt\frac{20+3x}{5}\leq2$$

On multiplying by 5, we get

⇒ – 60 < 20 + 3x ≤ 10

Subtracting 20 on both sides

⇒ – 80 < 3x ≤ – 10

On dividing by 3, we get

$$\Rarr\space\frac{-80}{3}\lt x\lt\frac{-10}{3}\\\text{All real numbers x greater than}\space\frac{-80}{3}\\\text{but less than or equal to}\space\frac{-10}{3}\\\text{are solution of given equality.}\\\therefore\space\text{The solution set is}\bigg(\frac{-80}{3},\frac{-10}{3}\bigg].$$

$$\textbf{6. 7}\leq\frac{\textbf{3x+11}}{\textbf{2}}\leq\textbf{11.}$$

$$\textbf{Sol.}\space\text{Given,}\space 7\leq\frac{3x+11}{2}\leq11.$$

On multiplying by 2, we get

14 ≤ 3x + 11 ≤ 22

Subtracting 11 on both sides

3 ≤ 3x ≤ 11

On dividing by 3, we get

$$1\leq3\leq\frac{11}{3}$$

All real numbers x greater than or equal to 1 but less than or equal to $$\frac{11}{3}$$ are solution of given equality.

$$\therefore\space\text{The solution set is}\bigg[1,\frac{11}{3}\bigg].$$

Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line:

7. 5x + 1 > – 24, 5x – 1 < 24.

Sol. Given, 5x + 1 > – 24 …(i)

and 5x – 1 < 24 …(ii)

So in equation (i),

Subtracting 1 on both sides, we get

5x > – 25

On dividing by 5, we get

x > – 5

and in equation (ii),

adding 1 on both sides, we get

5x < 25

On dividing by 5, we get

x < 5

∴ The solution set is (– 5, 5).

It is clear, x should less than 5.

The graphical representation of the solution of the given inequality is as follow:

8. 2(x – 1) < x + 5, 3(x + 2) > 2 – x.

Sol. Given, inequalities

2(x – 1) < x + 5

and 3(x + 2) > 2 – x

Now, 2(x – 1) < x + 5

2x – 2 < x + 5

Adding 2 on both sides, we get

2x < x + 7

Subtract x on both sides, we get

x < 7 …(i)

and 3(x + 2) > 2 – x

3x + 6 > 2 – x

Subtract 6, on oth sides

3x > – 4 – x

On adding x, on both sides

4x > – 4

On dividing by 4, we get

x > – 1 …(ii)

It is clear x should greater than – 1.

From (i) and (ii), we get

The solution set is (– 1, 7).

The graphical representation of the solution of the given inequality is as follow:

9. 3x – 7 > 2(x – 6), 6 – x > 11 – 2x.

Sol. Now, 3x – 7 < 2(x – 6)

3x – 7 ≥ 2x – 12

Subtracting 2x on both sides,

x – 7 > – 12

On add 7 both sides, we get

x > – 5 …(i)

and 6 – x > 11 – 2x

Adding 2x on both sides, we get

6 + x > 11

Subtracting 6 from both sides, we get

x > 5 …(ii)

It is clear, x should greater than 5.

From (i) and (ii), we get

The solution set is (5, ∞).

The graphical representation of the solution of the given inequality is as follow:

10. 5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47.

Sol. Now, 5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

Adding 44 on both sides, we get

⇒ 4x ≤ 44

On dividing by 4, we get

x ≤ 11 …(i)

and

2x + 19 ≤ 6x + 47

Subtracting 19 on both sides, we get

2x ≤ 6x + 28

Subtracting 6x on both sides,

– 4x ≤ 28

Dividing by – 4 on both sides,

x ≥ – 7 …(ii)

It is clear, x should greater than or equal to – 7.
From (i) and (ii), we get

The solution set is [– 7, 11].

The graphical representation of the solution of the given inequality is as follow:

11. A solution is to be kept between 68°F and 77°F. What is the range in temperature in degree Celsius (C) if the Celsius/Fahrenheit (F)

conversion formula is given by

$$\textbf{F}=\frac{\textbf{9}}{\textbf{5}}\textbf{C\space+\space32 ?}$$

Sol. Given that,

68° < F < 77°

$$\text{since}\space\text{F}=\frac{9}{5}\text{C}+32\degree\\\Rarr\space68\degree\lt\frac{9}{5}\text{C}+32\degree\lt77\degree$$

Taking LCM of 1 and 5 is 5,

$$\Rarr\space68\degree\lt\frac{\text{9C+160\degree}}{5}\lt77\degree$$

On multiplying by 5, we get

⇒ 340° < 9 C + 160° < 385°

Subtracting 160° on both sides, we get

⇒ 180° < 9 C < 225°

On dividing by 9, we get

⇒ 20° < C < 25°

∴ The range of temperature lies between 20°C to 25°C.

12. 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?

Sol. Let x litres is the 2% of boric acid solution.

∴ the mixture = (640 + x) litres

According to the question,

2% of x + 8% of 640 > 4% of (640 + x)

and 2% of x + 8% of 640 < 6% of (640 + x)

$$\Rarr\space\frac{2}{100}×x+\frac{640×8}{100}\gt\frac{4}{100}(640+x)\\\text{and}\space\frac{2}{100}×x+\frac{8×640}{100}\lt\frac{6}{100}(640+x)$$

$$\Rarr\space\frac{2x+5120}{100}\gt\frac{2560+4x}{100}\\\text{and}\space\frac{2x+5120}{100}\lt\frac{3840+6x}{100}$$

⇒ 2x + 5120 = 2560 + 4x

and 2x + 5120 < 3840 + 6x

On transferring, we get

2560 > 2x …(i)

and 1280 < 4x …(ii)

On dividing by 2 in inequation (i) and 4 in inequation (ii), we get

1280 > x and 320 < x

rewrite it as

320 < x < 1280

Therefore the number of litres of 20% of boric acid should be added greater than 320 litres and less than 1280 litres.

13. 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?

Sol. Let x litres water added in 1125 litres of 45% solution of acid.

According to the question, 25% of (1125 + x) < 45% of 1125 < 30% of (1125 + x)

$$\Rarr\space\frac{25}{100}×(1125+x)\lt\frac{45}{100}×1125\\\lt\frac{30}{100}(1125+x)$$

$$\text{On multiplying by}\frac{100}{5},\text{we get}$$

⇒ 5(1125 + x) < 9 × 1125 < 6(1125 + x)

⇒ 5625 + 5x < 10,125 < 6750 + 6x

Taking first two inequality,

5625 + 5x < 10,125

On transferring, we get

5x < 4500

x < 900 …(i)

and another two inequality,

10125 < 6750 + 6x

On transferring, we get

3370 < 6x

⇒ 562.5 < x …(ii)

From (i) and (ii)

x ∈ (562.5, 900)

Hence, the number of litres of water should be added greater than 562.5 litres and less than 900 litres.

14. IQ of a person is given by the formula

$$\textbf{IQ}=\frac{\textbf{MA}}{\textbf{CA}}\textbf{×100}$$

where, MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age.

Sol. Given, 80 ≤ IQ ≤ 140

$$\text{Since,}\space\text{IQ}=\frac{\text{MA}}{\text{CA}}×100\\\text{We get}\\\Rarr\space80\leq\frac{\text{MA}}{\text{CA}}×100\leq140\\\Rarr\space80\leq\frac{\text{MA}}{12}×100\leq140\space\\(\because kA=12)$$

On transferring, we get

960 ≤ MA × 100 ≤ 1680

9.6 ≤ MA ≤ 16.80

Hence, Range of mental age should be greater than 9.6 but less than 16.8.