Class XI CBSE-Mathematics Linear Inequalities

Class–XI–CBSE-Mathematics Linear Inequalities

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CBSE NCERT Solutions for Class 11 Mathematics Chapter 06

Back of Chapter Questions

Exercise 6.1

1. Solve 24𝑥 < 100, when

(i) 𝑥 is a natural number

(ii) 𝑥 is an integer

Solution: (i) Step 1:

Given, 24𝑥 < 100

Dividing both sides by 24

⇒24𝑥

24<

100

24

⇒𝑥 <25

6

1, 2, 3 and 4 are the only natural numbers less than 25

6

Therefore, when 𝑥 is a natural number, the solutions of the given inequality are 1, 2, 3 and 4. Thus, the solution set is {1, 2, 3, 4}.

Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

(ii) Step 1:

We know that the integers less than 25

6 are: … . . −3, −2, −1, 0, 1, 2, 3 and 4.

Therefore, when 𝑥 is an integer, the solutions of the given inequality are:

… . −3, −2, −1, 0, 1, 2, 3 and 4.

Therefore, the solution set is {… − 3, −2, −1, 0, 1, 2, 3, 4}. Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

2. Solve – 12𝑥 > 30, when

(i) 𝑥 is a natural number



(ii) 𝑥 is an integer

Solution:

Given, −12𝑥 > 30

−12𝑥 > 30

Dividing both sides by −12

⇒−12𝑥

−12<

30

−12

⇒𝑥 < −5

2

(i) Step1:

There is no natural number less than −5

2

Therefore when 𝑥 is a natural number, there is no solution of the given inequality. Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

(ii) Step1:

The integers less than −5

2 are … − 5, −4, −3.

Therefore, when 𝑥 is an integer, the solutions of the given inequality are … − 5, −4, −3.

Thus, the solution set is {… − 5, −4, −3} Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

3. Solve 5𝑥– 3 < 7, when

(i) 𝑥 is an integer

(ii) 𝑥 is a real number

Solution:

Given, 5𝑥 − 3 < 7

5𝑥 − 3 < 7

Adding 3 to both the sides we have,

⇒5𝑥 − 3 + 3 < 7 + 3

⇒5𝑥 < 10



Dividing both sides by 5 we get,

⇒5𝑥

5<

10

5

⇒𝑥 < 2 (i) Step1:

The integers less than 2 are … − 3, −2, −1,0 and 1.

Therefore, when 𝑥 is an integer, the solutions of the given inequality are … − 3, −2, −1, 0 and 1.

Thus, the solution set is {… − 3, −2, −1, 0, 1}. Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

(ii) Step1:

When 𝑥 is a real number then the solution of the given by 𝑥 < 2

That is all the real numbers 𝑥 which are less than 2.

Thus, the solution set (−∞, 2). Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

4. Solve 3𝑥 + 8 > 2, when

(i) 𝑥 is an integer

(ii) 𝑥 is a real number

Solution:

Given, 3𝑥 + 8 > 2

Subtracting 8 both the sides.

⇒3𝑥 + 8 − 8 > 2 − 8

⇒3𝑥 > −6

Dividing both sides by 3

⇒3𝑥

3>

−6

3

⇒𝑥 > −2 (i) Step1:

The integers greater than −2 are −1, 0, 1, 2, 3 …

Therefore, when 𝑥 is an integer, the solutions of the given inequality are −1, 0, 1, 2, 3 …



The solution set is given by {−1, 0, 1, 2, 3 … }. Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

(ii) Step1:

When 𝑥 is a real number then the solution of the given inequality is all the real numbers greater than −2.

The solution set is given by (−2, ∞). Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

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

Solution: Step1:

Given, 4𝑥 + 3 < 5𝑥 + 7

Subtracting 7 from both the sides.

⇒4𝑥 + 3 − 7 < 5𝑥 + 7 − 7

⇒4𝑥 − 4 < 5𝑥

Step2:

Subtracting 4𝑥 both the sides we have,

⇒4𝑥 − 4 − 4𝑥 < 5𝑥 − 4𝑥

⇒−4 < 𝑥

or 𝑥 > −4

Therefore, the solution of the given inequality is all the real numbers greater than −4.

Thus, the solution set of the given inequality is (−4, ∞) Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

6. Solve the given inequality for real 𝑥: 3𝑥 – 7 > 5𝑥 – 1



Solution: Step1:

Given, 3𝑥 − 7 > 5𝑥 − 1

Adding 1 both the sides we have,

⇒3𝑥 − 7 + 1 > 5𝑥 − 1 + 1

⇒3𝑥 − 6 > 5𝑥

Step2:

Subtracting 3𝑥 both the sides.

⇒3𝑥 − 6 − 3𝑥 > 5𝑥 − 3𝑥

⇒−6 > 2𝑥

⇒−6

2>

2𝑥

2

⇒ 𝑥 < −3

Therefore, the solution of the given inequality is all the real numbers less than−3.

Thus, the solution set of the given inequality is (−∞, −3) Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

7. Solve the given inequality for real 𝑥:

3(𝑥 – 1) ≤ 2 (𝑥 – 3)

Solution: Step1:

Given, 3(𝑥 − 1) ≤ 2(𝑥 − 3)

⇒3𝑥 − 3 ≤ 2𝑥 − 6

⇒3𝑥 − 3 + 3 ≤ 2𝑥 − 6 + 3

⇒3𝑥 ≤ 2𝑥 − 3

⇒3𝑥 − 2𝑥 ≤ −3

⇒ 𝑥 ≤ −3

Therefore, the solution of the given inequality is all the real numbers less than and equal to−3.

Thus, the solution set of the given inequality is (−∞, −3]





3(2 – 𝑥) ≥ 2(1 – 𝑥)

Solution: Step1:

Given, 3(2 − 𝑥) ≥ 2(1 − 𝑥)

⇒6 − 3𝑥 ≥ 2 − 2𝑥

Adding 3𝑥 both the sides we have,

⇒6 − 3𝑥 + 3𝑥 ≥ 2 − 2𝑥 + 3𝑥

⇒6 ≥ 2 + 𝑥

⇒6 − 2 ≥ 2 + 𝑥 − 2

⇒ 4 ≥ 𝑥

or 𝑥 ≤ 4

Therefore, the solution of the given inequality is all the real numbers less than and equal to 4.

Thus, the solution set of the given inequality is (−∞, 4] Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.


𝑥 +𝑥

2+

𝑥

3< 11

Solution:

Step1:

Given, 𝑥 +𝑥

2+

𝑥

3< 11

⇒6𝑥 + 3𝑥 + 2𝑥

6< 11

⇒11𝑥

6< 11



⇒6

11×

11𝑥

6<

6

11× 11

⇒ 𝑥 < 6

Therefore, the solution of the given inequality is all the real numbers less than 6.

Thus, the solution set of the given inequality is (−∞, 6). Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.


𝑥

3>

𝑥

2+ 1

Solution:

Step1:

Given, 𝑥

3>

𝑥

2+ 1

⇒𝑥

3−

𝑥

3>

𝑥

2−

𝑥

3+ 1

⇒ 0 > 3𝑥−2𝑥

6+ 1

⇒ 0 >𝑥

6+ 1

⇒ 0 − 1 >𝑥

6+ 1 − 1

⇒ −1 >𝑥

6

Step2:

⇒ −1 × 6 >𝑥

6× 6

⇒ −6 > 𝑥

or 𝑥 < −6

Therefore, the solution of the given inequality is all the real numbers less than −6.

Thus, the solution set of the given inequality is (−∞, −6). Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.




3(𝑥 − 2)

5≤

5(2 − 𝑥)

3

Solution:

Step1:

Given, 3(𝑥−2)

5≤

5(2−𝑥)

3

⇒ 9(𝑥 − 2) ≤ 25(2 − 𝑥)

⇒ 9𝑥 − 18 ≤ 50 − 25𝑥

⇒ 9𝑥 − 18 + 25𝑥 ≤ 50 − 25𝑥 + 25𝑥

⇒ 34𝑥 − 18 ≤ 50

Step2:

⇒ 34𝑥 − 18 + 18 ≤ 50 + 18

⇒ 34𝑥 ≤ 68

⇒34𝑥

34≤

68

34

⇒ 𝑥 ≤ 2


Thus, the solution set of the given inequality is (−∞, 2]. Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.


1

2(

3𝑥

5+ 4) ≥

1

3(𝑥 − 6)

Solution:

Step1:

Given, 1

2(

3𝑥

5+ 4) ≥

1

3(𝑥 − 6)

⇒ 3 (3𝑥

5+ 4) ≥ 2(𝑥 − 6)



⇒9𝑥

5+ 12 ≥ 2𝑥 − 12

⇒9𝑥

5+ 12 −

9𝑥

5≥ 2𝑥 − 12 −

9𝑥

5

⇒ 12 ≥ 2𝑥 −9𝑥

5− 12

⇒ 12 ≥10𝑥−9𝑥

5− 12

⇒ 12 ≥𝑥

5− 12

Step2:

⇒ 12 + 12 ≥𝑥

5− 12 + 12

⇒ 24 ≥𝑥

5

⇒ 24 × 5 ≥ 𝑥

⇒ 120 ≥ 𝑥

or 𝑥 ≤ 120




2(2 𝑥 + 3) − 10 < 6(𝑥 − 2)

Solution:

Step1:

Given, 2(2𝑥 + 3) − 10 < 6(𝑥 − 2)

⇒ 4𝑥 + 6 − 10 < 6𝑥 − 12

⇒ 4𝑥 − 4 < 6𝑥 − 12

⇒ 4𝑥 − 4 − 4𝑥 < 6𝑥 − 12 − 4𝑥

⇒ −4 < 2𝑥 − 12

⇒ −4 + 12 < 2𝑥 − 12 + 12



Step2:

⇒ 8 < 2𝑥

⇒8

2<

2𝑥

2

⇒ 4 < 𝑥

Or, 𝑥 > 4

Therefore, the solution of the given inequality is all the real numbers greater than 4.

Thus, the solution set of the given inequality is (7, ∞). Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.

14. Solve the given inequality for real 𝑥: 37 − (3𝑥 + 5) ≥ 9𝑥 − 8(𝑥 − 3)

Solution:

Step1:

Given, 37 − (3𝑥 + 5) ≥ 9𝑥 − 8(𝑥 − 3)

⇒ 37 − 3𝑥 − 5 ≥ 9𝑥 − 8𝑥 + 24

⇒ 32 − 3𝑥 ≥ 𝑥 + 24

⇒ 32 − 3𝑥 + 3𝑥 ≥ 𝑥 + 24 + 3𝑥

⇒ 32 ≥ 4𝑥 + 24

Step2:

⇒ 32 − 24 ≥ 4𝑥 + 24 − 24

⇒ 8 ≥ 4𝑥

⇒8

4≥

4𝑥

4

⇒ 2 ≥ 𝑥

Or, 𝑥 ≤ 2






𝑥

4<

(5𝑥 − 2)

3−

(7𝑥 − 3)

5

Solution:

Step1:

Given, 𝑥

4<

(5𝑥−2)

3−

(7𝑥−3)

5

⇒𝑥

4<

5(5𝑥−2)−3(7𝑥−3)

15

⇒𝑥

4<

25𝑥−10−21𝑥+9

15

⇒𝑥

4<

4𝑥−1

15

⇒ 15𝑥 < 4(4𝑥 − 1)

⇒ 15𝑥 < 16𝑥 − 4

⇒ 15𝑥 − 15𝑥 < 16𝑥 − 4 − 15𝑥

⇒ 0 < 𝑥 − 4

Step2:

⇒ 0 + 4 < 𝑥 − 4 + 4

⇒ 4 < 𝑥

Or, 𝑥 > 4

Therefore, the solution of the given inequality is all the real numbers greater than 4.

Thus, the solution set of the given inequality is (4, ∞). Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.


(2𝑥 − 1)

3≥

(3𝑥 − 2)

4−

(2 − 𝑥)

5

Solution:



Step1:

Given, (2𝑥−1)

3≥

(3𝑥−2)

4−

(2−𝑥)

5

⇒(2𝑥−1)

3≥

5(3𝑥−2)−4(2−𝑥)

20

⇒(2𝑥−1)

3≥

15𝑥−10−8+4𝑥

20

⇒(2𝑥−1)

3≥

19𝑥−18

20

⇒ 20(2𝑥 − 1) ≥ 3(19𝑥 − 18)

⇒ 40𝑥 − 20 ≥ 57𝑥 − 54

⇒ 40𝑥 − 20 − 40𝑥 ≥ 57𝑥 − 54 − 40𝑥

Step2:

⇒ −20 ≥ 17𝑥 − 54

⇒ −20 + 54 ≥ 17𝑥 − 54 + 54

⇒ 34 ≥ 17𝑥

⇒ 34

17≥

17𝑥

17

⇒2 ≥ 𝑥

Or, 𝑥 ≤ 2

Therefore, the solution of the given inequality is all the real numbers less than or equal to 2.


17. Solve the given inequality and show the graph of the solution on number line: 3𝑥 − 2 < 2𝑥 + 1

Solution:

Step1:

Given, 3𝑥 − 2 < 2𝑥 + 1

⇒ 3𝑥 − 2 − 2𝑥 < 2𝑥 + 1 − 2𝑥

⇒ 𝑥 − 2 < 1

⇒ 𝑥 − 2 + 2 < 1 + 2



⇒ 𝑥 < 3

Step2:

Graphical representation on the number line is shown as:


18. Solve the given inequality and show the graph of the solution on number line:

5𝑥 − 3 ≥ 3𝑥 − 5

Solution:

Step1:

Given, 5𝑥 − 3 ≥ 3𝑥 − 5

⇒ 5𝑥 − 3 − 3𝑥 ≥ 3𝑥 − 5 − 3𝑥

⇒ 2𝑥 − 3 ≥ −5

⇒ 2𝑥 − 3 + 3 ≥ −5 + 3

⇒ 2𝑥 ≥ −2

⇒2𝑥

2≥

−2

2

⇒ 𝑥 ≥ −1

Step2:


Overall Hint: Convert the given inequality into literal inequality and solve for 𝑥.and use the concepts of number line representation of the solutions.




3(1 − 𝑥) < 2(𝑥 + 4)

Solution:

Step1:

Given inequality is 3(1 − 𝑥) < 2(𝑥 + 4)

⇒ 3 − 3𝑥 < 2𝑥 + 8

⇒ 3 − 3𝑥 + 3𝑥 < 2𝑥 + 8 + 3𝑥

⇒ 3 < 5𝑥 + 8

⇒ 3 − 8 < 5𝑥 + 8 − 8

⇒ −5 < 5𝑥

Or, 5𝑥 > −5

⇒5𝑥

5>

−5

5

⇒ 𝑥 > −1

Step2:




𝑥

2≥

(5𝑥−2)

3−

(7𝑥−3)

5

Solution:

Step1:

Given inequality is, 𝑥

2≥

(5𝑥−2)

3−

(7𝑥−3)

5



⇒𝑥

2≥

5(5𝑥 − 2) − 3(7𝑥 − 3)

15

⇒𝑥

2≥

25𝑥 − 10 − 21𝑥 + 9

15

⇒𝑥

2≥

4𝑥 − 1

15

⇒ 15𝑥 ≥ 2(4𝑥 − 1)

⇒ 15𝑥 ≥ 8𝑥 − 2

⇒ 15𝑥 − 8𝑥 ≥ 8𝑥 − 2 − 8𝑥

⇒ 7𝑥 ≥ −2

⇒ 𝑥 ≥ −2

7

Step2:



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

Solution:

Step1:

Consider the marks obtained by Ravi in third unit test be 𝑥.

Given that the average marks of three unit test should be at least 60.

So,70+75+𝑥

3≥ 60

⇒ 145 + 𝑥 ≥ 180

⇒ 145 + 𝑥 − 145 ≥ 180 − 145

⇒ 𝑥 ≥ 35



Therefore, the minimum marks obtained by the student to have an average of 60 marks is 35. Overall Hint: Using the given data form an inequality and convert the given inequality into literal inequality and solve for 𝑥..

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.

Solution:

Step1:

Consider the marks obtained by Sunita in the fifth examination be 𝑥.

Given, Average marks of five examinations should be at least 90 in order to receive grade A in the course.

So, 87+92+94+95+𝑥

5≥ 90

⇒ 368 + 𝑥 ≥ 450

⇒ 368 + 𝑥 − 368 ≥ 450 − 368

⇒ 𝑥 ≥ 82

Therefore, the Sunita must obtain marks greater than or equal to 82 marks in the fifth examination. Overall Hint: Using the given data form an inequality and convert the given inequality into literal inequality and solve for 𝑥..

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

Solution:

Step1:

Consider the two consecutive odd positive integers be 𝑥 and 𝑥 + 2, where 𝑥 is the smaller of the two consecutive odd positive integers.

According to question we have.

𝑥 + 2 < 10

⇒ 𝑥 < 10 − 2



⇒ 𝑥 < 8………(1)

Also, the sum of the two integers is more than 11.

⸫ 𝑥 + (𝑥 + 2) > 11

⇒ 2𝑥 + 2 > 11

⇒ 2𝑥 + 2 − 2 > 11 − 2

⇒ 2𝑥 > 9

Step2:

⇒2𝑥

2>

9

2

⇒ 𝑥 >9

2…………..(2)

⇒ 𝑥 > 4.5

From equation (1) and (2), the possible values of 𝑥 are 5 and 7.

Therefore, the required possible pairs are (5, 7) and (7, 9). Overall Hint: Using the given data form an inequality and convert the given inequality into literal inequality and solve for 𝑥..

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

Solution:

Step1:

Consider the two consecutive even positive integers be 𝑥 and 𝑥 + 2, where 𝑥 is the smaller of the two consecutive even positive integers.

According to question we have,

𝑥 > 5………………(1)

Also, the sum of the two integers is less than 23.

⸫ 𝑥 + (𝑥 + 2) < 23

⇒ 2𝑥 + 2 < 23

⇒ 2𝑥 + 2 − 2 < 23 − 2

⇒ 2𝑥 < 21



Step2:

⇒2𝑥

2<

21

2

⇒ 𝑥 <21

2 …………..(2)

⇒ 𝑥 < 10.5

From equation (1) and (2), we get 5 < 𝑥 < 10.5. The possible values of 𝑥 are 6, 8 and 10.

Therefore, the required possible pairs are (6, 8), (8, 10) and (10, 12). Overall Hint: Using the given data form an inequality and convert the given inequality into literal inequality and solve for 𝑥..

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.

Solution:

Step1:

Consider the length of the shortest side of the triangle be 𝑥 cm

So, length of the longest side = 3𝑥 cm

Length of the third side= 3𝑥 − 2 cm

Given, the perimeter of the triangle is at least 61 cm.

Perimeter = (𝑥 + 3𝑥 + 3𝑥 − 2) cm = 7𝑥 − 2 cm

According to question,

7𝑥 − 2 ≥ 61

⇒ 7𝑥 ≥ 61 + 2

⇒ 7𝑥 ≥ 63

⇒ 𝑥 ≥ 9

Therefore, the minimum length of the shortest side is 9 cm. Overall Hint: Using the given data form an inequality and convert the given inequality into literal inequality and solve for 𝑥..



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 the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 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 ≤ 91and 2𝑥 ≥ (𝑥 + 3) + 5]

Solution:

Step1:

Consider the length of the shortest piece be 𝑥 cm.

And, the length of the second piece and third piece be (𝑥 + 3)cm and 2𝑥 cm respectively.

Since the three lengths are to be cut from a single piece of board of length 91 cm.

So, 𝑥 + (𝑥 + 3) + 2𝑥 ≤ 91

⇒ 4𝑥 + 3 ≤ 91

⇒ 4𝑥 ≤ 88

⇒ 𝑥 ≤ 22………………(1)

Step2:

And, the third piece is at least 5 cm longer than the second piece.

⸫2𝑥 ≥ (𝑥 + 3) + 5

⇒ 2𝑥 ≥ 𝑥 + 8

⇒ 2𝑥 − 𝑥 ≥ 8

⇒ 𝑥 ≥ 8 ………………..(2)

From (1) and (2)

8 ≤ 𝑥 ≤ 22

Therefore, the possible length of the shortest board is greater than or equal to 8 and smaller than or equal to 22. Overall Hint: Using the given data form an inequality and convert the given inequality into literal inequality and solve for 𝑥..

Exercise 6.2

1. Solve the given inequality graphically in two-dimensional plane:

𝑥 + 𝑦 < 5



Solution:

Step1:

Graphical representation of 𝑥 + 𝑦 = 5 is given as dotted line in the figure below.

That dotted line divides the 𝑥𝑦-plane in two half planes, I and II.

Pick a point, which doesn’t lie in one of the half planes but not on the line, to determine whether the point satisfies the given inequality or not.

Let’s select the point as (0, 0).

we know that,

0 + 0 < 5 or 0 < 5 which is true.

So, half plane II is not the solution region of the given inequality. Also, any point on the line does not satisfy the given strict inequality.

Therefore, the shaded half plane I excluding the points on the line is the solution region of the given inequality.

Step2:

Graphical representation can be given as follows:

Overall Hint: Use the concept of Graphical solution of linear inequalities in two variables:

The graph of the inequality ax + by > c will be one of the half plane (called solution region) and represented by shading in the corresponding half plane.




2𝑥 + 𝑦 ≥ 6

Solution:

Step1:

Graphical representation of 2𝑥 + 𝑦 = 6 is given as dotted line in the figure below.

The dotted line divides the 𝑥𝑦-plane in two half planes, I and II.

Pick a point, which lies in one of the half planes but not on the line, to determine whether the point satisfies the given inequality or not.


We know that,

2 × 0 + 0 ≥6 or 0 ≥ 6 which is false.

Hence, half plane I is not the solution region of the given inequality. Also, any point on the line satisfy the given inequality.

Therefore, the shaded half plane II including the points on the line is the solution region of the given inequality.

Step2:







3𝑥 + 4𝑦 ≤ 12

Solution:

Step1:

Graphical representation of 3𝑥 + 4𝑦 = 12 is given as dotted line in the figure below.

The dotted line divides the 𝑥𝑦-plane in two half planes, I and II.



We know that,

3 × 0 + 4 × 0 ≤12 or 0 ≤ 12 which is true.

Hence, half plane II is not the solution region of the given inequality. Also, any point on the line satisfy the given inequality.

Therefore, the shaded half plane I including the points on the line is the solution region of the given inequality.

Step2:







𝑦 + 8 ≥ 2𝑥

Solution:

Step1:

Graphical representation of 𝑦 + 8 = 2𝑥 is given as dotted line in the figure below.

The dotted line divides the 𝑥𝑦-plane in two half planes.


We select the point as (0, 0).

We know that,

0 + 8 ≥ 2(0) or 8 ≥ 0 which is true.

Hence, the lower half plane is not the solution region of the given inequality. Also, any point on the line satisfy the given inequality.

Therefore, the shaded half plane containing (0, 0) including the line is the solution region of the given inequality.

Step2:







𝑥– 𝑦 ≤ 2

Solution:

Step1:

Graphical representation of 𝑥 − 𝑦 = 2 is given as dotted line in the figure below.




We know that,

0 − 0 ≤2 or 0 ≤ 2 which is true.


Therefore, the shaded half plane containing (0, 0) including the line is the solution region of the given inequality.



Step2:





2𝑥 − 3𝑦 > 6

Solution:

Step1:

Graphical representation of 2𝑥 − 3𝑦 = 6 is given as dotted line in the figure below.




We know that,

2(0) − 3(0) >6 or 0 > 6 which is true.

Hence, the upper half plane is not the solution region of the given inequality. Also, any point on the line does not satisfy the given inequality.



Therefore, the shaded half plane which does not contain (0, 0) including the line is the solution region of the given inequality.

Step2:





−3𝑥 + 2𝑦 ≥ −6

Solution:

Step1:

Graphical representation of −3𝑥 + 2𝑦 = −6 is given as dotted line in the figure below.

The dotted line divides the xy-plane in two half planes.



We know that,



−3(0) + 2(0) ≥ −6 or 0 ≥ −6 which is true.


Therefore, the shaded half plane containing the point (0, 0) including the line is the solution region of the given inequality.

Step2:





3𝑦 − 5𝑥 < 30

Solution:

Step1:

Graphical representation of 3𝑦 − 5𝑥 = 30 is given as dotted line in the figure below.






We know that,

3(0) − 5(0) ≤ 30 or 0 ≤ 30 which is true.


Therefore, the shaded half plane containing the point (0, 0) excluding the line is the solution region of the given inequality.

Step2:





𝑦 < – 2

Solution:

Step1:



Graphical representation of 𝑦 = −2 is given as dotted line in the figure below.

The dotted line divides the xy-plane in two half planes.



We know that,

0 < −2, which is false.


Therefore, every point below the line 𝑦 = −2, excluding the line is the solution region of the given inequality.

Step2:







𝑥 > – 3

Solution:

Step1:

Graphical representation of 𝑥 = −3 is given as dotted line in the figure below.




We know that,

0 > −3, which is true.

And, any point on the line does not satisfy the given inequality.

Therefore, every point on the right hand side of the line 𝑥 = −3, excluding the line is the solution region of the given inequality.

Step2:






Exercise 6.3

1. Solve the following system of inequalities graphically:

𝑥 ≥ 3, 𝑦 ≥ 2.

Solution:

Step1:

Given inequalities are:

𝑥 ≥ 3………..(1)

𝑦 ≥ 2………..(2)

The required lines of 𝑥 = 3 and 𝑦 = 2 are drawn in the figure below.

Inequality (1) represents the region on the right hand side of the line, 𝑥 = 3, including the line 𝑥 =3, and inequality (2) represents the region above the line 𝑦 = 2, including the line 𝑦 = 2.

Therefore, the solution of the given system of linear in equations is represented by the common shaded region including the points on the respective lines as follows

Step2:





2. Solve the following system of inequalities graphically: 3𝑥 + 2𝑦 ≤ 12, 𝑥 ≥ 1, 𝑦 ≥ 2

Solution:

Step1:


3𝑥 + 2𝑦 ≤ 12 ……..(1)

𝑥 ≥ 1………..(2)

𝑦 ≥ 2 ………..(3)

The required lines of 3𝑥 + 2𝑦 = 12, 𝑥 = 3 and 𝑦 = 2 are drawn in the figure below.

Inequality (1) represents the region below the line 3𝑥 + 2𝑦 = 12, including the line 3𝑥 + 2𝑦 = 12, inequality (2) represents the region on the right hand side of the line, 𝑥 = 1, including the line 𝑥 =1, and inequality (2) represents the region above the line 𝑦 = 2, including the line 𝑦 = 2.

Therefore, the solution of the given system of linear in equations is represented by the common shaded region including the points on the respective lines as follows





3. Solve the following system of inequalities graphically: 2𝑥 + 𝑦 ≥ 6, 3𝑥 + 4𝑦 ≤ 12

Solution:

Step1:


2𝑥 + 𝑦 ≥ 6 …………(1)

3𝑥 + 4𝑦 ≤ 12 ……..(2)

The required lines of 2𝑥 + 𝑦 = 6 and 3𝑥 + 4𝑦 = 12 are drawn in the figure below.

Inequality (1) represents the region above the line 2𝑥 + 𝑦 = 6, including the line 2𝑥 + 𝑦 = 6, inequality (2) represents the region below the line 3𝑥 + 4𝑦 = 12, including the line 3𝑥 + 4𝑦 = 12.

Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows



Step2:



4. Solve the following system of inequalities graphically 𝑥 + 𝑦 ≥ 4, 2𝑥 − 𝑦 < 0

Solution:

Step1:


𝑥 + 𝑦 ≥ 4 …………(1)

2𝑥 − 𝑦 < 0 ……..(2)

The required lines of 𝑥 + 𝑦 = 4 and 2𝑥 − 𝑦 = 0 are drawn in the figure below.

Inequality (1) represents the region above the line 𝑥 + 𝑦 = 4, including the line 𝑥 + 𝑦 = 4.It is observed that (1, 3) satisfies the inequality (1)

Inequality (2) represents the half plane corresponding to the line, 2𝑥 − 𝑦 = 0, excluding the line 2𝑥 − 𝑦 = 0.



Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the line 𝑥 + 𝑦 = 4 and excluding the points on the line 2𝑥 −𝑦 = 0 as follows

Step2:



5. Solve the following system of inequalities graphically: 2𝑥 – 𝑦 > 1, 𝑥 – 2𝑦 < – 1

Solution:

Step1:


2𝑥 − 𝑦 > 1…………(1)

𝑥 − 2𝑦 < −1……..(2)

The required lines of 2𝑥 − 𝑦 = 1 and 𝑥 − 2𝑦 = −1 are drawn in the figure below.



Inequality (1) represents the region below the line 2𝑥 − 𝑦 = 1, excluding the line 2𝑥 − 𝑦 = 1, and inequality (2) represents the region above the line, 𝑥 − 2𝑦 = −1, excluding the line 𝑥 − 2𝑦 = −1.

Therefore, the solution of the given system of linear inequalities is represented by the common shaded region excluding the points on the respective lines as follows

Step2:



6. Solve the following system of inequalities graphically: 𝑥 + 𝑦 ≤ 6, 𝑥 + 𝑦 ≥ 4

Solution:

Step1:


𝑥 + 𝑦 ≤ 6 …………(1)

𝑥 + 𝑦 ≥ 4 ………...(2)

The required lines of 𝑥 + 𝑦 = 6 and 𝑥 + 𝑦 = 4 are drawn in the figure below.



Inequality (1) represents the region below the line 𝑥 + 𝑦 = 6, including the line 𝑥 + 𝑦 = 6, and inequality (2) represents the region above the line, 𝑥 + 𝑦 = 4, including the line 𝑥 + 𝑦 = 4.


Step2:



7. Solve the following system of inequalities graphically: 2𝑥 + 𝑦 ≥ 8, 𝑥 + 2𝑦 ≥ 10

Solution:

Step1:


2𝑥 + 𝑦 ≥ 8 …………(1)

𝑥 + 2𝑦 ≥ 10 ………...(2)



The required lines of 2𝑥 + 𝑦 = 8 and 𝑥 + 2𝑦 = 10 are drawn in the figure below.

Inequality (1) represents the region above the line 2𝑥 + 𝑦 = 8, including the line 2𝑥 + 𝑦 = 8, and inequality (2) represents the region above the line, 𝑥 + 2𝑦 = 10, including the line 𝑥 + 2𝑦 = 10.


Step2:



8. Solve the following system of inequalities graphically: 𝑥 + 𝑦 ≤ 9, 𝑦 > 𝑥, 𝑥 ≥ 0

Solution:

Step1:


𝑥 + 𝑦 ≤ 9 …………(1)



𝑦 > 𝑥 ………………..(2)

𝑥 ≥ 0 …..…………..(3)

The required lines of 𝑥 + 𝑦 = 9 and 𝑦 = 𝑥 are drawn in the figure below.

Inequality (1) represents the region below the line 𝑥 + 𝑦 = 9, including the line 𝑥 + 𝑦 = 9.It is observed that (0, 1) satisfies the inequality (2) [1 > 0]

Inequality (2) represents the half plane corresponding to the line, 𝑦 = 𝑥, containing the point (0, 1) excluding the line 𝑦 = 𝑥. Inequality (3) represents the region on the right hand side of the line, 𝑥 =0 or 𝑦-axis including 𝑦-axis.

Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the line 𝑥 + 𝑦 = 9 and 𝑥 = 0, and excluding the points on the line 𝑦 = 𝑥 as follows

Step2:



9. Solve the following system of inequalities graphically: 5𝑥 + 4𝑦 ≤ 20, 𝑥 ≥ 1, 𝑦 ≥ 2



Solution:

Step1:


5𝑥 + 4𝑦 ≤ 20 …………(1)

𝑥 ≥ 1………………..(2)

𝑦 ≥ 2…..…………..(3)

The required lines of 5𝑥 + 4𝑦 = 20, 𝑥 = 1 and 𝑦 = 2 are drawn in the figure below.

Inequality (1) represents the region below the line 5𝑥 + 4𝑦 = 20, including the line 5𝑥 + 4𝑦 = 20. Inequality (2) represents the region on the right hand side of the line 𝑥 = 1, including the line 𝑥 = 1. Inequality (3) represents the region above the line 𝑦 = 2, including the line 𝑦 = 2.


Step2:





10. Solve the following system of inequalities graphically: 3𝑥 + 4𝑦 ≤ 60, 𝑥 + 3𝑦 ≤ 30, 𝑥 ≥ 0, 𝑦 ≥ 0 Solution:

Step1:


3𝑥 + 4𝑦 ≤ 60 …………(1)

𝑥 + 3𝑦 ≤ 30 …………(2)

The required lines of 3𝑥 + 4𝑦 = 60, and 𝑥 + 3𝑦 = 30 are drawn in the figure below.

Inequality (1) represents the region below the line 3𝑥 + 4𝑦 = 60, including the line 3𝑥 + 4𝑦 = 60. Inequality (2) represents the region below the line 𝑥 + 3𝑦 = 30, including the line 𝑥 + 3𝑦 = 30.

AS, 𝑥 ≥ 0 and 𝑦 ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities.

Step2:





11. Solve the following system of inequalities graphically: 2𝑥 + 𝑦 ≥ 4, 𝑥 + 𝑦 ≤ 3, 2𝑥 – 3𝑦 ≤ 6 Solution:

Step1:


2𝑥 + 𝑦 ≥ 4…………(1)

𝑥 + 𝑦 ≤ 3…………(2)

2𝑥 − 3𝑦 ≤ 6…………(3)

The required lines of 2𝑥 + 𝑦 = 4, 𝑥 + 𝑦 = 3 and 2𝑥 − 3𝑦 = 6 are drawn in the figure below.

Inequality (1) represents the region above the line 2𝑥 + 𝑦 = 4, including the line 2𝑥 + 𝑦 = 4. Inequality (2) represents the region below the line 𝑥 + 𝑦 = 3, including the line 𝑥 + 𝑦 = 3. Inequality (3) represents the region above the line 2𝑥 − 3𝑦 = 6, including the line 2𝑥 − 3𝑦 = 6.


Step2:





12. Solve the following system of inequalities graphically: 𝑥 – 2𝑦 ≤ 3, 3𝑥 + 4𝑦 ≥ 12, 𝑥 ≥ 0, 𝑦 ≥ 1

Solution:

Step1:


𝑥 − 2𝑦 ≤ 3 …………(1)

3𝑥 + 4𝑦 ≥ 12 …………(2)

𝑦 ≥ 1 …………(3)

𝑥 ≥ 0……(4)

The required lines of 𝑥 − 2𝑦 = 3, 3𝑥 + 4𝑦 = 12 and 𝑦 = 1 are drawn in the figure below.

Inequality (1) represents the region above the line 𝑥 − 2𝑦 = 3, including the line 𝑥 − 2𝑦 = 3. Inequality (2) represents the region above the line 3𝑥 + 4𝑦 = 12, including the line 3𝑥 + 4𝑦 = 12. Inequality (3) represents the region above the line 𝑦 = 1, including the 𝑙𝑖𝑛𝑒 𝑦 = 1. The inequality, 𝑥 ≥ 0, represents the region on the right hand side of y-axis, including y-axis.

Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines and y-axis as follows.

Step2:





13. Solve the following system of inequalities graphically: 4𝑥 + 3𝑦 ≤ 60, 𝑦 ≥ 2𝑥, 𝑥 ≥ 3, 𝑥, 𝑦 ≥ 0

Solution:

Step1:


4𝑥 + 3𝑦 ≤ 60 …………(1)

𝑦 ≥ 2𝑥 …………(2)

𝑥 ≥ 3 …………(3)

The required lines of 4𝑥 + 3𝑦 = 60, 𝑦 = 2𝑥 and 𝑥 = 3 are drawn in the figure below.

Inequality (1) represents the region below the line 4𝑥 + 3𝑦 = 60, including the line 4𝑥 + 3𝑦 = 60. Inequality (2) represents the region above the line 𝑦 = 2𝑥, including the line 𝑦 = 2𝑥. The inequality (3) represents the region on the right hand side of the line 𝑥 = 3, including the line 𝑥 = 3.

Therefore, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.

Step2:





14. Solve the following system of inequalities graphically: 3𝑥 + 2𝑦 ≤ 150, 𝑥 + 4𝑦 ≤ 80, 𝑥 ≤ 15, 𝑦 ≥ 0, 𝑥 ≥ 0

Solution:

Step1:


3𝑥 + 2𝑦 ≤ 150 …………(1)

𝑥 + 4𝑦 ≤ 80 …………(2)

𝑥 ≤ 15 …………(3)

The required lines of 3𝑥 + 2𝑦 = 150, 𝑥 + 4𝑦 = 80 and 𝑥 = 15 are drawn in the figure below.

Inequality (1) represents the region below the line 3𝑥 + 2𝑦 = 150, including the line 3𝑥 + 2𝑦 =150. Inequality (2) represents the region below the line 𝑥 + 4𝑦 = 80, including the line 𝑥 + 4𝑦 =80. The inequality (3) represents the region on the left hand side of the line 𝑥 = 15, including the line 𝑥 = 15.

As 𝑥 ≥ 0 and 𝑦 ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities.

Step2:





15. Solve the following system of inequalities graphically: 𝑥 + 2𝑦 ≤ 10, 𝑥 + 𝑦 ≥ 1, 𝑥 – 𝑦 ≤ 0, 𝑥 ≥ 0, 𝑦 ≥ 0

Solution:

Step1:


𝑥 + 2𝑦 ≤ 10 …………(1)

𝑥 + 𝑦 ≥ 1 …………(2)

𝑥 − 𝑦 ≤ 0 …………(3)

The required lines of 𝑥 + 2𝑦 = 10, 𝑥 + 𝑦 = 1 and 𝑥 − 𝑦 = 0 are drawn in the figure below.

Inequality (1) represents the region below the line 𝑥 + 2𝑦 = 10, including the line 𝑥 + 2𝑦 = 10. Inequality (2) represents the region above the line 𝑥 + 𝑦 = 1, including the line 𝑥 + 𝑦 =1.Inequality (3) represents the region above the line 𝑥 − 𝑦 = 0, including the line 𝑥 − 𝑦 = 0.

As 𝑥 ≥ 0 and 𝑦 ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities.



Step2:



Miscellaneous Exercise on Chapter 6

1. Solve the inequality 2 ≤ 3𝑥 – 4 ≤ 5.

Solution: Step1:

Given inequality is, 2 ≤ 3𝑥 − 4 ≤ 5

⇒ 2 + 4 ≤ 3𝑥 − 4 + 4 ≤ 5 + 4 [adding 4 both]

⇒ 6 ≤ 3𝑥 ≤ 9

⇒6

3≤

3𝑥

3≤

9

3

⇒ 2 ≤ 𝑥 ≤ 3

Therefore, the solution set of given inequality is [2, 3]. Overall Hint: Solve the given inequality simultaneously and find the range of the 𝑥.



2. Solve the inequality 6 ≤ – 3(2𝑥 – 4) < 12

Solution: Step1:

Given inequality is, 6 ≤ −3(2𝑥 − 4) < 12

⇒ 6 ≤ −6𝑥 + 12 < 12

⇒ 6 − 12 ≤ −6𝑥 + 12 − 12 < 12 − 12 [Subtracting 12]

⇒ −6 ≤ −6𝑥 < 0

⇒−6

−6≥

−6𝑥

−6>

0

−6 [dividing by −6]

⇒ 1 ≥ 𝑥 > 0

Or, 0 < 𝑥 ≤ 1

Therefore, the solution set of given inequality is (0, 1]. Overall Hint: Solve the given inequality simultaneously and find the range of the 𝑥.

3. Solve the inequality −3 ≤ 4 −7𝑥

2≤ 18

Solution: Step1:

Given inequality is, −3 ≤ 4 −7𝑥

2≤ 18

⇒ −3 − 4 ≤ 4 −7𝑥

2− 4 ≤ 18 − 4 [Subtracting from 4]

⇒ −7 ≤ −7𝑥

2≤ 14

⇒ 7 ≥7𝑥

2≥ −14

⇒ 7 ×2

7≥

7𝑥

2×

2

7≥ −14 ×

2

7

⇒ 2 ≥ 𝑥 ≥ −4

Or, ⇒ −4 ≤ 𝑥 ≤ 2

Therefore, the solution set of given inequality is [−4, 2]. Overall Hint: Solve the given inequality simultaneously and find the range of the 𝑥.

4. Solve the inequality −15 <3(𝑥−2)

5≤ 0



Solution: Step1:

Given inequality is, −15 <3(𝑥−2)

5≤ 0

⇒ −75 < 3(𝑥 − 2) ≤ 0

⇒ −75 < 3𝑥 − 6 ≤ 0

⇒ −75 + 6 < 3𝑥 − 6 + 6 ≤ 0 + 6 [Adding6 both the sides]

⇒ −69 < 3𝑥 ≤ 6

⇒−69

3<

3𝑥

3≤

6

3

⇒ −23 < 𝑥 ≤ 2

Therefore, the solution set of given inequality is (−23, 2]. Overall Hint: Solve the given inequality simultaneously and find the range of the 𝑥.

5. Solve the inequality −12 < 4 −3𝑥

−5≤ 2

Solution: Step1:

Given inequality is, −12 < 4 −3𝑥

−5≤ 2

⇒ −12 < 4 +3𝑥

5≤ 2

⇒ −12 − 4 < 4 +3𝑥

5− 4 ≤ 2 − 4 [Adding 4 both the sides]

⇒ −16 <3𝑥

5≤ −2

⇒ −16 ×5

3<

3𝑥

5×

5

3≤ −2 ×

5

3

⇒ −80

3< 𝑥 ≤ −

10

3

Therefore, the solution set of given inequality is (−80

3, −

10

3].

Overall Hint: Solve the given inequality simultaneously and find the range of the 𝑥.

6. Solve the inequality 7 ≤(3𝑥+11)

2≤ 11



Solution: Step1:

Given inequality is, 7 ≤(3𝑥+11)

2≤ 11

⇒ 14 ≤ 3𝑥 + 11 ≤ 22

⇒ 14 − 11 ≤ 3𝑥 + 11 − 11 ≤ 22 − 11

⇒ 3 ≤ 3𝑥 ≤ 11

⇒3

3≤

3𝑥

3≤

11

3

⇒ 1 ≤ 𝑥 ≤11

3

Therefore, the solution set of given inequality is [1,11

3]

Overall Hint: Solve the given inequality simultaneously and find the range of the 𝑥.

7. Solve the inequalities and represent the solution graphically on number line:

5𝑥 + 1 > – 24, 5𝑥 – 1 < 24

Solution: Step1:

Given inequalities are, 5𝑥 + 1 > −24, 5𝑥 – 1 < 24

⇒ 5𝑥 > −24 − 1

⇒ 5𝑥 > −25

⇒ 𝑥 > −5 ……………………(1)

Also, 5𝑥 − 1 < 24

Step2:

⇒ 5𝑥 < 24 + 1

⇒ 5𝑥 < 25

⇒ 𝑥 < 5 ……………………(2)

From equation (1) and (2), the solution set of given inequalities is (−5, 5).

Representation of the solution can be shown on number line as follows:



Overall Hint: Solve the given inequality simultaneously and solve for 𝑥.and use the concepts of number line representation of the solutions.


2(𝑥 – 1) < 𝑥 + 5, 3(𝑥 + 2) > 2 – 𝑥

Solution: Step1:

Given inequalities are2(𝑥 − 1) < 𝑥 + 5, 3(𝑥 + 2) > 2 – 𝑥

⇒ 2𝑥 − 2 < 𝑥 + 5

⇒ 2𝑥 − 𝑥 < 2 + 5

⇒ 𝑥 < 7 …………………(1)

Also, 3(𝑥 + 2) > 2 − 𝑥

Step2:

⇒ 3𝑥 + 6 > 2 − 𝑥

⇒ 3𝑥 + 𝑥 > 2 − 6

⇒ 4𝑥 > −4

⇒ 𝑥 > −1 …………………..(2)

From equation (1) and (2), the solution set of given inequalities is (−1, 7).

Representation of the solution can be shown on number line as fallows:


9. Solve the following inequalities and represent the solution graphically on number line: 3𝑥 – 7 >

2(𝑥 – 6), 6 – 𝑥 > 11 – 2𝑥



Solution: Step1:

Given inequalities are 3𝑥 − 7 > 2(𝑥 − 6), 6 – 𝑥 > 11 – 2𝑥

⇒ 3𝑥 − 7 > 2𝑥 − 12

⇒ 3𝑥 − 2𝑥 > 7 − 12

⇒ 𝑥 > −5 ……………………(1)

Step2:

Also, 6 − 𝑥 > 11 − 2𝑥

⇒ −𝑥 + 2𝑥 > 11 − 6

⇒ 𝑥 > 5 ………………………(2)

From equation (1) and (2), the solution set of given inequalities is (5, ∞).




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

Solution: Step1:

Given inequalities are 5(2𝑥 − 7) − 3(2𝑥 + 3) ≤ 0, 2𝑥 + 19 ≤ 6𝑥 + 47

⇒ (10𝑥 − 35) − (6𝑥 + 9) ≤ 0

⇒ 4𝑥 − 44 ≤ 0

⇒ 4𝑥 ≤ 44

⇒ 𝑥 ≤ 11 ……………………..(1)

Step2:



Also, 2𝑥 + 19 ≤ 6𝑥 + 47

⇒ 19 − 47 ≤ 6𝑥 − 2𝑥

⇒ −28 ≤ 4𝑥

⇒ −7 ≤ 𝑥

Or 𝑥 ≥ −7 ………………(2)

From equation (1) and (2), the solution set of given inequalities is [−7, 11].



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

Solution: Step1:

As the solution is to be kept between 68oF and 77oF, so, 68 < F < 77

On putting F =9

5𝐶 + 32, we obtain,

68 <9

5𝐶 + 32 < 77

⇒ 68 − 32 <9

5𝐶 < 77 − 32

⇒ 36 <9

5𝐶 < 45

⇒ 36 ×5

9<

9

5𝐶 ×

5

9< 45 ×

5

9

⇒ 20 < 𝐶 < 25

Therefore, the required range of temperature in degree Celsius is between 20oC and 25oC. Overall Hint: Solve the given inequality simultaneously and solve for 𝑥.and use the concepts of number line representation of the solutions.



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? Solution: Step1:

Consider 𝑥 litres of 2% boric acid solution is required to be added.

So, total mixture = (𝑥 + 640) litres

This resulting mixture is to be more than 4% but less than 6% boric acid.

2% 𝑥 + 8% of 640 > 4% of (𝑥 + 640) and 2% 𝑥 + 8% of 640 < 6% of (𝑥 + 640)

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

⇒2

100× 𝑥 +

8

100× 640 >

4

100× (𝑥 + 640)

⇒2𝑥 + 8 × 640 > 4(𝑥 + 640)

⇒2𝑥 + 8 × 640 > 4(𝑥 + 640)

⇒2𝑥 + 5120 > 4𝑥 +2560

⇒5120 − 2560 > 4𝑥 − 2𝑥

⇒25600 > 2𝑥

⇒𝑥 < 1280 …………………(1)

Step2:

And, 2% 𝑥 + 8% of 640 < 6% of (𝑥 + 640)

⇒2

100× 𝑥 +

8

100× 640 <

6

100× (𝑥 + 640)

⇒2𝑥 + 8 × 640 < 6(𝑥 + 640)

⇒2𝑥 + 5120 < 6𝑥 + 3840

⇒5120 − 3840 < 6𝑥 − 2𝑥

⇒5120 − 3840 < 6𝑥 − 2𝑥

⇒1280 < 4𝑥

⇒𝑥 > 320 ………………….(2)

From equation (1) and (2), we obtain

320 < 𝑥 < 1280

Therefore, the number of liters of 2% of boric acid solution that is to be added will have to be more than 320 litres but less than 1280 litres. Overall Hint: Using the given data form the inequality and solve for 𝑥.



13. How many liters of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content? Solution: Step1:

Consider 𝑥 litres of water is required to be added.

And, total mixture = (𝑥 + 1125) litres

Evident is that the amount of acid contained in the resulting mixture is 45% of 1125 litres. This resulting mixture will contain more than 25% but less than 30% acid content.

⸫30% of (1125 + 𝑥) > 45% of 1125 and 25% of (1125 + 𝑥) < 45% of 1125

30% of (1125 + 𝑥) > 45% of 1125

⇒30

100× (1125 + 𝑥) >

45

100× 1125

⇒ 30(1125 + 𝑥) > 45 × 1125

⇒ 33750 + 30𝑥 > 50625

⇒ 30𝑥 > 50625 − 33750

⇒ 30𝑥 > 50625 − 33750

⇒ 30𝑥 > 16875

⇒ 𝑥 > 562.5 …………………(1)

Step2:

And, 25% of (1125 + 𝑥) < 45% of 1125

⇒25

100× (1125 + 𝑥) <

45

100× 1125

⇒ 25 × (1125 + 𝑥) < 45 × 1125

⇒ 28125 + 25𝑥 < 50625

⇒ 25𝑥 < 50625 − 28125

⇒ 25𝑥 < 22500

⇒ 𝑥 < 900 ………… (2)

From equation (1) and (2), we have

562.5 < 𝑥 < 900

Therefore, the required number of litres of water that is to be added will have to be more than

562.5 but less than 900. Overall Hint: Using the given data form the inequality and solve for 𝑥.



14. IQ of a person is given by the formula IQ =MA

CA× 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.

Solution: Step1:

Given that for a group of 12 years old children,

80 ≤ IQ ≤ 140 ………………..(1)

And for a group of 12 years old children, CA = 12 years

IQ =MA

12× 100

Step2:

Substitute this value of IQ in equation (1)

⇒ 80 ≤MA

12× 100 ≤ 140

⇒ 80 ×12

100≤ MA ≤ 140 ×

12

100

⇒ 9.6 ≤ MA ≤ 16.8

Therefore, the range of mental age of the group of 12 years old children is [ 9.6, 16.8] Overall Hint: Using the given data form the inequality and solve for 𝑥.

Class XI CBSE-Mathematics Linear Inequalities

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Transcript of Class XI CBSE-Mathematics Linear Inequalities