NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 5 Linear Inequalities help students revise and apply the concepts of linear inequalities in a comprehensive and exam-oriented manner. This exercise includes a variety of problems that test understanding of inequalities in one variable and their graphical representation.
Prepared according to the latest CBSE Class 11 Maths syllabus, this exercise focuses on solving different types of linear inequalities, representing solution sets on the number line, and interpreting results correctly. Practicing these questions helps students improve accuracy and strengthen their problem-solving skills.
NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 5 Linear Inequalities
Q.
Solve 24x < 100, when
(i) x is a natural number.
(ii) x is an integer.
Q.
Solve the following system of inequalities graphically:
Q.
Solve the following inequality graphically using two-dimensional plane : y < – 2
Q.
Solve the following inequality graphically using two-dimensional plane : x > – 3
Q.
Solve the following system of inequalities graphically:
x≥3,y≥2
Q.
Solve the following system of inequalities graphically:
3x+2y≤12,x≥1,y≥1
Q.
Q.
Q.
Solve the following system of inequalities graphically:
Q.
Solve the following system of inequalities graphically:
Q.
Solve the following system of inequalities graphically:
Q.
Q.
Solve the following inequality graphically using two-dimensional plane :
−3x+2y≥−6
Q.
Q.
Q.
Q.
Q.
Q.
Q.
Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.
7. 5x + 1 > – 24, 5x – 1 < 24
8. 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x
9. 3x – 7 > 2(x – 6), 6 – x >11–2x
10.
Q.
Q.
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?
Q.
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?
Q.
Solve the following inequality graphically using two-dimensional plane :
3y−5x<30
Q.
Solve the following inequality graphically using two-dimensional plane :
2x−3y>6
Q.
Solve – 12x > 30, when
(i) x is a natural number.
(ii) x is an integer.
Q.
Solve 5x – 3 < 7, when
(i) x is an integer.
(ii) x is a real number.
Q.
Solve 3x + 8 > 2, when
(i) x is an integer.
(ii) x is a real number.
Q.
Solve the following inequality graphically using two-dimensional plane :
x−y≤2
Q.
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.
Q.
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.
Q.
Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Q.
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Q.
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.
Q.
A man wants to cut three lengths from a single piece of board of length 91cm.
The second length is to be 3cm longer than the shortest and the third length is to 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?
Q.
Solve the following inequality graphically using two-dimensional plane :
2x+y≥6
Q.
Solve the following inequality graphically using two-dimensional plane :
3x+4y≤12
Q.
Solve the following inequality graphically using two-dimensional plane :
y+8≥2x
Q.
NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 5 Linear Inequalities
The solutions are explained in a structured, step-by-step format so students can easily understand the approach and avoid common mistakes in exams.
- Solve the inequality:
2 ≤ 3x – 4 < 5
Solution:
2 ≤ 3x – 4 < 5
⇒ 2 + 4 ≤ 3x – 4 + 4 < 5 + 4
⇒ 6 ≤ 3x < 9
⇒ 2 ≤ x < 3
Solution set: [2, 3)
- Solve the inequality:
6 ≤ –3(2x – 4) < 12
Solution:
6 ≤ –3(2x – 4) < 12
⇒ 2 ≤ –(2x – 4) < 4
⇒ –2 ≥ 2x – 4 > –4
⇒ 2 ≥ 2x > 0
⇒ 1 ≥ x > 0
Solution set: (0, 1]
- Solve the inequality:
–3 ≤ 4 – 7x/2 ≤ 18
Solution:
–3 ≤ 4 – 7x/2 ≤ 18
⇒ –7 ≤ –7x/2 ≤ 14
⇒ 7 ≥ 7x/2 ≥ –14
⇒ 2 ≥ x ≥ –4
Solution set: [–4, 2]
- Solve the inequality:
–15 < 3(x – 2)/5 ≤ 0
Solution:
–15 < 3(x – 2)/5 ≤ 0
⇒ –75 < 3(x – 2) ≤ 0
⇒ –25 < x – 2 ≤ 0
⇒ –23 < x ≤ 2
Solution set: (–23, 2]
- Solve the inequality:
–12 < 4 – 3x/(–5) ≤ 2
This becomes:
–12 < 4 + 3x/5 ≤ 2
Solution:
⇒ –16 < 3x/5 ≤ –2
⇒ –80 < 3x ≤ –10
⇒ –80/3 < x ≤ –10/3
Solution set: (–80/3, –10/3]
- Solve the inequality:
7 ≤ (3x + 11)/2 ≤ 11
Solution:
7 ≤ (3x + 11)/2 ≤ 11
⇒ 14 ≤ 3x + 11 ≤ 22
⇒ 3 ≤ 3x ≤ 11
⇒ 1 ≤ x ≤ 11/3
Solution set: [1, 11/3]
- Solve the inequalities and represent the solution graphically on number line:
5x + 1 > –24, 5x – 1 < 24
Solution:
5x + 1 > –24
⇒ 5x > –25
⇒ x > –5 ...(1)
5x – 1 < 24
⇒ 5x < 25
⇒ x < 5 ...(2)
From (1) and (2):
–5 < x < 5
Solution set: (–5, 5)
- Solve the inequalities and represent the solution graphically on number line:
2(x – 1) < x + 5, 3(x + 2) > 2 – x
Solution:
2(x – 1) < x + 5
⇒ 2x – 2 < x + 5
⇒ x < 7 ...(1)
3(x + 2) > 2 – x
⇒ 3x + 6 > 2 – x
⇒ 4x > –4
⇒ x > –1 ...(2)
From (1) and (2):
–1 < x < 7
Solution set: (–1, 7)
- Solve the following inequalities and represent the solution graphically on number line:
3x – 7 > 2(x – 6), 6 – x > 11 – 2x
Solution:
3x – 7 > 2(x – 6)
⇒ 3x – 7 > 2x – 12
⇒ x > –5 ...(1)
6 – x > 11 – 2x
⇒ –x + 2x > 11 – 6
⇒ x > 5 ...(2)
From (1) and (2):
x > 5
Solution set: (5, ∞)
- Solve the inequalities and represent the solution graphically on number line:
5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47
Solution:
5(2x – 7) – 3(2x + 3) ≤ 0
⇒ 10x – 35 – 6x – 9 ≤ 0
⇒ 4x – 44 ≤ 0
⇒ x ≤ 11 ...(1)
2x + 19 ≤ 6x + 47
⇒ 19 – 47 ≤ 6x – 2x
⇒ –28 ≤ 4x
⇒ –7 ≤ x ...(2)
From (1) and (2):
–7 ≤ x ≤ 11
Solution set: [–7, 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 conversion formula is:
F = 9C/5 + 32
Solution:
68 < F < 77
Substitute F = 9C/5 + 32
⇒ 68 < 9C/5 + 32 < 77
⇒ 36 < 9C/5 < 45
⇒ 20 < C < 25
Required temperature range: 20°C to 25°C
- A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?
Solution:
Let x litres of 2% solution be added.
Total mixture = (x + 640) litres
Condition 1: mixture > 4%
2% of x + 8% of 640 > 4% of (x + 640)
⇒ 2x + 5120 > 4x + 2560
⇒ 2560 > 2x
⇒ x < 1280
Condition 2: mixture < 6%
2% of x + 8% of 640 < 6% of (x + 640)
⇒ 2x + 5120 < 6x + 3840
⇒ 1280 < 4x
⇒ x > 320
So,
320 < x < 1280
Required quantity: More than 320 litres but less than 1280 litres
- How many litres of water will have to be added to 1125 litres of 45% acid solution so that the resulting mixture will contain more than 25% but less than 30% acid content?
Solution:
Let x litres of water be added.
Total mixture = (1125 + x) litres
Acid amount remains = 45% of 1125
Condition 1: acid content < 30%
30% of (1125 + x) > 45% of 1125
⇒ 30(1125 + x) > 45 × 1125
⇒ 30x > (45 – 30) × 1125
⇒ x > 562.5
Condition 2: acid content > 25%
25% of (1125 + x) < 45% of 1125
⇒ 25(1125 + x) < 45 × 1125
⇒ 25x < (45 – 25) × 1125
⇒ x < 900
So,
562.5 < x < 900
Required quantity of water: More than 562.5 litres but less than 900 litres
- IQ of a person is given by:
IQ = (MA/CA) × 100
Where:
- MA = Mental Age
- CA = Chronological Age
If 80 < IQ < 140 for a group of 12-year-old children, find the range of their mental age.
Solution:
CA = 12 years
IQ = (MA/12) × 100
Substitute into inequality:
80 < (MA/12) × 100 < 140
⇒ 80 × 12/100 < MA < 140 × 12/100
⇒ 9.6 < MA < 16.8
Mental age range: 9.6 years to 16.8 years
FAQs – Class 11 Maths Miscellaneous Exercise Chapter 5 Linear Inequalities
Q1. What is the focus of the Miscellaneous Exercise in Chapter 5?
The Miscellaneous Exercise focuses on applying the concepts of linear inequalities and solving a variety of problems based on them.
Q2. What topics are covered in this exercise?
This exercise includes problems related to solving inequalities, writing solution sets, and representing solutions on the number line.
Q3. Why is this exercise important for exams?
This exercise helps in revising the entire chapter and includes mixed questions that are important for exam preparation.
Q4. How can students prepare effectively for this exercise?
Students should practice different types of inequalities, understand the rules for solving them, and carefully represent the solutions to improve accuracy.