*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 5.
Chapter 6 Linear Inequalities of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. The last exercise of the chapter, the miscellaneous exercise, contains 14 questions that cover the topics from the entire chapter. The Miscellaneous Exercise of NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities is based on the following topics:
- Introduction to Linear Inequalities
- Inequalities
- Algebraic Solutions of Linear Inequalities in One Variable and Their Graphical Representation
The latest CBSE guidelines are followed while solving the questions in the textbook, and the students can enhance their confidence by solving these questions again and again. Hence, students who aim to get a high score in the Maths board examination should practise solving the NCERT Solutions for Class 11 Maths problems without fail.
Download the PDF of NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities Miscellaneous Exercise
Solutions for Class 11 Maths Chapter 6 – Miscellaneous Exercise
Solve the inequalities in Exercises 1 to 6.
1. 2 ≤ 3x – 4 ≤ 5
Solution:
According to the question,
The inequality given is
2 ≤ 3x – 4 ≤ 5
⇒ 2 ≤ 3x – 4 ≤ 5
⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4
⇒ 6 ≤ 3x ≤ 9
⇒ 6/3 ≤ 3x/3 ≤ 9/3
⇒ 2 ≤ x ≤ 3
Hence, all real numbers x greater than or equal to 2 but less than or equal to 3 are solutions of given equality.
x ∈ [2, 3]
2. 6 ≤ –3 (2x – 4) < 12
Solution:
According to the question,
The inequality given is
6 ≤ –3 (2x – 4) < 12
⇒ 6 ≤ -3 (2x – 4) < 12
Dividing the inequality by 3, we get
⇒ 2 ≤ – (2x – 4) < 4
Multiplying the inequality by -1,
⇒ -2 ≥ 2x – 4 > -4 [multiplying the inequality with -1 changes the inequality sign.]
⇒ -2 + 4 ≥ 2x – 4 + 4 > -4 + 4
⇒ 2 ≥ 2x > 0
Dividing the inequality by 2,
⇒ 0 < x ≤ 1
Hence, all real numbers x greater than 0 but less than or equal to 1 are solutions of given equality.
x ∈ (0, 1]
3. – 3 ≤ 4 – 7x/2 ≤ 18
Solution:
According to the question,
The inequality given is
– 3 ≤ 4 – 7x/2 ≤ 18
⇒ – 3 – 4 ≤ 4 – 7x/2 – 4 ≤ 18 – 4
⇒ – 7 ≤ – 7x/2 ≤ 18 – 14
Multiplying the inequality by -2,
⇒ 14 ≥ 7x ≥ -28
⇒ -28 ≤ 7x ≤ 14
Dividing the inequality by 7,
⇒ -4 ≤ x ≤ 2
Hence, all real numbers x greater than or equal to -4 but less than or equal to 2 are solutions of given equality.
x ∈ [-4, 2]
4. – 15 ≤ 3(x – 2)/5 ≤ 0
Solution:
According to the question,
The inequality given is
– 15 ≤ 3(x – 2)/5 ≤ 0
⇒ – 15 < 3(x – 2)/5 ≤ 0
Multiplying the inequality by 5,
⇒ -75 < 3(x – 2) ≤ 0
Dividing the inequality by 3, we get
⇒ -25 < x – 2 ≤ 0
⇒ – 25 + 2 < x – 2 + 2 ≤ 0 + 2
⇒ – 23 < x ≤ 2
Hence, all real numbers x greater than -23 but less than or equal to 2 are solutions of given equality.
x ∈ (-23, 2]
5. – 12 < 4 – 3x/ (-5) ≤ 2
Solution:
According to the question,
The inequality given is
Hence, all real numbers x greater than -80/3 but less than or equal to -10/3 are solutions of given equality.
x ∈ (-80/3, -10/3]
6. 7 ≤ (3x + 11)/2 ≤ 11
Solution:
According to the question,
The inequality given is
⇒ 14 ≤ 3x + 11 ≤ 22
⇒ 14 – 11 ≤ 3x + 11 – 11 ≤ 22 – 11
⇒ 3 ≤ 3x ≤ 11
⇒ 1 ≤ x ≤ 11/3
Hence, all real numbers x greater than or equal to -4 but less than or equal to 2 are solutions of given equality.
x ∈ [1, 11/3]
Solve the inequalities in Exercises 7 to 11 and represent the solution graphically on the number line.
7. 5x + 1 > – 24, 5x – 1 < 24
Solution:
According to the question,
The inequalities given are
5x + 1 > -24 and 5x – 1 < 24
5x + 1 > -24
⇒ 5x > -24 – 1
⇒ 5x > -25
⇒ x > -5 ……… (i)
5x – 1 < 24
⇒ 5x < 24 + 1
⇒ 5x < 25
⇒ x < 5 ……….(ii)
From equations (i) and (ii),
We can infer that the solution of given inequalities is (-5, 5).
8. 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x
Solution:
According to the question,
The inequalities given are
2 (x – 1) < x + 5 and 3 (x + 2) > 2 – x
2 (x – 1) < x + 5
⇒ 2x – 2 < x + 5
⇒ 2x – x < 5 + 2
⇒ x < 7 ……… (i)
3 (x + 2) > 2 – x
⇒ 3x + 6 > 2 – x
⇒ 3x + x > 2 – 6
⇒ 4x > -4
⇒ x > -1 ………. (ii)
From equations (i) and (ii),
We can infer that the solution of given inequalities is (-1, 7).
9. 3x – 7 > 2(x – 6), 6 – x > 11 – 2x
Solution:
According to the question,
The inequalities given are
3x – 7 > 2(x – 6) and 6 – x > 11 – 2x
3x – 7 > 2(x – 6)
⇒ 3x – 7 > 2x – 12
⇒ 3x – 2x > 7 – 12
⇒ x > -5 ………… (i)
6 – x > 11 – 2x
⇒ 2x – x > 11 – 6
⇒ x > 5 ……….(ii)
From equations (i) and (ii),
We can infer that the solution of given inequalities is (5, ∞).
10. 5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47
Solution:
According to the question,
The inequalities given are
5(2x – 7) – 3(2x + 3) ≤ 0 and 2x + 19 ≤ 6x + 47
5(2x – 7) – 3(2x + 3) ≤ 0
⇒ 10x – 35 – 6x – 9 ≤ 0
⇒ 4x – 44 ≤ 0
⇒ 4x ≤ 44
⇒ x ≤ 11 ……(i)
2x + 19 ≤ 6x +47
⇒ 6x – 2x ≥ 19 – 47
⇒ 4x ≥ -28
⇒ x ≥ -7 ……….(ii)
From equations (i) and (ii),
We can infer that the solution of given inequalities is (-7, 11).
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 F = (9/5) C + 32?
Solution:
According to the question,
The solution has to be kept between 68° F and 77° F.
So, we get 68° < F < 77°
Substituting,
⇒ 20 < C < 25
Hence, we get
The range of temperature in degree Celsius is 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?
Solution:
According to the question,
8% of solution of boric acid = 640 litres
Let the amount of 2% boric acid solution added = x litres
Then, we have
Total mixture = x + 640 litres
We know that
The resulting mixture has to be more than 4% but less than 6% boric acid.
∴ 2% of x + 8% of 640 > 4% of (x + 640) and
2% of x + 8% of 640 < 6% of (x + 640)
2% of x + 8% of 640 > 4% of (x + 640)
⇒ (2/100) × x + (8/100) × 640 > (4/100) × (x + 640)
⇒ 2x + 5120 > 4x + 2560
⇒ 5120 – 2560 > 4x – 2x
⇒ 2560 > 2x
⇒ x < 1280 ….(i)
2% of x + 8% of 640 < 6% of (x + 640)
⇒ (2/100) × x + (8/100) × 640 < (6/100) × (x + 640)
⇒ 2x + 5120 < 6x + 3840
⇒ 6x – 2x > 5120 – 3840
⇒ 4x > 1280
⇒ x > 320 ……….(i)
From (i) and (ii),
320 < x < 1280
Therefore, the number of litres of 2% of boric acid solution that has to be added will be more than 320 litres but 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?
Solution:
According to the question,
45% of solution of acid = 1125 litres
Let the amount of water added = x litres
Resulting mixture = x + 1125 litres
We know that,
The resulting mixture has to be more than 25% but less than 30% acid content.
Amount of acid in the resulting mixture = 45% of 1125 litres
∴ 45% of 1125 < 30% of (x + 1125) and 45% of 1125 > 25% of (x + 1125)
45% of 1125 < 30% of (x + 1125)
⇒ 45 × 1125 < 30x + 30 × 1125
⇒ (45 – 30) × 1125 < 30x
⇒ 15 × 1125 < 30x
⇒ x > 562.5 ………..(i)
45% of 1125 > 25% of (x + 1125)
⇒ 45 × 1125 > 25x + 25 × 1125
⇒ (45 – 25) × 1125 > 25x
⇒ 25x < 20 × 1125
⇒ x < 900 …..(ii)
∴ 562.5 < x < 900
Therefore, the number of litres of water that has to be added will have to be more than 562.5 litres but less than 900 litres.
14. IQ of a person is given by the formula,
, 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:
According to the question,
Chronological age = CA = 12 years
IQ for the age group of 12 is 80 ≤ IQ ≤ 140.
We get that
80 ≤ IQ ≤ 140
Substituting,
⇒ 9.6 ≤ MA ≤ 16.8
∴ Range of mental age of the group of 12 years old children is 9.6 ≤ MA ≤ 16.8
Access other exercise solutions of Class 11 Maths Chapter 6 – Linear Inequalities
Exercise 6.1 Solutions 26 Questions
Exercise 6.2 Solutions 10 Questions
Exercise 6.3 Solutions 15 Questions
Also explore – NCERT Class 11 Solutions
Comments