• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar

Class Notes

Free Class Notes & Study Material

  • Class 1-5
  • Class 6
  • Class 7
  • Class 8
  • Class 9
  • Class 10
  • Class 11
  • Class 12
  • NCERT SOL
  • Ref Books
Home » NCERT Solutions » Class 11 » Maths » NCERT Solutions for Chapter 5, Miscellaneous Exercise, Class 11, Maths

NCERT Solutions for Chapter 5, Miscellaneous Exercise, Class 11, Maths

Last Updated on July 29, 2023 By Mrs Shilpi Nagpal

Miscellaneous Exercise
Linear Equations
Question and Answers
Class 11 – Maths

Class Class 11
Subject Mathematics
Chapter Name Linear Equations
Chapter No. Chapter 5
Exercise Miscellaneous Exercise
Category Class 11 Maths NCERT Solutions

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

Answer 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]

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

Answer 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]

Question 3 Solve the inequality :  – 3 ≤ 4 – 7x/2 ≤ 18

Answer

– 3 ≤ 4 – 7x/2 ≤ 18

Miscellaneous Exercise-5 , Answer 3

As a result, the set of solutions for the given inequality is [−4,2].

Question 4 Solve the inequality :

Miscellaneous Exercise-5 , Question 4

Answer 4)  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]

Question 5 Solve the inequality :

Miscellaneous Exercise-5 , Question 5

Answer 

Miscellaneous Exercise-5 , Answer 5

Question 6 Solve the inequality :

Miscellaneous Exercise-5 , Question 6

Answer 6) 

Miscellaneous Exercise-5 , Answer 6

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]

Question 7 Solve the inequalities and represent the solution graphically on number line: 5x + 1 > – 24, 5x – 1 < 24

Answer 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).

Miscellaneous Exercise-5 , Graphical representation , Answer-7

Question 8 Solve the inequalities and represent the solution graphically on number line:

2 (x – 1) < x + 5, 3 (x + 2) > 2 – x

Answer

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).

Miscellaneous Exercise-5 , Answer 7, Graphical representation

Question 9 Solve the inequalities and represent the solution graphically on number line: 3x – 7 > 2 (x – 6) , 6 – x > 11 – 2x

Answer

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, ∞).

Miscellaneous Exercise-5 , Answer 9, Graphical representation

Question 10 Solve the inequalities and represent the solution graphically on number line: 5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

Answer 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).

Miscellaneous Exercise-5 , Answer 10, Graphical representation

Question 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?

Answer  The solution has to be kept between 68° F and 77° F. So, we get 68° < F < 77°

Miscellaneous Exercise-5 , Answer 11

The required temperature range in degrees Celsius is between 20° C to 25° C.

Question 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?

Answer 8% of solution of boric acid = 640 litres

Let the amount of 2% boric acid solution added = x litres

Total mixture = x + 640 litres

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.

Question 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?

Answer 45% of solution of acid = 1125 litres

Let the amount of water added = x litres

Resulting mixture = x + 1125 litres

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)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 57

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)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 58

⇒ 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.

Question 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.

Answer  Chronological age = CA = 12 years

IQ for the age group of 12 is 80 ≤ IQ ≤ 140.

We get that

80 ≤ IQ ≤ 140

Substituting,

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 60

⇒ 9.6 ≤ MA ≤ 16.8

∴ The range of mental age of the group of 12 years old children is 9.6 ≤ MA ≤ 16.8

Filed Under: Class 11, Maths

About Mrs Shilpi Nagpal

Author of this website, Mrs. Shilpi Nagpal is MSc (Hons, Chemistry) and BSc (Hons, Chemistry) from Delhi University, B.Ed. (I. P. University) and has many years of experience in teaching. She has started this educational website with the mindset of spreading free education to everyone.

Reader Interactions

Leave a Reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

  • Facebook
  • Pinterest
  • Twitter
  • YouTube

CATEGORIES

  • —— Class 6 Notes ——
  • —— Class 7 Notes ——
  • —— Class 8 Notes ——
  • —— Class 9 Notes ——
  • —— Class 10 Notes ——
  • —— NCERT Solutions ——

© 2016 - 2025 · Disclaimer · Privacy Policy · About Us · Contact Us