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Exercise 6.1 Class 11 maths 6. Linear Inequalities - ncert solutions - Toppers Study

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Exercise 6.1 Class 11 maths 6. Linear Inequalities - ncert solutions - Toppers Study

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Exercise 6.1 class 11 Mathematics Chapter 6. Linear Inequalities

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Exercise 6.1 Class 11 Maths 6. Linear Inequalities - Ncert Solutions - Toppers Study
Class 11 Ncert Solutions
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Chapter 6. Linear Inequalities Exercise 6.1 Class 11

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6. Linear Inequalities

| Exercise 6.1 |

Exercise 6.1 Class 11 maths 6. Linear Inequalities - ncert solutions - Toppers Study

Exercise 6.1

Q1. Solve 24x < 100, when

(i) x is a natural number

(ii) x is an integer

Solution:

The given inequality is 24x < 100.

(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than

∴ when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.

Hence, in this case, the solution set is {1, 2, 3, 4}.

(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than

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

Hence, in this case, the solution set is {1, 2, 3, 4}.

Q2. Solve –12x > 30, when

(i) x is a natural number

(ii) x is an integer

Solution:

The given inequality is –12x > 30.

(i) There is no natural number less than

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

Hence, in this case, the solution set is {…, –5, –4, –3}.

Q3. Solve 5x– 3 < 7, when

(i) x is an integer

(ii) x is a real number

Soluution:

The given inequality is 5x– 3 < 7.

Q5. Solve the given inequality for real x: 4x + 3 < 5x + 7

Solution :

4x + 3 < 5x + 7

⇒ 4x + 3 – 7 < 5x + 7 – 7

⇒ 4x – 4 < 5x

⇒ 4x – 4 – 4x < 5x – 4x

⇒ –4 < x

Thus, all real numbers x, which are greater than –4, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–4, ∞).

Q23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Solution:

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

Then, the other integer is x + 2.

Since both the integers are smaller than 10, 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 > 11 – 2

⇒ 2x > 9

⇒ x > 9/2

⇒ x > 4.5 ....... (2)

From (1) and (2), we get .

Since x is an odd number, x can take the values, 5 and 7.

Therefore, the required possible pairs are (5, 7) and (7, 9).

Other Pages of this Chapter: 6. Linear Inequalities

1. Chapter 6. Linear Inequalities | Exercise 6.1

2. Chapter 6. Linear Inequalities | Exercise 6.2

3. Chapter 6. Linear Inequalities | Exercise 6.3

4. Chapter 6. Linear Inequalities | Miscellaneous Exercise on Chapter - 6

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