# NCERT Solutions for Class 11th Mathematics

## Chapter 6. Linear Inequalities

### Exercise 6.1

**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).