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).