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).