NCERT Solutions for Class 11 Maths Chapter 6

Given that: 3x + 8 > 2
⇒ 3x > -6
⇒ x > (-6)/3
⇒ x > -2
x is an integer. The integers greater than –2 are –1, 0, 1, 2, …
Thus, when x is an integer, the solutions of the given inequality are –1, 0, 1, 2 …
Hence, in this case, the solution set is {–1, 0, 1, 2, …}.

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, ∞).

3x – 7 > 5x – 1
⇒ 3x – 7 + 7 > 5x – 1 + 7
⇒ 3x > 5x + 6
⇒ 3x – 5x > 5x + 6 – 5x
⇒ – 2x > 6
⇒ -x > 6/2
⇒ -x > 3
⇒ x < -3 Thus, all real numbers x, which are less than –3, are the solutions of the given inequality. Hence, the solution set of the given inequality is (–∞, –3).

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, therefore x + 2 < 10 ⇒ x < 10 – 2 ⇒ x < 8 … (i) 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 …(ii)
From (i) and (ii), we obtain that the value of x can be 4,5,6 or 7. .
Since x is an odd number, x can take the values, 5 and 7.
Hence, the required possible pairs of numbers are (5, 7) and (7, 9).

Let x be the smaller of the two consecutive even positive integers.
Then, the other integer is x + 2.
Since both the integers are larger than 5, x > 5 … (1)
Also, the sum of the two integers is less than 23. x + (x + 2) < 23 ⇒ 2x + 2 < 23 ⇒ 2x < 23 – 2 ⇒ 2x < 21 ⇒ x < 21/2 ⇒ x < 10.5 … (2) From (1) and (2), we obtain 5 < x < 10.5. Since x is an even number, so x can take the values, 6, 8 and 10. Hence, the required possible pairs are (6, 8), (8, 10) and (10, 12).

Let the length of the shortest side of the triangle be x cm.
Then, length of the longest side = 3x cm and the length of the third side = (3x – 2) cm
Since the perimeter of the triangle is at least 61 cm,
⇒ x + 3x + (3x – 2) ≥ 61
⇒ 7x – 2 ≥ 61
⇒ 7x ≥ 63
⇒ x ≥ 9
Hence, the minimum length of the shortest side is 9 cm.

Let the length of the shortest piece be x cm.
Then, length of the second piece and the third piece are (x + 3) cm and 2x cm respectively.
Since the three lengths are to be cut from a single piece of board of length 91 cm, x + (x + 3) + 2x≤ 91
⇒ 4x + 3 ≤ 91
⇒ 4x ≤ 91 – 3
⇒ 4x ≤ 88
⇒ x≤22 … (1)
Also, the third piece is at least 5 cm longer than the second piece.
∴ 2x ≥ (x + 3) + 5
⇒ 2x ≥ x + 8
⇒ x ≥ 8 … (2)
From (1) and (2), we obtain, 8 ≤ x ≤ 22
Thus, the possible length of the shortest board is greater than or equal to 8 cm but less than or equal to 22 cm.

2 ≤ 3x – 4 ≤ 5
⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4
⇒ 6 ≤ 3x ≤ 9
⇒ 2 ≤ x ≤ 3
Therefore, all the real numbers, x, which are greater than or equal to 2 but less than or equal to 3, are the solutions of the given inequality. The solution set for the given inequality is [2, 3].

Let x litres of water is required to be added.
Then, the total mixture = (x + 1125) litres
It is evident that the amount of acid contained in the resulting mixture is 45% of 1125 litres.
This resulting mixture will contain more than 25% but less than 30% acid content.
∴ 25% of (1125 + x) < 45% of 1125 < 30% of (1125 + x) ⇒ 25/100 (1125+x) < 45/100×1125<30/100 (1125+x) ⇒ 25(1125+x) < 45×1125<30(1125+x) ⇒ 5(1125+x) < 9×1125<6(1125+x) ⇒ 5625+5x < 10125<6750+6x ⇒ 5x<10125-5625 < 6750-5625+6x ⇒ 5x<4500 < 1125+6x ⇒ 5x<4500 and 4500<1125+6x ⇒ x<900 and 3375<6x ⇒ x<900 and 562.5