Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 6 Linear Inequalities Miscellaneous Exercise Questions and Answers.
PSEB Solutions for Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise
Question 1.
Solve the inequality 2 ≤ 3x – 4 < 5.
2 ≤ 3x – 4 < 5
⇒ 2x + 4 < 3x – 4 + 4< 5 + 4
⇒ 6 ≤ 3x ≤ 9
⇒ 2 ≤ x ≤ 3
Thus, 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].
Question 2.
Solve the inequality 6 ≤ – 3 (2x – 4) < 12.
Ans.
6 ≤ – 3 (2x – 4) < 12
or \(\frac{6}{3}\) ≤ – 1 (2x – 4) ≤ \(\frac{12}{3}\) [divide by 3]
or 2 ≤ – 1 (2x – 4) ≤ 4
or 2 ≤ (- 2x + 4) ≤ 4
or 2 – 4 ≤ – 2x + 4 – 4 ≤ 4 – 4
or – 2 ≤ – 2x ≤ 0
or \(\frac{-2}{-2} \geq \frac{-2 x}{-2}\) ≥ 0
or 1 ≥ x ≥ 0
Hence, x is less than or equal to 1 and is greater than 0 i.e., x ∈ [0, 1].
Question 3.
Solve the inequality – 3 ≤ 4 – \(\frac{7 x}{2}\) ≤ 18.
Answer.
– 3 ≤ 4 – \(\frac{7 x}{2}\) ≤ 18
⇒ – 3 – 4 ≤ – \(\frac{7 x}{2}\) ≤ 18 – 4
⇒ – 7 ≤ – \(\frac{7 x}{2}\) ≤ 14
⇒ 7 ≥ \(\frac{7 x}{2}\) ≥ – 14
⇒ 1 ≥ \(\frac{x}{2}\) ≥ – 2
⇒ 2 ≥ x ≥ – 4
Question 4.
Solve the inequality – 15 < \(\frac{3(x-2)}{5}\) ≤ 0.
Answer.
– 15 < \(\frac{3(x-2)}{5}\) ≤ 0
⇒ – 75 < 3 (x – 2) ≤ 0
⇒ – 25 < x – 2 ≤ 0
⇒ – 25 + 2 < x – 2 + 2 ≤ 0 + 2
⇒ – 23 < x ≤ 2
Thus, the solution set for the given inequality is (- 23, 2].
Question 5.
Solve the inequality – 12 < 4 – \(\frac{3 x}{-5}\) ≤ 2.
Answer.
We have, – 12 < 4 – \(\frac{3 x}{-5}\) ≤ 2
Question 6.
Solve the inequality 7 ≤ \(\frac{(3 x+11)}{2}\) ≤ 11.
Answer.
7 ≤ \(\frac{(3 x+11)}{2}\) ≤ 11
⇒ 14 ≤ 3x + 11 ≤ 22
⇒ 14 – 11 ≤ 3x ≤ 22 – 11
⇒ 3 ≤ 3x ≤ 11
⇒ 1 ≤ x ≤ \(\frac{11}{3}\).
Thus, the solution set for the given inequality is [1, \(\frac{11}{3}\)].
Question 7.
Solve the inequalities and represent the solution graphically on number line : 5x + 1 > – 24, 5x – 1 < 24.
Answer.
5x + 1 > – 24
5x > – 25
x > -5 …………..(i)
5x – 1 < 24
5x < 25
⇒ x < 5 ……………….(ii)
From inequalities (i) and (ii), it can be concluded that the solution set for the given system of inequalities is (- 5, 5).
The solution of the given system of inequalities can be represented on number line as
Question 8.
Solve the inequalities and represent the solution graphically on number line:
2(x – 1) < x + 5, 3 (x + 2) > 2x.
Solution.
2(x-l) < x+5
⇒ 2x – 2 < x + 5
⇒ 2x – x < 5 + 2
⇒ x< 7 …………(i) 3(x + 2) > 2 – x
⇒ 3x +6 > 2 – x
⇒ 3x + x > 2 – 6
⇒ 4x > – 4
⇒ x > – 1 ……………..(ii)
From inequalities (i) and (ii), it can be concluded that the solution set for the given system of inequalities is (- 1, 7).
The solution of the given system of inequalities can be represented on number line as
Question 9.
Solve the following inequalities and represent the solution graphically on number line : 3x – 7 > 2(x – 6), 6 – x > 11 – 2x.
Answer.
3x – 7 > 2 (x – 6)
⇒ 3x – 7 > 2x – 12
⇒ 3x – 2x > – 12 + 7
⇒ x > – 5
6 – x > 11 – 2x
⇒ – x + 2x > 11 – 6
⇒ x > 5 …………….(ii)
From inequalities (i) and (ii), it can be concluded that the solution set for the given system of inequalities is (5, ∞).
The solution of the given system of inequalities can be represented on number line as
Question 10.
Solve the inequalities and represent the solution graphically on number line : 5 (2x – 7) – 3(2x + 3) < 0, 2x + 19 < 6x + 47.
Answer.
5 (2x – 7) – 3(2x + 3) ≤ 0
⇒ 10x – 35 – 6x – 9 ≤ 0
⇒ 4x – 44 ≤ 0
⇒ 4x ≤ 44
⇒ x ≤ 11 …………….(i)
2x + 19 ≤ 6x + 47
19 – 47 ≤ 6x – 2x
⇒ – 28 < 4x
⇒ – 7 < x …………….(ii)
From inequalities (i) and (ii), it can be concluded that the solution set for the given system of inequalities is [- 7, 11]. The solution of the given system of inequalities can be represented on number line as
Question 11.
A solution is to be kept between 68°F and 77°F. What is the range in temperature in degree Celsius (C) if the Celsius/ Fahrenheit (F) conversion formula is given by F = \(\frac{9}{8}\) C + 32 ?
Answer.
Since the solution is to be kept between 68°F and 77°F,
68 < F < 77
Putting F = \(\frac{9}{5}\) C + 32 , we obtain
68 < \(\frac{9}{5}\) C + 32 < 77
⇒ 68 – 32 < \(\frac{9}{5}\) C < 77 – 32
⇒ 36 < \(\frac{9}{5}\) C < 45
⇒ 36 × \(\frac{5}{9}\) < C < 45 × \(\frac{5}{9}\)
⇒ 20 < C < 25
Thus, the required range of temperature in degree Celsius is between 20°C and 25°C.
Question 12.
A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?
Answer.
Let x be the number of litres of 2% boric acid solution.
Total mixture = (640 + x) L
∴ According to the question, 2% of x + 8% of 640 > 4% of (640 + x)
⇒ \(\frac{2 x}{100}\) + \(\frac{8}{100}\) × 640 > \(\frac{4}{100}\) (640 + x)
⇒ 2x + 5120 > 2560 + 4x [multiplying both sides by 100]
⇒ 2x + 5120 – 2x > 2560 + 4x – 2x [subtracting 2x from both sides]
⇒ 5120 > 2560 + 2x
⇒ 5120 – 2560 > 2560 + 2x – 2560
⇒ 2560 > 2x
⇒ 2x < 2560
⇒ x < 1280 ………………….(i)
[dividing bothsides by 2]
Also, 2% of x + 8% of 640 < 6% of (640 + x)
⇒ \(\frac{2}{100} \times x+\frac{8}{100} \times 640<\frac{6}{100} \times(640+x)\)
⇒ 2x + 5120 < 3840 + 6x [multiplying both sides by 100]
⇒ 2x + 5120 – 2x < 3840 + 6x – 2x . [subtracting 2x from both sides]
⇒ 5120 < 3840 + 4x
⇒ 5120 – 3840 < 3840 + 4x – 3840 [subtracting 3840 from both sides]
⇒1280 < 4x or 4x > 1280
⇒ \(\frac{4 x}{4}>\frac{1280}{4}\)
⇒ x > 320 ………….(ii)
On combining eqs. (i) and (ii), we get 320 < x < 1280 Thus, the number of litres to be added should be greater than 320 L and less than 1280 L.
Question 13.
How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
Answer.
Let x litres of water is required to be added.
Then,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.
∴ 30% of (1125 + x) > 45% of 1125
and, 25% of (1125 + x) < 45% of 1125
30% of (1125 + x) > 45% of 1125
\(\frac{30}{100}\) (1125 + x) < \(\frac{45}{100}\) × 1125 ⇒ 30 (1125 + x) > 45 × 1125
⇒ 30 × 1125 + 30x > 45 × 1125
⇒ 30x > 45 × 1125 – 30 × 1125
⇒ 30x > (45 – 30) × 1125
⇒ 30x > 15 × 1125
⇒ x > \(\frac{15 \times 1125}{30}\) = 5625
25% of (1125 + x) < 45% of 1125
⇒ \(\frac{25}{100}\) (1125 + x) < \(\frac{45}{100}\) × 1125
⇒ 25 (1125 + x) < 45 × 1125
⇒ 25 × 1125 + 25x < 45 × 1125
⇒ 25x < 45 × 1125 – 25 × 1125
⇒ 25x < (45 – 25) × 1125
⇒ x < \(\frac{20 \times 1125}{25}\) = 900
∴ 562.5 < x < 900
Thus, the required number of litres of water that is to be added will have to be more than 562.5 litres but less than 900 litres.
Question 14.
IQ of a person is given by the formula IQ = \(\frac{M A}{C A}\) × 100, where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental Age.
Ans.
Given, CA = 12 and 80 ≤ IQ ≤ 140 ………………..(i)
Also, IQ = \(\frac{M A}{C A}\) × 100
On putting this value in eq. (i), we get
80 ≤ \(\frac{M A}{C A}\) × 100 ≤ 140
⇒ 80 ≤ \(\frac{M A}{12}\) × 100 ≤ 140 [∵ CA = 12]
⇒ \(\frac{80 \times 12}{100}\) ≤ MA ≤ \(\frac{12}{100}\)
[multiplying bothsides by (\(\frac{12}{100}\))]
⇒ \(\frac{960}{100}\) ≤ MA ≤ \(\frac{1680}{100}\)
⇒ 9.6 ≤ MA ≤ 16.8.