Practicing the AP Board Solutions Class 11 Maths and Chapter 5 Inter 1st Year Maths Linear Inequalities Solutions Exercise 5b Pdf Download will help students to clear their doubts quickly.
Intermediate 1st Year Maths Linear Inequalities Solutions Exercise 5b
Linear Inequalities Exercise 5b Solutions
Linear Inequalities Class 11 Exercise 5b Solutions – Linear Inequalities 5b Exercise Solutions
I. Solve the inequalities in Exercises 1 to 6.
Question 1.
2 ≤ 3x – 4 ≤ 5
Solution:
According to the question,
The inequality given is 2 ≤ 3x – 4 ≤ 5
⇒ 2 ≤ 3x – 4 ≤ 5
⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4
⇒ 6 ≤ 3x ≤ 9
⇒ \(\frac{6}{3} \leq \frac{3 x}{3} \leq \frac{9}{3}\)
⇒ 2 ≤ x ≤ 3
Hence, all real numbers x greater than or equal to 2, but less than or equal to 3, are solutions of the given equality.
x ∈ [2, 3]
Question 2.
6 ≤ -3(2x – 4) < 12
Solution:
According to the question,
The inequality given is 6 ≤ -3(2x – 4) < 12
⇒ 6 ≤ -3 (2x – 4) < 12
Dividing the inequality by 3, we get
⇒ 2 ≤ -(2x – 4) < 4
Multiplying the inequality by -1,
⇒ -2 ≥ 2x – 4 > -4 [multiplying the inequality by -1 changes the inequality sign]
⇒ -2 + 4 ≥ 2x – 4 + 4 > -4 + 4
⇒ 2 ≥ 2x > 0
Dividing the inequality by 2,
⇒ 0 < x ≤ 1
Hence, all real numbers x greater than 0, but less than or equal to 1, are solutions of the given equality.
x ∈ (0, 1)
Question 3.
-3 ≤ 4 – \(\frac {7x}{2}\) ≤ 18
Solution:
According to the question,
The inequality given is -3 ≤ 4 – \(\frac {7x}{2}\) ≤ 18
⇒ -3 – 4 ≤ 4 – \(\frac {7x}{2}\) – 4 ≤ 18 – 4
⇒ -7 ≤ \(-\frac {7x}{2}\) ≤ 18 – 14
Multiplying the inequality by -2
⇒ (-7) × (-2) ≥ \(-\frac {7x}{2}\) × (-2) ≥ 14 × (-2)
⇒ 14 ≥ 7x ≥ -28
⇒ -28 ≤ 7x ≤ 14
Dividing the inequality by 7,
⇒ -4 ≤ x ≤ 2
Hence, all real numbers x greater than or equal to -4, but less than or equal to 2 are solutions of the given equality.
x ∈ [-4, 2]
Question 4.
-15 ≤ \(\frac{3(x-2)}{5}\) ≤ 0
Solution:
According to the question,
The inequality given is -15 < \(\frac{3(x-2)}{5}\) ≤ 0
⇒ -15 < \(\frac{3(x-2)}{5}\) ≤ 0
Multiplying the inequality by 5,
⇒ -15 × 5 < \(\frac{3(x-2)}{5}\) × 5 ≤ o × 5
⇒ -75 < 3(x – 2) ≤ 0
Dividing the inequality by 3, we get
⇒ \(-\frac{75}{3}<\frac{3(x-2)}{3} \leq \frac{0}{3}\)
⇒ -25 < x – 2 ≤ 0
⇒ -25 + 2 < x – 2 + 2 ≤ 0 + 2
⇒ -23 < x ≤ 2
Hence, all real numbers x greater than -23, but less than or equal to 2, are solutions of the given equality.
x ∈ (-23, 2)
Question 5.
-12 < 4 – \(\frac{3 x}{-5}\) ≤ 2
Solution:
According to the question,
The inequality given is -12 < 4 – \(\frac{3 x}{-5}\) ≤ 2
⇒ -12 < 4 – \(\frac{3 x}{-5}\) ≤ 2
⇒ -12 – 4 < 4 – \(\frac{3 x}{-5}\) – 4 ≤ 2 – 4
⇒ 16 < \(\frac{3 x}{5}\) ≤ -2
Multiplying the inequality by 5
⇒ -16 × 5 < \(\frac{3 x}{5}\) × 5 ≤ -2 × 5
⇒ -80 < 3x ≤ -10
⇒ \(-\frac{80}{3}\)
Hence, all real numbers x greater than \(-\frac {80}{3}\), but less than or equal to \(-\frac {10}{3}\) are solutions of the given equality.
x ∈ \(\left(\frac{-80}{3}, \frac{-10}{3}\right)\).
Question 6.
7 ≤ \(\frac{(3 x+11)}{2}\) ≤ 11
Solution:
According to the question,
The inequality given is, 7 ≤ \(\frac{(3 x+11)}{2}\) ≤ 11
Multiplying the inequality by 2.
⇒ 7 × 2 ≤ \(\frac{(3 x+11)}{2}\) × 2 ≤ 11 × 2
⇒ 14 ≤ 3x + 11 ≤ 22
⇒ 14 – 11 ≤ 3x + 11 – 11 ≤ 22 – 11
⇒ 3 ≤ 3x ≤ 11
⇒ 1 ≤ x ≤ \(\frac {11}{3}\)
Hence, all real numbers x greater than or equal to 1, but less than or equal to \(\frac {11}{3}\) are solutions of the given equality.
x ∈ (1, \(\frac {11}{3}\))
II. Solve the inequalities in Exercises 1 to 4 and represent the solution graphically on a number line.
Question 1.
5x + 1 > -24, 5x – 1 < 24
Solution:
According to the question,
The inequalities given are 5x + 1 > -24 and 5x – 1 < 24 5x + 1 > -24
⇒ 5x > -24 – 1
⇒ 5x > -25
⇒ x > -5 ………..(i)
5x – 1 < 24
⇒ 5x < 24 + 1
⇒ 5x < 25
⇒ x < 5 …………(ii)
From equations (i) and (ii),
We can infer that the solution of the given inequalities is (-5, 5).
Question 2.
2(x – 1) < x + 5, 3(x + 2) > 2 – x
Solution:
According to the question,
The inequalities given are 2(x – 1) < x + 5 and 3(x + 2) > 2 – x
2(x – 1) < 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 equations (i) and (ii),
We can infer that the solution of the given inequality is (-1, 7)
Question 3.
3x – 7 > 2(x – 6), 6 – x > 11 – 2x
Solution:
According to the question,
The inequalities given are, 3x – 7 > 2(x – 6) and 6 – x > 11 – 2x
3x – 7 > 2(x – 6)
⇒ 3x – 7 > 2x – 12
⇒ 3x – 2x > 7 – 12
⇒ x > -5 ……….(i)
6 – x > 11 – 2x
⇒ 2x – x > 11 – 6
⇒ x > 5 …………(ii)
From equations (i) and (ii),
We can infer that the solution of the given inequalities is (5, ∞)
Question 4.
5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47
Solution:
According to the question,
The inequalities given are 5(2x – 7) – 3(2x + 3) ≤ 0 and 2x + 19 ≤ 6x + 47
5(2x – 7) – 3(2x + 3) ≤ 0
⇒ 10x – 35 – 6x – 9 ≤ 0
⇒ 4x – 44 ≤ 0
⇒ 4x ≤ 44
⇒ x ≤ 11 ……….(i)
2x + 19 ≤ 6x + 47
⇒ 6x – 2x ≥ 19 – 47
⇒ 4x ≥ -28
⇒ x ≥ -7 ………(ii)
From equations (i) and (ii),
We can infer that the solution of the given inequalities is (-7, 11).
Question 5.
A solution is to be kept between 68°F and 77°F. What is the range in temperature in degrees Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by F = \(\frac {9}{5}\)C + 32?
Solution:
According to the question,
The solution has to be kept between 68°F and 77°F
So, we get, 68° < F < 77°
Substituting in F = \(\frac {9}{5}\)C + 32
⇒ 68 < \(\frac {9}{5}\)C + 32 < 77
⇒ 68 – 32 < \(\frac {9}{5}\)C + 32 – 32 < 77 – 32
⇒ 36 < \(\frac {9}{5}\)C < 45
⇒ \(36 \times \frac{5}{9}<\frac{9}{5} C \times \frac{5}{9}<45 \times \frac{5}{9}\)
⇒ 20 < C < 25
Hence, we get the range of temperature in degrees Celsius is between 20°C and 25°C.
Question 6.The
IQ of a person is given by the formula IQ = \(\frac {MA}{CA}\) × 100, where MA is mental age, and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12-year-old children, find the range of their mental age.
Solution:
According to the question,
Chronological age = CA = 12 years
IQ for the age group of 12 is 80 ≤ IQ ≤ 140.
We get that, 80 ≤ IQ ≤ 140
substituting in IQ = \(\frac {MA}{CA}\) × 100, We get
⇒ 80 ≤ \(\frac {MA}{CA}\) × 100 ≤ 140
⇒ 80 ≤ \(\frac {MA}{12}\) × 100 ≤ 140
⇒ 80 × \(\frac {12}{100}\) ≤ \(\frac {MA}{12}\) × 100 ≤ 140 × \(\frac {12}{100}\)
⇒ 9.6 ≤ MA ≤ 16.8
∴ The range of mental age of the group of 12 children is 9.6 ≤ MA ≤ 16.8
III.
Question 1.
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 must be added?
Solution:
According to the question,
8% of a solution of boric acid = 640 litres
Let the amount of 2% boric acid solution added = x litres
Then we have, Total mixture = x + 640 litres
We know that, the resulting mixture has to be more than 4% but less than 6% boric acid.
∴ 2% of x + 8% of 640 > 4% of (x + 640) and 2% of x + 8% of 640 < 60% of (x + 640) 2% of x + 8% of 640 > 4% of (x + 640)
⇒ \(\frac {2}{100}\) × x + \(\frac {8}{100}\) × 640 > \(\frac {4}{100}\) × (x + 640)
⇒ 2x + 5120 > 4x + 2560
⇒ 5120 – 2560 > 4x – 2x
⇒ 2560 > 2x
⇒ x < 1280 ………(i)
2% of x + 8% of 640 < 6% of (x + 640)
⇒ \(\frac {2}{100}\) × x + \(\frac {8}{100}\) × 640 < \(\frac {6}{100}\) × (x + 640)
⇒ 2x + 5120 < 6x + 3840 ⇒ 6x – 2x > 5120 – 3840
⇒ 4x > 1280
⇒ x > 320 …………(ii)
From (i) and (ii)
∴ 320 < x < 1280
Therefore, the number of litres of 2% boric acid solution that has to be added will be more than 320 litres but less than 1280 litres.
Question 2.
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?
Solution:
According to the question,
45% of the solution of acid = 1125 litres
Let the amount of water added = x litres
Resulting mixture = x + 1125 litres
We know that, the resulting mixture has to be more than 25% but less than 30% acid content.
Amount of acid in resulting mixture = 45% of 1125 liters
∴ 45% of 1125 < 30% of (x + 1125) and 45% of 1125 > 25% of (x + 1125)
45% of 1125 < 30% of (x + 1125)
⇒ \(\frac {45}{100}\) × 1125 < \(\frac {30}{100}\) × (x + 1125)
⇒ 45 × 1125 < 30x + 30 × 1125
⇒ (45 – 30) × 1125 < 30x
⇒ 15 × 1125 < 30x ⇒ x > 562.5 ……………(i)
45% of 1125 > 25% of (x + 1125)
⇒ \(\frac {45}{100}\) × 1125 > \(\frac {25}{100}\) × (x + 1125)
⇒ 45 × 1125 > 25x + 25 × 1125
⇒ (45 – 25) × 1125 > 25x
⇒ 25x < 20 × 1125
⇒ x < 900 ……….(ii)
From (i) and (ii)
∴ 562.5 < x < 900
Therefore, the number of litres of water that has to be added will have to be more than 562.5 litres but less than 900 litres.