Referring to the AP Inter 1st Year Maths Study Material Chapter 5 Linear Inequalities Exercise 5b Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Linear Inequalities Solutions Exercise 5b
I.
Question 1.
Solve the inequality 2 ≤ 3x – 4 ≤ 5.
Solution:
Given that 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].
Question 2.
Solve the inequality 6 ≤ -3 (2x – 4) ≤ 12.
Solution:
Given that 6 ≤ -3 (2x – 4) ≤ 12
⇒ 2 ≤ – (2x -4)<4
⇒ -2 ≥ 2x – 4 > -4
⇒ 4 – 2 ≥ 2x > -4 + 4
⇒ 2 ≥ 2x > 0
⇒ 1 ≥ x > 0
⇒ 0 < x ≤ 1
Hence, the solution set for the given inequality is (0, 1],
Question 3.
Solve the inequality -3 ≤ 4 – \(\frac{7x}{2}\) < 18.
Solution:
Given that -3 ≤ 4 – \(\frac{7x}{2}\) <18
⇒ -6 ≤ 8 – 7x ≤ 36
⇒ -14 ≤ -7x ≤ 28
⇒ 2 ≥ x ≥ -4
⇒ -4 ≤ x ≤ 2.
Hence, the solution set for the given inequality is [- 4, 2]
Question 4.
Solve the inequality -15 < \(\frac{3(x-2)}{5}\) ≤ 0.
Solution:
Given that -15 < \(\frac{3(x-2)}{5}\) ≤ 0
⇒ -75 < 3 (x – 2) ≤ 0
⇒ -25 < x-2 ≤ 0
⇒ -25 + 2 < x ≤ 2
⇒ -23 < x ≤ 2
Hence, the solution set for the given inequality is (-23, 2].
Question 5.
Solve the inequality -12 < 4 – \(\frac{3x}{-5}\) ≤ 2
Solution:
Given that -12 < 4 – \(\frac{3x}{-5}\) ≤ 2
⇒ -12-4 < \(\frac{3x}{-5}\) ≤ 2 – 4
⇒ —16 < \(\frac{3x}{-5}\) ≤ -2
⇒ -16 < \(\frac{3x}{-5}\) ≤ -2
⇒ \(\frac{-80}{3}\) < x ≤ \(\frac{-10}{3}\)
Hence, the solution set for the given inequality is \(\left(\frac{-80}{3}, \frac{-10}{3}\right]\)
Question 6.
Solve the inequality 7 ≤ \(\frac{(3 x+11)}{2}\) ≤ 11
Solution:
Given that 7 ≤ \(\frac{(3 x+11)}{2}\) ≤ 11
⇒ 14 ≤ (3x +11) ≤ 22
⇒ 14 – 11 ≤ 3x ≤ 22 – 11
⇒ 3 ≤ 3x ≤ 11
⇒ 1 ≤ x ≤ \(\frac{11}{3}\)
Hence, the solution set for the given inequality is [1, \(\frac{11}{3}\)]
II.
Question 1.
Solve the following inequalities and represent the solution graphically on number line.
5x + 1 > – 24, 5x – 1 < 24
Solution:
Given that 5x + 1 > – 24
⇒ 5x > – 25
⇒ x > -5 ……………..(1)
⇒ 5x – 1 < 24
⇒ 5x < 25
⇒ x < 5 ………(2)
From (1) and (2) 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 2.
Solve the following inequalities and represent the solution graphically on number line.
2 (x – 1) < x + 5, 3 (x + 2) > 2 – x
Solution:
Given that 2(x – 1) < x + 5
⇒ 2x – 2 < x + 5
⇒ 2x – x < 5 + 2
⇒ x < 7 ………(1)
⇒ 3(x + 2) > 2 – x
⇒ 3x + 6 > 2 – x
⇒ 3x + x > 2 – 6
⇒ 4x > – 4, x > -1 ………………..(2)
From (1) & (2), it can be concluded that the solution set for the given system of in¬equalities is (-1, 7).
The solution of the given system of inequalities can be represented on number line as
Question 3.
Solve the following inequalities and represent the solution graphically on number line.
3x – 7 > 2 (x – 6), 6 – x > 11- 2x.
Solution:
Given that 3x – 7 > 2 (x – 6)
⇒ 3x – 7 > 2x – 12
⇒ x > -5 (1)
⇒ 6 – x > 11 – 2x
⇒ -x + 2x > 11 – 6
⇒ x > 5 ……………(2)
From (1) & (2) it can be concluded that the solution set for the given system of inequalities is (5, ∞). Because x > 5 is a stricter condition than x > -5. Therefore the solution set is (5, ∞). The solution of the given system of inequalities can be represented on number line as
Question 4.
Solve the following inequalities and represent the solution graphically on number line.
5 (2x – 7) – 3 (2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47
Solution:
Given that 5 (2x -7) – 3 (2x + 3) ≤ 0
⇒ 10x – 35 – 6x – 9 ≤ 0
⇒ 4x – 44 ≤ 0
⇒ 4x ≤ 44
⇒ x ≤ 11 ……….(1)
⇒ 2x + 19 ≤ 6x + 47
⇒ 19 – 47 ≤ 6x – 2x
⇒ -28 ≤ 4x
⇒ -7 ≤ x ………(2)
From (1) & (2), 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 5.
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 = \(\frac{9}{5}\) C + 32?
Solution:
Since the solution is to be kept between 68°F and 77°F
68 < F < 77
Putting the value of F, we have
68 < \(\frac{9}{5}\) C + 32 < 77
⇒ 68 – 32 < \(\frac{9}{5}\)C < 77 – 32
⇒ 36 < \(\frac{9}{5}\) C < 45
⇒ \(\frac{36 \times 5}{9}\) < C < \(\frac{45 \times 5}{9}\)
⇒ 20 < C < 25
Hence, the required range of temperature in degree Celsius is between 20°C and 25°C.
Question 6.
IQ of a person is given by the formula IQ = \(\frac{\mathrm{MA}}{\mathrm{CA}}\) 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.
Solution:
It is given that for a group of 12 years old children, 80 ≤ IQ ≤ 140 …………(i)
For a group of 12 years old children, (CA =12 years)
Putting this value of IQ in (i), we obtain 80 ≤ \(\frac{\text { MA }}{12}\) × 100 ≤ 140
⇒ 960 ≤ MA × 100 ≤ 1680
⇒ \(\frac{960}{100}\) ≤ MA ≤ \(\frac{1680}{100}\)
⇒ 9.60 ≤ MA ≤ 16.80
Thus, the range of mean age of the group of 12 years old children is 9.60 ≤ MA ≤ 16.80.
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 will have to be added?
Solution:
Let “x’ litres of 2% boric acid solution is required to be added.
Then the total mixture = (x + 640) litres. This resulting mixture is to be more than 4% boric acid.
⇒ 2%x + 8% of 640 > 4% of (x + 640)
⇒ \(\frac{2}{100}\) x + \(\frac{8}{100}\) x 640 > \(\frac{4}{100}\)(x + 640)
⇒ 2x + 5120 > 4x + 2560
⇒ 5120 – 2560 > 4x-2x
⇒ 2560 > 2x
⇒ 1280 > x …………….. (1)
This resulting mixture is to be less than 6% boric acid.
2%x + 8% of 640 < 6% of (x + 640)
⇒ \(\frac{2}{100}\) x + \(\frac{8}{100}\) x 640 < \(\frac{6}{100}\) (x + 640)
⇒ 2x + 5120 < 6x + 3840
⇒ 5120 – 3840 < 6x – 2x
⇒ 1280 < 4x
⇒ 320 < x ……. (2)
∴ From (1) and (2)
⇒ 320 < x < 1280
Hence, From (1) and (2) the number of litres of 2% of boric acid solution that is to be added will have to 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:
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)
⇒ \(\frac{25}{100}\)(1125 + x) < \(\frac{45}{100}\) x 1125 < \(\frac{30}{100}\) (1125 + x)
⇒ 25 (1125 + x) < 45 x 1125 < 30(1125 + x)
⇒ 5 (1125 + x) < 9 x 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 < x
⇒ 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 but lessthan 900.