Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 6 Linear Inequalities Miscellaneous Exercise

Solve the inequalities in questions 1 to 6:

1. 2 â‰¤ 3x – 4 â‰¤ 5

2. 6 â‰¤ – 3(2x – 4) â‰¤ 12

3. – 3 â‰¤ 4 – \(\frac{7x}{2}\) â‰¤ 18

4. – 15 â‰¤ \(\frac{3(x – 2)}{5}\) â‰¤ 0

5. – 12 â‰¤ 4 – \(\frac{3x}{- 3}\) â‰¤ 2

6. 7 â‰¤ \(\frac{3x+11}{2}\) â‰¤ 11

Solutions to questions 1 – 6:

1. 2 â‰¤ 3x – 4 â‰¤ 5

Adding, we get

2 + 4 â‰¤ 3x – 4 + 4 â‰¤ 5 + 4

or 6 â‰¤ 3x â‰¤ 9.

Dividing by 3, we get

2 â‰¤ x â‰¤ 3.

Thus, the solution is {2, 3}

2. 6 â‰¤ – 3(2x – 4) â‰¤ 12

or \(\frac{6}{3}\) â‰¤ – (2x – 4) â‰¤ \(\frac{12}{3}\) [Divide by 3]

or 2 â‰¤ – (2x – 4) â‰¤ 4

or 2 â‰¤ – 2x + 4 â‰¤ 4

or 2 – 4 â‰¤ – 2x + 4 – 4 â‰¤ 4 – 4

or – 2 â‰¤ – 2x â‰¤ – 0

or \(\frac{- 2}{- 2}\) > \(\frac{- 2x}{- 2}\) â‰¥ 0 [Divide by – 2]

or 1 â‰¥ x â‰¥ 0.

Hence, x is less than or equal to ‘1’ and is greater than 0, i.e; x âˆˆ [0, 1].

3. – 3 â‰¤ 4 – \(\frac{7x}{2}\) â‰¤ 18

Subtracting 4, we get

– 3 – 4 â‰¤ 4 – \(\frac{7x}{2}\) – 4 â‰¤ 18 – 4

or – 7 â‰¤ – \(\frac{7x}{2}\) â‰¤ 14.

Dividing by – \(\frac{7}{2}\), we get

– 7 Ã— (- \(\frac{2}{7}\)) â‰¥ \(\frac{- 7}{2}\)x Ã— (- \(\frac{2}{7}\)) â‰¥ 14 Ã— (- \(\frac{2}{7}\)).

2 â‰¥ x â‰¥ – 4 or – 4 â‰¤ x â‰¤ 2 or [- 4, 2]

âˆ´ x is less than or equal to 2 and greater than or equal to – 4.

4. – 15 < \(\frac{3(x – 2)}{5}\) â‰¤ 0

– 15 Ã— 5 < \(\frac{3(x – 2)}{5}\) Ã— 5 â‰¤ 0 Ã— 5 [Multiplying by 5]

or – 75 < 3(x – 2) â‰¤ 0

or – \(\frac{75}{3}\) < x – 2 â‰¤ \(\frac{0}{3}\) [Divide by 3]

or – 25 < x – 2 â‰¤ 0

or – 25 < x – 2 â‰¤ 0

or – 25 + 2 < x – 2 + 2 â‰¤ 0 + 2 [Add 2]

or – 23 < x â‰¤ 2.

5. – 12 < 4 – \(\frac{3x}{- 5}\) â‰¤ 2.

Subtracting 4 from each side, we get

– 12 – 4 < 4 – \(\frac{3x}{- 5}\) – 4 â‰¤ 2 – 4.

or – 16 < \(\frac{3x}{5}\) â‰¤ – 2.

Multiplying each side by \(\frac{5}{3}\), we get

6. 7 â‰¤ \(\frac{3x+11}{2}\) â‰¤ 11.

Multiplying each side by 2, we get

7 Ã— 2 â‰¤ \(\frac{3x+11}{2}\) Ã— 2 â‰¤ 11 Ã— 2.

or 14 â‰¤ 3x + 11 â‰¤ 22.

Subtracting 11, we get

14 – 11 â‰¤ 3x + 11 – 11 â‰¤ 22 – 11.

or 3 â‰¤ 3x â‰¤ 11.

Dividing by 3, we get

1 â‰¤ x â‰¤ \(\frac{11}{3}\) or x âˆˆ [1, \(\frac{11}{3}\)].

i.e; x is greater than or equal to 1 and less than or equal to \(\frac{11}{3}\).

Solve the inequalities from questions 7 to 10 and represent the solutions graphically on the number line:

7. 5x + 1 > – 24, 5x – 1 < 24

8. 2(x – 1) < x + 5, 3x + 2) > 2 – x

9. 3x – 7 > 2(x – 6), 6 – x > 11 – 2x

10. 5(2x – 7) – 3(2x + 3) â‰¤ 0, 2x + 19 â‰¤ 6x + 47

Solutions to questions from 7 to 11:

7. (i) 5x + 1 > – 24, subtracting 1, 5x > – 24 – 1

or 5x > – 25

âˆ´ Dividing by 5, x > – 5.

(ii) 5x – 1 < 24.

âˆ´ Adding 1, 5x < 24 + 1 = 25.

âˆ´ Dividing by 5, x < 5. x > – 5 and x < 5 are shown graphically as follows:

i.e., x is greater than – 5 and less than 5.

8. The inequalities are

Graph of the inequalities are shown below

i.e; x âˆˆ (- 1, 3) or x is greater than – 1 and less than 3.

9. The inequalities are 3x – 7 > 2(x – 6) and 6 – x > 11 – 2x.

or 3x – 7 > 2x – 12

Graphs of these inequalities are shown below:

x > 5 satisfies both the inequalities x > – 5 and x > 5

âˆ´ Solution is x > 5.

10. The inequalities are

Graphs of these inequalities are shown below:

x â‰¤ 11 i.e; x âˆˆ [- 7, 11]

Solution of x is greater than or equal to – 7 and x less or equal to 11.

11. A solution is to be kept between 68Â° F and 77Â°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:

We have:

F = \(\frac{9}{5}\)C + 32

But 68Â° < F < 77Â°.

âˆ´ 68Â° < \(\frac{9}{5}\)C + 32 < 77Â°.

Subtracting 32 from each side

68 – 32 < \(\frac{9}{5}\)C < 77 – 32

36 < \(\frac{9}{5}\)C < 45

Multiplying by \(\frac{5}{9}\), we get

36 Ã— \(\frac{5}{9}\) < C < 45 Ã— \(\frac{5}{9}\)

20 < C < 25. 12. A solution of 8% boric acid is to be diluted by adding 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 8% solution, how many litres of the 2% solution will have to be added? Solution: Let the 2% boric acid solution be x litres. âˆ´ Mixture is (640 + x) litres. or \(\frac{2x}{100}\) + \(\frac{8 Ã— 640}{100}\) > \(\frac{4}{100}\) (640 + x)

or 2x + 5120 > 2560 + 4x

or 5120 – 2560 > 4x – 2x

or 2x < 2560.

or x < 1280

Also, 2% of x + 8% of 640 < 6% of (640 + x).

or \(\frac{2x}{100}\) + \(\frac{8Ã—640}{100}\) < \(\frac{6}{100}\) (640 + x)

or 2x + 5120 < 3840 + 6x

or 5120 – 3840 < 6x – 2x or 4x > 1280 or x > 320.

Thus, the number of litres to be added should be greater than 320 litres.

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?

Solution:

Let x litres water is added to 45% solution of acid.

(i) âˆ´ 25% of (1125 + x) < \(\frac{1125Ã—45}{100}\)

or \(\frac{1125Ã—x}{4}\) < \(\frac{1125Ã—9}{20}\).

Multiplying by 20, we get

5(1125 + x) < 10125

or 5625 + 5x < 10125.

Subtracting 5625 from both sides, we get

âˆ´ 5x < 10125 – 5625 = 4500.

or x < \(\frac{4500}{5}\) = 900 …………………. (1) (ii) Further 30% of (1125 + 4) > \(\frac{1125Ã—45}{100}\).

or \(\frac{3}{10}\) Ã— (1125 + x) > \(\frac{1125Ã—45}{100}\) > \(\frac{1125Ã—9}{20}\).

Multiplying both sides by 20, we get

6(1125 + x) > 1125 Ã— 9

or 6750 + 6x > 10125.

Subtracting 6750 from both sides, we get

6x > 10125 – 6750 > 3375

or x > \(\frac{3375}{6}\) = 562.5 ……………….. (2)

From (1) and (2), we have:

562.5 < x < 900.

14. IQ of a person is given by the formula

IQ = \(\frac{MA}{CA}\) Ã— 100,

where MA is mental age and CA chronological age. If 80 â‰¤ IQ â‰¤ 140 for a group of 12 years old children, find the range of their mental ages.

Solution:

We have:

IQ = \(\frac{MA}{CA}\) Ã— 100

We are given 80 â‰¤ IQ â‰¤ 140.

Putting the values of IQ in it, we get

80 â‰¤ \(\frac{MA}{CA}\) Ã— 100 â‰¤ 140.

Multiplying both sides by 12, we get

12 Ã— 80 â‰¤ MA Ã— 100 â‰¤ 140 Ã— 12

or 960 â‰¤ 100 Ã— MA â‰¤ 1680.

Dividing by 100, we get

9.6 â‰¤ MA â‰¤ 16.8.

Hence, mental age is greater than or equal to 9.6 and less than or equal to 16.8 years.