# Class 11 NCERT Solutions- Chapter 6 Linear Inequalities – Miscellaneous Exercise on Chapter 6

### Solve the inequalities in Exercises 1 to 6.

### Question 1. 2 â‰¤ 3x â€“ 4 â‰¤ 5

**Solution:**

In this case, we have two inequalities, 2 â‰¤ 3x â€“ 4 and 3x â€“ 4 â‰¤ 5, which we will solve simultaneously.

We have

2 â‰¤ 3x â€“ 4 â‰¤ 5or 2 â‰¤ 3x â€“ 4 and 3x â€“ 4 â‰¤ 5

â‡’ 2 + 4 â‰¤ 3x and 3x â‰¤ 5 + 4

â‡’ 6 â‰¤ 3x and 3x â‰¤ 9

â‡’ â‰¤ x and x â‰¤

â‡’ 2 â‰¤ x and x â‰¤ 3

â‡’ 2 â‰¤ x â‰¤ 3Hence, all real numbers x greater than or equal to 2 but less than or equal to 3 are solution of given 2 â‰¤ 3x â€“ 4 â‰¤ 5 equality.

x âˆˆ [2, 3]

### Question 2. 6 â‰¤ â€“ 3 (2x â€“ 4) < 12

**Solution:**

In this case, we have two inequalities, 6 â‰¤ â€“ 3 (2x â€“ 4) and â€“ 3 (2x â€“ 4) < 12, which we will solve simultaneously.

We have

6 â‰¤ â€“ 3 (2x â€“ 4) < 12or 6 â‰¤ â€“ 3 (2x â€“ 4) and â€“ 3 (2x â€“ 4) < 12

â‡’ â‰¤ -(2x â€“ 4) and -(2x â€“ 4) <

â‡’ 2 â‰¤ -(2x â€“ 4) and -(2x â€“ 4) < 4

â‡’ -2 â‰¥ (2x â€“ 4) and (2x â€“ 4) > -4 [

multiplying the inequality with (-1) which changes the inequality sign]â‡’ -2+4 â‰¥ 2x and 2x > -4+4

â‡’ 2 â‰¥ 2x and 2x > 0

â‡’ â‰¥ x >

â‡’ 1â‰¥ x > 0

â‡’ 0 < x â‰¤ 1Hence, all real numbers x greater than 0 but less than or equal to 1 are solution of given 6 â‰¤ â€“ 3 (2x â€“ 4) < 12 equality.

x âˆˆ (0, 1]

### Question 3. -3 â‰¤ 4- â‰¤ 18

**Solution:**

In this case, we have two inequalities, -3 â‰¤ 4- and 4- â‰¤ 18, which we will solve simultaneously.

We have

-3 â‰¤ 4-â‰¤ 18or -3 â‰¤ 4- and 4- â‰¤ 18

â‡’ -3-4 â‰¤ – and – } â‰¤ 18-4

â‡’ -7 â‰¤ – and – â‰¤ 14

â‡’ 7 â‰¥ and â‰¥ -14 [

multiplying the inequality with (-1) which changes the inequality sign]â‡’ 7Ã—2 â‰¥ 7x and 7x â‰¥ -14Ã—2

â‡’ â‰¥ x and x â‰¥

â‡’ 2 â‰¥ x â‰¥ -4

â‡’ -4 â‰¤ x â‰¤ 2Hence, all real numbers x greater than or equal to -4 but less than or equal to 2 are solution of given -3 â‰¤ 4- â‰¤ 18 equality.

x âˆˆ [-4, 2]

### Question 4. -15 < â‰¤ 0

**Solution:**

In this case, we have two inequalities, -15 < and } â‰¤ 0, which we will solve simultaneously.

We have -15 < â‰¤ 0

or -15 < and â‰¤ 0

â‡’ -15Ã— < (x-2) and (x-2) â‰¤ 0Ã—

â‡’ -25 < x-2 and x-2 â‰¤ 0

â‡’ -25+2 < x and x â‰¤ 0+2

â‡’ -23<x â‰¤ 2Hence, all real numbers x greater than -23 but less than or equal to 2 are solution of given -15 â‰¤ â‰¤ 0 equality.

x âˆˆ (-23, 2]

### Question 5. -12 < 4- () â‰¤ 2

**Solution:**

In this case, we have two inequalities, -12 < 4- and 4- â‰¤ 2, which we will solve simultaneously.

We have

-12 < 4-â‰¤ 2or -12 < 4- and 4- â‰¤ 2

â‡’ -12-4 < and â‰¤ 2-4

â‡’ -16 < and â‰¤ -2

â‡’ -16 < and â‰¤ -2

â‡’ -16Ã—5 < 3x and 3x â‰¤ -2Ã—5

â‡’ < x and x â‰¤

â‡’Hence, all real numbers x greater than -80/3 but less than or equal to -10/3 are solution of given -12 < 4- â‰¤ 2 equality.

x âˆˆ (-80/3, -10/3]

### Question 6. 7 â‰¤ â‰¤ 11

**Solution:**

In this case, we have two inequalities, 7 â‰¤ and â‰¤ 11, which we will solve simultaneously.

We have

7 â‰¤â‰¤ 11or 7 â‰¤ and â‰¤ 11

â‡’ 7Ã—2 â‰¤ 3x+11 and 3x+11 â‰¤ 11Ã—2

â‡’ 14 â‰¤ 3x+11 and 3x+11 â‰¤ 22

â‡’ 14-11 â‰¤ 3x and 3x â‰¤ 22-11

â‡’ 3 â‰¤ 3x and 3x â‰¤ 11

â‡’ 1 â‰¤ x and x â‰¤

â‡’ 1 â‰¤ x â‰¤Hence, all real numbers x greater than or equal to 1 but less than or equal to 11/3 are solution of given 7 â‰¤ â‰¤ 11 equality.

x âˆˆ [1, 11/3]

### Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.

### Question 7. 5x + 1 > â€“ 24, 5x â€“ 1 < 24

**Solution:**

So, from given data

5x + 1 > â€“ 24 ……………………(1)

5x â€“ 1 < 24 …………………….(2)

From inequality (1), we have

5x + 1 > â€“ 24

5x > â€“ 24-1

x > â€“ 25/5

x > -5 ………………………….(3)Also, from inequality (2), we have

5x â€“ 1 < 24

5x < 24+1

x < 25/5

x < 5 ……………………………(4)So, from (3) and (4), we can conclude that,

-5 < x < 5 ……………….(5)If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are

x âˆˆ (-5,5)Thus, solution of the system are real numbers x lying between -5 and 5 excluding -5 and 5.

### Question 8. 2 (x â€“ 1) < x + 5, 3 (x + 2) > 2 â€“ x

**Solution:**

So, from given data

2 (x â€“ 1) < x + 5 ……………………(1)

3 (x + 2) > 2 â€“ x …………………….(2)

From inequality (1), we have

2 (x â€“ 1) < x + 5

2x â€“ 2 < x + 5

2x â€“ x < 5+2

x < 7 ………………………….(3)Also, from inequality (2), we have

3 (x + 2) > 2 â€“ x

3x + 6 > 2 â€“ x

3x + x > 2 â€“ 6

4x > -4

x > -1 ……………………………(4)So, from (3) and (4), we can conclude that,

-1 < x < 7 ……………….(5)If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are

x âˆˆ (-1,7)Thus, solution of the system are real numbers x lying between -1 and 7 excluding -1 and 7.

### Question 9. 3x â€“ 7 > 2 (x â€“ 6) , 6 â€“ x > 11 â€“ 2x

**Solution:**

So, from given data

3x â€“ 7 > 2 (x â€“ 6) ……………………(1)

6 â€“ x > 11 â€“ 2x …………………….(2)

From inequality (1), we have

3x â€“ 7 > 2 (x â€“ 6)

3x â€“ 7 > 2x â€“ 12

3x â€“ 2x > â€“ 12+7

x > -5 ………………………….(3)Also, from inequality (2), we have

6 â€“ x > 11 â€“ 2x

â€“ x+2x > 11 -6

x > 5 ……………………………(4)So, from (3) and (4), we can conclude that,

5 < x ……………….(5)If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are

x âˆˆ (5,âˆž)Thus, solution of the system are real numbers x lying between 5 and âˆž excluding 5 .

### Question 10. 5 (2x â€“ 7) â€“ 3 (2x + 3) â‰¤ 0 , 2x + 19 â‰¤ 6x + 47

**Solution:**

So, from given data

5 (2x â€“ 7) â€“ 3 (2x + 3) â‰¤ 0 ……………………(1)

2x + 19 â‰¤ 6x + 47 …………………….(2)

From inequality (1), we have

5 (2x â€“ 7) â€“ 3 (2x + 3) â‰¤ 0

10x – 35 -6x – 9 â‰¤ 0

4x – 44 â‰¤ 0

4x â‰¤ 44

x â‰¤ 44/4

x â‰¤ 11 ………………………….(3)Also, from inequality (2), we have

2x + 19 â‰¤ 6x + 47

2x – 6x â‰¤ 47 – 19

-4x â‰¤ 28

4x â‰¥ -28 [

multiplying the inequality with (-1) which changes the inequality sign]x â‰¥ -28/4

x â‰¥ -7 ……………………………(4)So, from (3) and (4), we can conclude that,

-7 â‰¤ x â‰¤ 11 ……………….(5)

x âˆˆ [-7,11)Thus, solution of the system are real numbers x lying between -7 and 11 including -7 and 11 .

### Question 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 = (****)C + 32**

**Solution:**

According to the given data

The solution has to be kept between 68Â° F and 77Â° F

So, we have,

68Â° < F < 77Â°

Substituting,

F =C + 32â‡’ 68Â° < C + 32 < 77Â°

â‡’ 68Â°- 32Â° < C < 77Â°- 32Â°

â‡’ 36Â° < C < 45Â°

â‡’ 36Ã— < C < 45Ã—

â‡’ 20Â° < C < 25Â°Hence, here we get,

The range of temperature in degree Celsius is between

20Â° C to 25Â° 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?

**Solution:**

According to the given data,

Here, 8% of solution of boric acid = 640 litres

So, we can take the amount of 2% boric acid solution added as x litres

Hence, Total mixture = (x + 640) litres

As it is given,

The resulting mixture has to be

more than 4% but less than 6% boric acidâ‡’

(2% of x + 8% of 640) > (4% of (x + 640)) and (2% of x + 8% of 640) < (6% of (x + 640))â‡’ () Ã— (x + 640) < () Ã— x + () Ã— 640) < () Ã— (x + 640)

â‡’ 4(x + 640) < (2Ã—x + 8Ã— 640) < 6(x + 640)

â‡’ 4x + 2560 < 2x +5120 < 6x+3840

In this case, we have two inequalities,

â‡’ 4x + 2560 < 2x +5120

and2x +5120 < 6x+3840â‡’ 4x – 2x < 5120 – 2560

and5120-3840 < 6x-2xâ‡’ 2x < 2560

and1280 < 4xâ‡’ x < 2560/2

and1280/4 < xâ‡’ x < 1280

and320 < x

â‡’ 320 < x < 1280Therefore, the number of litres of 2% of boric acid solution that has to be added will be more than

320 litres but less than 1280 litres.

### 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?

**Solution:**

According to the given data,

Here, 45% of solution of acid = 1125 litres

Let the amount of water added in the solution = x litres

Resulting mixture = (x + 1125) litres

As it is given,

The resulting mixture has to be

more than 25% but less than 30% acid contentAmount of acid in resulting mixture = 45% of 1125 litres.

â‡’ 45% of 1125 < 30% of (x + 1125) and 45% of 1125 > 25% of (x + 1125)â‡’ 25% of (x + 1125) < 45% of 1125 < 30% of (x + 1125)

In this case, we have two inequalities,

â‡’ ( Ã— (x + 1125)) < ( Ã— 1125)

and( Ã— (x + 1125)) > ( Ã— 1125)â‡’ (25(x + 1125)) < (45Ã—1125)

and(30(x + 1125)) > (45Ã—1125)â‡’ (x + 1125) < (45Ã—1125)/25

and(x + 1125) > (45Ã—1125)/30â‡’ (x + 1125) < 2025

and(x + 1125) > 3375/2â‡’ 3375/2 < (x + 1125) < 2025

â‡’ (3375/2)-1125 < x < 2025-1125

â‡’ 1125/2 < x < 900

â‡’ 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.

### Question 14. IQ of a person is given by the formula, **IQ = (****) Ã— 100, **where MA is mental age and CA is chronological age. If 80 â‰¤ IQ â‰¤ 140 for a group of 12 years old children, find the range of their mental age.

**Solution:**

According to the given data, we have

Chronological age = CA = 12 years

IQ for age group of 12 is in the range,

80 â‰¤ IQ â‰¤ 140

Substituting,

IQ = () Ã— 100â‡’ 80 â‰¤ Ã— 100 â‰¤ 140

â‡’ 80 â‰¤ Ã— 100 â‰¤ 140

â‡’ 80Ã—12/100 â‰¤ MA â‰¤ 140Ã—12/100

â‡’ 96/10 â‰¤ MA â‰¤ 168/10

â‡’ 9.6 â‰¤ MA â‰¤ 16.8Hence, Range of mental age (MA) of the group of 12 years old children is

9.6 â‰¤ MA â‰¤ 16.8

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