# Chapter 6 Miscellaneous ncert math solution class 11

# Chapter 6(Miscellaneous Exercise) Class 11 Math

**Solve the inequalities in Exercises 1 to 6.I(Chapter 6 Miscellaneous ncert math solution class 11)**

**Question 1: 2 ≤ 3x – 4 ≤ 5**

**Solution: Given that,**

2 ≤ 3x – 4 ≤ 5

⇒ 2 ≤ 3x – 4 ≤ 5

⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4

⇒ 6 ≤ 3x ≤ 9

⇒ 6/3 ≤ 3x/3 ≤ 9/3

⇒ 2 ≤ x ≤ 3

Hence, The solutions of given inequality.

**x ∈ [2, 3]**

**Question 2: 6 ≤ –3 (2x – 4) < 12**

**Solution: Given that**

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 with -1 changes the inequality sign.]

⇒ -2 + 4 ≥ 2x – 4 + 4 > -4 + 4

⇒ 2 ≥ 2x > 0

⇒ 0 < 2x ≤ 2

Dividing the inequality by 2,

⇒ 0 < x ≤ 1

Hence, solutions of given inequality.

x ∈ (0, 1]

**Question 3: – 3 ≤ 4 – 7x/2 ≤ 18**

**Solution: Given that**

– 3 ≤ 4 – 7x/2 ≤ 18

⇒ – 3 – 4 ≤ 4 – 7x/2 – 4 ≤ 18 – 4

⇒ – 7 ≤ – 7x/2 ≤ 18 – 14

Multiplying the inequality by -2,

⇒ 14 ≥ 7x ≥ -28

⇒ -28 ≤ 7x ≤ 14

Dividing the inequality by 7,

⇒ -4 ≤ x ≤ 2

Hence, solutions of given inequality is

**x ∈ [-4, 2]**

**Solution 4: – 15 ≤ 3(x – 2)/5 ≤ 0**

**Solution: Given that**

– 15 ≤ 3(x – 2)/5 ≤ 0

⇒ – 15 < 3(x – 2)/5 ≤ 0

Multiplying the inequality by 5,

⇒ -75 < 3(x – 2) ≤ 0

Dividing the inequality by 3, we get

⇒ -25 < x – 2 ≤ 0

⇒ – 25 + 2 < x – 2 + 2 ≤ 0 + 2

⇒ – 23 < x ≤ 2

Hence, solutions of given inequality.

x ∈ (-23, 2]

**Question 5: – 12 < 4 – 3x/ (-5) ≤ 2**

**Solution: Given that**

** – 12 < 4 – 3x/ (-5) ≤ 2**

Multiplying by 5

Divide by 3

Hence, solutions of given inequality.

**x ∈ (-80/3, -10/3]**

**Question 6: 7 ≤ (3x + 11)/2 ≤ 11**

**Solution: Given that,**

Multiplying by 2

⇒ 14 ≤ 3x + 11 ≤ 22

⇒ 14 – 11 ≤ 3x + 11 – 11 ≤ 22 – 11

⇒ 3 ≤ 3x ≤ 11

⇒ 1 ≤ x ≤ 11/3

Hence, solutions of given equality.

**x ∈ [1, 11/3]**

**Solve the inequalities in Exercises 7 to 11 and represent the solution graphically on the number line.**

**Question 7: 5x + 1 > – 24, 5x – 1 < 24**

**Solution: Given that**

5x + 1 > -24 and 5x – 1 < 24

Taken equation 5x + 1 > -24

⇒ 5x > -24 – 1

⇒ 5x > -25

⇒ x > -5

Taken equation 5x – 1 < 24

⇒ 5x < 24 + 1

⇒ 5x < 25

⇒ x < 5 ……….(ii)

From equations (i) and (ii),

The solution of given inequalities is (-5, 5).

**Question 8: 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x**

**Solution: Given that**

2 (x – 1) < x + 5 and 3 (x + 2) > 2 – x

Taking first inequalities

2 (x – 1) < x + 5

⇒ 2x – 2 < x + 5

⇒ 2x – x < 5 + 2

⇒ x < 7 ……… (i)

Taking second inequalities

3 (x + 2) > 2 – x

⇒ 3x + 6 > 2 – x

⇒ 3x + x > 2 – 6

⇒ 4x > -4

⇒ x > -1 ………. (ii)

From equations (i) and (ii),

The solution of given inequalities is (-1, 7).

**Question 9: 3x – 7 > 2(x – 6), 6 – x > 11 – 2x**

**Solution: Given that**

3x – 7 > 2(x – 6) and 6 – x > 11 – 2x

Taking first equation

3x – 7 > 2(x – 6)

⇒ 3x – 7 > 2x – 12

⇒ 3x – 2x > 7 – 12

⇒ x > -5 ………… (i)

Taking second equation

6 – x > 11 – 2x

⇒ 2x – x > 11 – 6

⇒ x > 5 ……….(ii)

From equations (i) and (ii),

The solution of given inequalities is (5, ∞).

**Question 10: 5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47**

**Solution: Given that**

5(2x – 7) – 3(2x + 3) ≤ 0 and 2x + 19 ≤ 6x + 47

Taking first inequalities

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ 4x ≤ 44

⇒ x ≤ 11 ……(i)

Taking second inequalities

2x + 19 ≤ 6x +47

⇒ 6x – 2x ≥ 19 – 47

⇒ 4x ≥ -28

⇒ x ≥ -7 ……….(ii)

From equations (i) and (ii),

The solution of given inequalities is (-7, 11).

**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 = (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° ……(i)

Putting in (i)

Multiplying by 5

Divide by 9

⇒ 20 < C < 25

Hence, we get

The range of temperature in degree Celsius is between 20° C to 25° C.

**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 question,

8% of 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 < 6% of (x + 640)

Taking first inequality

2% of x + 8% of 640 > 4% of (x + 640)

⇒ (2/100) × x + (8/100) × 640 > (4/100) × (x + 640)

⇒ 2x + 5120 > 4x + 2560

⇒ 5120 – 2560 > 4x – 2x

⇒ 2560 > 2x

⇒ x < 1280 ….(i)

Taking second inequality

2% of x + 8% of 640 < 6% of (x + 640)

⇒ (2/100) × x + (8/100) × 640 < (6/100) × (x + 640)

⇒ 2x + 5120 < 6x + 3840

⇒ 6x – 2x > 5120 – 3840

⇒ 4x > 1280

⇒ x > 320 ……….(i)

From (i) and (ii),

320 < x < 1280

Therefore, 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.

**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 question,

45% of 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 the resulting mixture = 45% of 1125 litres

∴ 45% of 1125 < 30% of (x + 1125) and 45% of 1125 > 25% of (x + 1125)

Taking first inequality

45% of 1125 < 30% of (x + 1125)

⇒ 45 × 1125 < 30x + 30 × 1125

⇒ (45 – 30) × 1125 < 30x

⇒ 15 × 1125 < 30x

⇒ x > 562.5 ………..(i)

Taking second inequality

45% of 1125 > 25% of (x + 1125)

⇒ 45 × 1125 > 25x + 25 × 1125

⇒ (45 – 25) × 1125 > 25x

⇒ 25x < 20 × 1125

⇒ x < 900 …..(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.

**14. IQ of a person is given by the formula, ,
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: **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,

We get

Multiply by 12

Divide by 100

∴ Range of mental age of the group of 12 years old children is 9.6 ≤ MA ≤ 16.8

Ex 6.1 Linear Inequalities ncert math solution class 11

Ex 6.2 Linear Inequalities ncert math solution class 11

Ex 6.3 Linear Inequalities ncert math solution class 11

Chapter 6 Miscellaneous ncert math solution class 11