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,

$\Rightarrow (-7)\times (-2) \geq -\frac{7x}{2}\times (-2) \geq 14 \times (-2)$

⇒ 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,

$\Rightarrow -15\times 5 < \dfrac{3(x-2)}{5}\times 5\leq 0\times 5$

⇒ -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

$\Rightarrow -12 < 4-\frac{3x}{-5}\leq 2$

$\Rightarrow -12-4 < 4-\frac{3x}{-5} -4\leq 2-4$

$\Rightarrow -16 <\dfrac{-3x}{-5} \leq -2$

$\Rightarrow -16 < \dfrac{3x}{5} \leq -2$

Multiplying by 5

$\Rightarrow -80 < 3x \leq -10$

Divide by 3

$\Rightarrow \dfrac{-80}{3} < x \leq \dfrac{-10}{3}$

Hence, solutions of given inequality.

x ∈ (-80/3, -10/3]

Question 6: 7 ≤ (3x + 11)/2 ≤ 11

Solution: Given that,

$7 \leq \dfrac{(3x+11)}{2}\leq 11$

Multiplying by 2

$\Rightarrow 7\times 2 \leq (3x+11) \leq 11\times 2$

⇒ 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).

Chapter 6 Miscellaneous ncert math solution class 11

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, ∞).

Chapter 6 Miscellaneous ncert math solution class 11

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).

Chapter 6 Miscellaneous ncert math solution class 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)

$F=\frac{9}{5}C+32$

Putting in (i)

$68 <\frac{9}{5}C+32< 77$

$\Rightarrow 68-32<\frac{9}{5}C<77-32$

$\Rightarrow 36<\frac{9}{5}C<45$

Multiplying by 5

$\Rightarrow 180 <9C<225$

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)

$\Rightarrow \dfrac{45}{100}\times <\dfrac{30}{100}\times (x+1125)$

⇒ 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)

$\Rightarrow \dfrac{45}{100}\times 1125>\dfrac{25}{100}\times (x+1125)$

⇒ 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, $IQ = \dfrac{MA}{CA}\times 100$ ,
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.