**Miscellaneous Exercise**

Linear Equations

**Question and Answers**

**Class 11 – Maths**

Linear Equations

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Linear Equations |

Chapter No. | Chapter 5 |

Exercise | Miscellaneous Exercise |

Category | Class 11 Maths NCERT Solutions |

**Question 1 Solve the inequality 2 ≤ 3x – 4 ≤ 5**

**Answer** The inequality given is

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, all real numbers x greater than or equal to 2 but less than or equal to 3 are solutions of given equality.

x ∈ [2, 3]

**Question 2 Solve the inequality : 6 ≤ – 3 (2x – 4) < 12**

**Answer **The inequality given is

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

Dividing the inequality by 2,

⇒ 0 < x ≤ 1

Hence, all real numbers x greater than 0 but less than or equal to 1 are solutions of given equality.

x ∈ (0, 1]

**Question 3** **Solve the inequality : ****– 3 ≤ 4 – 7x/2 ≤ 18**

**Answer**

**– 3 ≤ 4 – 7x/2 ≤ 18**

As a result, the set of solutions for the given inequality is [−4,2].

**Question 4** **Solve the inequality :**

**Answer 4)** The inequality given is

– 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, all real numbers x greater than -23 but less than or equal to 2 are solutions of given equality.

x ∈ (-23, 2]

**Question 5** **Solve the inequality :**

**Answer **

**Question 6** **Solve the inequality :**

**Answer 6) **

All real numbers x greater than or equal to -4 but less than or equal to 2 are solutions of given equality.

x ∈ [1, 11/3]

**Question 7 Solve the inequalities and represent the solution graphically on number line:** **5x + 1 > – 24, 5x – 1 < 24**

**Answer** 5x + 1 > -24 and 5x – 1 < 24

5x + 1 > -24

⇒ 5x > -24 – 1

⇒ 5x > -25

⇒ x > -5 ……… (i)

5x – 1 < 24

⇒ 5x < 24 + 1

⇒ 5x < 25

⇒ x < 5 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-5, 5).

**Question 8** **Solve the inequalities and represent the solution graphically on number line:**

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

**Answer**

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

2 (x – 1) < x + 5

⇒ 2x – 2 < x + 5

⇒ 2x – x < 5 + 2

⇒ x < 7 ……… (i)

3 (x + 2) > 2 – x

⇒ 3x + 6 > 2 – x

⇒ 3x + x > 2 – 6

⇒ 4x > -4

⇒ x > -1 ………. (ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-1, 7).

**Question 9 Solve the inequalities and represent the solution graphically on number line: ****3x – 7 > 2 (x – 6) , 6 – x > 11 – 2x**

**Answer**

3x – 7 > 2(x – 6) and 6 – x > 11 – 2x

3x – 7 > 2(x – 6)

⇒ 3x – 7 > 2x – 12

⇒ 3x – 2x > 7 – 12

⇒ x > -5 ………… (i)

6 – x > 11 – 2x

⇒ 2x – x > 11 – 6

⇒ x > 5 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (5, ∞).

**Question 10 Solve the inequalities and represent the solution graphically on number line: 5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47**

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

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ 4x ≤ 44

⇒ x ≤ 11 ……(i)

2x + 19 ≤ 6x +47

⇒ 6x – 2x ≥ 19 – 47

⇒ 4x ≥ -28

⇒ x ≥ -7 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-7, 11).

**Question 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?**

**Answer ** The solution has to be kept between 68° F and 77° F. So, we get 68° < F < 77°

The required temperature range in degrees Celsius is between 20° C to 25° 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?**

**Answer** 8% of solution of boric acid = 640 litres

Let the amount of 2% boric acid solution added = x litres

Total mixture = x + 640 litres

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)

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)

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.

**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?**

**Answer** 45% of solution of acid = 1125 litres

Let the amount of water added = x litres

Resulting mixture = x + 1125 litres

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)

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

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

⇒ 45 × 1125 < 30x + 30 × 1125

⇒ (45 – 30) × 1125 < 30x

⇒ 15 × 1125 < 30x

⇒ x > 562.5 ………..(i)

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.

**Question 14 IQ of a person is given by the formula, IQ = (MA/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.**

**Answer ** Chronological age = CA = 12 years

IQ for the age group of 12 is 80 ≤ IQ ≤ 140.

We get that

80 ≤ IQ ≤ 140

Substituting,

⇒ 9.6 ≤ MA ≤ 16.8

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

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