## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4

**More Exercises**

- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test

**Question 1.**

**Solve the inequation 3x -11 < 3 where x ∈ {1, 2, 3,……, 10}. Also represent its solution on a number line**

**Solution:**

3x – 11 < 3 => 3x < 3 + 11 => 3x < 14 x < \(\\ \frac { 14 }{ 3 } \)

But x ∈ 6 {1, 2, 3, ……., 10}

Solution set is (1, 2, 3, 4}

Ans. Solution set on number line

**Question 2.**

**Solve 2(x – 3)< 1, x ∈ {1, 2, 3, …. 10}**

**Solution:**

2(x – 3) < 1 => x – 3 < \(\\ \frac { 1 }{ 2 } \) => x < \(\\ \frac { 1 }{ 2 } \) + 3 => x < \(3 \frac { 1 }{ 2 } \)

But x ∈ {1, 2, 3 …..10}

Solution set = {1, 2, 3} Ans.

**Question 3.**

**Solve : 5 – 4x > 2 – 3x, x ∈ W. Also represent its solution on the number line.**

**Solution:**

5 – 4x > 2 – 3x

– 4x + 3x > 2 – 5

=> – x > – 3

=> x < 3

x ∈ w,

solution set {0, 1, 2}

Solution set on Number Line :

**Question 4.**

**List the solution set of 30 – 4 (2.x – 1) < 30, given that x is a positive integer.**

**Solution:**

30 – 4 (2x – 1) < 30

30 – 8x + 4 < 30

– 8x < 30 – 30 – 4

– 8x < – 4 x > \(\\ \frac { -4 }{ -8 } \)

=> x > \(\\ \frac { 1 }{ 2 } \)

x is a positive integer

x = {1, 2, 3, 4…..} Ans.

**Question 5.**

**Solve : 2 (x – 2) < 3x – 2, x ∈ { – 3, – 2, – 1, 0, 1, 2, 3} .**

**Solution:**

2(x – 2) < 3x – 2

=> 2x – 4 < 3x – 2

=> 2x – 3x < – 2 + 4

=> – x < 2

=> x > – 2

Solution set = { – 1, 0, 1, 2, 3} Ans.

**Question 6.**

**If x is a negative integer, find the solution set of \(\\ \frac { 2 }{ 3 } \)+\(\\ \frac { 1 }{ 3 } \) (x + 1) > 0.**

**Solution:**

\(\\ \frac { 2 }{ 3 } \)+\(\\ \frac { 1 }{ 3 } \) x + \(\\ \frac { 1 }{ 3 } \) > 0

=> \(\\ \frac { 1 }{ 3 } \) x + 1 > 0

=> \(\\ \frac { 1 }{ 3 } \) x > – 1

x is a negative integer

Solution set = {- 2, – 1} Ans.

**Question 7.**

**Solve: \(\\ \frac { 2x-3 }{ 4 } \)≥\(\\ \frac { 1 }{ 2 } \), x ∈ {0, 1, 2,…,8}**

**Solution:**

\(\\ \frac { 2x-3 }{ 4 } \)≥\(\\ \frac { 1 }{ 2 } \)

=> 2x – 3 ≥ \(\\ \frac { 4 }{ 2 } \)

**Question 8.**

**Solve x – 3 (2 + x) > 2 (3x – 1), x ∈ { – 3, – 2, – 1, 0, 1, 2, 3}. Also represent its solution on the number line.**

**Solution:**

x – 3 (2 + x) > 2 (3x – 1)

=> x – 6 – 3x > 6x – 2

=> x – 3x – 6x > – 2 + 6

=> – 8x > 4

=> x < \(\\ \frac { -4 }{ 8 } \) => x < \(– \frac { 1 }{ 2 } \)

x ∈ { – 3, – 2, – 1, 0, 1, 2}

.’. Solution set = { – 3, – 2, – 1}

Solution set on Number Line :

**Question 9.**

**Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solve x – 3 < 2x – 1.**

**Solution:**

x – 3 < 2x – 1

x – 2x < – 1 + 3 => – x < 2 x > – 2

But x ∈ {1, 2, 3, 4, 5, 6, 7, 9}

Solution set = {1, 2, 3, 4, 5, 6, 7, 9} Ans.

**Question 10.**

**Given A = {x : x ∈ I, – 4 ≤ x ≤ 4}, solve 2x – 3 < 3 where x has the domain A Graph the solution set on the number line.**

**Solution:**

2x – 3 < 3 => 2x < 3 + 3 => 2x < 6 => x < 3

But x has the domain A = {x : x ∈ I – 4 ≤ x ≤ 4}

Solution set = { – 4, – 3, – 2, – 1, 0, 1, 2}

Solution set on Number line :

**Question 11.**

**List the solution set of the inequation**

**\(\\ \frac { 1 }{ 2 } \) + 8x > 5x \(– \frac { 3 }{ 2 } \), x ∈ Z**

**Solution:**

\(\\ \frac { 1 }{ 2 } \) +8x > 5x \(– \frac { 3 }{ 2 } \)

**Question 12.**

**List the solution set of \(\\ \frac { 11-2x }{ 5 } \) ≥ \(\\ \frac { 9-3x }{ 8 } \) + \(\\ \frac { 3 }{ 4 } \),**

**x ∈ N**

**Solution:**

\(\\ \frac { 11-2x }{ 5 } \) ≥ \(\\ \frac { 9-3x }{ 8 } \) + \(\\ \frac { 3 }{ 4 } \)

=> 88 – 16x ≥ 45 – 15x + 30

(L.C.M. of 8, 5, 4 = 40}

=> – 16x + 15x ≥ 45 + 30 – 88

=> – x ≥ – 13

=>x ≤ 13

x ≤ N.

Solution set = {1, 2, 3, 4, 5, .. , 13} Ans.

**Question 13.**

**Find the values of x, which satisfy the inequation : \(-2\le \frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le 1\frac { 5 }{ 6 } \), x ∈ N.**

**Graph the solution set on the number line. (2001)**

**Solution:**

\(-2\le \frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le 1\frac { 5 }{ 6 } \), x ∈ N

**Question 14.**

**If x ∈ W, find the solution set of**

**\(\frac { 3 }{ 5 } x-\frac { 2x-1 }{ 3 } >1\)**

**Also graph the solution set on the number line, if possible.**

**Solution:**

\(\frac { 3 }{ 5 } x-\frac { 2x-1 }{ 3 } >1\)

9x – (10x – 5) > 15 (L.C.M. of 5, 3 = 15)

=> 9x – 10x + 5 > 15

=> – x > 15 – 5

=> – x > 10

=> x < – 10

But x ∈ W

Solution set = Φ

Hence it can’t be represented on number line.

**Question 15.**

**Solve:**

**(i)\(\frac { x }{ 2 } +5\le \frac { x }{ 3 } +6\) where x is a positive odd integer.**

**(ii)\(\frac { 2x+3 }{ 3 } \ge \frac { 3x-1 }{ 4 } \) where x is positive even integer.**

**Solution:**

(i) \(\frac { x }{ 2 } +5\le \frac { x }{ 3 } +6\)

**Question 16.**

**Given that x ∈ I, solve the inequation and graph the solution on the number line :**

**\(3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \ge 2 \) (2004)**

**Solution:**

\(3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \) and \(3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \ge 2 \)

**Question 17.**

**Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2x – 1 < x + 4.**

**Solution:**

-3 < 2x – 1 < x + 4.

=> – 3 < 2x – 1 and 2x – 1 < x + 4

=> – 2x < – 1 + 3 and 2x – x < 4 + 1

=> – 2x < 2 and x < 5

=> – x < 1

=> x > – 1

– 1 < x < 5

x ∈ {1, 2, 3, 4, 5, 6, 7, 9}

Solution set = {1, 2, 3, 4} Ans.

**Question 18.**

**Solve : 1 ≥ 15 – 7x > 2x – 27, x ∈ N**

**Solution:**

1 ≥ 15 – 7x > 2x – 27

1 ≥ 15 – 7x and 15 – 7x > 2x – 27

**Question 19.**

**If x ∈ Z, solve 2 + 4x < 2x – 5 ≤ 3x. Also represent its solution on the number line.**

**Solution:**

2 + 4x < 2x – 5 ≤ 3x

2 + 4x < 2x – 5 and 2x – 5 ≤ 3x => 4x – 2x < – 5 – 2 ,and 2x – 3x ≤ 5

**Question 20.**

**Solve the inequation = 12 + \(1 \frac { 5 }{ 6 } x\) ≤ 5 + 3x, x ∈ R. Represent the solution on a number line. (1999)**

**Solution:**

12 + \(1 \frac { 5 }{ 6 } x\) ≤ 5 + 3x

**Question 21.**

**Solve : \(\\ \frac { 4x-10 }{ 3 } \)≤\(\\ \frac { 5x-7 }{ 2 } \) x ∈ R and represent the solution set on the number line.**

**Solution:**

\(\\ \frac { 4x-10 }{ 3 } \)≤\(\\ \frac { 5x-7 }{ 2 } \)

=> 8x – 20 ≤ 15x – 21

(L.C.M. of 3, 2 = 6)

**Question 22.**

**Solve \(\frac { 3x }{ 5 } -\frac { 2x-1 }{ 3 } \) > 1, x ∈ R and represent the solution set on the number line.**

**Solution:**

\(\frac { 3x }{ 5 } -\frac { 2x-1 }{ 3 } \) > 1

=> 9x – (10x – 5) > 15

=> 9x – 10x + 5 > 15

=> – x > 15 – 5

=> – x > 10

=> x < – 10

x ∈ R.

.’. Solution set = {x : x ∈R, x < – 10}

Solution set on the number line

**Question 23.**

**Solve the inequation – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line. (2000)**

**Solution:**

– 3 ≤ 3 – 2x < 9

– 3 ≤ 3 – 2x and 3 – 2x < 9

2x ≤ 3 + 3 and – 2x < 9 – 3

2x ≤ 6 and – 2x < 6 => x ≤ 3 and – x < 3 => x ≤ – 3 and – 3 < x

– 3 < x ≤ 3.

Solution set= {x : x ∈ R, – 3 < x ≤ 3)

Solution on number line

**Question 24.**

**Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on number line. (2003)**

**Solution:**

2 ≤ 2x – 3 ≤ 5 .

2 ≤ 2x – 3 and 2x – 3 ≤ 5

2 + 3 ≤ 2x and 2x ≤ 5 + 3

5 ≤ 2x and 2x ≤ 8.

**Question 25.**

**Given that x ∈ R, solve the following inequation and graph the solution on the number line: – 1 ≤ 3 + 4x < 23. (2006)**

**Solution:**

We have

– 1 ≤ 3 + 4x < 23 => – 1 – 3 ≤ 4x < 23 – 3 => – 4 ≤ 4x < 20 => – 1 ≤ x < 5, x ∈ R

Solution Set = { – 1 ≤ x < 5; x ∈ R}

**Question 26.**

**Solve tlie following inequation and graph the solution on the number line. (2007)**

**\(-2\frac { 2 }{ 3 } \le x+\frac { 1 }{ 3 } <3+\frac { 1 }{ 3 } \) x∈R**

**Solution:**

Given \(-2\frac { 2 }{ 3 } \le x+\frac { 1 }{ 3 } <3+\frac { 1 }{ 3 } \) x∈R

\(-\frac { 8 }{ 3 } \le x+\frac { 1 }{ 3 } <\frac { 10 }{ 3 } \)

Multiplying by 3, L.C.M. of fractions, we get

**Question 27.**

**Solve the following inequation and represent the solution set on the number line :**

**\(-3<-\frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le \frac { 5 }{ 6 } ,x\in R\)**

**Solution:**

\(-3<-\frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le \frac { 5 }{ 6 } ,x\in R\)

**Question 28.**

**Solve \(\frac { 2x+1 }{ 2 } +2(3-x)\ge 7,x\in R\). Also graph the solution set on the number line**

**Solution:**

\(\frac { 2x+1 }{ 2 } +2(3-x)\ge 7,x\in R\)

**Question 29.**

**Solving the following inequation, write the solution set and represent it on the number line. – 3(x – 7)≥15 – 7x > \(\\ \frac { x+1 }{ 3 } \), n ∈R**

**Solution:**

– 3(x – 7)≥15 – 7x > \(\\ \frac { x+1 }{ 3 } \), n ∈R

**Question 30.**

**Solve the inequation :**

**\(-2\frac { 1 }{ 2 } +2x\le \frac { 4x }{ 3 } \le \frac { 4 }{ 3 } +2x,\quad x\in W\). Graph the solution set on the number line.**

**Solution:**

\(-2\frac { 1 }{ 2 } +2x\le \frac { 4x }{ 3 } \le \frac { 4 }{ 3 } +2x,\quad x\in W\)

**Question 31.**

**Solve the inequation 2x – 5 ≤ 5x + 4 < 11, where x ∈ I. Also represent the solution set on the number line. (2011)**

**Solution:**

2x – 5 ≤ 5x + 4 < 11 2x – 5 ≤ 5x + 4

=> 2x – 5 – 4 ≤ 5x and 5x + 4 < 11

=> 2x – 9 ≤ 5x and 5x < 11 – 4

and 5x < 7

=> 2x – 5x ≤ 9 and x < \(\\ \frac { 7 }{ 5 } \)

=> 3x > – 9 and x< 1.4

=> x > – 3

**Question 32.**

**If x ∈ I, A is the solution set of 2 (x – 1) < 3 x – 1 and B is the solution set of 4x – 3 ≤ 8 + x, find A ∩B.**

**Solution:**

2 (x – 1) < 3 x – 1

2x – 2 < 3x – 1

2x – 3x < – 1 + 2 => – x < 1 x > – 1

Solution set A = {0, 1, 2, 3, ..,.}

4x – 3 ≤ 8 + x

4x – x ≤ 8 + 3

=> 3x ≤ 11

=> x ≤ \(\\ \frac { 11 }{ 3 } \)

Solution set B = {3, 2, 1, 0, – 1…}

A ∩ B = {0, 1, 2, 3} Ans.

**Question 33.**

**If P is the solution set of – 3x + 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find**

**(i) P ∩ Q**

**(ii) Q – P.**

**Solution:**

(i) – 3 x + 4 < 2 x – 3

– 3x – 2x < – 3 – 4 => – 5x < – 7

**Question 34.**

**A = {x : 11x – 5 > 7x + 3, x ∈R} and B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}**

**Find the range of set A ∩ B and represent it on a number line**

**Solution:**

A = {x : 11x – 5 > 7x + 3, x ∈R}

B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}

Now, A = 11x – 5 > 7x + 3

=> 11x – 7x > 3 + 5

=> 4x > 8

=>x > 2, x ∈ R

**Question 35.**

**Given: P {x : 5 < 2x – 1 ≤ 11, x∈R)**

**Q{x : – 1 ≤ 3 + 4x < 23, x∈I) where**

**R = (real numbers), I = (integers)**

**Represent P and Q on number line. Write down the elements of P ∩ Q. (1996)**

**Solution:**

P= {x : 5 < 2x – 1 ≤ 11}

5 < 2x – 1 ≤ 11

**Question 36.**

**If x ∈ I, find the smallest value of x which satisfies the inequation \(2x+\frac { 5 }{ 2 } >\frac { 5x }{ 3 } +2\)**

**Solution:**

\(2x+\frac { 5 }{ 2 } >\frac { 5x }{ 3 } +2\)

=>\(2x-\frac { 5x }{ 3 } >2-\frac { 5 }{ 2 } \)

=>12x – 10x > 12 – 15

=> 2x > – 3

=>\(x>-\frac { 3 }{ 2 } \)

Smallest value of x = – 1 Ans.

**Question 37.**

**Given 20 – 5 x < 5 (x + 8), find the smallest value of x, when**

**(i) x ∈ I**

**(ii) x ∈ W**

**(iii) x ∈ N.**

**Solution:**

20 – 5 x < 5 (x + 8)

⇒ 20 – 5x < 5x + 40

⇒ – 5x – 5x < 40 – 20

⇒ – 10x < 20

⇒ – x < 2

⇒ x > – 2

(i) When x ∈ I, then smallest value = – 1.

(ii) When x ∈ W, then smallest value = 0.

(iii) When x ∈ N, then smallest value = 1. Ans.

**Question 38.**

**Solve the following inequation and represent the solution set on the number line :**

**\(4x-19<\frac { 3x }{ 5 } -2\le -\frac { 2 }{ 5 } +x,x\in R\)**

**Solution:**

We have

\(4x-19<\frac { 3x }{ 5 } -2\le -\frac { 2 }{ 5 } +x,x\in R\)

Hence, solution set is {x : -4 < x < 5, x ∈ R}

The solution set is represented on the number line as below.

**Question 39.**

**Solve the given inequation and graph the solution on the number line :**

**2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.**

**Solution:**

2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.

(a) 2y – 3 < y + 1

⇒ 2y – y < 1 + 3

⇒ y < 4

⇒ 4 > y ….(i)

(b) y + 1 ≤ 4y + 7

**Question 40.**

**Solve the inequation and represent the solution set on the number line.**

**\(-3+x\le \frac { 8x }{ 3 } +2\le \frac { 14 }{ 3 } +2x,Where\quad x\in I\)**

**Solution:**

Given : \(-3+x\le \frac { 8x }{ 3 } +2\le \frac { 14 }{ 3 } +2x,Where\quad x\in I\)

**Question 41.**

**Find the greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer.**

**Solution:**

Let the greatest integer = x

According to the condition,

2x + 7 > 3x

⇒ 2x – 3x > – 7

⇒ – x > – 7

⇒ x < 7

Value of x which is greatest = 6 Ans.

**Question 42.**

**One-third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water is greater than or equal to 3 metres. Find the length of the shortest pole.**

**Solution:**

Let the length of the shortest pole = x metre

Length of pole which is burried in mud = \(\\ \frac { x }{ 3 } \)

Length of pole which is in the water = \(\\ \frac { x }{ 6 } \)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 are helpful to complete your math homework.

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