Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 - Linear Inequations (in one variable)

Selina Concise Mathematics Class 10 ICSE Solutions Linear Inequations (in one variable)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations (in one variable)

Linear Inequations in One Variable Exercise 4A – Selina Concise Mathematics Class 10 ICSE Solutions

Exercise 4(A)

Question 1.

Solution:

Question 2.

State, whether the following statements are true or false:

(i) a < b, then a – c < b – c (ii) If a > b, then a + c > b + c

(iii) If a < b, then ac > bc

(iv) If a > b, then
(v) If a – c > b – d, then a + d > b + c

(vi) If a < b, and c > 0, then a – c > b – c

Where a, b, c and d are real numbers and c ≠ 0.

Solution:

(i) a < b ⇒ a – c < b – c The given statement is true.

(ii) If a > b ⇒ a + c > b + c

The given statement is true.

(iii) If a < b ⇒ ac < bc The given statement is false.

(iv) If a > b ⇒
The given statement is false.

(v) If a – c > b – d ⇒ a + d > b + c

The given statement is true.

(vi) If a < b ⇒ a – c < b – c (Since, c > 0)

The given statement is false.

Question 3.

 If x ∈ N, find the solution set of inequations.

(i) 5x + 3 ≤ 2x + 18

(ii) 3x – 2 < 19 – 4x

Solution:

(i) 5x + 3 ≤ 2x + 18

5x – 2x ≤ 18 – 3

3x ≤ 15

x ≤ 5

Since, x ∈ N, therefore solution set is {1, 2, 3, 4, 5}.

(ii) 3x – 2 < 19 – 4x

3x + 4x < 19 + 2

7x < 21

x < 3

Since, x ∈ N, therefore solution set is {1, 2}.

Question 4.

Solution:

(i) x + 7 ≤ 11

x ≤ 11 – 7

x ≤ 4

Since, the replacement set = W (set of whole numbers)

⇒ Solution set = {0, 1, 2, 3, 4}

(ii) 3x – 1 > 8

3x > 8 + 1

x > 3

Since, the replacement set = W (set of whole numbers)

⇒ Solution set = {4, 5, 6, …}

(iii) 8 – x > 5

– x > 5 – 8

– x > -3

x < 3

Since, the replacement set = W (set of whole numbers)

⇒ Solution set = {0, 1, 2}

  

Since, the replacement set = W (set of whole numbers)

∴ Solution set = {0, 1, 2}

  

Since, the replacement set = W (set of whole numbers)

∴ Solution set = {0, 1}

(vi) 18 ≤ 3x – 2

18 + 2 ≤ 3x

20 ≤ 3x

x ≥
Since, the replacement set = W (set of whole numbers)

∴ Solution set = {7, 8, 9, …}

Question 5.

Solve the inequation:

3 – 2x ≥ x – 12 given that x ∈ N.

Solution:

3 – 2x ≥ x – 12

-2x – x ≥ -12 – 3

-3x ≥ -15

x ≤ 5

Since, x ∈ N, therefore,

Solution set = {1, 2, 3, 4, 5}

Question 6.

If 25 – 4x ≤ 16, find:

(i) the smallest value of x, when x is a real number,

(ii) the smallest value of x, when x is an integer.

Solution:

25 – 4x ≤ 16

-4x ≤ 16 – 25

-4x ≤ -9

x ≥
x ≥ 2.25

(i) The smallest value of x, when x is a real number, is 2.25.

(ii) The smallest value of x, when x is an integer, is 3.

Question 7.

Solution:

Question 8.

Solution:

Thus, the required smallest value of x is -1.

Question 9.

Find the largest value of x for which

2(x – 1) ≤ 9 – x and x ∈ W.

Solution:

2(x – 1) ≤ 9 – x

2x – 2 ≤ 9 – x

2x + x ≤ 9 + 2

3x ≤ 11

x ≤
x ≤ 3.67

Since, x ∈ W, thus the required largest value of x is 3.

Question 10.

Solution:

Question 11.

Given x ∈ {integers}, find the solution set of:

-5 ≤ 2x – 3 < x + 2

Solution:

Question 12.
Given x ∈ {whole numbers}, find the solution set of:
-1 ≤ 3 + 4x < 23

Solution:

Linear Inequations in One Variable Exercise 4B – Selina Concise Mathematics Class 10 ICSE Solutions

Exercise 4(B)

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

x ∈ {real numbers} and -1 < 3 – 2x ≤ 7, evaluate x and represent it on a number line.

Solution:

-1 < 3 – 2x ≤ 7

-1 < 3 – 2x and 3 – 2x ≤ 7

2x < 4 and -2x ≤ 4

x < 2 and x ≥ -2

Solution set = {-2 ≤ x < 2, x ∈ R}

Thus, the solution can be represented on a number line as:

  

Question 6.

List the elements of the solution set of the inequation

-3 < x – 2 ≤ 9 – 2x; x ∈ N.

Solution:

-3 < x – 2 ≤ 9 – 2x

-3 < x – 2 and x – 2 ≤ 9 – 2x

-1 < x and 3x ≤ 11

-1 < x ≤
 Since, x ∈ N

∴ Solution set = {1, 2, 3}

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Given x ∈ {real numbers}, find the range of values of x for which -5 ≤ 2x – 3 < x + 2 and represent it on a number line.

Solution:

-5 ≤ 2x – 3 < x + 2

-5 ≤ 2x – 3 and 2x – 3 < x + 2

-2 ≤ 2x and x < 5

-1 ≤ x and x < 5

Required range is -1 ≤ x < 5.

The required graph is:

  

Question 10.

If 5x – 3 ≤ 5 + 3x ≤ 4x + 2, express it as a ≤ x ≤ b and then state the values of a and b.

Solution:

5x – 3 ≤ 5 + 3x ≤ 4x + 2

5x – 3 ≤ 5 + 3x and 5 + 3x ≤ 4x + 2

2x ≤ 8 and -x ≤ -3

x ≤ 4 and x ≥ 3

Thus, 3 ≤  x ≤ 4.

Hence, a = 3 and b = 4.

Question 11.

Solve the following inequation and graph the solution set on the number line:

2x – 3 < x + 2 ≤ 3x + 5, x ∈ R.

Solution:

2x – 3 < x + 2 ≤ 3x + 5

2x – 3 < x + 2 and x + 2 ≤ 3x + 5

x < 5 and -3 ≤ 2x

x < 5 and -1.5 ≤ x

Solution set = {-1.5 ≤ x < 5}

The solution set can be graphed on the number line as:

Question 12.

Solve and graph the solution set of:

(i) 2x – 9 < 7 and 3x + 9 ≤ 25, x ∈ R (ii) 2x – 9 ≤ 7 and 3x + 9 > 25, x ∈ I

(iii) x + 5 ≥ 4(x – 1) and 3 – 2x < -7, x ∈ R

Solution:

Question 13.

Solve and graph the solution set of:

(i) 3x – 2 > 19 or 3 – 2x ≥ -7, x ∈ R

(ii) 5 > p – 1 > 2 or 7 ≤ 2p – 1 ≤ 17, p ∈ R

Solution:

(i) 3x – 2 > 19 or 3 – 2x ≥ -7

3x > 21 or -2x ≥ -10

x > 7 or x ≤ 5

Graph of solution set of x > 7 or x ≤ 5 = Graph of points which belong to x > 7 or x ≤ 5 or both.

Thus, the graph of the solution set is:

  

(ii) 5 > p – 1 > 2 or 7 ≤ 2p – 1 ≤ 17

6 > p > 3 or 8 ≤ 2p ≤ 18

6 > p > 3 or 4 ≤ p ≤ 9

Graph of solution set of 6 > p > 3 or 4 ≤ p ≤ 9

= Graph of points which belong to 6 > p > 3 or 4 ≤ p ≤ 9 or both

= Graph of points which belong to 3 < p ≤ 9

Thus, the graph of the solution set is:

Question 14.

Solution:

(i) A = {x ∈ R: -2 ≤ x < 5}

B = {x ∈ R: -4 ≤ x < 3}

(ii) A ∩ B = {x ∈ R: -2 ≤ x < 5}

It can be represented on number line as:

B’ = {x ∈ R: 3 < x ≤ -4}

A ∩ B’ = {x ∈ R: 3 ≤ x < 5}

It can be represented on number line as:

Question 15.

Use real number line to find the range of values of x for which:

(i) x > 3 and 0 < x < 6

(ii) x < 0 and -3 ≤ x < 1

(iii) -1 < x ≤ 6 and -2 ≤ x ≤ 3

Solution:

(i) x > 3 and 0 < x < 6

Both the given inequations are true in the range where their graphs on the real number lines overlap.

The graphs of the given inequations can be drawn as:

  

From both graphs, it is clear that their common range is

3 < x < 6

(ii) x < 0 and -3 ≤ x < 1

Both the given inequations are true in the range where their graphs on the real number lines overlap.

The graphs of the given inequations can be drawn as:

From both graphs, it is clear that their common range is

-3 ≤ x < 0

(iii) -1 < x ≤ 6 and -2 ≤ x ≤ 3

Both the given inequations are true in the range where their graphs on the real number lines overlap.

The graphs of the given inequations can be drawn as:

From both graphs, it is clear that their common range is

-1 < x ≤ 3

Question 16.

Illustrate the set {x: -3 ≤ x < 0 or x > 2, x ∈ R} on the real number line.

Solution:

Graph of solution set of -3 ≤ x < 0 or x > 2

= Graph of points which belong to -3 ≤ x < 0 or x > 2 or both

Thus, the required graph is:

Question 17.

Given A = {x: -1 < x ≤ 5, x ∈ R} and B = {x: -4 ≤ x < 3, x ∈ R}

Represent on different number lines:

(i) A ∩ B

(ii) A’ ∩ B

(iii) A – B

Solution:

Question 18.

P is the solution set of 7x – 2 > 4x + 1 and Q is the solution set of 9x – 45 ≥ 5(x – 5); where x ∈ R. Represent:

(i) P ∩ Q

(ii) P – Q

(iii) P ∩ Q’

on different number lines.

Solution:

Question 19.

Solution:

Question 20.

Given: A = {x: -8 < 5x + 2 ≤ 17, x ∈ I}, B = {x: -2 ≤ 7 + 3x < 17, x ∈ R}

Where R = {real numbers} and I = {integers}. Represent A and B on two different number lines. Write down the elements of A ∩ B.

Solution:

Question 21.

Solve the following inequation and represent the solution set on the number line 2x – 5 ≤ 5x +4 < 11, where x ∈ I

Solution:

Question 22.

Solution:

Question 23.

Given:

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 number line.

Solution:

A = {x: 11x – 5 > 7x + 3, x ∈ R}

= {x: 4x > 8, x ∈ R}

= {x: x > 2, x ∈ R}

B = {x: 18x – 9 ≥ 15 + 12x, x ∈ R}

= {x: 6x ≥ 24, x ∈ R}

= {x: x ≥ 4, x ∈ R}

A ∩ B = {x: x ≥ 4, x ∈ R}

It can be represented on number line as:

Question 24.

  Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Find three consecutive largest positive integers such that the sum of one-third of first, one-fourth of second and one-fifth of third is atmost 20.

Solution:

Let the required integers be x, x + 1 and x + 2.

According to the given statement,

  

Thus, the largest value of the positive integer x is 24.

Hence, the required integers are 24, 25 and 26.

Question 28.

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

⇒ 2y – 3 – y < y + 1 – y ≤ 4y + 7 – y

⇒ y – 3 < 1 ≤ 3y + 7

⇒ y – 3 < 1 and 1 ≤ 3y + 7

⇒ y < 4 and 3y ≥ 6 ⇒ y ≥ – 2

⇒ – 2 ≤ y < 4

The graph of the given equation can be represented on a number line as:

Question 29.

Solve the inequation:

3z – 5 ≤ z + 3 < 5z – 9, z ∈ R.

Graph the solution set on the number line.

Solution:

3z – 5 ≤ z + 3 < 5z – 9

3z – 5 ≤ z + 3 and z + 3 < 5z – 9

2z ≤ 8 and 12 < 4z

z ≤ 4 and 3 < z

Since, z R

∴ Solution set = {3 < z ≤ 4, x ∈ R }

It can be represented on a number line as:

  

Question 30.

  

Solution:

Question 31.

Solution:

Question 32.

Solution:

Question 33.

Solution:

Question 34.

Solve the following in equation and write the solution set:

13x – 5 < 15x + 4 < 7x + 12, x ∈ R

Solution:

Question 35.

Solve the following inequation, write the solution set and represent it on the number line.

-3(x – 7) ≥ 15 – 7x > x+1/3, x R.

Solution:

Question 36.

Solve the following inequation and represent the solution set on a number line.

  

Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 - Linear Inequations (in one variable)