Linear Inequation MCQ Type Questions Class-10 ICSE Maths

Linear Inequation MCQ Type Questions Class-10 ICSE Maths for Sem-1. These MCQ  / Objective Type Questions of Linear Inequation is based on latest reduced syllabus according 2021-22 session on bifurcated pattern. Main motto of MCQ Type Question is cracking the next upcoming exam of council. Visit official website CISCE for detail information about ICSE Board Class-10 Maths

MCQ Type Questions of Linear Inequation for ICSE Class-10

Board	ICSE
Class	10th ( x )
Subject	Maths
Chapter	Linear Inequation
Syllabus	on bifurcated syllabus (after reduction)
bifurcated pattern	Semester-1
Session	2021-22
Topic	MCQ / Objective Type Question

ICSE Maths Linear Inequation MCQ Type Questions 

Question 1  Find the largest value of x for which  2(x – 1)  9 – x and x ∈W.

(a) 1

(b) 2

(c) 3

(d) 0

Answer- (c) 3

Question 5   3 – 2x  ≥ x – 12 given that x  N. the solutions set  of x is

(a) {1, 2, 3, 4, -5}

(b) {- 1, 2, 3, 4, 5}

(c) {1, – 2, 3, 4, 5}

(d) {1, 2, 3, 4, 5}

Answer- (d) {1, 2, 3, 4, 5}

Hint

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  Solution set of  the inequation, 3x – 11 < 3 where x ∈ {1, 2, 3, ……, 10}.is

(a) {-1, 2, 3, 4,}

(b) {1, -2, 3, 4,}

(c) {1, 2, -3, 4,}

(d) {1, 2, 3, 4,}

Answer- (d) {1, 2, 3, 4, }

Hint

Given inequation, 3x – 11 < 3

3x < 3 + 11

3x < 14

⇒ x < 14/3

But, x ∈ {1, 2, 3,……, 10}

Hence, the solution set is {1, 2, 3, 4}.

Question 7  If x is a negative integer, find the solution set of 2/3 + 1/3 (x + 1) > 0.

(a) {-1, -2}.

(b) {-2, -3}.

(c) {-3, -4}.

(d) {-4, -5}.

Answer-(a) {-1, -2}.

Hint

Given inequation, 2/3 + 1/3 (x + 1) > 0.

2/3 + x/3 + 1/3 > 0

x/3 + 1 > 0

x/3 > -1

⇒ x > -3

As x is a negative integer

The solution set is {-1, -2}.

Question 8   Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solution set will be x – 3 < 2x – 1.

(a) {, 2, 3, 4, 5, 6, 7, 9}

(b) {1, 2, 3, 4, 5, 6, 7, 9}

(c) {1, 2, 3, 4, 5, 6, 7, }

(d) {1, 2, 3, 5, 6, 7, 9}

Answer-(b) {1, 2, 3, 4, 5, 6, 7, 9}

Hint

Given inequation, x – 3 < 2x – 1

x – 2x < – 1 + 3

-x < 2

⇒ x > -2

But, x ∈ {1, 2, 3, 4, 5, 6, 7, 9}

Hence, the solution set is {1, 2, 3, 4, 5, 6, 7, 9}

Question 9   In following inequation the solution set are

-2 + 10x ≤ 13x + 10 ≤ 24 + 10x, x ∈ Z

(a) {, -1, 0, 1, }

(b){ -2, -1, 0, 1, 2, }

(c){-4, -3, -2, -1, 0, 1, 2, 3, 4}

(d) { -3, -2, -1, 0, 1, 2, 3, }

Answer-(c){-4, -3, -2, -1, 0, 1, 2, 3, 4}

 Hint

Given inequation, -2 + 10x ≤ 13x + 10 ≤ 24 + 10x

So, we have

-2 + 10x ≤ 13x + 10 and 13x + 10 ≤ 24 + 10x

10x – 13x ≤ 10 + 2 and 13x – 10x ≤ 24 – 10

-3x ≤ 12 and 3x ≤ 14

x ≥ -12/3 and x ≤ 14/3

x ≥ -4 and x ≤ 14/3

So, -4 ≤ x ≤ 14/3

As x ∈ Z

Thus, the solution set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}

Question 10 . If x ∈ I, A is the solution set of 2 (x – 1) < 3x – 1 and B is the solution set of 4x – 3 ≤ 8 + x, set of A ∩ B.

(a) {0, 1, 2, 3, …)

(b) {0, 1, 2, 3)

(c)  {4,5,6,…. )

(d) {3, 2, 1, 0, -1, …)

Answer- {0, 1, 2, 3)

Hint

Given inequations,

2 (x – 1) < 3x – 1 and 4x – 3 ≤ 8 + x for x ∈ I

Solving for both, we have

2x – 3x < 2 – 1 and 4x – x ≤ 8 + 3

-x < 1 and 3x ≤ 11

x > -1 and x ≤ 11/3

Hence,

Solution set A = {0, 1, 2, 3, …}

Solution set B = {3, 2, 1, 0, -1, … }

Thus, A ⋂ B = {0, 1, 2, 3)

Question 11   A = {x : 11x – 5 > 7x + 3, x ∈ R} and B = {x : 18x – 9 ≥ 15 + 12x, x ∈ R}  the range of set A ∩ B is

(a) x < 4

(b)x < 4

(c) x ≤ 4

(d) x ≥ 4

Answer- (d) x ≥ 4

Hint

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

Solving for A,

11x – 5 > 7x + 3

11x – 7x > 3 + 5

4x > 8

x > 2

Hence, A = {x : x > 2, x ∈ R}

Next, solving for B

18x – 9 ≥ 15 + 12x

18x – 12x ≥ 15 + 9

6x ≥ 24

x ≥ 4

Hence, B = {x : x ≥ 4, x ∈ R}

Thus, A ∩ B = x ≥ 4

Question 12  Given 20 – 5 x < 5 (x + 8),  the smallest value of x, when (i) x ∈ I, (ii) x ∈ W, (iii) x ∈ N.

(a) {2,),(3) ,(4 )

(b)x {-2),(-1),( 0)

(c) x {-1), (0), (1,)

(d) x {1), (2,) (3,)

Answer- (c) x {-1), (0), (1,)

Hint

Given inequation, 20 – 5 x < 5 (x + 8)

20 – 5x < 5x + 40

-5x – 5x < 40 – 20

-10x < 20

-x < 20/10

x > -2

Thus,

(i) For x ∈ I, the smallest value = -1

(ii) For x ∈ W, the smallest value = 0

(iii) For x ∈ N, the smallest value = 1

Question -13 the solution set of   

  if x ∈ N

(a){1, 2, 3, 4, 5, .. , 13}

(b){, 2, 3, 4, 5, .. , 13}

(c) {, 3, 4, 5, .. , 13}

(d){, 4, 5, .. , 13}

Answer- (a) {1, 2, 3, 4, 5, .. , 13}

Hint

=> 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 -14 the values of x, which satisfy the inequation :

, x ∈ N.

(a) 15/4 ≥ x ≥ – 5

(b)15/4 ≥ x ≥ – 4

(c) 15/4 ≥ x ≥ – 3

(d) 15/4 ≥ x ≥ – 2

Answer-(d) 15/4 ≥ x ≥ – 2

Hint

, x ∈ N

⇒ – 2 – 1/2 ≤ 1/2 – 2x/3 – 1/2 ≤ 11/6 – 1/2

[By subtracting 1/2 on both sides of inequality]

⇒ – 5/2 ≤ 2x/3 ≤ 8/6

⇒ – 15 ≤ – 4x ≤ 8

⇒ 15 ≥ 4x ≥ – 8

⇒ 15/4 ≥ x ≥ – 2

Question- 15  The greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer

(a) greatest = 6

(b) greatest = 5

(c) greatest =4

(d) greatest = 3

Answer -(a) greatest = 6

Hint 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 -16 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.

(a) shortest in length 4 metres

(b) shortest in length5 metres

(c) shortest in length 6 metres

(d) shortest in length 7 metres

Answer- (c) shortest in length 6 metres

Hint Length of pole which is buried in mud = x/3

Length of pole which is in the water = x/6

According to this problem,

x – [x/3 + x/6] ≥ 3

⇒ x – (2x + x)/6 ≥ 3

⇒ x – x/2 ≥ 3

⇒ x/2 ≥ 3

⇒ x ≥ 6

∴ Length of pole (shortest in length) = 6 metres

Question- 17 If x ∈ { – 3, – 1, 0, 1, 3, 5}, then the solution set of the inequation 3x – 2 ≤ 8 is

(a) { – 3, – 1, 1, 3}
(b) { – 3, – 1, 0, 1, 3}
(c) { – 3, – 2, – 1, 0, 1, 2, 3}
(d) { – 3, – 2, – 1, 0, 1, 2}

Solution set = (b) { -3, -1, 0, 1, 3}

Hint x ∈ { -3, -1, 0, 1, 3, 5}
3x – 2 ≤ 8
….⇒ 3x ≤ 8 + 2
and ⇒ 3x ≤ 10
so ⇒ x ≤ 10/3
therefore ⇒ x < 10/3
Solution set = { -3, -1, 0, 1, 3}

Question -18 If x ∈ W, then the solution set of the inequation 3x + 11 ≥ x + 8 is

(a) { – 2, – 1, 0, 1, 2, …}
(b) { – 1, 0, 1, 2, …}
(c) {0, 1, 2, 3, …}
(d) {x : x∈R,x≥-3/2}

Answer -(c) {0, 1, 2, 3, …}

Hint x ∈ W
3x + 11 ≥ x + 8
⇒ 3x – x ≥ 8 – 11

⇒ 2x ≥ – 3

⇒ x ≥ -3/2

⇒ x ≥ -1.1/2

Solution set = {0, 1, 2, 3,…..}

Question -19 If x ∈ W, then the solution set of the inequation 5 – 4x ≤ 2 – 3x is

(a) {…, – 2, – 1, 0, 1, 2, 3}
(b) {1, 2, 3}
(c) {0, 1, 2, 3}
(d) {x : x ∈ R, x ≤ 3}

Answer -(c) {0, 1, 2, 3}

Hint x ∈ W
5 – 4x < 2 – 3x
⇒ 5 – 2 ≤ 3x + 4x
⇒ 3 ≤ x
Solution set = {0, 1, 2, 3,} (c)

Question -20  If x ∈ I, then the solution set of the inequation 1 < 3x + 5 ≤ 11 is

(a) { – 1, 0, 1, 2}
(b) { – 2, – 1, 0, 1}
(c) { – 1, 0, 1}
(d) {x : x ∈ R, -4/3 < x ≤ 2}

Answer- (a) { – 1, 0, 1, 2}
Hint
x ∈ I
1 < 3x + 5 ≤ 11
⇒ 1 < 3x + 5
⇒ 1 – 5 < 3x

⇒ – 4 < 3x

⇒ -4/3 < x

And 3x + 5 ≤ 11 ⇒ 3x ≤ 11 – 5

⇒ 3x ≤ 6

⇒ x ≤ 6/3

⇒ x ≤ 2

∴ -4/3 < x ≤ 2

Solution set = {- 1, 0, 1, 2}

Question -21  If x ∈ R, the solution set of 6 ≤ – 3 (2x – 4) < 12 is

(a) {x : x ∈ R, 0 < x ≤ 1}
(b) {x : x ∈ R, 0 ≤ x < 1}
(c) {0, 1}
(d) none of these

Answer- (a) {x : x ∈ R, 0 < x ≤ 1}

Hint x ∈ R
6 ≤ – 3(2x – 4) < 12
⇒ 6 ≤ – 3(2x – 4)
⇒ 6 ≤ – 6x + 12

⇒ 6x ≤ 12 – 6

⇒ 6x ≤ 6

⇒ x ≤ 6/6

⇒ x ≤ 1

And -3(2x – 4) < 12

⇒ – 6x + 12 < 12

⇒ – 6x < 0

⇒ x < 0 ………(ii)

From (i) and (iii),

∴ 0 < x ≤ 1

Solution set = {x : x ∈ R, 0 < x ≤ 1}

Question -22 Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is at least 3.

(a) {18, 19, 20}
(b) {17, 18, 19}
(c) {16, 17, 18}
(d) none of these

Answer- (a) {18, 19, 20}

Hint

Let first least natural number = x
then second number = x + 1
and third number = x + 2

According to the condition 1/3(x + 2) – 1/5 (x) ≥ 3

5x + 10 – 3x ≥ 45

(Multiplying by 15 the L.C.M. of 2 and 5)

2x ≥ 45 – 10

⇒ 2x ≥ 35

x ≥ 35/2

⇒ x ≥ 17.1/2

∵ x is a natural least number

∴ x = 18

∴ find least natural number = 18

Second number = 18 + 1 = 19

And third numbers = 18 + 2 = 20

Hence, least natural numbers are 18, 19, 20

–: End of Linear Inequation MCQ Type Questions Class-10 ICSE Maths :-

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