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### Maths MCQ Class 11 Ch-6 | Linear Inequalities

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__Mathematics__

## MCQ | Class 11 | Chapter 06

__Linear Inequalities__

__Multiple Choice Questions (MCQ)__

__Multiple Choice Questions (MCQ)__

**MCQ Based on the Linear inequalities.**

**MCQ Based on the Linear inequalities.****MCQ Based on the Graph of Linear Inequalities.**

**MCQ Based on the Graph of Linear Inequalities.****MCQ Based on the average of numbers.**

**MCQ Based on the average of numbers.**__Features__

__Features__

**In this post given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations.****Solution Hints are also given to some difficult problems.****Each MCQ contains four options from which one option is correct.**

__Action Plan__

__Action Plan__

**First of all students should Learn and write all basic points and Formulas related to the Linear Inequalities.****Start solving the NCERT Problems with examples.****Solve the important assignments on the Linear Inequalities Chapter 6 Class XI.****Then start solving the following MCQ.**

# MCQ | CHAPTER 6 | CLASS 11

LINEAR INEQUALITIES

**Question 1) **

**The length of a rectangle is three times the breadth. If the
minimum perimeter of the rectangle is 160 cm, then**

(a) breadth >
20 cm

(b) length <
20 cm

(c) breadth x ≥
20 cm

(d) length ≤ 20
cm

Answer: (c)

**Solution:**** **Let x be the breadth of a rectangle.

So, length = 3x

Given that the minimum perimeter of a rectangle is 160 cm. Thus,

⇒ 2 (3x + x) ≥ 160

⇒ 4x ≥ 80 ⇒ x ≥ 20

**Question 2):**7 > 5 is

a) linear inequality

b) quadratic inequality

c) numerical inequality

d) literal inequality

Explanation: Since here numbers are compared with inequality sign so, it is called numerical inequality.

**Question 3):**If x is a whole number and 10x ≤ 50 then find solution set of x.

a) {0, 1, 2, 3, 4, 5}

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

c) {1, 2, 3, 4}

d) {0, 1, 2, 3, 4}

Explanation: 10x ≤ 50

Dividing by 10 on both sides, x ≤ (50/10) ⇒ x ≤ 5

Since x is a whole number so x = 0, 1, 2, 3, 4, 5.

**Question 4):**

**If – 3x + 17 < – 13, then**

(a) x ∈
(10, ∞)

(b) x ∈
[10, ∞)

(c) x ∈
(– ∞, 10]

(d) x ∈
[– 10, 10)

Answer: (a)

**Solution:
**Given, -3x + 17
< -13

Subtracting 17
from both sides,

-3x + 17 – 17
< -13 – 17

⇒ -3x < -30

⇒ x > 10 {since the division by
negative number inverts the inequality sign}

⇒ x ∈ (10, ∞)

**Question 5):**If 2x + 1 > 5 then which is true?

a) x > 4

b) x < 4

c) x > 2

d) x < 2

**Answer c**

**Question 6):**x > 5 is

a) double inequality

b) quadratic inequality

c) numerical inequality

d) literal inequality

Explanation: Since a variable ‘x’ is compared with number ‘5’ with inequality sign so it is called literal inequality.

**Question 7):**If x – 1 > - x + 7 then which is true?

a) x > 4

b) x < 4

c) x > 2

d) x < 2

**Answer a**

**Question 8):**ax

^{2 }+ bx + c > 0 is

a) double inequality

b) quadratic inequality

c) numerical inequality

d) linear inequality

**Answer b**

**Question 9):**Rahul obtained 20 and 25 marks in first two tests. Find the minimum marks he should get in the third test to have an average of at least 30 marks.

a) 60

b) 35

c) 180

d) 45

**Question 10):**

**If |x −1| > 5, then**

(a) x ∈
(– 4, 6)

(b) x ∈
[– 4, 6]

(c) x ∈
(– ∞, – 4) U (6, ∞)

(d) x ∈
[– ∞, – 4) U [6, ∞)

Answer: (c)

**Solution:**** **|x – 1| > 5

x – 1 < – 5
and x – 1 > 5

x < -4 and x
> 6

Therefore, x ∈
(-∞, -4) U (6, ∞)

**Question 11):**Find all pairs of consecutive odd positive integers both of which are smaller than 8 such that their sum is more than 10.

a) (5, 7)

b) (3, 5), (5, 7)

c) (3, 5), (5, 7), (7, 9)

d) (5, 7), (7, 9)

**Question 12):**

**If**

**, then**

(a) x ∈
[7, ∞)

(b) x ∈
(7, ∞)

(c) x ∈
(– ∞, 7)

(d) x ∈
(– ∞, 7]

Answer: (b)

**Solution:**** **

Given, |x –
7|/(x – 7) ≥ 0

This is possible
when x − 7 ≥ 0, and x – 7 ≠ 0.

Here, x ≥ 7 but
x ≠ 7

Therefore, x
> 7, i.e. x ∈ (7, ∞).

**Question 13):**The longest side of a triangle is 2 times the shortest side and the third side is 4 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

a) 7

b) 9

c) 11

d) 13

**Question 14): ** If
Ram has x rupees and he pay 40 rupees to shopkeeper then find range of x if
amount of money left with Ram is at least 10 rupees is given by inequation is

a) x ≥ 10

b) x ≤ 10

c) x ≤ 50

d) x ≥ 50

Answer
d

x - 40 ≥ 10 ⇒ x ≥ 50.

**Question 15):**

**If |x + 3| ≥ 10, then**

(a) x ∈
(– 13, 7]

(b) x ∈
(– 13, 7]

(c) x ∈
(– ∞, – 13] ∪ [7, ∞)

(d) x ∈
[– ∞, – 13] ∪ [7, ∞)

Answer: (d)

**Solution:
**Given, |x + 3| ≥
10

⇒ x + 3 ≤ – 10 or x + 3 ≥ 10

⇒ x ≤ – 13 or x ≥ 7

⇒ x ∈ (– ∞, –
13] ∪ [7, ∞)

**Question 16):**2x + y > 5. Which of the following will satisfy the given equation?

a) (1, 1)

b) (1, 2)

c) (2, 1)

d) (2, 2)

**Answer d**

**Question 17):**

**If 4x + 3 < 6x +7, then x belongs to the interval**

(a) (2, ∞)

(b) (-2, ∞)

(c) (-∞, 2)

(d) (-4, ∞)

Answer: (b)

**Solution:**** **Given, 4x + 3 < 6x + 7

Subtracting 3
from both sides,

4x + 3 – 3 <
6x + 7 – 3

⇒ 4x < 6x + 4

Subtracting 6x
from both sides,

4x – 6x < 6x
+ 4 – 6x

⇒ – 2x < 4 or

⇒ x > – 2 i.e., all the real numbers
greater than –2, are the solutions of the given inequality.

Hence, the
solution set is (–2, ∞), i.e. x ∈ (-2, ∞)

**Question 18):**A solution is to be kept between 77° F and 86° 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° ?

a) [15°, 20°]

b) [20°, 25°]

c) [25°, 30°]

d) [30°, 35°]

**Answer c**

**Question 19):**If Rohit has x rupees and he pay 40 rupees to shopkeeper then find range of x if amount of money left with Rohit is at most 10 rupees is given by inequation

a) x ≥ 10

b) x ≤ 10

c) x ≤ 50

d) x ≥ 50

Explanation: Amount left is at most 10 rupees i.e. amount left ≤ 10.

x - 40 ≤ 10 ⇒ x ≤ 50.

**Question 20):**

**Solving – 8 ≤ 5x – 3 < 7, we get**

(a) –1/2 ≤ x ≤ 2

(b) 1 ≤ x < 2

(c) –1 ≤ x <
2

(d) –1 < x ≤
2

Answer: (c

**Solution:**** **Given,
– 8 ≤ 5x – 3 and 5x – 3 < 7

Let us solve
these two inequalities simultaneously.

– 8 ≤ 5x – 3 and
5x – 3 < 7 can be written as:

– 8 ≤ 5x –3 <
7

Adding 3, we get

– 8 + 3 ≤ 5x – 3
+ 3 < 7 + 3

⇒ –5 ≤ 5x < 10

Dividing by 5,
we get

–1 ≤ x < 2

**Question 21):**If x is a positive integer and 20x<100 then find solution set of x.

a) {0, 1, 2, 3, 4, 5}

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

c) {1, 2, 3, 4}

d) {0, 1, 2, 3, 4}

Explanation: 20x < 100

Dividing by 20 on both sides, x < (100/20) ⇒ x < 5

**Question 22):**If x is a natural number and 20x ≤ 100 then find solution set of x.

a) {0, 1, 2, 3, 4, 5}

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

c) {1, 2, 3, 4}

d) {0, 1, 2, 3, 4}

Explanation: 20x ≤ 100

Dividing by 20 on both sides, x ≤ (100/20) ⇒ x ≤ 5

**Question 23):**

**Observe the figure given below.**

The interval at
which the value of x lies is

(a) x ∈
(– ∞, – 2)

(b) x ∈
(– ∞, – 2]

(c) x ∈
(– 2, ∞]

(d) x ∈
[– 2, ∞)

Answer (b)

**Solution:**** **In the given figure, the circle is
filled with dark color at -2 which means -2 is included and the highlighted is
towards the left of -2.

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