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Maths MCQ Class 11 Ch-5 | Linear Inequalities
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MCQ | Class 11 | Chapter 05
Linear Inequalities
MCQ | CHAPTER 5 | 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
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.
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.
(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, ∞)
a) x > 4
b) x < 4
c) x > 2
d) x < 2
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.
a) x > 4
b) x < 4
c) x > 2
d) x < 2
a) double inequality
b) quadratic inequality
c) numerical inequality
d) linear inequality
a) 60 b) 35 c) 180 d) 45
(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, ∞)
a) (5, 7)
b) (3, 5), (5, 7)
c) (3, 5), (5, 7), (7, 9)
d) (5, 7), (7, 9)
(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, ∞).
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.
(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, ∞)
a) (1, 1) b) (1, 2) c) (2, 1) d) (2, 2)
(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, ∞)
a) [15°, 20°]
b) [20°, 25°]
c) [25°, 30°]
d) [30°, 35°]
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.
(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
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
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
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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