Assertion & Reason Questions For Math Class 10 | Arithmetic Progression

Top of Form ASSERTION & REASON QUESTIONS CLASS 10  CHAPTER 1  REAL NUMBERS Competency based questions on ARITHMETIC PROGRESSION chapter 5 , Assertion and Reason based questions for class 10 ARITHMETIC PROGRESSION chapter 5

Mathematics

Multiple Choice Questions (MCQ)

MCQ Based on the average of numbers.

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

• 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 11LINEAR INEQUALITIES

Question 1)

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

(b) length < 20 cm

(c) breadth x ≥ 20 cm

(d) length ≤ 20 cm

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
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)

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

Question 6): x  > 5 is

a) double 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

Question 8):  ax2 + bx + c > 0 is

a) double inequality
c) numerical inequality
d) linear inequality

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, ∞)

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 $\frac{|x-7|}{(x-7)}\geq 0$ , then

(a) x [7, ∞)

(b) x (7, ∞)

(c) x (– ∞, 7)

(d) x (– ∞, 7]

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

Explanation: Amount left is at least 10 rupees i.e. amount left ≥ 10.
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, ∞)

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)

Question 17): If 4x + 3 < 6x +7, then x belongs to the interval

(a) (2, ∞)

(b) (-2, ∞)

(c) (-∞, 2)

(d) (-4, ∞)

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°]

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

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

– 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
Since x is a positive integer so x = 1, 2, 3, 4.

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
Since x is a natural number so x = 1, 2, 3, 4, 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, ∞)

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.