Quiz-summary
0 of 10 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
Information
Hello Aspirants,
Welcome to our official website http://www.studyvirus.com for important study materials,quizzes,current affairs,etc.We are providing the most important quizzes for all bank exams-SBI PO,SBI Clerk,IBPS PO,IBPS Clerk,IBPS RRB exams,Insurance and many more.All the quizzes are based on latest pattern.
In this series of quizzes,we will be providing most important questions on Quadratic Equations (Inequalities) which are usually asked in the exams.In Prelims Examination,usually 5-10 questions are asked from this topic.
All the best !
Get the best selling Data Interpretation and Puzzle Pen Drive Course (350+ videos) along with PDF and online quizzes.Click here
Click here for Real Exam Pattern Based Test Series for Banking exams (SBI,IBPS,Insurance,etc) at the most affordable price.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
- Not categorized 0%
⇒ Please share your marks in comment section.
⇒Ask your doubts in too in comment section,we will try our best to clear all your doubts.
Keep Working Hard and Never Give Up
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- Answered
- Review
- Question 1 of 10
1. Question
Directions: 1-5) In the following questions, two equations numbered I and II are given with variables a and b. Solve both the equations and find out the relationship between a and b. Then give answer accordingly.
I. 3a2 – a – 10 = 0
II. 3b2 – 11b + 6 = 0Correct3a2 – a – 10 = 0
3a2 – 6a + 5a – 10 = 0
3a (a-2) + 5 (a-2) = 0
(3a+5) (a – 2) = 0
Gives a = -5/3, 2
3b2 – 11b + 6 = 03b2 – 9b – 2b +6 =0
3b (b-3) – 2(b-3) =0
(3b -2) (b-3) =0
Gives b = 2/3, 3
Relationship cannot be determinedIncorrect3a2 – a – 10 = 0
3a2 – 6a + 5a – 10 = 0
3a (a-2) + 5 (a-2) = 0
(3a+5) (a – 2) = 0
Gives a = -5/3, 2
3b2 – 11b + 6 = 03b2 – 9b – 2b +6 =0
3b (b-3) – 2(b-3) =0
(3b -2) (b-3) =0
Gives b = 2/3, 3
Relationship cannot be determined - Question 2 of 10
2. Question
I. 3a2 – (3 – 2√2)a – 2√2 = 0
II. 3b2 – (1 + 3√3)b + √3 = 0Correct3a2 – (3 – 2√2)a – 2√2 = 0
(3a2 – 3a) + (2√2a – 2√2) = 0
3a (a – 1) + 2√2 (a – 1) = 0
So a = 1, -2√2/3 (-0.9)
3b2 – (1 + 3√3)b + √3 = 0
(3b2 – b) – (3√3b – √3) = 0
b (3b – 1) – √3 (3b – 1) = 0
So, b = 1/3, √3 (1.7)Relationship cannot be determined
Incorrect3a2 – (3 – 2√2)a – 2√2 = 0
(3a2 – 3a) + (2√2a – 2√2) = 0
3a (a – 1) + 2√2 (a – 1) = 0
So a = 1, -2√2/3 (-0.9)
3b2 – (1 + 3√3)b + √3 = 0
(3b2 – b) – (3√3b – √3) = 0
b (3b – 1) – √3 (3b – 1) = 0
So, b = 1/3, √3 (1.7)Relationship cannot be determined
- Question 3 of 10
3. Question
I. a2+ (4 + √2)a + 4√2 = 0
II. 5b2+ (2 + 5√2)b + 2√2 = 0Correcta2 + (4 + √2)a + 4√2 = 0
(a2 + 4a) + (√2a + 4√2) = 0
a (a + 4) + √2 (a + 4) = 0
So a = -4, -√2 (-1.4)
5b2 + (2 + 5√2)b + 2√2 = 0
(5b2 + 2b) + (5√2b + 2√2) = 0
b (5b + 2) + √2 (5b + 2) = 0
So, b = -2/5 (-0.4), -√2 (-1.4)a ≤ b
Incorrecta2 + (4 + √2)a + 4√2 = 0
(a2 + 4a) + (√2a + 4√2) = 0
a (a + 4) + √2 (a + 4) = 0
So a = -4, -√2 (-1.4)
5b2 + (2 + 5√2)b + 2√2 = 0
(5b2 + 2b) + (5√2b + 2√2) = 0
b (5b + 2) + √2 (5b + 2) = 0
So, b = -2/5 (-0.4), -√2 (-1.4)a ≤ b
- Question 4 of 10
4. Question
I. 6a2– (3 + 4√3)a + 2√3 = 0,
II. 3b2– (6 + 2√3)b + 4√3 = 0Correct6a2 – (3 + 4√3)a + 2√3 = 0
(6a2 – 3a) – (4√3a – 2√3) = 0
3a (2a- 1) – 2√3 (2a – 1) = 0,
So a = 1/2, 2√3/3 (1.15)
3b2 – (6 + 2√3)b + 4√3 = 0
(3b2 – 6b) – (2√3b – 4√3) = 0
3b (b – 2) – 2√3 (b – 2) = 0
So, b = 2, 2√3/3a ≤ b
Incorrect6a2 – (3 + 4√3)a + 2√3 = 0
(6a2 – 3a) – (4√3a – 2√3) = 0
3a (2a- 1) – 2√3 (2a – 1) = 0,
So a = 1/2, 2√3/3 (1.15)
3b2 – (6 + 2√3)b + 4√3 = 0
(3b2 – 6b) – (2√3b – 4√3) = 0
3b (b – 2) – 2√3 (b – 2) = 0
So, b = 2, 2√3/3a ≤ b
- Question 5 of 10
5. Question
I. 8a2 + (4 + 2√2)a + √2 = 0
II. b2 – (3 + √3)b + 3√3 = 0Correct8a2 + (4 + 2√2)a + √2 = 0
(8a2 + 4a) + (2√2a + √2) = 0
4a (2a + 1) + √2 (2a + 1) = 0
So a = -1/2 (-0.5), -√2/4 (-0.35)
b2 – (3 + √3)b + 3√3 = 0
(b2 – 3b) – (√3b – 3√3) = 0
b (b – 3) – √3 (b – 3) = 0
So b = 3, √3 (1.73)a < b
Incorrect8a2 + (4 + 2√2)a + √2 = 0
(8a2 + 4a) + (2√2a + √2) = 0
4a (2a + 1) + √2 (2a + 1) = 0
So a = -1/2 (-0.5), -√2/4 (-0.35)
b2 – (3 + √3)b + 3√3 = 0
(b2 – 3b) – (√3b – 3√3) = 0
b (b – 3) – √3 (b – 3) = 0
So b = 3, √3 (1.73)a < b
- Question 6 of 10
6. Question
Directions:6-10) In each of the following questions, two equations are given. You have to solve these questions and find out the values of p and q.
I.16p² + 20p + 6 = 0
II.10q² + 38q + 24 = 0CorrectI. 16p² + 20p + 6 = 0
or, 8p² + 10p + 3 = 0
or, (4p + 3 ) ( 2p + 1) = 0
Therefore, p = -3/4 = -0.75 or p = – 1/2 = – 0.5II. 10q² + 38q + 24 = 0
or, 5q² + 19q + 12 = 0
or, ( q + 3) (5q + 4 ) = 0
Therefore, q = – 3 or q = – 4/5 = -0.8
Hence p > qIncorrectI. 16p² + 20p + 6 = 0
or, 8p² + 10p + 3 = 0
or, (4p + 3 ) ( 2p + 1) = 0
Therefore, p = -3/4 = -0.75 or p = – 1/2 = – 0.5II. 10q² + 38q + 24 = 0
or, 5q² + 19q + 12 = 0
or, ( q + 3) (5q + 4 ) = 0
Therefore, q = – 3 or q = – 4/5 = -0.8
Hence p > q - Question 7 of 10
7. Question
I.18p² + 18p + 4 = 0
II.12q² + 29q + 14 = 0Correct- 18p² + 18p + 4 = 0
or, 9p² + 9p + 2 = 0
or, ( 3p + 2 ) ( 3p + 1) = 0
Therefore, p = – 2/3 = -0.67 or p = – 1/3 = -0.33 - 12q² + 29q + 14 = 0
or, (3q + 2) ( 4q + 7 ) = 0
Therefore, q = – 2/3 = -0.67 or q = -7/4 = -1.75
Hence p ≥ q
Incorrect- 18p² + 18p + 4 = 0
or, 9p² + 9p + 2 = 0
or, ( 3p + 2 ) ( 3p + 1) = 0
Therefore, p = – 2/3 = -0.67 or p = – 1/3 = -0.33 - 12q² + 29q + 14 = 0
or, (3q + 2) ( 4q + 7 ) = 0
Therefore, q = – 2/3 = -0.67 or q = -7/4 = -1.75
Hence p ≥ q
- Question 8 of 10
8. Question
I.8p² + 6p = 5
II.12q² – 22q + 8 = 0CorrectI. 8p² + 6p – 5 = 0
or, (4p + 5) (2p – 1) = 0
Therefore, p = – 5/4 = -1.25 or p = 1/2 = 0.5II. 12q² – 22q + 8 = 0
or, 6q² -11q + 4 = 0
or, (2q -1) (3q – 4) = 0
Therefore, q = 1/2 = 0.5 or q = 4/3 = 1.33
Hence p ≤ qIncorrectI. 8p² + 6p – 5 = 0
or, (4p + 5) (2p – 1) = 0
Therefore, p = – 5/4 = -1.25 or p = 1/2 = 0.5II. 12q² – 22q + 8 = 0
or, 6q² -11q + 4 = 0
or, (2q -1) (3q – 4) = 0
Therefore, q = 1/2 = 0.5 or q = 4/3 = 1.33
Hence p ≤ q - Question 9 of 10
9. Question
I.17p² + 48p – 9 = 0
II.13q² = 32q – 12CorrectI. 17p² + 48p – 9 = 0
or, (p + 3) (17p – 3) = 0
Therefore, p = -3 or p = 3/17 = 0.18II. 13q² – 32q + 12 = 0
or, (q -2) (13q – 6) = 0
Therefore, q = 2 or q = 6/13 = 0.46
Hence p < qIncorrectI. 17p² + 48p – 9 = 0
or, (p + 3) (17p – 3) = 0
Therefore, p = -3 or p = 3/17 = 0.18II. 13q² – 32q + 12 = 0
or, (q -2) (13q – 6) = 0
Therefore, q = 2 or q = 6/13 = 0.46
Hence p < q - Question 10 of 10
10. Question
I.821p² – 757p² = 256
II.√196 q³ – 12q³ = 16Correct64p² = 256
or, p² = 4
or, p = ±2
AND
14q³ – 12q³ = 16
or, 2q³ = 16
or, q³ = 8
or, q = 2
Hence p ≤ qIncorrect64p² = 256
or, p² = 4
or, p = ±2
AND
14q³ – 12q³ = 16
or, 2q³ = 16
or, q³ = 8
or, q = 2
Hence p ≤ q
