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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.
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Question 1 of 10
1. Question
Directions:1-5)In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,
√324 x + √11664 = 0
(16807)1/5 y + (9261)1/3 = 0
Correct
√324 x + √11664 = 0
18x + 108 = 0
18x = -108
x = – 6
II. (16807)1/5 y + (9261)1/3 = 0
(75)1/5y + (213)1/3 = 0
7y + 21 = 0
7y = -21
y = -3
Hence x < y
Incorrect
√324 x + √11664 = 0
18x + 108 = 0
18x = -108
x = – 6
II. (16807)1/5 y + (9261)1/3 = 0
(75)1/5y + (213)1/3 = 0
7y + 21 = 0
7y = -21
y = -3
Hence x < y
Question 2 of 10
2. Question
2x – y = 31
y = ∛19683
Correct
y = ∛19683
y = 27
II. 2x – y = 31
2x – 27 = 31
2x = 58
x = 29
Hence x > y
Incorrect
y = ∛19683
y = 27
II. 2x – y = 31
2x – 27 = 31
2x = 58
x = 29
Hence x > y
Question 3 of 10
3. Question
x2 – 10x + 21 = 0
y2 – 10y + 16 = 0
Correct
x2 – 10x + 21 = 0
(x – 7) (x – 3) = 0
x = 7, 3
II. y2 – 10y + 16 = 0
(y -8) (y – 2) = 0
y = 8, 2
Hence the relationship can’t be determined.
Incorrect
x2 – 10x + 21 = 0
(x – 7) (x – 3) = 0
x = 7, 3
II. y2 – 10y + 16 = 0
(y -8) (y – 2) = 0
y = 8, 2
Hence the relationship can’t be determined.
Question 4 of 10
4. Question
12x2 – 17x – 57 = 0
4y2 -7y – 36 = 0
Correct
12x2 – 17x – 57 = 0
12x2 – 36x + 19x – 57 = 0
12x(x – 3) + 19(x – 3) = 0
(12x + 19) (x – 3) = 0
x = -19/12, 3
x = -1.58, 3
II.4y2 -7y – 36 = 0
4y2 -16y + 9y -36 = 0
4y(y – 4) + 9(y – 4) = 0
(4y + 9) (y – 4) = 0
y = -9/4, 4
y= -2.25, 4
Hence the relationship can’t be determined.
Incorrect
12x2 – 17x – 57 = 0
12x2 – 36x + 19x – 57 = 0
12x(x – 3) + 19(x – 3) = 0
(12x + 19) (x – 3) = 0
x = -19/12, 3
x = -1.58, 3
II.4y2 -7y – 36 = 0
4y2 -16y + 9y -36 = 0
4y(y – 4) + 9(y – 4) = 0
(4y + 9) (y – 4) = 0
y = -9/4, 4
y= -2.25, 4
Hence the relationship can’t be determined.
Question 5 of 10
5. Question
3x + 5y = 34
5x – 2y = 5
Correct
3x + 5y = 34 –à (1)
5x – 2y = 5—à (2)
By solving the equation (1) and (2), we get,
x = 3, y = 5
Hence x < y
Incorrect
3x + 5y = 34 –à (1)
5x – 2y = 5—à (2)
By solving the equation (1) and (2), we get,
x = 3, y = 5
Hence x < y
Question 6 of 10
6. Question
Directions:6-10) In the following questions, two equations numbered I and II are given. You have to solve both the equations and Give Answer If
x2 – 452 = 874 -√900
y = (24+√(121+√(500+29)))
Correct
x2 – 452 = 874 -√900
x2 = 1326 – 30 = 1296
x =± 36
y = (24+√(121+√(500+29)))
y = (24+√(121+23))
y = (24+12)
y = 36
x ≤ y
Incorrect
x2 – 452 = 874 -√900
x2 = 1326 – 30 = 1296
x =± 36
y = (24+√(121+√(500+29)))
y = (24+√(121+23))
y = (24+12)
y = 36
x ≤ y
Question 7 of 10
7. Question
3 + 18/x + 15/x2 =0
II. 2 + 4/y+2/y2 =0
Correct
3 + 18/x + 15/x2 =0
3x2 + 18x + 15 =0
3x2 + 3x + 15x +15 =0
3x (x+1) +15 (x+1) =0
(3x+15) (x+1) =0
x =-1, -15/3 = -1, -5
2 + 4/y+2/y2 =0
2y2 + 4y + 2 =0
2y2 +2y +2y +2 =0
2y (y+1) +2 (y+1) =0
(2y+2) (y+1) =0
y = -1, -2/2 = -1, -1
x ≤ y
Incorrect
3 + 18/x + 15/x2 =0
3x2 + 18x + 15 =0
3x2 + 3x + 15x +15 =0
3x (x+1) +15 (x+1) =0
(3x+15) (x+1) =0
x =-1, -15/3 = -1, -5
2 + 4/y+2/y2 =0
2y2 + 4y + 2 =0
2y2 +2y +2y +2 =0
2y (y+1) +2 (y+1) =0
(2y+2) (y+1) =0
y = -1, -2/2 = -1, -1
x ≤ y
Question 8 of 10
8. Question
x3 = 512
II. y2 – 17y + 72 =0
Correct
x3 = 512
x = 8
y2 – 17y + 72 =0
y2 – 8y – 9y+ 72 =0
(y-8) (y-9) =0
y = 8, 9
x ≤ y
Incorrect
x3 = 512
x = 8
y2 – 17y + 72 =0
y2 – 8y – 9y+ 72 =0
(y-8) (y-9) =0
y = 8, 9
x ≤ y
Question 9 of 10
9. Question
9x+7y = 25
II. 5x+14y = 24
Correct
9x+7y = 25 —– (1)
5x+14y = 24 —– (2)
Simplify the above equation, we get x = 2 and y =1
x > y
Incorrect
9x+7y = 25 —– (1)
5x+14y = 24 —– (2)
Simplify the above equation, we get x = 2 and y =1
