Practice Quantitative Aptitude Questions For SBI PO Prelims& NIACL 2017 (Quadratic Equation)

Dear Readers, Important Practice Aptitude Questions for Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this.

 
Directions (Q. 1-5): In each question, two equations numbered I and II are given. You have to solve both the equations and mark an appropriate answer.
a)  x< y
b)  x ≥ y
c)  x> y
d)  x ≤ y
e)  Relation between x and y can’t be established
 
1). I.9x2 – 29x + 22 = 0
II.6y2 – 31y + 39 = 0
 
2). I.x2 = 11449
II.y = √11449
 
3). I. 6x2 – 25x – 14 = 0
II.8y2 – 35y + 12 = 0
 
4). I.5x – 4y = 83
II.6x + 3y = 45
 
5). I.9x2 + 6√7 x + 7 = 0
II.4y2 + 4√3 y + 3 = 0
 
Directions (Q. 6-10): In each question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
6). I.14x2+27x+9=0
II.2y2+ 9y +10 = 0
a)  x ≥ y
b)  x<y
c)  x = y or no relation can be established
d)  x ≤ y
e)  x> y
 
7). I. 20x2— 53x + 18=0
II.5y2— 37y + 32 = 0
a)  x ≤ y
b)  x>y
c)  x≥y
d)  x<y
e)  x = y or no relation can be established
 
8). I.12x2— 25x+ 13 = 0
II.5y2— 12y + 4 = 0
a)  x<y
b)  x≥y
c)  x≤y
d)  x = y or no relation can be established
e)  x>y
 
9). I.2x2-7x-22=0
II.6y2-13y-19 = 0
a)  x>y
b)  x = y or no relation can be established
c)  x≤y
d)  x≥y
e)  x<y
 
10). I.10x2 – (15 + 8√3)x + 12√3 = 0
II.6y2 + (8 + 3√3)y + 4√3 = 0
a)  x≤y
b)  x≥y
c)  x<y
d)  x>y
e)  x = y or no relation can be established
 
Explanation:
 
1). A)I. 9×2 – 29x + 22 = 0
x = (-18/9), (-11/9)
x = 2, 11/9
II. 6y2 – 31y + 39 = 0
y = (-18/6), (-13/6)
y = 3, 13/6
hence x< y
 
2). D)I. x2 = 11449
x = √11449 = ±107
II. y = √11449 = 107
Hence x ≤ y
 
3). E)I. 6×2 – 25x – 14 = 0
x = (3/6), (-28/6)
x = -1/2, 14/3
II. 8y2 – 35y + 12 = 0
y = (-3/8), (-32/8)
y = 3/8, 4
hence no relation can be established.
 
4). C)I. 5x – 4y = 83     ..(i)
II. 6x + 3y = 45      …(ii)
Solving equation (i)×3 + (ii)×4, we get
(15x – 12y = 249) + (24x + 12y = 180)
x = (429/39) = 11
putting the value of x in equation (i), we get
55 – 4y = 83
-4y = 83-55
y = -28/4 = -7
hence x> y
 
5). A)I. 9×2 + 6√7 x + 7 = 0
Or, (3x + √7)2 = 0
Or, 3x = -√7
x = -3/√7
II. 4y2 + 4√3 y + 3 = 0
Or, (2y – √3)2 = 0
Or, 2y = √3
y = √3/2
hence x< y
 
6). E)I. 14×2+27x+9=0
x = (21/4), (6/14)
x = -3/2, -3/7
II. 2y2+ 9y +10 = 0
y = (5/2), (4/2)
y = -5/2, -2
hence x> y
 
7). E)I. 20×2— 53x + 18=0
x = (-45/20), (-8/20)
x = 9/4, 2/5
II. 5y2— 37y + 32 = 0
y = (-32/5), (-5/5)
y = 6.4, 1
Hence no relation can be established.
 
8). B)I.12×2— 25x+ 13 = 0
x = (-13/12), (-12/12)
x = 13/12, 1
II. 5y2— 12y + 4 = 0
y = (-10/5), (-2/5)
y = 2, 2/5
Hence relationship can’t be established.
 
9). B)I. 2×2-7x-22=0
x = (-11/2), (4/2)
x = 5.5, -2
II.6y2-13y-19 = 0
y = (-19/6), (6/6)
y = 19/6, -1
Hence relation can’t be established.
 
10). D)I. 10×2 – (15 + 8√3)x + 12√3 = 0
x = (-15/10), (-8√3 /10)
x = 3/2, 4√3/5
II. 6y2 + (8 + 3√3)y + 4√3 = 0
y = (-4/3), (-√3 /2)

hence x> y

 

 

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