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

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

Dear Readers, Important Practice Aptitude Questions forย Upcoming Exams was given here with Solutions. Aspirants those who areย preparingย for the Bank Examination and other Competitive Examination canย use this.

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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)ย ย x< y
b)ย ย x โฅ y
c)ย ย x> y
d)ย ย x โค y
e)ย ย Relation between x and y can’t be established
ย
1). I.9x2 – 29x + 22 = 0
II.6y2 โ 31y + 39 = 0
ย
2). I.x2 = 11449
II.y = โ11449
ย
3). I. 6x2 – 25x โ 14 = 0
II.8y2 โ 35y + 12 = 0
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4). I.5x โ 4y = 83
II.6x + 3y = 45
ย
5). I.9x2 + 6โ7 x + 7 = 0
II.4y2 + 4โ3 y + 3 = 0
ย
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)ย ย x โฅ y
b)ย ย x<y
c)ย ย x = y or no relation can be established
d)ย ย x โค y
e)ย ย x> y
ย
7). I. 20x2โ 53x + 18=0
II.5y2โ 37y + 32 = 0
a)ย ย x โค y
b)ย ย x>y
c)ย ย xโฅy
d)ย ย x<y
e)ย ย x = y or no relation can be established
ย
8). I.12x2โ 25x+ 13 = 0
II.5y2โ 12y + 4 = 0
a)ย ย x<y
b)ย ย xโฅy
c)ย ย xโคy
d)ย ย x = y or no relation can be established
e)ย ย x>y
ย
9). I.2x2-7x-22=0
II.6y2-13y-19 = 0
a)ย ย x>y
b)ย ย x = y or no relation can be established
c)ย ย xโคy
d)ย ย xโฅy
e)ย ย x<y
ย
10). I.10x2 – (15 + 8โ3)x + 12โ3 = 0
II.6y2 + (8 + 3โ3)y + 4โ3 = 0
a)ย ย xโคy
b)ย ย xโฅy
c)ย ย x<y
d)ย ย x>y
e)ย ย x = y or no relation can be established
ย
Explanation:
ย
1). A)I. 9×2 – 29x + 22 = 0
x = (-18/9), (-11/9)
x = 2, 11/9
II. 6y2 โ 31y + 39 = 0
y = (-18/6), (-13/6)
y = 3, 13/6
hence x< y
ย
2). D)I. x2 = 11449
x = โ11449 = ยฑ107
II. y = โ11449 = 107
Hence x โค y
ย
3). E)I. 6×2 – 25x โ 14 = 0
x = (3/6), (-28/6)
x = -1/2, 14/3
II. 8y2 โ 35y + 12 = 0
y = (-3/8), (-32/8)
y = 3/8, 4
hence no relation can be established.
ย
4). C)I. 5x โ 4y = 83ย ย ย ย  ..(i)
II. 6x + 3y = 45ย ย ย ย ย  โฆ(ii)
Solving equation (i)ร3 + (ii)ร4, we get
(15x โ 12y = 249) + (24x + 12y = 180)
x = (429/39) = 11
putting the value of x in equation (i), we get
55 โ 4y = 83
-4y = 83-55
y = -28/4 = -7
hence x> y
ย
5). A)I. 9×2 + 6โ7 x + 7 = 0
Or, (3x + โ7)2 = 0
Or, 3x = -โ7
x = -3/โ7
II. 4y2 + 4โ3 y + 3 = 0
Or, (2y – โ3)2 = 0
Or, 2y = โ3
y = โ3/2
hence x< y
ย
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
ย
7). E)I. 20×2โ 53x + 18=0
x = (-45/20), (-8/20)
x = 9/4, 2/5
II. 5y2โ 37y + 32 = 0
y = (-32/5), (-5/5)
y = 6.4, 1
Hence no relation can be established.
ย
8). B)I.12×2โ 25x+ 13 = 0
x = (-13/12), (-12/12)
x = 13/12, 1
II. 5y2โ 12y + 4 = 0
y = (-10/5), (-2/5)
y = 2, 2/5
Hence relationship canโt be established.
ย
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 canโt be established.
ย
10). D)I. 10×2 – (15 + 8โ3)x + 12โ3 = 0
x = (-15/10), (-8โ3 /10)
x = 3/2, 4โ3/5
II. 6y2 + (8 + 3โ3)y + 4โ3 = 0
y = (-4/3), (-โ3 /2)

hence x> y