# 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. 1-10): In each of the following questions two equations are given. You have to solve the equations and give answer.

a)ย ย ifx< y
b)ย ย ifx โค y
c)ย ย ifx = y
d)ย ย ifx> y
e)ย ย ifx โฅ y
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1). I.4x2 โ 8x + 3 = 0
II.2y2 โ 7y + 6 = 0
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2). I.x2 + x – 6 = 0
II.2y2 โ 13y + 21 = 0
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3). I.x2โ x – 6 = 0
II.2y2 + 13y + 21 = 0
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4). I. x2= 4
II.y2 + 6y + 9 = 0
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5). I.2x + 3y = 4
II.3x + 2y = 11
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6). I.16x2 + 20x + 6 = 0
II.10y2 + 38y + 24 = 0
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7). I.18x2 + 18x + 4 = 0
II.12y2 + 29y + 14 = 0
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8). I.8x2 + 6x = 5
II.12y2โ 22y + 8 = 0
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9). I.17x2 + 48x = 9
II.13y2 = 32y โ 12
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10). I.4x + 7y = 209
II.12x – 14y = -38
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Solution:
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1). B)ย From equation I:
4x2-8x+3 = 0
or, 4x2-6x-2x+3=0
or, 2x(2x -3)-1(2x -3) = 0
or, (2x -1)(2x -3) = 0
If, 2x -1 = 0ร ย x =1/2
and if 2x – 3 =0ร x = 3/2
x =1/2 or 3/2
From equation II:
2y2– 7y +6 = 0
or,2y2-4y-3y+6 = 0
or, (2y-3) (y-2)=0
If2y-3 =0ร  y=3/2
and if y โ 2 = 0ร y=2
ย y =3/2 or 2
Hence, x โค y
ย
2). A) From equation I:
ย x2+ x – 6 = 0
or, x2 โ 2x + 3x โ 6 = 0
or, x(x -2) +3(x – 2) = 0
(x – 2) (x + 3) = 0
If, x – 2 =0ร ย x= 2 and x + 3 = 0ร  x = -3 or 2
From equation II:
2y2โ 13y + 21 = 0
or, 2y2 โ 6y โ 7y + 21 = 0
or, 2y(y-3) -7(y-3) = 0
or, (2y – 7) (y – 3) = 0
ย If, 2y- 7= 0ร y = 7/2 and if y-3 = 0ร y = 3
ย y = 3 or 7/2 Hence, x<y
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3). E) From equation I:
x2โ x โ 6 = 0
or, x2 โ 3x + 2x โ 6 = 0
or, x(x-3) + 2(x-3) = 0
or, (x-3)(x+2) = 0
if, x-3 = 0ร  x =3 and if x+2 = 0ร x = -2
x = -2 or 3
From equation II:
ย 2y2 +13y+21=0
or, 2y2 +6y+7y+21 = 0
or,2y(y+3)+7(y+3)=0
or, (2y+7)(y+3) = 0
If, 2y + 7 = 0ร y = -(7/2)and if y + 3 =0ร y =-3
Hence x> y
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4). E) From equation I:
x2= 4
x = โ4ร  x = ยฑ2
from equation II:
y2+ 6y + 9 = 0
or, y2 + 3y + 3y + 9 = 0
or,y (y + 3)+3(y+3) = 0
ย or, (y+3) (y+3)=0
or, y+3 = 0
Hence x> y
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5). E)From I and II we get y =-2 and x = 5
x> y
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6). D) I.16x2 + 20x + 6 = 0
or, 16x2 +12x+8x+6 = 0
or, 4x(4x+3)+y(4x+3)=0
ย or, (4x + 3)(4x + 2) = 0
x = -(3/4) or โ(1/2)
II.10y2 +38y+24 = 0
or, 5y2 +19y+12 = 0
or, 5y2+15y+4y+12=0
or, (5y + 4) (y + 3) = o
y = -(4/5) or -3
Hence,ย  x> y
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7). E) I.18x2 +18x + 4 = 0
or, 9x2 +9x+2 =0
or, 9x2+6x+3x+2 =0
or, 3x (3x +2) + 1(3x + 2) = 0
x = -(1/3) or โ(2/3)
II.12y2 +29y+14=0
or, 12y2 +21y+8y+14 = 0
or, 3y(4y 7) + 2(4y + 7) =0
or, (3y + 2) (4y + 7) = 0
y= -(2/3) or โ(7/4)
Hence, xโฅy
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8). B) I.8x2+6x-5=0
or,2x(4x+5)-1( 4x+5) = 0
or, (2x-1)(4x+5) = 0
x = (1/2) or โ(5/4)
II.12y2-22y+8=0
or, 12y2 โ 16y โ 6y + 8 = 0
or, 4y(3y – 4) โ 2 (3y -4) = 0
or, (4y – 2) (3y- 4) = 0
y = (1/2) or (4/3)
Hence, x โค y
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9). A) I. 17x2 + 48x โ 9 = 0
or, 17x2 + 51x โ 3x โ 9 = 0
or, (17x -3)(x+3) = 0
x = (3/17) or -3
II.13y2 โ 32y + 12 = 0
Or, 13y2 โ 26y โ 6y + 12 = 0
Or, (13y – 6)(y -2) = 0
Y = (6/13) or 2
Hence x< y.
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10). C)
I.4x + 7y = 209
Or, 8x+14y=418ย ย ย ย  โฆ(i)
II.12x – 14y = -38
From (i) and (ii), we get
20x = 380
x = 19
putting value of x in equation (i),
we get y = 19
x=y