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

Practice Quantitative Aptitude Questions For SBI PO Prelims& NIACL 2017

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.

 
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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)  ifx< y
b)  ifx ≤ y
c)  ifx = y
d)  ifx> y
e)  ifx ≥ y
 
1). I.4x2 – 8x + 3 = 0
II.2y2 – 7y + 6 = 0
 
2). I.x2 + x – 6 = 0
II.2y2 – 13y + 21 = 0
 
3). I.x2– x – 6 = 0
II.2y2 + 13y + 21 = 0
 
4). I. x2= 4
II.y2 + 6y + 9 = 0
 
5). I.2x + 3y = 4
II.3x + 2y = 11
 
6). I.16x2 + 20x + 6 = 0
II.10y2 + 38y + 24 = 0
 
7). I.18x2 + 18x + 4 = 0
II.12y2 + 29y + 14 = 0
 
8). I.8x2 + 6x = 5
II.12y2– 22y + 8 = 0
 
9). I.17x2 + 48x = 9
II.13y2 = 32y – 12
 
10). I.4x + 7y = 209
II.12x – 14y = -38
 

 

 

 

Solution:
 
1). B) 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 = 0à x =1/2
and if 2x – 3 =0àx = 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 =0à y=3/2
and if y – 2 = 0ày=2
 y =3/2 or 2
Hence, x ≤ y
 
2). A) From equation I:
 x2+ x – 6 = 0
or, x2 – 2x + 3x – 6 = 0
or, x(x -2) +3(x – 2) = 0
(x – 2) (x + 3) = 0
If, x – 2 =0à x= 2 and x + 3 = 0à x = -3 or 2
From equation II:
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= 0ày = 7/2 and if y-3 = 0ày = 3
 y = 3 or 7/2 Hence, x<y
 
3). E) From equation I:
x2– x – 6 = 0
or, x2 – 3x + 2x – 6 = 0
or, x(x-3) + 2(x-3) = 0
or, (x-3)(x+2) = 0
if, x-3 = 0à x =3 and if x+2 = 0àx = -2
x = -2 or 3
From equation II:
 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 = 0ày = -(7/2)and if y + 3 =0ày =-3
Hence x> y
 
4). E) From equation I:
x2= 4
x = √4à x = ±2
from equation II:
y2+ 6y + 9 = 0
or, y2 + 3y + 3y + 9 = 0
or,y (y + 3)+3(y+3) = 0
 or, (y+3) (y+3)=0
or, y+3 = 0
Hence x> y
 
5). E)From I and II we get y =-2 and x = 5
x> y
 
6). D) I.16x2 + 20x + 6 = 0
or, 16x2 +12x+8x+6 = 0
or, 4x(4x+3)+y(4x+3)=0
 or, (4x + 3)(4x + 2) = 0
x = -(3/4) or –(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,  x> y
 
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 –(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 –(7/4)
Hence, x≥y
 
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 –(5/4)
II.12y2-22y+8=0
or, 12y2 – 16y – 6y + 8 = 0
or, 4y(3y – 4) – 2 (3y -4) = 0
or, (4y – 2) (3y- 4) = 0
y = (1/2) or (4/3)
Hence, x ≤ y
 
9). A) I. 17x2 + 48x – 9 = 0
or, 17x2 + 51x – 3x – 9 = 0
or, (17x -3)(x+3) = 0
x = (3/17) or -3
II.13y2 – 32y + 12 = 0
Or, 13y2 – 26y – 6y + 12 = 0
Or, (13y – 6)(y -2) = 0
Y = (6/13) or 2
Hence x< y.
 
10). C)
I.4x + 7y = 209
Or, 8x+14y=418     …(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
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