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. 1-5):In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
Give answer
a)  If x< y
b)  If x> y
c)  If x ≥ y
d)  If x ≤ y
e)  If relationship between x and y cannot be determined
 
1). I.4x2 —15x + 14=0
II.6y2 —10y — 4=0
 
2). I.3x2 + 10x + 3 =0
II.2y2 + 15 y + 27 =0
 
3). I.7x2 + 12x + 5 =0.
II.3y2+ 7y + 2 =0
 
4). I.16x2 —14x + 3 = 0
II.6y2—19y + 15 =0
 
5). I.x2 + 11x + 18 =0
II.y2 – √81 = 0
 
Directions (Q. 6-10) :In the following questions, two equations are given. You have to solve both the equations and give answer.
 
6). I.x2 +3x-28 = 0
II.y2 – y -20 = 0
a)  x = y or relationship cannot be decided
b)  x> y
c)  x< y
d)  x ≥ y
e)  x ≤ y
 
7). I.5 x2 + 11x+ 6 =0
II.y2 – 34y – 336 = 0
a)  x> y
b)  x ≤ y
c)  x = y or relationship cannot be decided
d)  x<y
e)  x≥y
 
8). I.2x2+ 18x+ 40 = 0
II.2y2 +15y+ 27 = 0
a)  x<y
b)  x≥y
c)  x≤y
d)  x> y
e)  x = y or relationship cannot be decided
 
9). I.6x2-29x+35 = 0
II.3y2-11y+10=0
a)  x ≥ y
b)  x = y or relationship cannot be decided
c)  x ≤ y
d)  x>y
e)  x<y
 
10). I.x2 + x – 20 = 0
II.y2-y-30=0
a)  x = y or relationship cannot be decided
b)  x>y
c)  x<y
d)  x ≤ y
e)  x ≥ y

 

 
Explanation with answers key:
1). E)I. 4x2 -15x + 14 = 0
4x2-8x-7x+ 14 = 0
4x (x -2) – 7(x -2) = 0
(x-2) (4x – 7) = 0
x=2, 7/4
II. 6y2 -10y-4=0
6y2 -12y+ 2y-4= 0
6y(y-2)+ 2 (y-2)=0
 (y-2) (6y+2) =0
y=2, -1/ 3
If relationship between x and y cannot be determined
 
2). C)
 I. 3x2 +10x+3 =0
3x2+9x+ x+3 =0.
3x(x+3)+1(x+3)=0
(x+3)(3x+1)=0
x = -3, -1/3
II. 2y2 + 15y + 27 = 0
2y2+6y+9y+ 27 =0
2y(y+ 3)+ 9 (y+ 3)=0
(y+3)(2y+9)=0
y= -3, -9/2
 
3). E)
 I. 7x2+ 12x + 5 = 0
7x2+7x+5x+5 = 0
7x(x+1)+5(x+1)=0
(x+1)(7x+5)=0
x = -1, -5/7
II. 3y2 + 7y + 2 = 0
3y2+6y+ y+ 2 =0
3y(y+2)+1 (y+ 2)=0
 (y+2)(3y+ 1)= 0
y= -2, -1/3
x and y relationship cannot be decided
 
4). A)
I. 16x2-14x+3=0
16x2-8x-6x+3 =0
8x (2x-1) -3(2x-1) = 0
(2x -1) (8x -3) =0
 x= 1/2, 3/8
II. 6y2-19y+15=0
6y2-10y-9y+ 15 =0
2y (3y -5) -3(3y-5) =0
(2y-3) (3y-5) = 0
y = 3/2, 5/3
x< y
 
5). E)I. x2 +11x+18 =0
x2+9x+2x+18 =0
x (x + 9)+2 (x +9) =0
(x+2) (x+ 9)=0
x = -2, -9
II. y2 — √81 = 0
y2=√81
y2 = 9
y = ± 3
If relationship between x and y cannot be determined
 
6). A)
I. x2+3x-28 = 0
x2+ 7x+ 4x – 28 =
x(x+7)-4(x+7) =.0
(x- 4) (x+ 7) = 0
X = 4, -7
II. y2 -y-20 = 0
y2-5y+ 47-20 = 0
y(y-5)+ 4(y-5) = 0
(y + 4) (y – 5) = 0
y = -4, 5
Relationship cannot be decided.
 
7). C)I. 5x2+11x+6 =0
5x2+6x+5x+6 =0
x(5x+6)+1(5x+6)=0
(5x+ 6)(x+1)=0
x= -1, -6/5
II. y2 -34y-336 =0
y2-427+8y-336 =0
y(y-42)+8(y-42)=0
(y-42)(y+8)=0
y = 42, -8
Relationship cannot be decided.
 
8). E)I. 2x2 +18x+40 = 0
2 x2+10 x+8x+ 40 = 0
2 x(x+5)+8(x+5) = 0
(2x+8) (x+5) = 0
(x+5)(2x+8)=. 0
x = -4 , -5
II.  2y2 +15y -27 = 0
2y2+ 9y + 6y + 27 =0
y (2y + 9) + 3 (2y + 9) =0
(2y + 9) (y + 3) = 0
y = – 4.5, -3
Relationship cannot be decided.
 
9). D)
 I. 6x2—29x + 35 = 0
6x2—15x —14x + 35 = 0
3x (2x — 5) —7 (2x —5) = 0
(2x —5)(3x —7) = 0
x = 2.3, 2.5
II. 3y2 —11y +10 = 0
3y2— 6y — 5y + 10 = 0
3y(y— 2) — 5 (y — 2) = 0
(7-2) (3y —5) = 0
y = 1.6, 2
x> y
 
10). A)I. x2 +x-20 =0
x2+ 5x —4x-20 =0
x(x+5)-4(x+5)=0
(x+ 5)(x-4)=0
x = 4,-5
II. y2 — y-30 =0
y2—6y+ 5y —30 =0
y(y-6)+5(y-6) = 40
(y – 6)(y + 5) =0
y = —5, 6
If relationship between x and y cannot be determined
More Practice Aptitude Questions for Upcoming Exams 2017 –Click Here    

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